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INTRODUCTION CQ WORLD! CQ WORLD! CQ WORLD! ... This is W----; W----; W----; ... Over!

If you've read this far, you may have already started on the road toward making that "call" a reality for yourself!

Yes! Amateur radio is fascinating! You can talk to the world from your "hamshack". You will meet new friends in Radio Clubs and in "Amateur Radio Nets". You will "handle traffic" and relay messages via the magic of radio. You may sometime be involved in saving lives by providing emergency communications during a natural disaster, or after an accident. In addition, it is almost a "sure-fire" thing that once you have your Ham ticket, you'll frequently enjoy the art of "rag -chewing" (just talking with other amateur operators from all over, on the air). Perhaps you'll be the type who enjoys the aspect of experimentation and building your own electronic equipment. Whatever "your thing" . . . you'll find that Amateur Radio is truly fun! fascinating! and a fantastic hobby.

In this PROGRAMMED STYLE book, you will find that learning is easy! We don't want to imply that learning comes without effort . . . but we know that you'll find that learning in "digestible bits" and getting virtually "instant feedback" as to whether you are grasping the key ideas, will be very helpful to your learning process.

The basic philosophy of the book is to have you UNDERSTAND CONCEPTS RATHER THAN MEMORIZE A GROCERY LIST OF FACTS! Wherever it is possible to have you learn to understand key principles and concepts, rather than have you simply memorize a fact .. .

we will do it!

Before we show you "how to use" the book ... take a look at the pictures below of just a few of the many available Amateur Radio activities that you will be able to participate in after you've studied this book. We hope these will inspire you to jump into the learning experiences of this book ... get your Amateur Radio license ... and truly travel the road "from 5 watts ... to ... 1000 watts!

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®Copyright 1974, Radio Shack, A Tandy Corporation Company, Fort Worth, Texas 76109 PRINTED IN U.S.A.

2

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)IOOB 3H1 1fO8d

USING THE "PROGRAMMED PROGRESS COVER"

Below is a sample of the "three -column" approach used in this book. An example is given

of how the "cover" is used to "hide the answer" for a checkpoint until you've had time

to give your answer ... then, of course, you would simply move the cover downward to reveal the correct answer . . . and to check yourself. To make a Programmed

Progress Cover, just cut a thin piece of cardboard, about 11" wide by 2-4" high;

then use it as illustrated.

INFORMATION REFERENCE

Ohm's Law states that electrical current is I)Illl:(:'I'I,Y related to Voltage. 'Phis means that if the applied voltage In a given

electrical circuit is doubled . . . Ihen, the circuit current will double.

VOLTAGE SOURCE

"PROGRAMMED PROGRESS COVER"

MOVE COVER DOWN AS YOU PROGRESS!

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CHECKPOINTS

CHECKPOINT:

If the voltage input In the circuit shown in column 2 were tripled ... what would happen Io the circuit current.?

ANSWERS:

ll would triple! Note: "increase" would also be a good answer, but "triple" is a more precise answer.

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5

Chapter 1

LEARNING THE INTERNATIONAL MORSE CODE

A NEW LANGUAGE

Radiotelegraph code is a different language ... but it's a universal language! Fortunately, code

is not as difficult to learn as other languages. Thank goodness, there's no parts of speech, idioms, or conjugation of verbs to worry about. It really boils down to simply learning some

"sound patterns."

Because it is necessary to learn to identify "sound patterns" ... THE ONLY WAY TO PROP- ERLY LEARN CODE IS BY SOUND . . . AND BY SOUND ONLY! DON'T MEMORIZE WRITTEN CODE CHARTS!

For this reason we highly recommend that you practice the sounds AT EVERY OPPOR- TUNITY by ...

INFORMATION

SIMULATING THE SOUNDS OF CODE Before we begin to show you these "patterns in sound" mentioned earlier-let's look at a simple method of simulating code -like sounds.

Incidentally, we want to emphasize that you should learn sound patterns ... that is, complete character sounds, rather than individual dots or dashes. In fact it is very interesting to note that as one becomes more and more proficient in code, he not only hears complete characters (letters, numbers or punctuation marks), hut, whole -word sound patterns. Really expert code men can hear patterns far in excess of single words. This last level of proficiency is really not needed for passing Amateur License exams, but many hams do like code so much that they achieve this by on -the -air experience.

For the time being, let's learn how we simulate dots and dashes.

A series of dots can be simulated by voice by a rapid repetition of the syllable "di" ... pronounced like the first part of the word did. This series of "di's" should all be run together in a staccato string. Try saying "di-di-di-dit." Notice that the last syllable has a "t" attached, denoting the end of the series. Try a string of 10, and keep trying until you can say them smoothly run together with a staccato, even repetition of each syllable. Don't get discouraged, you'll have a "trained tongue" with just a little practice.

Try some more "di's" - "di-di-di-di-di-di-drt". Don't worry about sounding silly-this is one of the best ways to get used to the "sound of code". Practice some more-OUT LOUD!

6

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INFORMATION To help you gel started. here are some code exercises for you. Try lo accent the dahs ... allow 3 units of time between characters ... and 7 units of lime between "groups''.

If you can gel a frien(I \alto can "wart'. these letters anti groups to you; either by the di-dah method. or with a code practice oscillator, after you have thoroughly learned the letters by "didahing'' to yourself. you will find it quite helpful as receiving practice.

For most people. receiving code takes much more initial practice than sending code. We'll teach you to send ;i little later. Right now. concentrate on learning Io hear the sound patterns for each character . . . so no matter xvhere it "appears" in a group of code characters . . . you automatically say . or write down the correct letter when you hear its sound pattern.

REFERENCE & DATA Initial Exercise: (sound it . . . then say the leper-out loud.)

dah T; dah T; dah T; dah T; dah T; dah 'I'; dah T; dah T: dah T; dah T.

di-di-di-dit H; di-di-di-dit H; di-di-di-dit H; di-di-di-dit H; di-di-di-dit H; di-di-di-dit I -I; di-di-di-dit I -I; di-di-di-dit H: di-di-di-dit II; di-di-di-dil I -I.

dit E; dit E: dit E; dit E; dit dit E: dit E; dit I?; dit I?; dit E.

di-dah-dii N; di-dah-dit R: di-dah-dii R; di-dah-dit N: di-dah-dit R: di-dah-dil N: di-dah-dit R: di-dah-dil It; di-dah-dit R: di-dah-dil R.

dah-dah-dah O: dah-dah-dah O; dah-dah-dah O: dah-dah-dah O; dah-dah-dah O;

dah-dah-dah O: dah-dah-dah O; dah-dah-dah O; dah-dah-dah O: dah-dah-dah O.

Second Exercise: (say the sound . . . then the letter. etc.)

dah dah 'I': di-dah-dil N: di-dah-dil R: dit I?; dil E:

di-dah-dil R; dab T; di-dah-dit R; dil E: dah 'I'; dil E: dit E: dah-dah-dah O: dah-dah-dah O: di-di-di-dii H: di-di-di-dii I -I; dah-dah-dah O; di-di-di-dil II; dah T; di-di-di-dil 1-I; dan-dah-dah O: di-dah-dil N; di-dah-dii N: di-di-di-dii II; (lit I;: dil E; di-dah-dit R; dab 'I': di-dah-dil N: dah-dah-dah O; di-di-di-dil H; dit K.

CHECKPOINTS

CHECKPOINT

Letter Groups Exercise: Say the sound you read for each character . . . then, write down the letter you think it is, and when you've finished writing down the groups of letters . . . check yourself using the "Instant Feedback Cover" as appropriate.

1st:

2nd:

3rd:

4th:

5th:

6th:

7th:

8th:

9th:

1uth:

dit dab di-dah-dit dah-dah-dah dil di-di-di-dit di-di-di-dit dil dah dah-dah-dah dit di-dah-dit dah-dah-dah dab di-di-di-dil dah-dah-dah di-dah-dit dit di-di-di-dit dab dit dab di-dah-dit di-dah-dit dah dah-dah-dah dit di-di-di-dit di-di-di-dit dab dab dab dah-dah-dah dah-dah-dah dit di-dah-dit dah di-di-di-dit dah-dah-dah di-dah-dit dah di-di-di-dit dit di-dah-dit dah-dah-dah di-di-di-dit di-dah-dii dah-dah-dah dii dit

ANSWERS: 1s1: ETROE

2nd: HHETO 3rd: ERO')'H 4th: OREHT 5th: ETRRT

6th: OEHHT 7th: TTOOF. 8th: RTHOR 9th: THERO

10th: HROEE

8

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riai e Iu!lrin(1!uetti lo sa!ur.ya.nu a111 1111)(111 ñ.1.111N1

1,11011 nori I11111 .ía11)(1 si! lulucl sn(1 111 T1'1 u1 -Á.1>I e litlisn ua\a lnntll1.11 .1s1a.1.1$ a(Ll-,Yulpuas 11 no, amtY, pm: pu(11)du a)u)nllos sttll u.1nl sasmdaxa .u1.<1-1fu!n!))a.1 e s! tulue.nu si! Limo!) Iu11!.1.\i u,(11l ptinns at11 Iu1.i11.n1 s1t11 ltassan a.\ 1a1 ñ11u1 nori sv

s1u.lsa.t(la.1 pinos .ull .1.(Ia11.ietl) ,(yl lo yu!tIl ñ11r.)!Ieulolne nuri . .1a1)11.letla 1!

.lo punus all Y,ui.n!a11 u,(y.tt 1111. WO 1:11!.111 01 51 ;tu1t11 1.111:t0 .111 31a . 1a111 .1111 111 .111110 01 .011 .u( I1,10.11 11111 01 >1)11(1 .1t11 t110.11 Y,U1)I.1unt Ail 1111-111:1 1111 . . .11dun:xa .ío,1 unrli11 .(.1I 10.11! asla,laxa aa11))1.t(i ay1 aYuel.nua.1 uI 1111.1111 .111 1111)1) ))11))1.1(1 a.nuu p.(.1:1 !pis nori Iaa.l no. II i.(s1a.1a.1 sclno.l,) dal l('I.. I5.111 .inu, uu up nori in!) A1oII

INFORMATION Another practice hint (if you are fortunate enough to own a shortwave receiver) is to listen to a "code station" and see if you can pick out the first group characters you have learned. If you are careful to choose a station that is sending good "clean" code (proper spacing and timing), you will find this great sport and excellent practice. Incidentally, if you are considering buying a SW receiver, you might go down to the nearest Radio Shack store anti look over the variety they have. Generally you will find that they have an economical receiver that will be excellent for learning code and as a first ham station receiver'. If this is not practical at this time, perhaps you can find a friend who has a receiver that you can listen to.

We'll say it again . . the secret of learning code is practice, practice, practice.

If you've got the first group of letters clown so there is no difficulty recognizing their sound patterns-then proceed with the next group we'll introduce you to now. If you still need practice on the first group-then "recycle" yourself as required.

All right! Let's try a "letter groups" exercise for this second group of characters.

REFERENCE & DATA

SECOND GROUP OF RANDOM CHARACTERS

dah-dah = M dah-dit = N di-dah-di-dit = I. di-di-dah-dit = F di-dah = A di-dit = I

dah-di-dah-dit = C

Note: Remember lo practice each character over and over saying the sound first, then writing and/or pronouncing the letter until you know its sound pattern . . . then move on to the next one. To refresh your memory on how this is done, refer back to the initial exercise for the "FIRST CROUP OF RANDOM CHARAC'T'ERS" on page 7.

CHECKPOINTS

CHECKPOINT

Say the sound you read for each character . . then write down the letter you think it is. When yuu'\e finished writing down the groups of letters . . . check yourself by using the "Instant Feedback Cover" as appropriate.

LETTER GROUPS EXERCISE 1st: dah-dah dah-dah dah-dit

dah-dit di-dah-di-dit 2nd: di-di-dah-dit di-di-dah-dit

di-dah-di-dit di-dah di-dit 3rd: dah-di-dah-dit dah-di-dah-dit

di-dah di-dah di-dit 4th: di-dah-di-dit di-dah-di-dit

dah-dit di-dah dah-dit 5th: dah-di-dah-dit dah-di-dah-dit

dah-di-dah-dit dah-di-dah-dit dah-di-dah-dit

6th: di-dit di-dit di-di-dah-dit di-dah-di-dit dah-dit

7th: dah-dah di-dah dah-di-dah-dit di-dit di-dah-di-dit

10

11

I!P'!p-4eP-!P I!P-'p-yeP-!p yep -yep -yep pp -yep -q3

pp-yep-lp yep -up I!p pp-yep-lp-w 1!p 1p 1!p -yep -p 1!p-tplplp tip :418

yep yep -!P 1!P I!p'!P'!P'!P I!P-yep-'P l!P

l!P-!P-!P-!P yep yep -!p pp'4eP'!P-!p pp'4ep-!p 1!P

I!p-'p-'p-!p yep yep -yep -yep 4ep-L ep :41G pp -yep yep -yep -yep yep -yep -yep yep

pp'4eP-!P yep -yep -yep 1!p-4ep'!p 1!P -!P I!P 1!P'!P !P'!P yep :419

I!P 4eP'4eP 4eP-!P !p'4ep l!P'!P'4eP-!P up'11-yeP'!P 4eP'!P 4eP

I!P yep -yep yep -yep -yep pp -!p -'p -'p :418 pp -!p -yep -!p I!p I!p I!p-yep-!p-lp

1!P I!P-4eP iP up I!P'!P'!P !P yep 1p 4ep'4ep yep -yep -yep 1!p -yep -!p :4117

yep I!P'!P 1!P -!P -'P -!P

I!P yep yep -!p pp -!P -yep -!p yep 4ep-!3 yep -yep

4ep 4eP-!P up-4eP-!p 4eP-4eP 4eP-!P up'4eP'!P-yeP

1!P -yep yep -!P up'yep-!p-yep :pul I!p'4ep yep-ip yep -yep

4ep yep -!P pp'4eP-!P-!p I!P 1!P -!P -!P -!P yep :Isi

3sI0a3x3 saaonn s11O3Nt11130s1w

1NIOd)103H0

WDHNI :4101 33D33 :41s WNdNW :416 NHN'l1 :416

113V:1 : 4143 1b'HDD :ptE 1DHW :41L 'VW :pul

Mldll :419 'iNNLNW :1St :Sli3MSNt1

yep'4ep pp'yep-!P'4ep yep !p I!p yep 111-!1) :4101

yep -yep pp -yep 4ep-!p pp -yep yep -yep :416

1!P'!I)-4e13'!P !p -!p lup-yep-lp-yep 4eP-!P I!3 -4e3 -!P -!P :419

{; utunlua u! asl:l.r,lxa swam sn(1autrlla:lslul au 1 Á.i1 '11au.nral s.ua11111 ,lo sdno.tll omI 1s.tlj ay l amey n(1A Ieyl 111alnuuo:1 a,utr noA ua4M t11a41 1ali.tn 1.00A1 nnA !Is a:1ll:nr.td .to] 7 huir 1 sdnn.ili liu!uuluto:l Á.tl ua41 .. 11(11111 ut Ilam (Ino.t:i Ituu:ras a41 amry nu,i :Luis :.10 nuA uatlM tlnlLtil SiI :yl .n11 It11) nnÁ se 1:11ryltao:l Iuelstll., ;nuliu!sn a:1il:ur.ul liutituas lawliutAia:rl.t aAtli Innt 01 axed uu Yu!Is!I ..c1(101 -I:) (IN(YMS.. ay1 n1 1:11r4 l.taA:N r.nli 1! 101) MoII '>I.O

INFORMATION

How did your first encounter with "real words" go? As you can see, with a little ingenuity, you can develop other word lists using the letters from the first two groups you've learned for further practice.

Let's move on to our third group of random letters now.

Time to try a "letter groups" exercise for this third group of characters.

12

REFERENCE & DATA

THIRD GROUP OF RANDOM CHARACTERS

di-di-dit = S di-dah-dah = W dah-di-di-dit = 13

dah-di-dit = I) dah-dah-dit = G dah-di-dah-dah = Y dah-dah-di-dah = Q

Note: Remember to say the sound (out loud), then the letter-until you know its sound pattern . . . then move on to the next . .

and so on.

CHECKPOINTS 9th: di-di-dah-dit di-dah di-dit dah

di-di-di-dit dah-di-dah-dit di-dah-di-dit dit di-dah di-dah-dit di-dah-dit di-dah dah dit

10th: dah-dah dit di-dah dah-dit di-dah-di-dit di-dah dah dit di-dah-dit dah-di-dah-dit di-dah-dit dit di-dah dah dit

ANSWERS: 1st: THE FAT MAN

2nd: CAN CAM RAT 3rd: MAT LATE HIT 4th: 5th: 6th: 7th: 8th: 9th:

10th:

ROME THERE FEEL HOME TALL NAME THEIR OR MOON MOTHER FATHER HEAT THREE FEAR ROLL FAITH CLEAR RATE MEAN LATER CREATE

CHECKPOINT

Use the usual procedure here! That is: say the sound . . . then write clown the letter you think it is. After completing the exercise. check yourself with the "Instant Feedback Cover".

THIRD LETTER GROUPS EXERCISE

1st: di-di-dit di-dah-dah di-di-dit dah-dah-dit di-di-dit

2nd: di-dah-dah dah-dah-dit di-dah-dah di-di-dit dah-di-di-dit

3rd: dah-di-di-dit dah-di-dit dah-di-di-dit dah-di-dit dah-di-di-dit

4th: dah-di-dit dah-di-di-dit dah-di-dah-dah dah-dah-di-dah dah-di-dah-dah

51 h: di-dah-dah dah-dah-dit dah-dah-dit di-dah-dah dah-dah-di-dah

Cl

¡1! le )ali slal put! 'uwnlo:) prly) ay) 01

paa:wd ).)yeyd)e ay) roJ turn)) ol paau noA srallal jo dnorW Ise! ayi un as!:).iaxa sdno.ix rallal e .ioJ Apea] a.i no,k ¡suo!leln)erWuoa

¡paa:)ord uayl llaM dno.i1 ylJnoJ s)yi weal s,lal -ND magi Mou)I noA l!lun asayl an!lae.)d :aloN

Z = 1!P-!P-4eP-4ep X = 4'P-!P-!P-4eP

A = 4eP-'p-!P-!p I7 = 4eP1P1P

d = 1!p-yep-4ep1p X = 4eP-iP-4eP

= 4e13-4EP-4PP-!P

Sa313vavH3 woaNva 3o dnoaJ H1ano3

ADsMS :4101 bMwM :41s BbASD :416 AÓABu :4i6 AuDsu :418 tunas TrE aSbMA :416 tSMDM :puz ADatS :419 SaSMS :1s1

:Sa3MSNV

4e13-4eP1P-4eP 1!p -yep -yep I!3-!3-!p-4ep

yep-yep-lp 1!P -!P -!P :4101 pp-LP-!P'Lipp

4ep-!p-4ep-4ep yep-4ep-!p'4ep 1!p -!p -1p 14) -yep -yep :416

4ep-yep-!P-4eP I!P-!P-4BP 1!P-4eP-4eP

I!p-IP-IP-yep 1!P -!P -yep :419 1!p -!p -yep

pp -'P -'P 4ep-!P-4eP-4"P yeP-4ep-10 yep-4ep-)p..yep :416

yep-4ep-!p-4ep 1!P-yep-4eP pp -!p -yep

I!P-!P-!p-yep 1!p -431p :919

dnaiá ay) Jo.' sasi:).)axa a.)!Iaerd ay) oI paa:)orcl 'dnorli ayI U! srailal ay) Ile 1lu!uaeal .ia)Jd dnoil ay) u! sralal a4) Ile n)u)I nori )llun 'up os pue dnor;l ay) u! auo Ixau ay) o) uo anon) 'uay) 'Ham I! Xu!ureal ra1JH .r)I:)e.ir.y:) ley) .ioJ uralled punos ay) Mou)I noA l!lun :saw!) Auow siyl liUiloaclar puo raila) ay) xu!:)unouodd ro/pue UMop ñUIIWM uayl raloe.)ey:) ayi .)oJ punos ay) lu!ries sueaw s!y,I, sdnorU railaea ay) roJ aney noA se a.mpanord awes ay) as11 layeydle ay) .ioJ s.)allal jo dnorW Ise) ay) Uu!ureal Jo )Ise) ay) Ie la;i s,lal 'os

apo:) Wu!ureal Jo ired llna!J.IIp 1sUu1 ay) payslldwo:)ae AI)ear ane9 noA . uoiyseJ wopuer u ui wayl reay noA uayM -layeydle ayI u! sial la' ay1 Ile .ioJ suralled punos ay) Moui noA uayM layeyd)e ay) alaldwo:) II!M yo!yM 'dnorx ylrnoJ ay) .ioJ ripear Mou are noA uMop s.iallal lo sdnorl aaryl Isr!) ay) aney noA 11 ¡learg

INFORMATION

Ilope you did well on this exercise. You have now learned the twenty six letters in the alphabet. just for good measure, we suggest that you try the sending and receiving practice using the "Instant Feedback Cover" technique on letter groups 3 and 4, just as you did for groups 1 and 2. After you are sure you have all twenty six letter sound patterns pretty svell in mind, we want you to try the "mixed code groups" practice in the exercise shown in the third column. When you're ready ... give it a try.

14

REFERENCE & DATA CHECKPOINTS

CHECKPOINT FOURTH LETTER GROUPS

EXERCISE 1st: di-dah-dah-dah di-dah-dah-dit

di-dah-dah-dit di-dah-dah-dah dah-di-dah

2nd: dah-di-dah di-dah-dah-dit dah-di-di-dah dah-di-di-dah di-dah-dah-dit

3rd: di-dah-dah-dah di-dah-dah-dit dah-di-dah dah-di-di-dah dah-di-dah

4th: di-di-dah di-di-di-dah di-di-dah di-di-di-dah di-di-dah

5th: di-di-di-dah di-dah-dah-dit dah-di-dah di-dah-dah-dah di-di-dah

6th: di-di-di-dah di-di-dah dah-dah-di-dit dah-dah-di-dit dah-dah-di-dit

7th: dah-dah-di-dit dah-di-di-dah dah-di-di-dah dah-dah-di-dit di-dah-dah-dah

8th: di-dah-dah-dit dah-di-dah di-di-di-dah di-dah-dah-dit dah-di-dah

9th: di-dah-dah-dit dah-dah-di-dit di-di-dah di-di-dah di-dah-dah-dah

10th: dah-di-dah di-dah-dah-dit di-di-dah di-di-di-dah dah-di-di-dah

ANSWERS: 1st: JPPJK 6th: VUZZZ

2nd: KPXXP 7th: ZXXZJ 3rd: JPKXK 8th: PKVPK 4th: UVUVU 9th: PZUUJ 5th: VPKJU 10th: KPUVX

.LHSI3 :4191 333O.L :4151

NIN/1f1V :4161 XM/1fl.L :41E1 DBVZA :41ZL IHOd3 :41LL

NW DII :4101 S2IbdO :416

aDflNZ :418 AASdIN :41L Í9aHfl :419 adH[4 :415

Nd2i.L!\ :416 XZAMII :p1£ Sbowx :puZ

IO33H :lsl :Sti3MSNV

4eP I!P-!P-!P-!P I!P-'P-!P 1!P -!P 1!P 1!P 1!13-40P-IP-4eP I!P-4eP-!p-4eP

yep -yep -yep 4eP

1!P-4eP 4eP-4eP 4eP-iP-!13-!p 4ep-113-!P 4eP1P

49)-!P-!P-4eP 4"P-4eP-!P 4eP-!P-!P-!P 4eP-!P-!P 4eP

1!p-4ep-!p-4ep 11P-!p-1p-4eP 4eP-!P

1!p-!P-4eP-4ep 4ep-4ep1p-4ep 1!p -!p /!p -!P -!P -'P

l!P-4e3-4eP 11P-4eP-!P-!P 1!P PP -4°13 4eP-11°P l!P-!P-4eP-!P

4eP-!P-9eP 4eP-4ep-4eP-IP I!P-!P-!P

I!P-4 eP-!I) 4eP-!P-4eP-4 eP 1!P-4eP-4eP-!p 4RP-4BP-4"P

1!p-!p-4ep 1!13-4eP-!p-4eP 1!P-!P1P-4RP

4e131P 1!P-!p-4e13-4eP 4eP-4RP1P-4IP

4eP-!p-!p-!P l!P-'P-!P 1!p-4ep-49)-!p 4e13-4eP

4eP-4°P-4eP-IP 1!P-4eP-4eP 1!P-!P-4BP 4eP1P 1!P-!P1P-4eP

1!p-!p-4ep 1!P-4RP-1P-!P 1!P -!P -'P -!P

4eP-4BP-4IP-!P 1!P-!P-4eP-!P 1!P-4eP 1!P-4e3-4eP-!p

I!P-4eP1P 4eP 4"P-!P1P-!P 4eP-!P-!P-4eP

1!P-!p-4BP-4RP 4e13-4eP1P-4eP 9PP-4°P-!P 4eP1P-!P

1!P -!p -!P 4eP-!P-4eP-11eP 4eP-4eP-4eP

4e13-4eP 4eP-!P-4eP l!P-!P 1!P-4eP-4eP

1!P 1!P-4e13-!P-4eP 4eP1P

:14191

:4191

:4161

:41EL

:41Z1

:411 L

:4101

:416

:419

:411

:419

:415

:4 ft

:puZ

:1st

3SI31:13X3 Sdf1O1:19 3a03 a3XI1A1 138VHd1V 1t11O1

0101S )IoeyS O!pe}J 1e301 JnoA le Olqel!e+1e s,ll enoge peleAlsnll! pioo0l:i 01)03 e4l se Lions me Bu!weel e esn ol noA e6in em Jellel 43e0 ez!u6000u pinoys noA pue epoo;o spunos e4l LiIIM peo en,no/l MoN

LUL UU'hRlUn ulur, 1I11IN hiSIOIlllldl SMOSS3101

35Hl03 9:111Wf)38 3003 ~ bVlM1VNN31N1. v> >r

INFORMATION

How did it go? You should have detected from this exercise the letters you are weak on. You should, of course, work on your weak ones until you have them as well as the others. Let's say again, the secret is PRAC'T'ICE! PRAC'T'ICE! PRAC'T'ICE! "Off the air" practice in receiving is really the best . . . providing you listen to a station sending "clean" code. The best vvay Io do this at this point in your learning process is to listen to some of station W1AW's code practice transmissions.

If you don't have a receiver that will receive these frequencies you might want to run down to your nearest Radio Shack and look over their selection. As we said before . . . the hest practice in the world is listening to the sounds . . . and listening to them being well sent . . . so get a receiver, or the use of one if al all possible. Take advantage of the premium practice available from W IAW. You'll find that you will progress much faster in code if you'll do this. Good luck!

As you listen to radio stations sending code, you will discover that they send other characters than just the 26 letters in the alphabet. Pur that reason, let's move into the area of learning the numbers and important punctuation. By the time you learn these if you have been faithfully practicing . . . you should be ready In take the Novice Class License Code lest.

Here's the numbers study list. See if you can determine the unique pattern of sounds that were developed for communicating the numbers via code.

Did you catch the logical pattern that was developed for the numbers? WATCH OUT 't'I-IOl1Gl-l! The temptation is going to be to "count" the dits or dabs . . . then try and think out which of the numbers it is. You must learn to hear the TO'T'AL CHARAC'I'I;R SOUND as one unique sound you recognize, and do no start counting the elements which make up a character.


W1AW CODE PRACTICE SCHEDULE

Practice is transmitted three times a day during the weekdays and twice a day on the weekends. On weekdays, the times are 9:00 A.M. NS'I', 7:30 P.M. EST and 9:30 P.M. EST, and on weekends the P.M. schedules are maintained. For the beginner, the best times for listening are: 9:00 A.M. Monday. Wednesday. and Friday and 9:30 P.M. on Sundays, Tuesdays, Thursdays, and Satur- days. At these times W1AW sends practice at speeds of 5, 7'/2, 10, 13, 20 and 25 WPM (words per minute). As you progress you will find that the 7:30 schedule EVERY night is a good one as they send al 10. 13 and 15 WPM for practice purposes. All these transmissions are made simultaneously on the following frequencies: 1805, 3580, 7080, 14,080, 21,080, 28,080, 50,080 and 145,588 kHz. (If your receiver dial is calibrated in Megacycles, Mc., or Megahertz, MHz., simply put the decimal point in the above named frequencies BACK three places from the right. Poi' example 14,080 kHz = 14.08 Mc. or MHz.)

NUMBERS

di-dah-dah-dah-dah = 1

di-di-dah-dah-dah = 2 di-di-di-dah-dah = 3 di-di-di-di-dah = 4 di-di-di-di-dil = 5

dah-di-di-di-dit = 6 dah-dah-di-di-dit = 7

dah-dah-dah-di-dit = 8 dah-dah-dah-dah-dit = 9 dah-dah-dah-dah-dah = (this is the way

we write zero tu avoid confusion with the letter 0)

16

Ll

IfeMr\Ue way) MOM! O) paau IIno/f ssetJ leJauaJ JOJ

a:)uls ÁeM at1 JO inn watll 'al pue 'Wall UJeal pUe l)eelll'. nti 01 hilly 1.u~ I! OS ' Isa1 ato JO uO!IJOII JNIQNBS at) xu!.nlp uo!I -enl:lund puas 01 paJIl1I)aJ al if nu! noif ',1.1111

u011nniutind anlaaa.l 01 pa.I!nbaJ al IOU

II!M nOrf Isal iiU!n!aDa.I awn) asuaall SSV IJ íIJI!\ON 1)41 Un into JanaMOy 'iftlJOMaloti SI II IfeM Il:nsn ayI asayl a:l!oaeJd :aloN

OI.IeUI 'tints leuolie!p) 2IV11

NOI,I.3V21.4 = I!P-IIPP-!P'!1)-4PP UnISSIwSUnJ1

JO) >INOM :IO (IN:I = 4eP1P-IIPP-!P1P1P

)2IV Sl'. t1:)Il1.IM -SOW !I:111105)

:)VSS:I :10 (IN;1 = I!P-4eP-!P-4PP-!1)

),LÚ SP. UaIIIJM MU] I iaUlI1S)

I I5V(I :I'I1t110(1 = 4eP-!P-!P-!P-4PP

>2IVIN NOI.LS:11 = I!3-!P-4eP-40P-!P-!1)

VW1N0J = 4PP-4eP-!P-!P-4PP-4PP (I0I21:IcI = 4PP1P-40P-!13-4PIP-!P

S3Jt/SS3W 3003 NI a3Sf1 JIINOWW00 N011vfllONfld

asuani o!peg JnaoewV IsJ!J ley) du!1)a1 o1 ÁeM Jno/f uo !lam 'auo slyl gpV UOÁ Jl Ja)deya apoJ asJoiN teuo!leu.lalui ay1 Wu!uJeai alp JOJ lno)iaa4J leu!d éU!n!aUaJ Jnolf sop Jap!suoJ uo!lenlaund pue 'sJagwnu 'sJa)Ial JO as!aJaxa

..JaylaWol IIP I! Ind., s!41 J,L ,.iJaylalol IIP ! Ind.. s,1a'I i>I.O

vOI oenlaund pasn flluanhe.IJ aiow ayo JO awns 1ulurnal 'sI ono puny in (1uI lxau ayI Uo paaaoJd '(IN(lOS A8 ,,.Ialndwo:) le:)!Woloty !u0/< Ui p.)liotr.Ie:) Ham sJagwnu ay' aney n0Á a.lns ale noñ uatM

weyl .IuJ punos ale!JdOJdde ayI atos!yM JO 'tes 'Uo!ysei wopueJ e UI s.laywnu ato lP Wu!)lool al!y.nn pun=. Jann,) )iiegpaaa oueosul ayl y1!M spunos ayl ñu!Jan03 ñy a:)!i:n'Jd ,liwpuas 01 paa:)uJd uayl sueaw I! noonuMop a1iJM uayo ')unos at Ifes aI Ilann sJa1:)eJey:) ayo Ile Mou)i nnA t!Iun 'IsJ!J a:)!o:)eJd xuinla:)a.I,. Jo a.Inpa:w.ui lensn ayl asf1

Code Receiving Checkout-Conglomerate of Straight Text, Mixed Ciphers, Numbers and Punctuation

Line 1-dah di-di-di-dit di-dit di-di-dit di-dit di-di-dit

Line 2-di-dah di-di-dah dah-dit di-dit dah-dah-di-dah di-di-dah dit

Line 3-dah-di-dah-dit dah-dah-dah dah-dit dah-dah-dit di-dah-di-dit

Line 4-dah-dah-dah dah-dah dit di-dah-dit di-dah dah di-dit

Line 5-dah-dah-dah dah-dit dah-dah-dah di-di-dah-dit

Line 6-di-dah-dah dah-dah-dah di-dah-dit dah-di-dit di-di-dit

Line 7-dah-dah-di-di-dah-dah dah-dah di-dit dah-di-di-dah

Line 8-dit dah-di-dit dah-di-dah-dit di-dit di-dah-dah-dit

Line 9-di-di-di-dit dit di-dah-dit di-di-dit dah-dah-di-di-dah-dah

Line 10-dah-dit di-di-dah dah-dah dah-di-di-dit dit di-dah-dit di-di-dit

Line 11-di-dah dah-dit dah-di-dit di-dah-dah-dit

Line 12-di-di-dab dah-dit dah-di-dah-dit dah di-di-dah di-dah

Line 13-dah di-dit dah-dah-dah dah-dit di-dah-di-dah-di-dah

Line 14-di-dit di-di-dah-dit dah-di-dah-dah dah-dah-dah di-di-dah

Line 15-dah-dah-dit dit dah dah-di-dah-dit dah-di-dah-dit dah-dah-dit

Line 16-dit dah di-dah-dah-dah-dah dah-dah-dah-dah-dah dah-dah-dah-dah-dah

Line 17-di-dah-dah-dit dit di-dah-dit dah-di-dah-dit dit dah-dit dah

Line 18-dah-dah-di-di-dah-dah dah-di-dah-dah dah-dah-dah di-di-dah

18

Line 19-di-di-di-dit di-dah di-di-di-dah dit

Line 20-dah-di-dit dah-dah-dah dah-dit dit di-dah dah-dit

Line 21-dit dah-di-di-dah dah-di-dah-dit dit di-dah-di-dit di-dah-di-dit

Line 22-dit dah-dit dah di-dah-dah-dah dah-dah-dah dah-di-di-dit

Line 23-dah-dah-dah di-di-dah-dit di-dah-di-dit dit di-dah di-dah-dit

Line 24-dah-dit di-dit dah-dit dah-dah-dit dah di-di-di-dit dit

Line 25-dah-di-dah-dit dah-dah-dah dah-di-dit dit di-dah-di-dah-di-dah

Line 26-dah-dit dah-dah-dah di-dah-dah dah-dah-di-di-dah-dah

Line 27-di-dah di-dah-di-dit di-dah-di-dit dah-di-dah-dah dah-dah-dah

Line 28-di-di-dah dah-dit dit dit dah-di-dit dah dah-dah-dah

Line 29-dah-di-dit dah-dah-dah di-dit di-di-dit dah-di-dah-dit

Line 30-dah-dah-dah dah-dit dah di-dit dah-dit di-di-dah dit

Line 31-di-dah-dah-dit di-dah-dit di-dah dah-di-dah-dit dah di-dit

Line 32-dah-di-dah-dit di-dit dah-dit dah-dah-dit di-dit da h -di t

Line 33-dah-dah-dah di-dah-dit dah-di-dit dit di-dah-dit

Line 34-dah dah-dah-dah di-dit dah-dit dah-di-dah-dit di-dah-dit

Line 35-dit dí-dah di-di.-dít dit dah-di-dah-dah dah-dah-dah

Line 36-di-di-dah di-dah-dit di-di-dit di-dah-dah-dit dit dit dah-di-dit

Line 37-di-dah-di-dah-di-dah dah-dah-di-dah di-di-dah-dah-dah

Line 38-di-di-di-di-dit dah-dah-dah-dah-dah dah-di-di-dit dab

Line 39-dah-dah di-dah-dah-dit di-di-dit di-di-di-dah-dah dah-dah

Line 40-dah-di-di-di-dit di-dah-dah-dah dah-di-di-di-dah

Line 41-di-dah-di-dah-dit dah dah-dit dah-di-di-dah

Line 42-di-di-dah-dit dit di-dah-dit dab di-di-di-dit dit

Line 43-dah-dah-di-dah di-di-dit dah-dah-dah dah-dah-dah dah-dah

Line 44-di-dah-di-dah-di-dah dah-dah dah-di-dah-dah

Line 45-di-di-dit di-dit dah-dah-dit di-di-dit

Line 46-di-dah-dah dit di-dah-dit dit dah-dit dah-dah-dah dah

Line 47-di-dah-dah-dah-dah dah-di-di-dah-dit di-di-dah-dah-dah

Line 48-di-dah di-di-dit dah-dah-dit dah-dah-dah dah-dah-dah dah-di-dit

Line 49-di-dah di-di-dit dah-di-dah-dah dah-dah-dah di-di-dah

Line 50-di-dah-dit di-di-dil dah-dah-di-di-dah-dah

Line 51-di-dah-dah dit di-dah-dit dit dah di-di-di-dit dit

Line 52-dah-di-dah-dah di-di-dah-dah-di-dit

Line 53-di-di-di-dit dah-dah-dah di-dah-dah-dit dit

Line 54-dah-di-dah-dah dah-dah-dah di-di-dah

Line 55-dah-di-dah-dit dah-dah-di-dah dah-di-dah-dit dah-dah-di-dah

Line 56-dah-di-dah-dit dah-dah-di-dah dah-dah di-dah dah-di-dit dit

Line 57-dah-dah-dah-dah-dit dah-dah-dah-di-dit dah-dah-di-di-dit

Line 58-dah-dah-dah di-di-dah dah di-dah-dah dit di-dah-di-dit

Line 59-di-dah-di-dit dah-dah-dah dah-dit dah di-di-di-dit di-dit

Line 60-di-di-dit dah-di-dah-dit di-di-di-dit dit dah-di-dah-dit

Line 61-dah-di-dah dah-dah-dah di-di-dah dah di-dah-di-dah-di-dah

Line 62-di-dit di-di-dah-dit dah-di-dah-dah dah-dah-dah di-di-dah

Line 63-dah-dah di-dah dah-di-dit dit dah-dah-dah-dah-dit dah-dah-dah-dah-dah

Line 64-di-dah-dah-dit dit di-dah-dit dah-di-dah-dit dit dah-dit dah

Line 65-dah-dah-dah di-dah-dit di-dah-di-dit dit di-di-dit di-di-dit

Line 66-dah-dah-di-di-dah-dah dah-di-dah-dah dah-dah-dah di-di-dah

Line 67-di-dah-dah-dit di-dah-dit dah-dah-dah dah-di-di-dit di-dah

Line 68-dah-di-di-dit di-dah-di-dit dah-di-dah-dah

Line 69-dah-dit dit dit dah-di-dit dah-dah dah-dah-dah di-dah-dit dit

Line 70-dah-di-di-di-dah di-dah-dah-dit di-dah-dit di-dah

Line 71-dah-di-dah-dit dah di-dit dah-di-dah-dit dit

Line 72-dah-di-di-di-dah dah-dah-di-di-dah-dah

Line 73-dah-di-di-dah-dit di-di-di-dah-di-dah

6-

(mom jo pua '.lo) (uolsslwsuejl Jo pua) xS /=EL aun

.LH =ZL aun aala =LL aun

ejd ,lg =0L aun a.low paau =69 aun

AIg =89 aun egojd =G9 aun noA ' =g9 aun

ssal jo =Sg aun luan jad =179 aun 06 apptu =E9 auto

noA J1 =Zg aun Ino)J =19 aun

saya S =09 aun lyl uo l =6S aun lana )no =g! aun

L96 =LS aun apew b3 =99 aun

b3 b:) =SS aun noA =179 aun

adoH =ES aun ¿A =ZS aun

ay! ajan =LS aun

1!P-4eP-4BP-!P yep -yep -yep

yep )!P -!P -!P

sj =09 aun no/1 se =617 aun

pooS se =917 aun Z/1 =Lt' aun

lou ajaM =9t' aun s8!s =St aun

Ayv =bb aul3 ¡NO OSb =E6 aun

ayl jaj =Zt aun XN.L 21V =C6 aun

1,819 =017 aun ¡NE SdW =6E aun

.L13 OS =g£ auto Zb - =L£ aun

paads Jn =g£ aun oil asea =ge aun .loul ol =b£ aun

apio =E£ aun ul Sula =Z£ aun llpejd =Le aun

anulluo =0£ aun a sl op =6Z aun

ol paau n =9Z aun oil !le =zz aun 'moN =9Z aun

jalseJ ynnw 'Llanto leox jnoA ol 1a1 norf sdlay a:!l:tt:.id ut-pa.yeaus snll Ieyl pasi.td.ins atl II,nuA

ñliunl.tuddo /liana pue Aun le aailae.td paapul (Al.ladojd pallslynt .tu papunos jaylla) spunus apta olul pea.l noA sults ayI üutlelsue.tl Ay ;tupllem .lo jea e ut üulpl.t a.t nnA uaym 301.1.7H21d

i331.L3H21d s! apoa .loJ p.luM ,ssed jnu .laywautaN loll JO SulAdoa aje nnA ]l llal AllejattaY Upa noA

jantaaa.l e o! Yulualsll uaynn uan3 Jlas.lnoA >laaya ue:t noA ley sl 'janannuy 'a.tay aSeluenpe ay" .tanlaaat e U() .11e ayl .IJo.. 1iutAdoa pue apta 000D01 xulualsll se pool se Anead l.usl Inonlnayn sly! asjnun .10 ¿op noA mi) nnoH ilnon:tay:t autos SeM ley,l ¡MOM

apoa =SZ aun ayl Sulu =17z aun

.leal Jo =£Z aun gol lua =zz aun ilaaxa =LZ aun

ue auop =OZ aun aney =61 aun noA =91 aun

Iuaa jad =LL aun 00113 =91 aun

9 33 IaS =SL aun

noA J1 =iiL aun uoll =EL aun

enloun =ZL aun d pue =LL aul']

s.lagwnu =0L aun 's.tay =6 aug

dta pa =g aun xitu ' =L aun

sp.lo.M =9 aun Jo uo =S aun

Ile.lauto =17 aun =E aun

anblunlSuoa e =Z aulrl sl sltu, =L aun

Sli3MSNV

INFORMATION Now we want lo give you some ...

HINTS FOR SENDING CODE WITH A KEY We have delayed until now getting you involved with a code key because it is an absolute necessity that you KNOW "1'11E SOUND OF CORRECTLY SENT CODE before you can "mimic." these sounds with your own sending.

Three things are of paramount importance in regard to the "mechanics" of using a code key well. These are:

1. Proper adjustment of the key action. (spring tension, side play and up and down movement, i.e., how far the key knob moves from the unpressecl position to the fully depressed position).

2. Proper body posture and position.

3. A comfortable "fluid" movement of the wrist and hand while sending, rather than a tense and tiring use of he muscles involved.

Let's look at these items one al a time.

Note: Adjust side adjustment pivots for good contact alignment and so the key does not hind; but no so loose that the key can have "slop" sideways. Generally. the best way to do this is to tighten pivots until "snug", then back off just a "tad". Adjust the Spring tension adjustment for a moderately heavy tension. In other words

. . . don't have "hair trigger" action when pushing the key knob . . . but don't have tension so hard it will tire you quickly when repeatedly depressing the key.

Note: Generally. when beginning to learn use of the key, it is best to have a wider spacing between the contacts than later un, when you become more proficient. A good rule of thumb for a beginner might be about 1/16" contact gap. As you increase in proficiency. you can decrease this gap slightly.

20


SIDE ADJUSTMENT PIVOTS

SPRING TENSION SCREW

CIRCUIT SCREWS

GAP SPRING ADJUSTMENT SCREW

SPRING TENSION

CONTACTS

SHORTING LEVER

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-1AIo3 Si .LHH.L uolllsod e la8 oI s! way (eon au algel alp uo 8u! -isad s! wdeaJoJ alp pue 'gaJe iy8tls 'lednieu e two.' IS[JM pue pueq 'sJa8ulJ alp Imp aalloN ÁeM at{i Jo Ino Álasool palJna Ja8ulJ alill pue glJnoJ JnoÁ pue low! aq! Jo a8pa wadi ayi pJeMoI rflgelJo}woa SulisaJ Ja8ulJ alppiw pue xaput JnoÁ 'Áan ayi jo a8pa pal ayi uo qwnyi JnoÁ mum qoun ifan ay! dseJO algei alp uo mow Jno/f !sal oI wood aney noÁ os noÁ jo IuoJJ u! y8noua JeJ Áa>¡ ay! uo!ISod algei 8uliedado ayi yIlM adenbs pue Jleya JnoÁ u! iysiJdn I!S

uoli!sod pue adnlsod Ápog Jadodd seM is!! Jno uo ixaN

INFORMATION After you practice series of (fits and (lahs until you think they sound right, and you don't lire because of the wrong position or wrist action, then you are ready to move on to some specific character practice.

The ARRL booklet entitled. "Learning the Radiotelegraph Code" gives some good hints along these lines. In fact. you will find that many useful ideas are given in this booklet for both receiving and sending.

This booklet suggests the following groups for sending practice. The reason they are grouped as shown is obvious when you analyze the relationships of the character sounds included in each group.

Try these exercises until you can comfortably send all the characters.

Relieve it or not . . . saving the sounds for signs along the road, etc., can help you in learning to send. Make a habit of translating on sight, any letters ,you see, into code sounds. Even though this will not help you with the mechanics of "pushing the key down"

. . . it will help develop a proper sense of rhythm and timing which is imperative for good sending of code.

Again . . NOTHING WILL TAKE THE PLACE OF PRACTICE! Practice with the key until it is second nature to translate written letters, words and sentences into natural wrist, hand ant! arm movements to produce the appropriate sounds. Look at some of the sending practice aids shown in the next column. Get sonic of these aids and use them! Use them! Use them!

WHAT THIS CHAPTER SAYS .. .

IN "CAPSULE" FORM 1 You can only learn code by sound, if you want to learn it

properly.

2. You can only learn code by practice, practice, practice! 3. Learn to receive code and recognize good code before trying

to use a code key.

4. If at all possible . . . listen to good code (copy it) "off the air." Note some of the "receiving" system equipment Radio Shack carries shown in the next column. Also, refer back to page 16 for practice schedules from W IAW.

5. Practice at everN opportunity-translating signs. etc., into code sounds.

6. When sending code with as key. observe.: I. Proper key adjustment, 2. Proper body position and posture, 3. Comfortable gripping and moving of the key for a fluid sending movement.

7. Don't gel impatient and give. up! Code conies easier to some than others. but it will come to everyone who will continue to practice.

22

REFERENCE & DATA CHECKPOINTS Group 1: E, I. S, I -I and 5.

Group 2: T, M, O and 0 (remember this is the way we write zero)

Group 3: A, R, I., W, I, I and P

Group 4: U, F, 2, V. 3 and 4

Group 5: N, D, B, 6, 8, 9 and X

Group 6: G. Q, Z, 7, K, C and Y

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INFORMATION REFERENCE & DATA CHECKPOINTS

Radio Frequency voltage and current complete many many cycles in one second. In fact we designate the frequency of a radio wave as being so many cycles per second. If you rode your bicycle so that the wheel rotated 2 times in 1 second, we might say that frequency of rotation was 2 cycles per second.

Now, since Mr. Hertz, a German scientist, was instrumental in early knowledge about radio waves and their propagation, the scientific world has honored him by naming the "cycle" of electrical energy in his name. Nowadays it is correct to say our house current has a frequency of 60 Hertz, rather than saying 60 cycles per second ... and it means the same thing.

All right, if an RF wave has a frequency of 1 million Hertz; how long do you think one cycle takes? If you figured 1 millionth of a second . . . you were right! Since you were so smart on that one . . . try this one in your "biological computer". If Radio energy travels at a rate of speed of 300,000,000 meters per second . . . how far would our 1,000,000 Hertz signal gel in the lime it takes for one cycle? You should be thinking to yourself ... If one cycle takes one millionth of a second . . . and the energy travels at a rate of 300 million meters per second, then, in one millionth of a second the RF "signal" should have traveled one millionth of 30t) million meters . . . or 300 meters. This signal then might be called a 300 meter signal.

Now, hack to our 80 meter, 40 meter, etc. signals. When we say that the Novice Band frequencies between 3700-3750 thousand Hertz (kHz for kilo -Hertz) are in the 80 meter band . . . it is just an approximation of how far that frequency signal will travel in the time it takes for one cycle of that signal, traveling at the speed of light (300,000,000 meters/second) of course, this is not an exact statement, because one, and only one precise frequency will Irakel that exact distance during the time of one cycle (or Hertz) of itself. Anyway, you get the idea. The "40 meter band" are frequencies whose RF energy will travel about 40 meters in the time it takes for one cycle (or Hertz) of that energy.

This brings us to another new term ... wavelength. A wavelength is also the distance a wave will travel during the time of one cycle. Look at the formula in the next column which shows the relationship of frequency and wavelength.

Now, what this says to us is that the higher the frequency. the shorter the wavelength, and vice -versa. Try the problem in the third column to see if you have the idea.

24

RF VOLTAGE AND CURRENT "WAVEFORM"

BETTMANN ARCHIVE. INC.

HEINRICH HERTZ

Formula

Distance = Speed x Time

Example: Distance RF energy of 1,000,000 Hertz frequency travels during time for I Hertz

(Speed) x (Time) I I-Iz = 300,000,000 Meters/Sec x 0.000,001

Second = 300 Meters

Formula

Wavelength = 300.000,000 Frequency in Hertz

AS "F" (frequency) A (wavelength) 4.

Note: This type of relationship is often referred to as an inverse relationship; that is, if one factor increases, the other decreases, and vice -versa. Remember this term: inverse.

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INFORMATION Try this last practice exercise to see if you have the idea.

Did you make out all right? May we suggest that you don't memorize the conversion chart, but rather UNDERSTAND what is happening when you convert. In fact, let's make a general statement here regarding all the "theory" and formulas you'll be learning as you go through this book. It is always better to study a formula, or point of theory until you UNDERSTAND THE CONCEPT . . .

rather than using "rote memory" methods and depending upon your ability to recall needed information. If you really do understand the concept ... you can then "figure out" the formula, or conversion factor, etc., as required. If you understand that a Megahertz is 1000 times greater than a Kilohertz, you can figure out that 1

MHz must equal 1000 kHz, and so on.

Now that we have gotten some preliminary information out of the way, and you have a general idea what the Novice Bands mean in terms of frequency and wavelength, and know the general requirements for, and privileges of being a Novice . . let's get to the "nitty gritty", and begin learning some Electronic theory.

First, let's find out "what in the world the matter is"! By this poor bit of humor I simply want to say, let's lake a quick look at the atom . . . of which all matter is composed. The chair you are sitting on, the book you are reading, etc., are all made up of tiny particles called atoms. Now the atom is a little galaxy all its own. In the center is the greatest mass, called the nucleus. The nucleus is composed of particles called protons . . . which have a positive electrical charge, and neutrons which are neutral electrically.

26


CHECKPOINT Convert the following as required:

2 MHz =

1000 kHz= 1000 kHz =

1,540,000 Hz =

1,540,000 Hz =

2.546 MHz =

1.65 kHz =

ANSWERS: 2 MHz = 2,000 kHz 1000 kHz = 1 MHz 1000 kHz = 1,000,000 Hz 1,540,000 Hz = 1.54 MHz 1,540,000 Hz = 1,540 kHz 2.546 MHz = 2,546 kHz 1.65 kHz = 1,650 Hz

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Reminder . . . working on your code? Don't forget it. A Code Practice Oscil- lator like the one illustrated above is a great aid to learn code. Available at your Radio Shack store.

kHz MHz. Hz MHz kHz kHz Hertz (Hz)

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INFORMATION The fact Ihat we can by various forms of energy cause Ihis electric.:íl unbalance allows the production of electricity and various electrical phenomena. To "separate these charges in an atom (or group of atoms) requires work . . . and that is why we mentioned Ihr use of various forms of energy. The battery or cell utilizes chemical energy to separate charges and produce an excess of electrons ¿it its negative terminal and an excess of protons al its positive terminal. The generator. or alternator in our automobile uses mechanical energy to perform this tvork of "generating electricity' by moving charges. Other devices may use heal, or light energy. and so on.

(),I<.. lo \v dues all This relate to electricity and electronics! Let's start by seeing what causes some materials to be good electrical conductors, while others are poor conductors. but good insulators: and still others are not real good conductors or insulators. lint are classified as "semiconductors

You may have noticed in our previous sketches of atoms. Ihe

orbiting electrons \very not all in one ring. or shell (orbiting path). It is beyond the scope of this book to go into the chemical or atomic reasons for this, but we do want to touch on a couple of important farts regarding this.

The "electrical stability" of an atom is determined by this orbital ..layout" of electrons. The most intporl:ant group of electrons. for our consideration. are the Ines in Ihe. "ring" farthest from the nucleus . . . or the 'outer shell". For any :atom . . . Ihe "magic: number" of electrons that will cause complete electrical "stability.. is 8 in this outermost ring. All alums also have the stipulation Ihat no more than 2 electrons can appear in Ihe "innermost ring. Thus, the carbon atom we showed Nino is nil completely stable. or inert". since it has only 4 outer ring electrons. OI course, Ihe other 2 electrons appear in the inner ring. If a material has 4 electrons in its outer ring. it is categorized as a senticontluctur. If the material's atomic structure is such that there are fewer than 4 electrons in its puler ring. it is usually classed as a conductor: and if more than 4 electrons are in its muter shell, it is generally an insulator, or nun -conductor.

Look at the diagram of the copper atom. Notice that is has 29

protons in its nucleus: therefore must normally have 29 orbiting electrons. I)ue to the laws of nature, we find that the innermost ring is considered "filled" when it has two electrons. the next ring out takes 8 electrons to he "filled", the next ring 18, and the outer ring of any atom is filled when it has 8 electrons.

REFERENCE & DATA

--0--

COPPER ATOM

CHECKPOINTS

CHECKPOINT

Several forms of energy that can be used to "move electrons" and produce electricity are: and energy.

ANSWERS: Several forms of energy that can be used to "move electrons" and produce electricity are: (chemical), (mechanical), (heat) or (Ther- mal) and (light) energy.

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INFORMATION

Now, let's amplify and explain the terms resistance and current, and at the same time throw in one more important electrical term, electromotive force, abbreviated emf. We have already stated that an electrical current is an organized, progressive movement or "flow" of electrons. As you might guess.

in an electrical circuit it is possible to cause "electron flow". or

current to flow through the circuit by applying force, or energy which can cause the ''free" electrons to move from atom to atom

through the material. Also, as you might suspect . . the lower the resistance of the circuit, the more electrons will he moved

by a given force in this case, electromotive force (emf.)

These three electrical quantities or parameters are of utmost importance in discussing electrical circuits. YOU MUST LEARN THESE, and the laws of electrical circuits related to them in order to

obtain meaningful knowledge of electronics and Ham Radio. so let's

discuss these terms. a little more fully. EMF might be considered similar to pressure in a water system. Notice in the diagram. the source of water pressure is the pump.

In the electrical circuit we'll discuss, the source of electrical pressure will be the battery. (ust as water pressure has units of measure

"pounds of pressure", electrical pressure, or emf has a unit of measure called the volt, just as the amount of water "moved" by the pressure can be measured in gallons per minute, the amount of electrical current through the electrical circuit can be measured in amperes (number of electrons moved past a given point in

a unit of time). (An ampere = 6.28 x 10'" electrons past a given point in one second). In the "water circuit", the smaller the water pipes, the more resistance they offer to the flow of water, and therefore, the less water will flow, per unit time, with any given water pressure. In the

electrical circuit, the higher the resistance (that is, electrical resistance) of the circuit . . . the less the current (in amperes) that will flow for any given emf (or voltage). This relationship of current (electron flow), voltage (emf) and resistance was studied by a

scientist named George. Simon Ohm and he set forth these relationships in the now fatuous OHM'S LAW, which we will discuss shortly. First, though, review the concepts we've been talking about by carefully studying the illustrations in the adjacent column.

Try this checkpoint to make sure you're getting the idea.

REFERENCE & DATA

(FORCE) (RESISTANCE) WATER

PIPE :FAUCET

CONTAINER

WATER "CIRCUIT"

(FORCE) RESISTANCE

BATTERY

(FLOW)

ELECTRICAL CIRCUIT

CHECKPOINTS

CHECKPOINT

Electrical pressure is called , abbreviated

The unit of measure for this electrical quantity is the

Electrical current consists of the flow of through an electrical circuit.

The amount of current is determined by the number of flowing past

a given point in the circuit in a unit of time. The basic unit of measure of electrical current flow is the A current of 1 indicates a

flow of 6.28 x 10'" electrons flowing past

a given point in the circuit in a time of

ANSWERS: Electrical pressure is called (electromotive force), abbreviated (emf). The unit of measure for this electrical quantity is the (volt).

Electrical current consists of the flow of (electrons) through an electrical circuit. The amount of current is determined by the number of (electrons) flowing past a given point in the circuit in a unit of time. The basic unit of measure of electrical current flow is the (ampere). A current of 1 (ampere) indicates a flow of 6.28 x 10'" electrons flowing past a given point in the circuit in a time of (one second).

30

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INFORMATION REFERENCE & DATA CHECKPOINTS

How did you fare? (.earn these terms well, for Ihey are important!

O.K. Let's begin to see how these three important electrical parameters are related by learning OilTS LAW!

By experimentation in his laboratory, hlr. Ohm was able to define the interrelationships of the three electrical quantities we have been talking about.

As logic would tell us, he discovered that the amount of current through an electrical circuit was directly related to voltage . .

which means, for a given resistance circuit, the higher the applied emf, or voltage ... the higher the current, in amperes.

He also noted that current in an electrical circuit is inversely related to the resistance . . . which means, for a given emf applied to the circuit, the higher the resistance . . . the lower the current. This certainly makes sense, for when you think of emf as a force which is trying to move electrons (current) through a circuit against some electrical resistance . . . it is obvious . . .

the higher the resistance . . . the less the number of electrons Current (in amps)

moved with a given force applied. Stating the above in formula form, we have the relationships shown in the next column.

Let's restate that formula using the standard abbreviations . . . or "shorthand" letters used to represent these terms.

The shorthand way of writing emf is E

The short hand way of 1vriting current is I

The shorthand way of writing resistance is R I (in amperes) = E (in volts) divided by R

If it will help . . . think of E for emf; I for intensity of electron (in ohms) flow (or current) ... and R for resistance. Using these abbreviations, or the basic statement showing the relationships of I, E and R is E

shown in the next column. I R

Voltage (volts)

Resistance (ohms)

ANSWERS: 1. The resistance of a conductor is: low. 2. The resistance of an insulator is: high. 3. The unit of resistance is called the (ohm). 4. All fixed resistors used the same elec-

trical symbol. (true). 5 Potential difference between two points

is measured in (volts). 6. The terminal of a "voltage source"

where there is an excess of electrons would be called the (negative) terminal of the source.

7. One ohm of resistance will limit the current through an electrical circuit to one ampere of current when one (volt) of emf is applied.

8. The unit of current is the (ampere). 9. The unit of emf is the (volt).

10. The ohm is the unit of (resistance).

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INFORMATION

O.K. You've used all three forms of the Ohm's Law formula now ... but we want you to begin learning how to apply this knowledge in a practical way by giving you some problems using circuit "schematic diagrams" to convey information to you.

Before we do that, however, let's talk for just a minute about what "instruments" are used to measure the three important electrical parameters we've been discussing. The instruments used are

really quite obvious: To measure emf in volts use a voltmeter. For our purposes we will use the following symbol to represent a voltmeter "schematically":

To measure current in amperes, use an ammeter. For our purposes. we will use the following symbol to represent an ammeter "schematically":

Incidentally . . . many times we will be dealing in thousandths of an ampere rather than amperes. As we mentioned earlier . . .

the metric "prefix" for one thousandth of a unit is "milli". Therefore. we use the term milliampere to indicate one thousandth of an

ampere ... and we abbreviate this mA.

The symbol we'll use to show a millammeter ín a schematic diagram is:

To measure resistance, an ohmmeter is used. We'll use the symbol:

While we're on the subject of resistance and ohms . . . we should point out that in many electronic circuits, we'll be dealing with thousands of ohms, and sometimes millions of ohms; and, again, you will find the metric prefixes come in handy. Remember that the term kilo means 1000. Our earlier example was that a kilometer is 1000 meters. Therefore, a kilohm is 1000 ohms. The common letter symbol, or abbreviation for kilo is K. 'Thus, if I am talking about a resistor which has 10,000 ohms of resistance, it would be general practice to say a 10K ohm resistor. For designating millions, we use the prefix Mega. Hence, a 10 Meg (we leave off the

a) resistor has 10 million ohms of resistance.

Now, let's get "down to brass tacks" in becoming familiar with schematic diagrams and using Ohm's Law to analyze these. circuits.

Initially, let's define a circuit as being a voltage source and a

complete path for electron flow from one terminal of the source to the other. Observe the diagram in the next column.


NOTE: The Greek letter omega, 11, is often used to symbolize the word ohm (or ohms).

BATTERY

RESISTOR

AMMETER

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INFORMATION

Do you have those formulas well in mind? I hope so. You will use them the rest of your life as an Amateur or technician.

We've been speaking of moving electrons through a resistance by applying an electromotive force. As you can imagine . . . when this work is clone, heal is generated. Even though we are talking about moving things as microscopic as electrons, heal is generated. In fact, a measure of how much work has liven clone in a given unit of time is called power. Electrical power is measured in units called watts . . named in honor of the scientist who studied work, power and energy. For your purposes at this time, it is

only necessary that you learn that the amount of power dissipated by a circuit (like we've been studying) is related to the three electrical quantities-E, I and R. For the lime being, you'll only use one form of the power formulas, namely:

Try the checkpoint!

We have been dealing with some "pretty heavy" electronics fundamentals so far in this chapter. We felt that it was important to

begin to lay the "ground -work" and give you a good "foundation" upon which to build in later chapters. Because the Novice Class License test sloes not require "in depth" electronic theory knowledge, we'll just introduce you to terms and rules and regulations which you may encounter in the Novice Class examination. In other words, we will not go into detailed explanation at this lime, but simply will familiarize you with them for "exam" purposes. Later in the

book, many of these topics will be taken up in detail.

Most all electronic circuits are made op of a combination of some,

if not all, of the following six basic electronic components: Resistors, Capacitors, Inductors, Tubes, 'I ransistors and Semiconductor Diodes.

You should learn the symbols for these, and some of their basic functions. Let's go through the list, one at a time.

The resistor is used to limit current, and in some cases to provide a desired voltage drop across itself. You have already seen pictures of resistors and the proper schematic symbol, so let's move on

to the Capacitor.

The symbol for a Capacitor is:


P (power in watts) = E x I, where E is in volts and I is in amperes.

CHECKPOINT

If a transmitter circuit has an applied voltage of 100 volts and a circuit current of 0.75 amperes . . . what is the power "input" to the stage?

P = waits.

ANSWER: P=ExI P = 100 x 0.75 P = (75) watts

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INFORMATION The next component on our list is the Inductor, sometimes called a coil. The capacitor stores electrical energy in its "electric" field between plates of the capacitor. The inductor also stores energy in a field . . . but the inductor stores energy in the form of a "magnetic field", which we'll learn more about later.

Whereas the capacitor has the characteristic of "blocking" DC, and "passing" AC with little opposition . . . the inductor, or coil does just about the opposite. This means the coil allows DC to pass with little opposition, but tends to offer opposition to the passage of AC. In fact, the higher the "frequency" of the alternating current . . . the more the opposition to its passage through the inductor. You do remember our "old friend" frequency don't you? Recall that "house current" is probably at a frequency of 60 Hertz . . . and it is, of course, alternating current . . . abbreviated AC.

There are two basic types of coils that you need symbols for: namely, the "air core" type . . . a around a form with nothing but air in the middle iron -core type. Note the pictures of these in the

to know the coil wrapped

. . . and the next column.

The symbol for the air -core coil is:

The symbol for the iron -core coil is:

The unit of inductance is the henry, named in honor of another scientist.

Iron -core inductors are used in circuits which are operating at relatively low AC frequencies. For example, at power line frequency (60 Hz) up through what is called "audio" frequencies (from about 20 Hz - 20 kHz).

Air -core inductors are used at higher frequencies ... often called RF (radio frequency). Many circuits in your TV set may use these RF coils. Almost all electronic equipment requires power supplies of various sorts, where Iron -core inductors are used frequently. So, you see, inductors are very important electronics component.

Verify your learning by doing the checkpoint.


CHECKPOINT

1. A capacitor tends to block but to pass signals.

2. An inductor tends to allow to pass with little opposition, but to impede signals. The higher the frequency of signal, the (greater, smaller) will be the inductor's opposition to its passage.

3. The unit of capacitance is the 4. The unit of inductance is the 5. The schematic symbol for a capacitor

is:

6. The schematic symbol for an air -core inductor is:

7. The schematic symbol for an iron -core inductor is:

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INFORMATION

Now, what do those terms mean? For the time being, let's give you a brief definition of each function.

A rectifier changes AC voltages or currents to DC voltage or current. One example of the need for "rectification" is your radio or TV. If there weren't a rectifier and filter circuit In the power supply, you'd hear terrible 60 Hertz hum all the time. In a very simple statement ... a rectifier changes AC to DC.

To amplify means to make larger or build up. A tube used to amplify simply builds up an input signal to a larger output signal. For example, one might put in a signal to an amplifier that is only 1 volt . . . but the "amplified" output voltage could be 100 volts ... etc.

When a tube is used to generate an AC signal, it is often called an Oscillator. The tube's function here can simply be stated as changing the DC voltages into an AC signal of desired frequency-this becomes the output.

Most vacuum tubes have a filament . . . called a heater. The heater is used to heat up a tube element called the cathode. The job of the cathode is to "emit" electrons when it is heated. These electrons are then available to he "collected" by the tube's plate. The plate is the tube element which attracts the electrons which have been emitted by the cathode. The tube we have just described is called the Diode tube, because it has two active elements . . .

the cathode . . . which emits electrons, and the plate . . . which collects them. The prefix "di" means two. For example . . . when you "disect" something, you cut it into two parts.

Look at the schematic symbol for the tube we have described is in the next column.

We've mentioned several functions of vacuum tubes as being rectification, amplification and their use as oscillators. The diode tube we just discussed often finds application as a rectifier. We will discuss how this is accomplished later in the book.

A tremendous "giant -step" forward was made in the world of electronics when Dr. Lee DeForest introduced a third major element in the vacuum tube. This element was physically placed between the cathode and the plate, and this third element was called the control -grid. Instead of being a solid piece of metal, such as the cathode or plate, the control grid has evolved to be a thin "spiral" of wire.

Because the tube now had three active elements, it became known as the -Triode -tube. The prefix-tri-means 3. Many examples of this prefix being used to designate 3 are available, but perhaps a very common one would be the common toy the "tricycle", which has three wheels.


PLATE

HEATER] IICATHODE

DIODE TUBE

. . .

, .

Dr. Lee DeForest

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INFORMATION

You may have noticed that we have changed our style in giving the answers. This is to conserve space and allow us to give you more teaching in a given space.

Now that you have had a chance to get used to the format of the book, you can see how to easily refer hack to the CHECKPOINTS when checking your Answers.

In the General Class section, we will expand on the vacuum tube, and discuss some other tubes that have been derived from the basic diodes and triodes. For now, we need to move ahead and briefly mention the device which has revolutionized the electronics industry ... the transistor.

One of the tremendous advantages of the transistor is its small size. Also, because it doesn't need "heater" power, the associated circuitry is less bulky. Because of the low power requirements to operate transistors, most all the associated components have also been "miniaturized".

Transistors are frequently referred to as semiconductors, and as solid-state devices. They are called semiconductors because they are made of materials that "atomically" fall into the semiconductor category. The two most popular semiconductor materials for transistors are: germanium and silicon.

Both of these semiconductor materials have four "valence" electrons. (Remember our discussion of conductors, insulators, and semiconductors?) They are called solid-state devices because the various materials that make the active part of the transistor are all in one solid piece, as opposed to vacuum tubes, where the various elements are separated by space. etc.

For now, you should learn the symbols for the two basic types of transistors, and briefly, what is meant by "N" type material, and "P" type material.

REFERENCE & DATA

171

Transistors vs Tube (size comparison)

BASE

EMITTERO COLLECTOR

Pictorial Concept of Transistor Materials

"Sandwich" for "NPN" Transistor

3. Triode 4. Three 5.

CONTROL GRID

CHECKPOINTS

PLATE -TUBE ENVELOPE,

GLASS OR METAL

HEATER) IICATHODE

6. To emit electrons 7. To control how many of the emitted elec-

trons reach the plate. 8. To "collect" the emitted electrons.

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INFORMATION

Two other transistor terms you ought to know (but don't need to know the "theory"): Alpha and Beta. Transistors have the capability of amplification, as many tubes do. These terms Alpha and Beta are used to indicate the amount of "gain" a transistor has. Alpha is simply the comparison of the "collector" current to the "emitter" current. The value of alpha for most transistors ranges between 0.95 and 0.99.

Beta, on the other hand, is the relationship of the collector current to the base current. The value of beta for transistors can range from 10 or 20 up to many hundreds. Beta is a very meaningful term for the design engineer, and is term that you ought to at least have heard.

Checkpoint!


While we are on the subject of semiconductors, we should mention that semiconductor diodes have replaced many vacuum tube diodes in application. Just as you learned the symbol for the plate of the tube (collector of electrons emitted by the cathode), and the cathode (emitter of electrons), you should learn the symbol for the semiconductor diode. The anode does the same job for this diode that the plate does in the tube . . . and the cathode basically

CATHODE ANODE performs the function of the cathode in the vacuum tube diode. Diodes, either semiconductor, or tube, perform quite a number of useful electronic functions. We've already mentioned rectification (changing alternating current to direct current). Diodes also are used in receivers as detectors. You've heard the term AM radio. This refers to the method whereby the "intelligence" (sound) is put on the transmitted radio signal. A detector is often called a "demodulator", because the detector circuit (diode and several associated filter components) recovers the audio intelligence from the complex "modulated RF" signal, and sends this audio signal on to the audio amplifier circuitry in the receiver (to be built up to a level to drive a speaker or earphones).

Another function that diodes are sometimes used for is to help "modulate" an RF wave . . . in other words . . . as a "modulator". Just learn the term for now.

Still other functions are electronic switching and limiting or clipping. (Limiting the amplitude of a signal . . . or clipping off part of the signal).

SEMICONDUCTOR DIODE

CHECKPOINT

1. The ratio of collector current to emitter current is called the transistor's

2. The ratio of the collector current to the base current is called the transistor's

ANSWERS:

1. Alpha 2. Beta

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INFORMATION

O.K., let's move on to some other terms that might appear on the FCC test. It is not neccessary at this time to know all the detailed theory for each term, but you should be aware of the basic concepts of each term.

In Amateur Radio and in Commercial Radio, many transmitters have stages referred to as frequency multipliers . . . or simply multipliers. The function of a multiplier stage is to produce some multiple of the input signal frequency at its output. (See the illustration). In a practical sense, multipliers may produce useable output at 2, 3, or 4 times the input signal frequency. Efficiency factors prevent a single stage of "multiplication of frequency" to go much above 4 times. If more multiplication is needed, the output of one multiplier stage is fed to another multiplier stage as required. Incidentally, a multiple of any given frequency is called a harmonic. ri' For example, the 2nd harmonic of 100 Hz is 200 Hz; the 4th harmonic of 300 Hz is 1200 Hz ... and so on.


Another term related to transmitters is parasitics. You are probably familiar with the word parasite ... referring to the animal kingdom. Generally, this refers to something attached to "whatever" and drains off part of the resources of the main object to sustain itself. In the transmitter . . . a parasitic oscillation is one which is not at the desired frequency . . . and is often far removed in frequency. To sustain this "undesired" parasitic . . . some of the stage's useful energy is used.

Still another transmitter term is the word overmodulation.

Think back a little and you will remember that we mentioned the term modulation. Transmitters accomplish the feat of putting the audio "intelligence" on the transmitted wave by one of several methods. AM signals are those using Amplitude changes for doing this . . . hence, AM = Amplitude Modulation. 1'M signals use either frequency changes or phase changes to perform this function. In any case, overmodulation refers to a transmitter signal in which the audio signal modulates the radio frequency signal MORE THAN 100%. When this happens ... "spurious" (unwanted) new frequencies are generated . . . and the FCC frowns on this.

While on the subject of transmitters, let's quickly discuss what A-1 transmission means. This is the type transmission you will be using as a Novice Class operator. In essence ... it is "amplitude - modulated telegraphy" without use of audio . . . in other words "on -off" keying of an RF carrier. The "A" shows that it is Amplitude modulation; whereas the 1 inciates telegraphy by "on -off" keying.

Another term you might see, which is related to keying, is the term key -click filter. The purpose of the key -click filter is to reduce radiation of undesired, or spurious frequencies, which can be caused when the RF signal is turned "on and off" abruptly. The filler, in essence, smooths the beginning and ending of each "pulse" to reduce this effect.

46

INPUT

INPUT

01

FREQUENCY MULTIPLIER

MULTIPLIER

V V

OUTPUT r rl.11l-VV

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MULTIPLIER

OUTPUT

RF BURSTS (Keyed RF) WITHOUT KEY CLICK FILTER

i

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INFORMATION O.K., let's move on now and talk about some fairly common problems related to Amateur transmitters . . . and what precautions can be used to reduce or prevent them.

First, let's mention the term chirp as it related to c.w. (telegraph) transmissions. Chirp is a quick change in transmitter frequency which can occur as the transmission is keyed on and off. II is due usually to the oscillator being keyed, and sounds to the person at the receiving end of the transmission like a bird's chirping . . .

hence the term "chirp". Common preventative measures include: Don't use a circuit which will key the oscillator (key a stage after the oscillator); use an oscillator with good frequency stability; use a stage between the oscillator and final transmitter stage (called a buffer stage).

Another problem in some transmitters is the radiation of harmonic frequencies (undesired). To hell) reduce this possibility, the following measures should be used: 1. Be sure final stage is being operated with proper electrical parameters. 2. The transmitter should lw designed with tuned circuits which are selective. 3. Ilse a good coupling circuit between the final amplifier stage and the antenna. (One which has good frequency selectivity).

A third problem in transmitters for which YOU should be very aware is SHOCK HAZARD. Most transmitters use voltages and currents which can be lethal. It behooves any user of a transmitter to know SAFETY PRECAUTIONS you should use ANY TIME you operate or work on a transmitter. In fact, these precautions should be used when working with ANY type electrical equipment. Some of these precaution are: I. Keep your body away from open electronic circuitry which is

on. (Usually, good equipment has enclosures for this purpose). 2. Be sure equipment is properly grounded. 3. Be sure antenna lead-in wires, etc., are kept away from power -

lines. 4. Use equipment which has good design (Such as power supply

bleeder resistors to discharge capacitors when equipment is turned off ... etc.)

5. COMMON SENSE

When an Amateur "puts" a transmitter "on the air", he must he sure he knows what the transmitter output frequency is-lo he sure he is "legal".

There are several methods of determining the operating frequency of a transmitter . . . hut, basically they all involve comparing indirectly the transmitter frequency with broadcasts of the National Bureau of Standards stations. (WWV, WWVH, WWVI3, etc).


48

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REFERENCE & DATA CHECKPOINTS INFORMATION

Just before we leave the electronics portion of this chapter and proceed into the "Rules & Regulations" part . . . we need to teach you how to calculate the "input power" to a transmitter's final stage (the stage which sends the signal to the antenna). Since a Novice is limited to a maximum of 75 watts of input power to the "final stage"; you need to know how to determine what the input power is to YOUR transmitter's final stage.

Here's the simple procedure:

Power input = DC plate voltage (to the final tube/s) times the final stage plate current.

For example, if a transmitter final stage consists of two tubes in parallel, and: Their plate voltage = 450 volts, and each tube current is 50 mA . . . this means the total final plate current is 100 mA ... therefore: Pin = 450 x 0.1 = 45 watts. (Note: 100 mA = 0.1 amperes ... remember the units and milliunits study earlier in the chapter?)

Now, you try one! Solve the checkpoint problem.

Did you do alright? In most transmitters, the plate voltage to the final stage is virtually the same as the transmitters's high voltage B+ supply ... so don't let the term "B+" throw you.

O.K. Let's proceed to a section of the chapter where we will go somewhat against our basic philosophy of NOT MEMORIZING BUT UNDERSTANDING.

F.C.C. RULES AND REGULATIONS

When learning the F.C.C. "Rules and Regulations" regarding Amateur Radio . . . it is just about impossible to do it in any other way than "rote memory". So for this reason, we have "lumped" the rote memory learning necessary for the Novice test into this one section. The FCC test is sure to have many of these items included ... so study them well.

Let's deal first with information regarding the transmitter and transmissions from a Novice station.

As mentioned earlier, the maximum input power permitted to the final stage of a transmitter operated by a Novice is 75 watts. In other words, the product of the final plate voltage and the final plate current must not exceed 75 watts.

50

Pin = E (plate) x I (plate)

CHECKPOINT

An amateur transmitter utilizes a B+ value of 500 volts which is fed to the plate of the final amplifier stage. The final stage is being operated at a plate current of 120 mA. What is the input power to the final stage of the transmitter? Use column 2 space for your calculations.

ANSWER:

Pin = E (plate) x I (plate)

Pin = 500 (volts) x 0.12 (amps)

Pinput = 60 watts

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INFORMATION OPERATIONAL PROCEDURES

Moving ahead . . . let's consider some operational rules and regulations.

You should know that the maximum penalty for a violation of FCC rules and regulations is a fine of up to $500 for each day of the offense, suspension of operator license and revocation of the station license.

One of the infringements of rules and regulations would be the transmission of obsene or profane language. Also, transmission of false signals, or malicious interference are prohibited.

In regard to who may operate a transmitter licensed by a holder of the Novice Class license, the FCC says that any U.S. Amateur operator may do so. (Licensed operator, that is). There is a stipulation, however, if the operator operates other than within the limitation of the Novice Class license (let's say he's a General Class operator) . . . Both your station call letters and HIS must be used for identification purposes on the air.

A very important document to all amateur operators is the Station Log. The "log" is the station's written record of transmissions, and should contain the following basic information: 1. Callsign of station and written signature of the licensee. 2. Signature of the control operator (and his callsign, if not the S -

licensee). 3. Location of station. (If a "mobile" station operated within 100

miles of station address, use the term "local" . . . otherwise enter location for first and last transmission of the day.)

4. Input power 5. Type of emission (A-1 for Novice Class) 6. Frequency band used 7. Dates of operation 8. Information regarding any "3rd party" messages. 9. Time of starting and ending each given series of transmissions.

A Novice Class license is valid for two (2) years, and normally MAY NOT BE RENEWED. The exceptions are certain military service- men cases, and one may take a new test and obtain a new Novice license 12 months after expiration of any FCC amateur license. Hopefully . . . you won't let your license expire, and you won't take two years before obtaining a higher than Novice Class license.

When operating an amateur radio station (of any Class), the FCC requires that you identify your station (by giving the call letters) a minimum of once each ten minutes. Also, when ending a transmission or series of transmissions, the callsign shall l given for the station with which you have been in contact.

52

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INFORMATION REFERENCE & DATA CHECKPOINTS When you want to operate an amateur station in the "portable", or "mobile" operation modes, you don't have to give written notice to the FCC "Engineer -in -charge" in your inspection district-unless you intend to operate in this manner for more than 15 days. In this case, you must give notification in writing to him.

There are antenna height restrictions on amateur stations, however, we will not go into detail on these at this time. You just need to know that height restrictions are to prevent the antenna from becoming a hazard to air navigation.

There are only certain specific situations where "one-way" amateur communication is allowed. These are: 1. emergency communications. 2. information bulletins consisting of subject matter of direct

interest to amateurs . . . such as code practice transmissions, or "round -table" discussions involving more than two stations.

however, normally 2 -way communications is expected by FCC.

Regarding the proper procedure for calling or answering other amateur stations, the following is an example:

1. Give callsign of station being called. 2. Follow this by word "de" (meaning "from") 3. Then give call sign of station transmitting.

Here's an example:

"W2XYZ W2XYZ de W4ABC AR."

Whereupon W2XYZ. answers ...

"W4ABC de W2XYZ K."

This example means:

"W2XYZ W2XYZ from W4A13C W4ABC end of transmission (AR)"

Whereupon W2XYZ answered:

"W4ABC from W2XYZ invitation to transmit (K)"

As you get on the air after getting your license, you will find that many abbreviations are used by telegraphy stations in order to conserve time. As a result, a series of "Q" signals has been developed which are commonly used by Amateurs. A few of the more common ones are listed below, and may be found on the FCC exam.

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INFORMATION

Also, to help you, we are now going to summarize in "distilled" form most of the important terms and formulas. This will make a good study of the electronics part of the Novice lest in "Capsule form".

The rules and regulations will not be repeated here. . . so recycle yourself as required to get them down. Good luck!

REFERENCE & DATA

IMPORTANT TERMS: EMF = electromotive force

Unit of measure is volt

I = current Unit of measure is ampere

R = resistance Unit of measure is ohm

P = power Unit of measure is watt

Hertz = cycle kHz = kilohertz (1000 cycles) MHz = megahertz (1,000,000 cycles)

J1= lamda (greek letter) meaning "wavelength" (in meters)

for F = frequency (number of hertz per second)

C.W. = continuous wave (or code type transmission)

RF = radio frequency signal DC = direct current AC = alternating current Cycle = one completed chain of events out of

a series of repetitive recurring events Wavelength = distance an electrical wave can

travel during the time of one cycle (or Hertz).

Atom = smallest particle of matter which contains the characteristics of the element

Electron = smallest particle of negative charge (revolving about nucleus)

Proton = a positively charged particle (contained in nucleus of an atom)

Neutron = neutral (electrically) particle Valence Electrons = those in outer ring of

at om

REFERENCE & DATA

Ion = atom which has lost or gained electrons

Conductor = material with many "free" electrons ... and consequently has low electrical resistance to current flow.

Insulator = material with few "free" electrons .. high resistance.

Semiconductor = material with 4 valence electrons per atom.

Resistor = component designed to limit current flow

Capacitor = component that stores electrical energy in form of an electric field

Inductor = component that stores electrical energy in form of an electromagnetic field

Milliammeter = electrical measuring device that measures "thousandths" of an ampere of current.

Ohmmeter = measuring device for measuring resistance

Voltmeter = measuring device for measuring emf

Henry = the unit of measure of inductance Farad = the unit of measure of capacitance Rectifier = device that changes AC to pulsat-

ing DC Filter = device (circuit) that changes pulsat-

ing DC to smooth DC Diode = a vacuum tube or semiconductor with

two active elements Cathode = emitter of electrons Anode or Plate = collector of electrons Control Grid = element in tube that controls

flow of electrons from cathode to plate Filament or Heater = part of a tube that heats

the cathode to a temperature that will cause "emission" of electrons

REFERENCE & DATA

Amplifier = device to enlarge signals elec- tronically

Oscillator = electronic device which generates an AC signal (when supplied with appropriate DC element voltages)

Triode = tube with three active elements Transistor = three element "solid state" device Emitter = element of transistor roughly simi-

lar to cathode Base = element of transistor which helps

control electron flow Collector = element in transistor somewhat

similar to plate in tube. NPN = transistor where emitter and collec-

tor are made of "N" type semiconductor material and base is "P" type material.

PNP = transistor where emitter and collector are made of "P" type semiconductor material and base is "N" type material.

Alpha = ratio of collector current to emitter current in a transistor

Beta = ratio of collector current to base current in a transistor ("current gain" of transistor)

Frequency Multiplier = Stage that amplifies a signal and produces harmonics in its output, from which it selects a desired multiple of the input signal as its output.

Sky wave = RF transmission "bounced" off of ionosphere

Skip = distance between origination of sky wave signal and its reception point.

Harmonic = multiple of a given frequency Log = document in which Amateur's keep

record of transmissions, etc., of their radio station

56

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1. The schematic symbol for an NPN transistor is:

a.

b. -1E-

c.

d.

e. -141- 2. A conductor has:

a. 8 outer ring electrons per atom b. 4 outer ring electrons per atom c. 0 outer ring electrons per atom d. Many free electrons e. High resistance

3. Ohm's Law states that:

a.I=ExR b. R = I/E c. I = E/R d. R=IxE e. I = E + R

4. To measure current, use a/an:

a. Voltmeter b. Wattmeter c. Ohmmeter d. Watt-hour meter e. Ammeter

5. The unit of capacitance is the:

a. Ohm h. Henry c. Farad d. Beta e. Hertz

SAMPLE NOVICE TEST

6. An insulator is a:

a. Good conductor of electricity b. Poor conductor of electricity c. Material with many free electrons d. Material with no electrons e. None of the above

7. An "amplifier"

a. Changes AC to pulsating DC b. Changes DC to AC c. Increases the amplitude of a signal d. Changes the frequency of a signal e. Changes pulsating DC to smooth DC

8. An oscillator:

a. Changes AC to pulsating DC b. Changes DC to AC c. Increases the amplitude of a

signal d. Changes the frequency of a

signal e. Changes pulsating DC to

smooth DC

9. A rectifier:

a. b. c. d. e.

Changes AC to pulsating DC Changes DC to AC Increases the amplitude of a signal Changes the frequency of a signal Changes pulsating DC to smooth DC

10. A filter:

a. b. c. d. e.

Changes AC to pulsating DC Changes DC to AC Increases the amplitude of a signal Changes the frequency of a signal Changes pulsating DC to smooth DC

11. A multiplier:

a. Changes AC to pulsating DC b. Changes DC to AC c. Increases the amplitude of a signal d. Changes the frequency of a signal e. Changes pulsating DC to smooth DC

12. A sky wave:

a. Is an RF signal that never returns to earth.

b. Is an RF signal that is reflected off the Ionosphere

c. Is an audio signal used to modulate d. Is a DC wave e. None of the above

13. A-1 Transmission means:

a. I).

c. d. e.

Radiotelephone FM Continuous wave telegraphy TV None of the above

14. A modulator:

a. b.

c. d. e.

Generates the transmitter RF Varies the frequency, amplitude, or phase of a transmitter in accordance with the intelligence. Varies the height of the antenna Filters the pulsating DC None of the above

15. A key -click filter:

a. b. c. d. e.

Helps prevent spurious radiation Softens the key action mechanically Isolates the oscillator from the load Injects the intelligence None of the above

16. A detector is sometimes called:

a. b. c. d. e.

A modulator A demodulator A harmonic A plainclothesman None of the above

17. The symbol for a triode is:

a.

b.

PC?

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32. One-way communications from an amateur station are allowed:

a. Never b. Anytime c. For music broadcasts d. For code practice e. None of the above

INFORMATION

SAMPLE NOVICE TEST(Cont.) 33. The frequency of a Novice transmitter

may be controlled:

a. By crystal only b. By VFO only c. By either crystal or VFO d. By the National Bureau of Standards e. None of the above

34. A novice station may be operated by:

a. Only Novices b. Any U.S. Licensed amateur c. Any foreign licensed amateur d. Relatives of the Novice only e. None of the above

Hope you made out well on the 'Sample Novice Test'. It is very similar to the one you will take for the FCC to get your license. If you did well on our sample test, then chances are you won't have trouble with the FCC test.

The procedure for applying for a Novice Class license and for obtaining the necessary test is as follows:

1. Obtain a copy of Form No. 610 from the nearest District FCC office. (See the list of district offices & addresses shown in columns 2 and 3)

2. Find a volunteer examiner who meets the following requirements: a. Is holder of an unexpired FCC amateur license of General,

Advanced, or Extra Class level,

b. or, Is holder of an unexpired FCC commercial RADIO -TELE- GRAPH, license,

c. or, Is currently the operator of a manually -operated radiotelegraph station in the service of the United States.

When your volunteer gets the necessary papers, he will administer the Code tests first . . . which will consist of five minutes of text at 5 words (25 characters) per minute. One minute of copying without error is required to pass . . . and, likewise, one flawless minute of sending is required at 5 WPM to pass.

After passing the code test, the examiner will give you the written test.

Within ten days after passing the code tests, the applicant must send the Form No. 610 to the FCC Licensing Unit ... Gettysburg, Pa. 17325. Be sure the examiner completes the proper certification on the reverse side of Form No. 610 before mailing it in.

Good luck!

Dist. No. 1

1600 Customhouse India & State Streets Boston, Mass. 02109

Dist. No. 2

748 Federal Bldg. 641 Washington St. New York, N.Y. 10014

Dist No. 3

1005 Customhouse Second & Chestnut Sts. Philadelphia, Pa. 19106

Dist. No. 4

819 Geo. M. Fallon Fed. Bld 31 Hopkins Plaza Baltimore, Md. 21201

Dist No. 5 Military Circle 870 Military Highway Norfolk, Va. 23502

Dist. No. 6 1602 Gas Light Tower 235 Peachtree St., N.E. Atlanta, Ga. 30303

Dist. No. 7

51 S.W. First Ave. Miami, Fla. 33130

Dist. No. 8 600 South St. New Orleans, La. 70130

g

35. When checked against NBS broadcast station(s), the following may be used to measure your transmitted signal frequency:

a. wave -meter b. frequency meter c. receiver d. All of the above e. None of the above

FCC DISTRICT OFFICES

Dist. No. 9 5636 Federal Bldg. 515 Rusk Ave. Houston, Tex. 77002

Dist. No. 10 Rm. 13E7; Fed. Bldg. 1100 Commerce St. Dallas, Tex. 75202

Dist. No. 11 Rm. 1754 312 N. Spring St. Los Angeles, Calif. 90012

Dist. No. 12 323 A Customhouse 555 Battery St. San Francisco, Calif. 94111

Dist. No. 13 314 Multnomah Bldg. 319 S.W. Pine St. Portland, Ore. 97204

Dist. No. 14 8012 Federal Office Bldg. 900 1st Ave. Seattle, Wash. 98104

Dist. No. 15 504 New Customhouse 19th St. btwn Calif. & Stout Denver, Colo. 80202

Dist. No. 16 691 New Federal Bldg. 4th and Robert Sts. St. Paul, Minn. 55101

Dist. No. 17 1703 Federal Bldg. 601 E. 12th St. Kansas, City, Mo. 64106

Dist. No. 18 1872 U. S. Courthouse 219 S. Dearborn Chicago, Ill. 60604 Dist. No. 19 1054 New Federal Bldg. Washington Blvd. & Lafayette St. Detroit, Mich. 48226

Dist. No. 20 905 Fed. Bldg. 111 W. Huron St. Buffalo, N.Y. 14202

Dist. No. 21 502 Federal Bldg. P.O. Box 1021 Honolulu, Hawaii 96808

Dist. No. 22 P.O. Box 2987 322 U.S.P.O. & Courthouse San Juan, P.R. 00903 Dist. No. 23 Room G-63 U.S.P.O. 4th & G Streets P.O. Box 644 Anchorage, Alaska 99501

Dist. No. 24 Room 216 1919 M Street N.W. Washipgtop. n . 20554

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CHAPTER 3 THE TECHNICIAN CLASS LICENSE

Again, congratulations on coming this far! You are now beginning to personally experience some of the excitement of "ham" radio, if you have gotten on the air with your Novice license. The ground work that has been laid by your diligent work in learning code and some basic electronic theory will pay great dividends as you continue to work toward the General Class License. In this chapter, however, we simply wish to give you an overview of the Technician Class license privileges, requirements and the like. May we repeat ourselves once more here and say that even though the Technician Class license does not require any more code proficiency than the Novice class . . . we strongly urge you to "continuously continue" your code practice . . . working toward the General Class 13 WPM requirement. The reason we keep telling you to practice, practice, practice is because in order to learn code it takes practice, practice, practice! Now, let's take a quick look at some of the features of the Technician Class license.

INFORMATION

The basic reason the FCC opened this class of license was to en- courage radio amateurs to experiment in the "higher frequency" ranges. Radio amateurs, over the years, have made great contri- butions to technology by being first to perform meaningful experimentation and operation in much of the radio spectrum.

The requirements for acquiring a Technician License boil down to the following:

1. Must be a U.S. citizen, or a resident alien who has filed "first papers" for citizenship.

2. Pass a sending and receiving code test at 5 words per minute (same as Novice).

3. Pass a written test on theory and regulations. This is the same as the General class.

The Technician tests are generally taken by mail under the direct supervision of a General Class or higher licensee. The reason we used the word "generally", is that there is an exception to this. The exception is-if you go to an FCC office to try for a General Class license and you should fail the 13 WPM test for code, but you get 25 characters in a row (equivalent of copying 5 consecutive words at the 13 WPM rate), you may take the written test and have the code credited toward the Technician Class license. If you pass the written theory test, and have met the above 25 character requirement, you will be issued a Technician Class license. This successful passing of the written test at the FCC office can be credited toward your General license, so that to get the General, all that's left for you to do is to pass the 13 WPM code tests at the FCC office. (Sending and receiving).

REFERENCE & DATA

, ... ....

TL t c _ = S

Code Reminder . . . Hope you have a

receiver now and are getting daily "on the air code practice". Keep up the good work. Stop by your local Radio Shack store to select a receiver, antenna and other helpful accessories.

62

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é3AtIH f1OA Oa S3J311lt114d .LHHM

CHAPTER 4 THE GENERAL CLASS LICENSE

Well, this is it! This is the chapter you've been working toward. When you have mastered

the information in this chapter, you should be well prepared for the Technician and General

Class written examination. In order to get that coveted General Class license, it will require continued diligence, on your own, to reach that 13 WPM code proficiency .. .

but, we have confidence in you ... you can do it!

In early pages of this book, we tried to "spoon feed" you. Now that you have some back-

ground, we will begin to treat you like an electronic technician in this chapter. By this, we simply mean that there will be less "spoon feeding". We will expect you to "recycle" yourself when required, without having to be told. We will want you to carefully study any figures, charts or drawings associated with the text in order to get the full significance-and we won't spell out every single detail in words. In other words ... we're going to force you to do some reasoning on your own ... just like you will have to do on

the General test.

Also, in this chapter, there will be a number of "schematic" diagrams given in which you

can either use a "pencil and paper" method of analyzing the parameters, and/or, you

can connect the circuits, if you have the parts, and make actual measurements to verify your "theory" calculations. Virtually all the parts and supplies can be purchased at your Radio Shack Store.

Generally, we will try to give the theory in this chapter by topic, so if you need more

study in a particular area of theory you can easily find the segment of the chapter devoted to that area for restudy (since information on a particular topic, for the most part, will not be scattered throughout the chapter.) Of course, there will be some cases of "overlapping" information.

Some of the basic topics include: Privileges and requirements of General Class; DC

and AC circuit theory; Active and passive electronic components; Various electronic sys-

tems; Applications of various electronics components and circuits; Significant electronic terms definitions; Antennas and feedlines; Rules and regulations; and various operating procedures.

Bon voyage through this interesting and challenging chapter!

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INFORMATION

Hope you did all right on that checkpoint! Knowing these frequency limits is not only important for the test, but later for operating on the air. Knowing these limits and abiding by them will help prevent you from getting any "pink tickets" (notices of violation) from the FCC. Study them until you know them!

As a holder of a General Class license, you will also have the privilege or right to conduct and or supervise code tests and written exam- inations (if you are over 21 years old) for Novice, Technician and Conditional class licenses which are available via the "mail method".


7. For 10 meters: a. CW is from to

MHz.

b. Phone is from to MHz.


MHz.



MHz.


ANSWERS:

1. 3.525 to 3.775 MHz and 3.890 to 4.0 MHz.

2. 3.89 to 4.0 MHz.

3. 7.025 to 7.150 MHz and 7.225 to 7.3 MHz.

4. 7.225 to 7.3 MHz.

5. 14.025 to 14.2 MHz, and 14.275 to 14.350 MHz.

6. 21.025 to 21.250 MHz and 21.350 to 21.450 MHz.

7. 28.0 to 29.7 MHz. 28.5 to 29.7 MHz.

8. a. 50.1 to 54 MHz. b. 50.1 to 54 MHz.

9. a. 144 to 148 MHz b. 144.1 to 148 MHz.

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INFORMATION

Now that you know the main colors used in color coding ... you must learn the meaning of the various bands and their placement.

Carefully study the "BAND SIGNIFICANCE CHART" shown in the next column.

Now let's mention the Gold and Silver colors again. You will only find them used as either the third band (decimal multiplier band), or in the fourth band (tolerance band). Look at the description for these colors in the next column.

68

REFERENCE & DATA

B

C BAND SIGNIFICANCE

Band A-lst significant figure of resistance value in ohms. Band B -2nd significant figure of resistance value in ohms. Band C-Decimal multiplier Band D-Tolerance in percentage (If no fourth band, tolerance is 20 per cent. Gold = 5% tolerance, when used as fourth band (band D) Gold = 0.1 multiplier, when used as third band (band C)

SILVER = 10% tolerance, when used as fourth band (band D) SILVER = 0.01 multiplier, when used as third band (band C)

CHECKPOINTS

CHECKPOINT: WITHOUT LOOKING BACK AT THE COLOR SIGNIFICANCE CHART ... fill in the blanks below with either the appropriate number, or color as required.

1. Green =

2. The number 2 is represented by the color

3. Brown =

4. 3 = what color?

5. Black =

6. 9 = what color?

7. Blue =

8. Gray =

9. 4 = what color?

10. 7 = what color? ANSWERS:

1. 5 6. White 2. Red 7. 6 3. 1 8. 8 4. Orange 9. Yellow 5. 0 10. Violet

69

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INFORMATION REFERENCE & DATA CHECKPOINTS Did you catch that tricky one with Black as the third band? When- ever black is used as a multiplier (third band), it means to add "zero" zeroes . . . and simply read the first two significant digits . . . in our sample the value is 10 ohms. Remember this one, for it appears quite often in real life. Also, we could have tricked you by putting gold or silver in the third band position. All you have to do is move the decimal in accordance with the color. Example: if the first two numbers were decoded as 15, and the third band was Gold (a 0.1 multiplier) . then the resistor would be a 1.5 ohm resistor. If Silver were the 3rd band . . . the value would be 0.15 ohms.

THE ONLY WAY TO LEARN THE COLOR CODE IS TO USE IT! We realize that we don't have room enough in this book to give you all the practice you need . . . so it's up to you! Every time you see a resistor . . . "decode it". It's a lot of fun to see how quickly and accurately you can decode it. You'll be amazed at how quickly those colors will almost look like numbers to you, if you will practice. EA RI 10K CHECKPOINT: Look at the diagram in column 2 and fill in the appropriate blanks in If the current meter is reading 2 mA . .

the checkpoint in column 3. HINT: Start by writing the OHM'S LAW What is the value of the "applied voltage"? formula for voltage . . . then substitute the appropriate values of current and resistance into the formula. Use the blank space on these EA = volts. pages for your calculations, etc.

That wasn't bad was it? Here's another little hint ... whenever you multiply milli -units times kilo -units ... the answer is in units. The reason is that a milli -unit is "one -thousandth" of a unit and the kilo - unit is "one thousand" units ... so when you multiply one -thousandth by one thousand you "end up" with 1.

Let's see if you remember the power formula. For the same circuit .. .

how much power does R dissipate? Don't forget, in order to prevent simple errors ..

1. Write down the appropriate formula, then . . .

2. Substitute the appropriate values ... keeping track of the decimals.

Work out the answer and fill in the Checkpoint blank.

ANSWER:

EA = 20 volts

Here's the procedure:

Ohm's Law formula for voltage is: E=IxR Substituting, we get: E = 0.002 A. x 10,000 ohms = 20 volts.

CHECKPOINT: The power dissipated by Ri (PR1) =

mw. (milliwatts)

ANSWER:

40 mw.

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INFORMATION

Let's take a closer look at our three -element series circuit and "pickup" some other very important facts regarding series circuits.

Fact: THE TOTAL RESISTANCE OF A SERIES CIRCUIT IS EQUAL TO THE SUM OF ALL THE INDIVIDUAL RESISTANCES IN SERIES! For our sample circuit then: RT (total res.) = R1 + R2 + R3 Thus: RT = 18K + 12K + 10K = 40,000 ohms (40K).

Another way we could have solved for R total is by using Ohm's Law as follows:

RT = ET/IT which simply means total resistance = total voltage divided by total current ... so for our sample circuit:

RT = 40v/ImA = 40K ohms.

Many, many times in working on electronics calculations you will be using this version of Ohm's Law . . . so here's another quickie hint at no extra charge! Whenever you divide units by milli -units . . . the answer is in K units. Most of the time when you are solving for resistance in electronic circuits you will be working with volts and milliamps . . . so remember this statement . . .

volts divided by milliamps = Kilohms.

Let's look at our sample circuit again and make another important observation. FOR SERIES CIRCUITS: THE APPLIED VOLTAGE (EA OR ET) IS EQUAL TO THE SUM OF THE INDIVIDUAL VOLTAGE DROPS!

For our sample circuit then, EA (E applied) must equal ER1 + ER2 + ER3, or in other words: 18v + 12v + 10v ... hence E applied or ET = 40 volts.

Putting these three important facts together (regarding series circuits) ... you have tremendous flexibility in solving series circuits. Those 3 key facts again are:

1. Current is the same through all parts of a series circuit.

2. Total resistance is equal to the sum of all the resistances.

3. Total voltage is equal to the sum of each of the individual voltage drops.

O.K. Here's a "brain twister" problem to see if you can begin to apply these facts to a practical problem. Give it your best effort! Try not to look ahead to the answer until you've really given it "a college try".

72

EA 40V

EA 60V


I= 2 mA

10K

10K

CHECKPOINT: For the circuit shown in column 2 find the following:

ER1 = volts ER3 = volts ER2 = volts

R2 = K ohms IR, = mA.

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INFORMATION

O.K. Let's make mention of a couple more facts that are very practical . . . even though you may not be called upon to analyze a circuit with these conditions on an FCC exam.

Fact number one relates to the effect of an "OPEN" in a series circuit. By an "open" we mean that for some reason or another, the current path is broken through a given part of the circuit. For example ... if a resistor were to physically break in half, it would create an open. If ANY PART OF A SERIES CIRCUIT IS OPEN ... THE ONLY CURRENT PATH THROUGH THE CIRCUIT IS BROKEN ... HENCE CURRENT WILL DECREASE TO ZERO THROUGHOUT THE SERIES PATH. Of course, an open can be caused by a wire becoming disconnected from a terminal or component, etc. When current is zero, then the IxR drop across all other components, except the open one, is zero. Across the open portion of the circuit, one would measure E applied, since there are no "IxR" drops to sub- tract from EA . . . THE POTENTIAL DIFFERENCE ACROSS THE OPEN WILL BE EA.

Fact number two relates to the effect of a "SHORT" in a series circuit. If any resistance is shorted out (becomes close to zero ohms), then R total will decrease ... causing I total to increase. If the current through the circuit increases ... then the IxR (voltage) drops across the "unshorted" components will increase proportionately ... and the voltage drop across the shorted part of the circuit will decrease to zero . . . since E = IxR, and R (for the shorted portion) has become virtually zero . . . anything times zero is still zero, hence E = zero.

Of course there are many adverse effects from a "shorted" part, or all, of a circuit. Everything from blown fuses to destroyed circuit elements can (and often do) happen when shorts occur. How much damage is done usually depends on how much "unshorted" resistance is left in the circuit to limit current.

Here's a summary of key facts regarding series circuits ... remember these and understand them, and you will be able to "think your way" through almost any problem involving series circuits.

Facts: (for series circuits)

1. Total resistance (RT) equals the sum of the individual resistances ...or: RT = R,+R2+...and so on.

2. Total voltage equals the sum of the individual voltage drops or:

ET = ER1 + ER2 + . . . and so on.

3. Total power equals the sum of the individual power dissipations,or: Pr = PR, + PR2 + ... and so on.


74

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INFORMATION

Let's see if you can "put it all together" and solve an interesting series -circuit problem. Get the "info" from the circuit in column 2,

REFERENCE

FROM

& DATA

and determine the answers asked for in the checkpoint. Calculate as required in the blank portions of these pages ... if possible when through . . . verify by actual circuit hookup and measurements.

VOLTAGE SOURCE +

Rl 4.7K

Good luck! EA R2 2.2K

=1.5V

ANSWERS:

As a form of review . let's "logic" you through this problem step-by-step.

RT, or total resistance, was the first electrical parameter asked for. By remembering that in a series circuit the total resistance simply equals the sum of the individual resistances ... you should have simply added 4.7K + 2.2K + 1.0K and arrived at 7.9K as your answer.

Next parameter asked for was total current. Here's a key step in solving a series circuit. Once you know the current through any part of a series circuit . . . you automatically know the current through all the components in series, since the current is the same through all parts of a series circuit. Just about everytime you have a circuit problem . . . the first thing to look for is any part of that circuit ... or any component about which you know . . . or are given TWO KNOWNS (or parameters). In this case we have such a condition for R3. We know its resistance value, AND we know the voltage dropped across it ... so we can easily solve for the current through it by: la. = ER3/R3. From this we get: IR3 = 1.5v divided by 1.0K ... and thus IR3 = 1.5 mA. Since the current is the same through all parts of a series circuit ... this is also the value for I total ... therefore, IT = 1.5 mA.

76

CHECKPOINTS

CHECKPOINT: Find the Following:

RT =

IT =

ET =

PT =

ER, =

PR2 =

What fractional part of EA is ER,?

Now that we know the current through all the components, and we were given each component's resistance value . . . we have TWO KNOWNS about each component and the rest is easy.

To solve for total voltage (ET) we simply use "good ole Ohm's Law" again: ET = IT

x RT, OR, ET = 1.5 mA. x 7.9K = 11.85 volts.

Next on the agenda was total power calcula- tion. You should have written down the formula: PT = ET x IT and substituted: PT =

11.85v x 1.5 mA to get the answer: 17.77 mW. (milliwatts). Incidentally, you could have used: IT2 x RT, OR ET2/RT and gotten the same results. Usually it's easier though . . . if you know both E and I, to use the version of the power formula that doesn't require any "squaring" . . . because then you have to perform two mathematical functions ... "squaring" and then either multiplication or division (depending on which formula you used).

O.K. To solve for ER,, just multiply the current through it by its resistance value to get its "IR" drop. Answer is: 7.05 volts.

To get PR2 requires two operations no matter what approach you use . . . so let's use IR22

x R2 here, and get: PR2 = 1.5 mA2. x 2.2K = 4.95 mW. (Watch out for those decimals whenever you square with them.)

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INFORMATION

THE PARALLEL CIRCUIT

The reason we took so much time looking at principles and concepts in series circuits was because you will be using the same general approach in solving all types of circuit problems, and we wanted to give you a good foundation to work from.

Remember that a series circuit could be defined as a circuit where all the components were connected "in tandem", and there was only one path for current flow ... well, a parallel circuit differs in these two basic points. First, a parallel circuit is one in which there are TWO OR MORE CURRENT PATHS. Second, the components ARE NOT connected "in tandem". Study the pictorial and schematic diagrams in column 2 illustrating these facts.

Notice that when total current leaves the negative side of the source, there is one path . . . but soon it arrives at a "junction" where R, and R2 join ... and the current divides into TWO paths (I, and 12).

I, passes through Ri . .. 12 passes through R2 ... then the two currents JOIN again at the top of R, and R2 at the junction of those two resistors, and we now have total current (II + 12) traveling from the junction back to the source through the single -wire path. As you see ... the parallel circuit offers more than one current path. Also, the components are not connected "in tandem", but rather "in a parallel configuration" with "ends common."

Since their ends are connected together, another valuable fact to remember about components "in parallel" is that the VOLTAGE ACROSS ALL PARALLEL COMPONENTS IS THE SAME! In the case of a "simple" parallel circuit, the voltage across all "branches" is equal to E applied . . . since they are each in parallel with the source as well as being in parallel with each other.

Notice, we used the term "branch" to indicate one path.

In our earlier illustration then, I, represents the "branch current" through R, and 12 is the symbol we used to indicate the "branch current" through R2.

This brings us to a third important fact to remember about parallel circuits. Each branch current is INVERSE to its branch resistance. This simply means that the higher the branch resistance . . . the lower the branch current. Makes sense doesn't it!

Ohm's Law told us that current is inverse to resistance when it says that I = E/R. This means that if branch No. ] has twice the resistance of branch No. 2 . . . then the current through branch No. 1 will be half that of branch No. 2's current . . .

since both R's have the same voltage across them.


CELL Pictorial of a

Parallel Circuit

Schematic of same Parallel Circuit

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INFORMATION

OK ... here's another practice problem for parallel circuits.

é

Hope you did well on this. We'll think through it here so you can check your method of solution.

First ... we know that EA is going to be the same as the voltage across any of the branches. Since we have "2 knowns" about branch No. 1, we can solve for ER1 easily. 3mA. x 10K = 30v. Hence, we know that EA is 30 volts.

We can see that branch No. 2's current is half that of branch No. 1 . . . so we know that branch No. 2's resistance must be TWICE that of branch No. 1. This means that R2 = 20 K ohms.

We know that branch No. 3's current must be the same as branch No. l's . . . since it has the same E and R . . . thus, 13 = 3 mA.

Total current for a parallel circuit must equal THE SUM OF THE BRANCH CURRENTS! This gives us 7.5 mA. for the total current.

The power dissipated by each branch resistance can be calculated simply by using the product of the branch E and the branch I. This gave us 90 mW. for branches No. 1 and No. 3, and 45 mW. for branch No. 2.

80


IT

IT

CHECKPOINT Solve the circuit shown in column 2 for the following:

R3 10K

1.

2.

EA = V.

R2 = K ohms.

3. 13 = mA.

4. IT = mA.

5. PR, = mW.

6. PR2 = mW.

7. PR3 = mW.

8. PT = mW.

ANSWERS:

1. 30 volts. 2. 20K ohms 3. 3 mA. 4. 7.5 mA..

5. 90 mW. 6. 45 mW, 7. 90 mW. 8. 225 mW.

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INFORMATION

Let's show you how to use this to advantage . . . even in a

circuit where NOT ALL the branches are the same value but some of them are.

Look at the circuit in column 2, and follow our reasoning below.

If R2 were not present . . . the total resistance of the circuit would be 5K . . . correct? Sure, two 10Ks in parallel equal an equivalent resistance of 5K . . . even when you work it out by the "product over the sum" formula. Now . . . here's a nifty trick. We CAN say that the equivalent total resistance of R, and R in parallel IS 5K. Let's use the Re (R equivalent) of R, & R3 and combine it with R2's resistance to get R total. HOW ABOUT THAT! We now can easily see the equivalent total resistance of R2's 5K in parallel with R, & R3's 5K will yield a circuit total resistance of 2.5K.

All we've said is this. You can combine equal resistance branches . . using this technique, and then apply the product over the

sum formula, or any other means, to solve for RT. If R2 had not been 5K (for convenience sake); then, you would have had to apply the product over the sum formula to find the final results (R total for circuit). In fact, let's try a problem where you must use your "tricky technique" of combining equal branches . . . plus the product over the sum formula to get the answer. Solve the problem at right for total resistance.

Here's the pattern of thinking you should have used:

1. R, and R3 are equal resistance branches whose equivalent resistance would equal 0.5K.

2. R2 and R4 are equal resistance branches whose equivalent resistance would equal 5K.

3. Total resistance will equal:

0.5K x 5K divided by 0.5K + 51( Giving: 2.5/5.5 = 0.45K

Now, let's suppose Nou had a circuit where none of the branches were equal . . . now what? All right . . . one approach that will always work is to use the product over the sum formula to solve any two of the branches . . . then take that equivalent resistance (called Re) of the two branches just solved and work it with the next branch using the same formula. Look at the example in column 2.


R3 10K

R2 R4 10K 10K CHECKPOINT

Solve for RT for the circuit shown:

RT =

ANSWER:

RT = 0.45K ohms

R3 Son

SOLUTION

R,xR2 10x20 6.662 Step 1: RI + R2 10 + 20 =

Re x R3 6.66 x 30 199.8 - 5.452 Step 2:

Re + R3 6.66 + 30 36.66

RT = 5.4552

K ohms.

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INFORMATION REFERENCE & DATA

It's about time that we summarize the key things we've told you about parallel circuits. These are things you should remember for sure!

1. Voltage across all parallel components is the same.

2. Current through any given branch is inverse to that branch's resistance. (This means the smallest R branch has the most current ... the highest branch R the least, etc.)

3. Total current equals the sum of the branch currents.

4. Total resistance is always LESS THAN THE LEAST BRANCH R.

5. The smallest resistance branch dissipates the most power. (And, the highest R branch, the least, etc.)

6. Total power equals the sum of all the individual power dissipations, and/or equals EA x IT.

7. Ohm's Law can be used to solve parameters anytime you have enough knowns to use it.

8. Several methods for solving for RT are: Ohm's Law; Product over the Sum formula; Equal branches trick; and the Assumed Voltage technique.

To bring into focus all the things that we've said about SERIES circuits and PARALLEL circuits . . . look at the COMPARISON CHART. If you will study AND UNDERSTAND . . . NOT MEM- ORIZE these facts ... you will be "money ahead" for the FCC exam.

THE SERIES-PARALLEL CIRCUIT

The final circuit configuration we want to discuss is the Series -Parallel circuit. As the name implies . . . parts of this circuit have series elements and other parts of the circuit have parallel elements. Sometimes you hear this type circuit called a "combination" circuit. We'll use the term series -parallel circuit most of the time . .. and for convenience, we'll sometimes abbreviate it an "S -P" circuit.

You have already learned the most important concepts needed to analyze S -P circuits. The key here is to use the rules for series circuits in analyzing the portions of the circuit where elements are in series . . . and use the rules for parallel circuits for the parts of the circuit for which that is appropriate. To get the final results . . . you merely combine the results, as you will see in our next example.

COMPARISON CHART FOR SERIES & PARALLEL CIRCUITS

Item being considered Series Ckt. Parallel Ckt.

Circuit connection Components connected "end -to -end".

Components connected "side -by -side".

Total Resistance Sum of all resistances Less than least branch resistance

Voltage Distribution ET = sum of all individual voltage drops; and largest R drops most voltage.

Voltage is the same across all parallel components. (= EA)

Current Distribution Same current through all components in series (_ IT)

Current splits or divides at "branching points"; inverse to branch R. Total current equals SUM OF ALL BRANCH CURRENTS.

Power Distribution Largest R dissipates most power. Total P equals sum of individual P dissipations or EA X IT.

Smallest R dissipates most power. Total power = sum of all individual power dissipations, or, EA X IT.

CIRCUIT COMMON FACTOR

Current Voltage

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INFORMATION

If you had any trouble with that problem, go back and review the pertinent information regarding series, parallel and S -P circuits until you understand the answers.

AC VOLTAGES AND CURRENTS

Now that you've had quite a bit of practice using "glorified Ohm's Law" circuit analysis of DC circuits ... it's time to begin learning how this same approach can be used for AC circuit analysis.

Let's quickly review the difference between DC and AC. Look at

the illustrations of DC voltage and AC voltage.

Observe that the DC voltage is at one level . . . and of one

polarity (in this case +). The AC voltage is continuously changing level (amplitude), and alternately is going in a positive direction ... then a negative direction (polarity).

As a radio amateur ... you will be working with both DC voltages and currents from DC sources and with AC voltages and currents from AC sources. A car battery is an example of a DC source. The outlet in your home is an example of a 60 Hz, AC source.

Probably the first thing we should do is to familiarize you with the important values of a "sine -wave" AC voltage . . . since this can appear on your FCC test . . . and besides, it will make you a

better technician if you learn it.

Study the sketch of the various sine -wave values . . . then we'll discuss some of the meanings.

The sketch shows one "cycle" (Hertz) of a sine -wave AC voltage. The reason it is sometimes called a sine -wave is that its value or amplitude at any given instant of time during the cycle can

be computed from the mathematical function called the "sine". For example, if we consider the cycle starting at 0, at 0 degrees

. the sine of an angle of 0 degrees is equal to zero. Since the complete cycle takes 360 degrees . . . then you can see from the diagram that after 90 degrees (1/4th cycle), the amplitude is at peak value . . . and sure enough . . . the sine of 90 degrees equals 1 (maximum value) ... and so on.


+10V

0 - DC VOLTAGE

)

AC VOLTAGE

"PEAK VALUE"(=1.414 X EFF VALUE) EFF.VALUE"(=0.707 X PK)

-"AVERAGE VALUE" (=0.637 X PK; OR 0 0.9 X E F F. VALUE)

"PEAK TO PEAK" VALUE

ANSWERS:

1. RT = 15K ohms 2. I, = 0.5 mA. 3. 13 = 0.5 mA. 4. IT = 1.0 mA. 5. ER3 = 5 volts 6. ER1 = 15 volts 7. PR, = 7.5 mW. 8. PR3 = 2.5 mW. 9. PT = 15.0 mW.

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INFORMATION

How did you do? If you missed any of these ... better review ... .

As you may have gathered from the discussion we just finished probably the most important value with reference to sine -wave

AC voltages and currents is the EFFECTIVE value. Be aware that virtually all AC measurement instruments, such as voltmeters and current meters are designed so that they measure effective value. Some do have provision for measuring peak or peak -to -peak . . .

however, these are special scales on the instrument. When you use an AC voltmeter or current meter . . . what you read is rms, or effective value, unless otherwise stipulated.

Now, you may be wondering . . . can I use Ohm's Law to analyze AC circuits, the same way I do for analyzing DC circuits. In a broad sense, the answer is YES! For purely resistive circuits . . .

you use the same exact procedures when using the EFFECTIVE VALUES of voltage or current. In other words . . . for all those problems we had you solve for series circuits, parallel circuits, and S -P circuits . . . if the values of voltage and/or current were given as effective values . . . your answers would be exactly the same as for DC. If the values were stated as peak values . .

you would have simply had to convert them to effective values by multiplying by 0.707 ... and then plug the effective values into the formulas. The same rules apply to a resistive circuit with AC applied, as with DC applied. That simply means that the rules for current distribution, voltage distribution, etc. hold.

Rather than waste your time working more problems with purely resistive circuits . . . let's move on now and begin to consider some components and circuits that operate a little differently with AC applied than does the simple resistor element.

Remember our old friend Frequency? Well, the elements we are about to discuss show different oppositions to current flow for different frequencies of AC applied to them. Because of this . . .

these components open up a whole new "vista" of controlling electron flow in electronic circuits.

REFERENCE & DATA

EA 20V DC SORCE

(AC SOURCE SYMBOL)

EA 20V EFFECTIVE AC

EA 28.28V PEAK VALUE AC

I= 1mA

10K 10V

10K 10V

IrrA=EFFECTIVE AC CURRENT

10V (RMS OR EFFECTIVE)

10V (RMS OR EFFECTIVE)

EFFECTIVE I = 1 mA

EFFECTIVE E=10V

EFFECTIVE E=10V

CHECKPOINTS

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INFORMATION

Now for the formula for XL.

When you think about it . . . that formula makes a lot of sense. As you recall . . . we said that the faster the rate of change of current . . . the higher the opposition, or XL That obviously relates directly to the frequency. Look at the diagram of two frequencies . . . as you can see, the higher the frequency . . . the faster the rate of change of voltage and/or current . . . so "sure enough" if frequency is doubled ... XL will double.

The same DIRECT relationship is true regarding inductance and inductive reactance. If we double the inductance of a coil . . .

it will give double the XL for any given frequency.

Try the little XL problem at this checkpoint.

Did you use the XL formula correctly? In this case the way you should have done it is: 1. Write down the formula:

XL = 2 77f

2. Then substitute the values: XL = 6.28 x 100 x 10 = 6,280 ohms.

To see if you understand the concept of how inductive reactance changes with frequency and inductance values . . . try the next checkpoint.

As you can see from the results on the checkpoint . . . Inductive Reactance is DIRECTLY PROPORTIONAL TO FREQUENCY AND TO INDUCTANCE.

REFERENCE & DATA

XL = 2nfL

Where: XL = inductive reactance in ohms 2 = 6.28 f = frequency in Hertz L = inductance in Henrys

+10Vr- I

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I

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f 2=2Xf 1

f= 1000 Hz

TIME -1>

CHECKPOINTS

f= 100 Hz

L= 10Hy

CHECKPOINT Refer to the diagram above and solve for the inductive reactance of the coil.

XL = ohms.

ANSWER: XL = 6,280 ohms

CHECKPOINT 1. Solve for XL for the circuit shown in col-

umn 2. XL = ohms.

2. If the frequency were doubled and the inductance were the same value, what would be the XL? ohms.

3. If the frequency were cut to 1/2 and the inductance were doubled, what would be the resultant XL? XL = ohms.

4. If the frequency were doubled and the inductance tripled . . . the resultant XL

would be times as great as the original circuit shown.

ANSWERS: 1. 9,420 ohms 2. 18,840 ohms 3. 9,420 ohms 4. 6 times

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INFORMATION

So far we've talked about inductances combined with other inductances . . . and discussed how to find total inductance and how to find total inductive reactance. Let's move ahead now into a

"generalized" understanding of what happens when you combine resistors and inductors in one circuit. We do not need to go into "deep AC theory" and mathematics to give you the concepts you need ... so we won't!

The fact that an inductor "opposes a change in current" causes the current through an inductor to LAG BEHIND THE VOLTAGE ACROSS THE INDUCTOR by 1/4th cycle ... which is 90 degrees. A whole cycle represent 360 degrees, as you already know. Now a crutch to help you remember the fact that current LAGS voltage in an inductor . . . remember the man's name "ELI". The E in the name stands for the voltage ... the L stands for the inductance ... and the I stands for current. So to use this "crutch" ... think: E comes before I in an inductor (L). As you can see with the name ELI ... the E is before the I!

Because maximum current does not occur at the same time maximum voltage does . . . we say that the two are out of phase. In this case . . . they are out of phase by approximately 90 degrees. If the inductor were perfect . . . and only had inductance with no resistance . . . it would be 90 degrees . . . but since there is some resistance in the wire of the coil . . . the phase difference ís really less than 90 degrees for a practical coil.

Look at one way this 90 degree out of phase condition can be represented (showing the phase difference between E and I in

an inductor). Notice that when one parameter is at maximum amplitude ... the other is at zero ... and vice versa.

THE RL CIRCUIT

Now let's think about what will happen when we put a resistor and an inductor together in a single circuit. Look at the schematic in column 2, and follow our reasoning below:

We haven't really said it before, but in a purely resistive circuit, the voltage and current are in phase. This means that when E

across a resistor is maximum . . . the current through it is also at maximum. We have just finished telling you that in an inductor, there is approximately 90 degrees phase difference between the Et, and the IL. When we combine the two elements . . . logic tells us that the resultant total circuit voltage (ET or EA) will NOT be perfectly in phase with total circuit current (IT) since the circuit is not purely resistive . . . nor will they be 90 degrees out of phase . . . since the circuit is not purely inductive. The result will be the ET and IT are out of phase by some angle between 0 degrees and 90 degrees. Does that make sense? Sure it does ... and that is actually what takes place.


E

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VOLTAGE AND CURRENT RELATIONSHIP IN AN INDUCTOR

E IS MAX. WHEN I IS MAX.

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INFORMATION

Vectorially is a big word. You don't need to worry about it because we are not going to go into a lot of vector math . . . but simply want you to get a concept.

If two fellows were tugging on two ropes that are tied at the other ends to a common "load" . . . and the two fellows are tugging with 10 pounds force each . . . but the ropes were aimed in directions that were 90 degrees apart, the resultant force on the "common load" point would not be 20 pounds . . . nor 10 pounds . . . but 14.14 pounds . . . similar in concept to our example of the voltage drops. You get the idea . . . the resultant is not as large as the sum of the two individual vectors . . .

but is greater than the largest one of the vectors. So for a series RL circuit, the applied voltage does not equal the simple sum of the two voltage drops ... but the vector sum.

Time to consider some facts about a parallel RI, circuit. As we have already hinted . . . the component in an RL circuit that has the most control over the circuit current is the one which has the greatest effect on the resultant phase angle between E

applied and circuit total current. Look at our simple parallel RL circuit diagram-which element do you think is having the greatest effect on the circuit phase angle? If you said the resistor in this case, you were correct. Why? Because there is more "resistive branch current", than there is "inductive branch current". What this says to us is that in PARALLEL RL CIRCUITS, THE BRANCH WITH THE SMALLEST OHMIC VALUE HAS THE MOST EFFECT! Notice this is just the opposite from the series RL circuit, where, the largest ohmic value component has the greatest effect on the circuit.

In parallel RL circuits, the voltage is the same across all parallel branches . . . and, of course is equal to E applied . . . so the voltages across all the branches are in phase . . . BUT THE CURRENTS THROUGH THE RESISTIVE AND INDUCTIVE BRANCHES ARE OUT OF PHASE. This means that the total circuit current, (IT) must not be equal to the simple sum of all the branch currents, like they were for DC circuits . . . or purely resistive AC circuits . . . but total current is the vector sum of the branch currents.

If the XL branch = 10K ohms, and the R branch = 10K ohms, then the circuit phase -angle will be 45 degrees. (Each branch current has an equal influence on IT). If Xi. is smaller than R . . . then the circuit will tend to act "more inductive" . . . since Ii. will be greater than la. Of course, if R is smaller than Xt., the opposite is true. Let's see if you have the general idea of series and parallel RL circuits. Try the Checkpoint!


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NET ( FORCE

LOAD( o

CHECKPOINT: 1. For a series RL circuit with a resistor of

20K and an inductor whose reactance is 30K, the phase angle between EA and IT

would be (more than 45 degrees; less than 45 degrees)?

IR

2. For a parallel RL circuit with a resistor of EA

20K an an inductor whose reactance is

IT 30K, thé phase angle between EA and IT

IL would be (more than 45 degrees: less than 45 degrees?

ANSWERS:

1. More than 45 degrees. 2. Less than 45 degrees.

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INFORMATION A very interesting point regarding the charging and discharging of the capacitor is that the "charge or discharge" current must flow in order to change the potential between the plates . . .

that is ... current "leads" voltage changes on a capacitor. Remember our old friend "ELI" for inductors. Now we want to complete the "crutch" for you. REMEMBER THE SENTENCE . . . "ELI THE ICE MAN". Interpreting the "crutch" sentence . . . E Leads I

for an inductor and I Leads E for a capacitor. In other words . . .

E is before I in our crutch word ELI (where L stands for inductance); and I is before E in our crutch word ICE (where C stands for capacitance).

In our charge and discharge discussion, we showed a fixed DC voltage source. Once the capacitor charged to the source voltage, there was no more charging current in the circuit. (No more current flow). Now ... if we were to continuously vary the source voltage

. . guess what would happen. You're right . . . there would be a constant flow of charging and discharging current as the capacitor attempted to charge to the source voltage and discharge, as appropriate. In effect, that is what happens when AC is applied to a circuit with capacitance in it. Look at the diagram at the right.

Note two things from the diagram. 1. Current continues to flow as long as the source voltage is chang-

ing ... thus even though a capacitor BLOCKS DC ... it effectively PASSES AC!

2. The current and voltage are 90 degrees out of phase for the capacitor . . . with current LEADING voltage. (Our old crutch ... "ICE", I before E.)

Before we move on to look at series and parallel RC circuits in the same manner as we studied series and parallel RI, circuits, let's discuss total capacitance of capacitors in series and parallel. And the resultant total capacitive reactances, in series and parallel.

To help you understand this, we'd better tell you what PHYSICAL factors of a capacitor determine its capacitance value. You recall that one farad of capacitance was that amount of capacitance which would store one coulomb (6.28 x 10'" electrons) of charge with one volt applied. The THREE physical factors which determine the amount of capacitance for any given capacitor are: 1. Total area of plates facing each other. 2. Spacing between the plates (dielectric thickness). 3. Characteristic of the material, or dielectric between the plates.

Some examples of dielectric materials commonly used are: air, mica, waxed paper . . . and other non-conductive materials.

REFERENCE & DATA

Ec "LAGS" I BY 90°

ER "IN PHASE" WITH I

CHECKPOINTS

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INFORMATION As a reminder, let's mention that because a farad is a huge amount of capacitance . . . (a capacitor with this value might be the physical size of a small room) . . . capacitors generally come in values of micro farads (millionths of a farad); and micro -micro farads (trillionths of a farad). A more modern term for the later is "pico farads" . . . meaning the same thing. A 10 microfarad capacitor is normally written as 10 uf or 10 mfd. A 10 micro -micro farad (sometimes written uuf) can he found frequently written as 10 pf (for pico farad).


PLATE N0. 2 PLATE NO.2

O.K. We've seen that capacitors in series add like resistors in parallel. - C1 C2

As you might already have anticipated . capacitors in parallel, add like resistances in series. This means that if I have a 10 uf capacitor in parallel with another 10 uf capacitor . . . the total capacitance of the circuit is 20 uf. Look at the circuit in column 2, and let's play our little game of pretending we are the source again. Now, the source sees the following;

T PLATE Il NO. 1

PLATE NO.1

1. The negative side of the source sees CI's plate No. 1 plate area AND C2's plate No. 1 plate area.

2. The positive side of the source sees Ci's No. 2 plate area PLUS C2's plate No. 2 area.

3. The spacing between all these plate areas is effectively ONLY ONE DIELECTRIC THICKNESS.

Therefore, following the concept mentioned earlier that the capacitance is directly related to plate area and inversely related to the thickness of the dielectric . . . it is obvious that C total will be the sum of the capacitances. (We have twice the plate area and the other two factors are the same: namely, spacing between plates and dielectric constant.)

Try the checkpoint!

As you can see . . . you simply add up the values of all the capacitances in parallel to get the total capacitance. So, summarizing: CAPACITANCES IN SERIES ADD LIKE RESISTORS IN PARALLEL . . CAPACITANCES IN PARALLEL ADD LIKE RESISTORS IN SERIES! (Just the opposite from the way finding total inductance works ... right? Right!)

CHECKPOINT: What is the total capacitance in the circuit shown?

Cl IC2 C3

-plow 5}tf 5f

ANSWER: CT = 20uf(10+5+5)

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INFORMATION

Now then . . . let's relate the Xc and total capacitance concepts into one package.

Remember that capacitors in series added like resistors in parallel, and, capacitances in parallel added like resistors in series. This means when we put capacitors in series the total capacitance decreases ... when we put capacitors in parallel, the total capacitance increases. Since the reactance (or opposition to current) that a

capacitor shows is inverse to its capacitance value . . . it follows that when we series capacitors, we will increase the total reactance, since CT decreases. THIS SIMPLY MEANS REACTANCES IN SERIES ADD LIKE RESISTANCES IN SERIES. This is true for any type reactance . . . either capacitive, or inductive. It is also true that REACTANCES IN PARALLEL ADD LIKE RESISTANCES IN PARAL- LEL. (In other words . . . the total reactance of parallel capacitive reactances will be less than the least branch reactance).

Try the checkpoint!

If you didn't get the answers, here's a hint, so you can try again until you understand. You should have used a form of the "product over the sum" formula to find the total capacitance in question 1. (CT = CI x C2/Ci + C2) ... and total reactance in problem 4. You should have simply added the series reactances in question 2 ... and added the capacitance values to find CT for question 3. If you missed these . . . recycle yourself through the reading material which is pertinent.

Now we've talked about factors that influence capacitance factors that influence capacitive reactance . . . and we have seen how total capacitance and total Xc can be solved for series and parallel circuits.

THE RC CIRCUIT

Let's move on now to a quick analysis of how current and/or voltage distributes itself in circuits that have both resistance and capacitance. We'll confine ourselves to AC voltage and current . . . since you already understand that in DC, once the capacitor is charged to EA, current ceases (since the capacitor "blocks DC").

Let's look at the simple RC series circuit first.

100

REFERENCE & DATA

TOTAL c = 5yt TOTAL Xc = 200 I1

SERIES C CIRCUIT

Xc2=10011

TOTAL c = 20 Pf TOTAL Xc = 5011

PARALLEL C CIRCUIT

CHECKPOINTS

CHECKPOINT:

Xc1 =30011

Xc2=1001l

1. What is the total capacitance of the circuit shown above? uf.

2. What is Xc total? ohms.

3. What is the total capacitance of the circuit shown below? uf.

C2 15 pf

Xc1= 300f1 Xc2=100II

4. What is the total Xc of this circuit? ohms.

ANSWERS:

1. 3.75 uf 2. 400 ohms 3. 20 uf 4. 75 ohms

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INFORMATION We can conclude then that if R = Xc in a parallel RC circuit

. . . the circuit phase angle will be 45 degrees. If R is greater than Xc . . . then more current will flow through the capacitive branch . . . and the circuit will act more capacitively .. . hence, phase', angle will be between 45 and 90 degrees. If Xc is greater than R . . . then the opposite is true and the circuit phase angle will be less than 45 degrees . . . and be somewhere between 0 degrees (a pure resistive circuit) and 45 degrees (a circuit where R= Xc. )

To see if you understand these principles . . . try the next checkpoint.

Trust you got those answers correct. If not . .. here's the thinking you should have used. For question 1 . . . we can make a general statement which is true in virtually all cases . . . If there is any "net" capacitive reactance in the circuit ... the resultant circuit total current will lead E applied by some amount. THIS IS TRUE FOR BOTH SERIES AND PARALLEL RC CIRCUITS. (By the way . . . if there is a net "inductive reactance" in the circuit . . . the voltage will lead the current by some amount). It may help you to remember the concept that inductors oppose changes in current . . . thus voltage leads . . . capacitors oppose a change in voltage . . . thus current leads. For question 2 . . . since Xc was greater than R . . . then there will be more current through the R branch (since E is the same across both components in a parallel circuit) . .. hence, the resistive branch has more effect on total current . . . therefore, the total current will lead Ea by less than 45 degrees. The thinking on question 3 is: if capacitance were decreased . . .

then Xc would increase (remember the inverse relationship?). If Xc increased . . . then the current through the capacitor branch would decrease . . . and the resistive branch current is even more in control of IT ... hence the phase angle would decrease.

REFERENCE & DATA Ic

Ic

Ic

IT

IT

IR

IR

or EA

CHECKPOINTS

CHECKPOINT: 1. Refer to the circuit below and determine

if the circuit current leads or lags Ea. IT Ea.

Y,c=20K

EA 2. Will the phase angle be less than 45 de- li

grees . . . more than 45 degrees . . . or exactly 45 degrees?

1.1 EA IR

3. If the value of capacitance were decreased . . would the phase angle increase, de-

crease or remain the same?

4. In order to achieve a phase angle of 45 degrees for this circuit, would the applied FREQUENCY have to be increased, decreased or kept the same?

ANSWERS:

1. Leads 2. Less than 45 degrees 3. Decrease 4. Increased

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INFORMATION Impedance then, is the total opposition the source "sees" to current flow. Another way of stating this is to say that the ratio of Er to IT is the impedance . . . or opposition to current. Ohm's Law states that ohms of opposition equals E/I. Because the presence of any reactive component (I. or C) in an AC circuit causes circuit current/s to be out of phase with circuit voltage/s ... it is logical to expect that we cannot just simply add up the reactances and resistances to find total circuit opposition (impedance).

For our purposes . . . it is only necessary that you realize that the circuit impedance (commonly represented by the symbol Z) equals the circuit voltage divided by the circuit total current. The only difference between Z and R is that Z is made up of a combination of R and X (reactance). For PURELY resistive AC circuits . . . you can deal with R -total as always. For PURELY REACTIVE circuits (All C's or All L's) you can deal with the reactances just like you do resistances to find total opposition. When both resistive and reactive components are present in the same circuit . .

it takes vector type analysis, or other math to analyze . . . but in ALL cases . . . the total opposition equals E applied divided by I total.

Try the checkpoint!

How'd we get those answers? Well, Z = h/I ... thus. 100v/ 10 mA = OK ohms. If frequency is increased ... then Xc will decrease (look at the formula). If Xc decreases and R stays the same . . . then it is logical that the total opposition must go down. This brings about a

very "nifty" general statement regarding circuit total resistance and/or total impedance.

NO MATTER WHAT KIND OF CIRCUIT .. . SERIES, PARALLEL OR SERIES -PARALLEL . . . IF ANY ONE OR MORE OF THE ELEMENTS CHANGES VALUE OF OPPOSITION ... THE TOTAL CIRCUIT OPPOSITION WILL BE CHANGED.

REFERENCE & DATA

ZT = ÍT IN ALL CASES

R2 RT = RI x R2

R1+R2

ZT DOES NOT EQUAL

R x Xc R + Xc BUT RATHERIT

CHECKPOINTS

CHECKPOINT: 1. What is the Z (impedance) of the circuit

shown below? ohms.

I=10mA

EA= 100 V

2. Would the circuit impedance increase, decrease, or remain the same if frequency were increased?

ANSWERS:

1. 10K ohms. 2. Decrease

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INFORMATION

THE RESONANT CIRCUIT

Since we have now studied inductance, inductive reactance and

RL circuits . . . capacitance, capacitive reactance and RC circuits . . let's put all these components together and study a very interesting phenomonon called Resonance.

We'll start with a series RLC circuit and the phenomenon of series resonance.

Look at the circuit in column 2 and note the values of reactances and the resistance value.

Because XL (inductive reactance) and Xc (capacitive reactance) affect circuit current and voltages in "opposite ways" ... some interesting things happen in this circuit. In effect . . . the XL and the Xc cancel each other out . . . so the source sees a circuit where the only effective limitation on current is the R value within the

circuit. (This is assuming a perfect coil with no resistance and a perfect capacitor, with no resistance . . . in which case they would perfectly cancel.) Assume the theoretically "perfect" condition ... what would be the current in this circuit?

Well, using common sense and Ohm's Law . . if the reactances cancel each other, the circuit impedance is equal to the R value of 1K ohm. Current = voltage divided by resistance (or impedance) ... thus 10v/1K = 10 mA of circuit current.

That was easy wasn't it? Nov, here is where the fun begins. Since the current is the same through all parts of a series circuit . . . then the 10 mA of current must be passing through the Xc and the XL as well as the R. Using Ohm's Law to solve for voltage drops we find that 10 mA times XL of 10K = 100 volts . . .

AND, 10 mA times Xc of 10K = 100 volts. Wow! With only 10 volts of E applied, we're getting 100 volts across the coil, and 100 volts the capacitor ... THAT'S RIGHT! The secret is that these reactive voltages are equal. . . BUT OPPOSITE IN PHASE . . . thus, as

far as the total circuit is concerned EA is dropped across the

R, and the reactive voltages cancel each other. Nevertheless, there REALLY IS 100 volts across each reactance, individually. If we

were able to have a meter (voltmeter) which did not "load" down the circuit ... we could actually measure that voltage. This unusual circumstance occurs in series resonant circuits. In fact, one way of judging the "Quality" of the reactive components is to note how many times greater their voltage drops are than E applied. This is called the Q of the circuit. It's obvious then that this ratio is directly related to how many times greater the XL or Xc:

is, than the R.

106


EA= 10V

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INFORMATION For question 1 ... I = E/Z. In this case Z=R because the reactances cancel each other, since they are equal in value and opposite in type. I = 10v/1K = 10 mA.

2.-EL = I x XL = 10 mA x 20K = 200v

3.-ER = I x R = 10 mA x 1K = 10v.

4.-Q = XL/R (or EL/EA) = 20

5.-XL is directly related to frequency and/or inductance. In this case L stayed the same and frequency was doubled . . . then XL would double. (Remember Xi. = 2 ft. formula?) Since the 2 pi and the L were constant, but f doubled . . . the product, or answer is twice as great.

6.-Explanation given in answers column.

7.-Since the circuit impedance would now be made up of 1K of R and 30K of "left -over" XL . . . rather than just the R, as at resonance . . . obviously the circuit Z would increase.

8.-Since circuit current is inverse to circuit impedance, and impedance increased ... then I must decrease.

In column 2 we have put a little "Summary Chart" for series resonance. Try to learn these key facts!

Moving on ... let's talk for a few minutes about Parallel Resonance.

Again . . . resonance (for our purposes) occurs when Xt. = Xc. We will only consider the theoretical "perfect" coil and capacitor (no internal resistance). In a practical sense, all these components are not perfect . . . but the characteristics we mention are true in a

general sense for parallel resonant circuits . . . and you need to know these "generalizations" for your FCC test.

Look at the circuit in column 2. If XL really does equal Xc . . .

then both "branches" should have equal currents . . . because in parallel circuits E is the same across all branches. Each branch current = E/opposition in ohms. In our example I = 10 volts/10K = 1 mA for each branch.

108


CIRCUIT CHARACTERISTICS AT SERIES RESONANCE

1. XL = Xc (net reactance = 0) 2. Z = R (thus Z is minimum) 3. I = maximum 4. Circuit Phase Angle = 0° 5. EL or Ec = Q times E applied 6. Q = XL orXcp EL Ec

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INFORMATION Another way of looking at the resonant frequency is to imagine that you are pushing a swing with someone in it. If you push "at just the right moment", the person will continue to swing at the same level. If you push at the wrong moment . . . you will change the level of the swing . . . or may even slop it. The "frequency" at which you push, is the natural, or resonant frequency of the swing. In the electrical circuit . . . the resonant frequency is the frequency at which new energy "injected" into the circuit will sustain the level of voltage and current and overcome :he losses. It's not too important that you know any deep theoretical knowledge here . . . but if you have the concept that there is one natural frequency . . . the resonant frequency . . . for any given LC combination.

Try using the "resonant frequency formula" on the problem. HINT:

fr = 0.159/ C Where fr= the resonant frequency in Hertz.

L = inductance in Henrys C = capacitance in farads.

Watch out for your decimal places. Good luck with the checkpoint!

By the way, you can find the resonant frequency of a series I.C. or RLC circuit using the same formula.

THE ELECTROLYTIC CAPACITOR

Time now to move ahead from the DC and AC circuit theory portion of study to look at several components you will need to have a basic knowledge of for the FCC test. These components include some information about Electrolytic capacitors, semiconductor diodes (including Zener diodes), basic vacuum tube information and transistors.

You have already learned that a capacitor is composed of conductive plates separated by a non-conductor material, called the "dielectric". There is one particular type of capacitor that is unique; you need to know some of its characteristics. This is the Electrolytic capacitor.

110


"Photo Courtesy of Centraleb, the Electronics Division of Globe -Union, Inc."

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CHECKPOINT: Solve for the resonant frequency of the circuit shown in column /.

fr Hertz.

ANSWER:

fr = 53 Hertz.

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INFORMATION Let's see if you have picked up the most important points regarding the Electrolytic capacitor and the Zener diode. Try the checkpoint!

VACUUM TUBES

Moving on now . . . let's give you a quick refresher regarding vacuum tubes. You remember that in the Novice chapter we

talked about the diode and triode tubes. Here, we'll briefly review and then we'll add basic knowledge regarding the tetrode and pentode tubes.

At this point, it may be good to "flat out" state a law of electricity that we have implied over and over again in our discussions . . .

but have never really said in so many words. The "law of electric charges" is that LIKE CHARGES REPEL . . . UNLIKE CHARGES ATTRACT.

112


CHECKPOINT: 1. Two key characteristics of electrolytic ca-

pacitors are: they have (high. low) capacity for their size? they (can, cannot) be connected into AC circuits?

2. True or False? Probably the most common usage of electrolytic capacitors is as power supply "filter elements".

3. The most important characteristic of the Zener diode is (hat in operation it has a

relatively constant (voltage drop across it; current through it)?

4. This characteristic becomes evident for the Zener diode when the voltage applied to it reaches a level to cause "breakdown", "avalanche", or, its "zener" point. True or False?

ANSWERS:

1. High capacity for size; and Cannot be connected into AC circuit.

2. True 3. Voltage drop across it 4. True

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INFORMATION To reduce this plate -to -grid capacitance, so that higher frequencies could be amplified, someone got the idea of introducing a fourth key element into the tube, called a screen grid. The screen grid is located between the plate and the control grid. This in effect put two capacitors in series . .. thus reducing the total capacitance between plate and control grid. (Output circuit and input circuit.) note the schematic symbol for the TETRODE.

To make a long story short . . . this idea of /the extra grid, the screen grid, worked! It was now possible to amplify higher frequencies. By the way . . . the screen grid normally operated at a DC positive potential with respect to the cathode. Because the screen grid is positive, it was found that under certain conditions, the screen might "drag away" some of the plate current. This was generally undesirable, for the plate circuit is the output circuit. The electrons that were being drained from the plate circuit, were electrons which "bounced off" the plate due to high impact electrons coming from the cathode. These secondary emission electrons tended to be attracted to the positive screen. To reduce this undesired effect, a third grid was added to the tube, called the suppressor grid. It suppressed this undesired effect. Because the suppressor grid is electrically at the same potential as the cathode ... negative with respect to the plate . . . electrons bounced off the plate are "repelled" back to it by the negative suppressor grid.

Note the "cutaway view" of a typical tube; it shows the internal construction of these elements we've been talking about (plus others).

This' really completes the BASIC family of vacuum tubes. The diode. triode, tetrode and pentode are really the key types of vacuum tubes. There are some special-purpose tubes which have been in- vented for particular jobs . . . but generally, all tubes are derived from the basic types just mentioned. The advantages of the tetrode and pentode over the triode are that they can amplify at higher frequencies . . . and there is good isolation between the input circuit (control grid) and output circuit (plate circuit). We look a little more deeply at how these tubes are used to amplify later in the book. Try the checkpoint to see if you picked up the needed terms and idea.

114

REFERENCE & DATA

CONTROL

GRID

PLATE

SCREEN GRID

TETRODE TUBE CATHODE

PENTODE TUBE

Courtesy of RCA

CHECKPOINTS

CHECKPOINT: 1. A "tetrode" is a tube with four main ele-

ments. True or False?

2. Name the five elements in a pentode tube ... not counting the heater.

3. The screen grid of a tube is normally operated at a (+, -) potential with respect to the cathode? potential.

4. The suppressor grid in a pentode is physically located between the grid and the

5. Which can amplify higher frequencies .

a triode or a pentode?

ANSWERS:

1. True 2. Cathode, Control Grid, Screen Grid, Sup-

pressor Grid & Plate. 3. Positive 4. Screen grid and Plate 5. A Pentode

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INFORMATION

Up to this point, we have generally gone into quite a bit of detail on

each subject covered ... mainly because we felt the subjects were significant and foundational. If a person learns his "fundamentals" well ... he can learn the rest far more easily. We felt that you, as a

hobbyist or student, would find the understanding of Ohm's Law, basic circuits, and the fundamental components of which all electronics circuits are made up, to be well worth learning. After all, ít's from these components (resistors, inductors, capacitors, tubes and transistors) that single function stages, such as amplifiers and oscillators, are made up . .. and from these stages, systems are built (con- taining combinations of these stages and components to perform a

useful task).

Obviously, it is beyond the scope of this book to delve deeply into each and every one of the myriad topics associated with Amateur Radio.

Our approach from here on out will be to try to organize the many topics into general categories. We will not attempt too detailed a study of any given item, but will endeavor to give you the information you

need to know to procure that coveted General Class License.

Good Luck! Stick with it . . . you're closing in on your goal!

APPLICATIONS OF COMPONENTS

RECTIFIERS

Remember us mentioning the term rectification (changing AC to pulsating DC) back in the Novice chapter? There are three basic rectifier circuits that you should he able to recognize, or draw, in schematic form, for the FCC test.

These are: the Half -wave rectifier, the Full -wave rectifier and the

Bridge rectifier. In the next column you will see a diagram of each

of these types ... both in vacuum -tube type, and semiconductor diode type. Learn to recognize these circuits ... or, better yet, learn to draw them yourself without looking (use the 3rd Column).

The reason the half -wave rectifier is called that is during only one half cycle (one alternation) is the plate positive with respect to the cathode (or anode positive with respect to cathode for the semiconductor). As you know, the only time these diodes will conduct is when the plate, or anode is positive with respect to the cathode ... therefore, no current flows during the other alternation of the AC signal ...and there is no output for that half cycle ... hence, a half -wave rectifier.


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INFORMATION

RF FILTERS

While on the subject of filters, we might as well mention a couple of

names, or types, that may appear on the FCC test. These are not low

frequency filters such as the power supply filters we just looked at. This, you can tell by the fact that the coils are not iron -core, such as

those used at power frequencies, or audio frequencies, but rather are air -core inductors.

So you won't be caught off -guard, these filters are sometimes referred to as the Constant -K type.

The most important thing to remember about these filters Is that the

reactive elements layout identifies whether its pi- section, or T -

section ... and whether it's a high-pass, or a low-pass filter. As you

can see . . . if the reactive elements form the shape of the term pi

. . . it's called a pi -section filter ... íf they form the shape of a T

... it's called a T -section filter. Now, for the "high-pass" and "low- pass" terms ... let's quickly explain. High-pass means that it passes the high frequencies along ... Low pass means that it passes the

low frequencies along. With your knowledge that inductor's opposition increases with frequency, while capacitor's opposition to AC

decreases as frequency increases, you can reason why the circuits are as shown. In the low-pass filters the inductors are "in series with the signal path", and offer low opposition to low frequencies . .

but high opposition to high frequencies . . . therefore, the low frequencies are passed along.

In the high-pass filters, the capacitor/s are in series with the signal path, and offer high opposition to low frequencies . . . but low

opposition to high frequencies . . . therefore, high frequencies are passed along.

THE AMPLIFIER

The next application of components we'd like to talk about is the

"tying together" of resistors, capacitors and the vacuum tube to create an amplifier. Really, what you're going to study here is just a "hot

rod" application of Ohm's Law, which you've already learned. (Now, you can see why we worked you over so thoroughly back there .. it makes this easier.)

Look at the diagram of a typical triode amplifier stage. First let's

analyze the DC voltages and currents. As you can see by looking at

the Equivalent DC circuit diagram . . . R2 and tube and R3 form a

simple series circuit. We've shown the tube as a variable resistance,

since that is the way it acts. When the grid is made more negative . . . it's resistance increases. When the grid is made less negative

(with respect the cathode), the tube conducts more ... so it's resistance is effectively lower.

INPUT SIGNAL

REFERENCE & DATA

L

C 1,--

o C j o

"PI-TYPE" LOW-PASS FILTER (or Pi -section)

"PI -TYPE" HIGH-PASS FILTER (or "Pi -section")

o T` o

"T -TYPE" LOW-PASS FILTER (or "T -section")

"T -TYPE" HIGH-PASS FILTER (or "T -section")

C2

R3 47K AMPLIFIED

SOUTPUT SIGNAL

=200V

TRIODE AMPLIFIER

DC EQUIVALENT CIRCUIT

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INFORMATION

Without going into great detail, we'd like to mention that the

amount of bias (negative voltage on the grid with respect to the

cathode) used for any given amplifier tube stage . . . determines its general operating characteristics. This is termed its class of operation. There are three main classes of operation for amplifiers and you should know these for our friend "Frank Charlie Charlie" (FCC). We've implied that these classes of operation only refer to

vacuum tube operation . . . Let's clear that up . . . these classifica- tions apply to BOTH TUBE AND TRANSISTOR AMPLIFIERS!

First, there is class A operation, in which plate (or collector) current flows all the time. (During.100% of the input signal cycle). This class operation produces the "highest fidelity" output . . . or, the most

faithful reproduction in the output of the input signal. Because DC

power is being used at all times, with, or without signal, Class A operation is the least efficient when comparing the DC power supplied to the circuit with the AC signal output power.

Class B is amplifier operation where the tube plate current (or transistor collector current) flows only during 1/2 the input cycle

. . because the tube or transistor are biased at cutoff. In other words, only the half cycle (or alternation) of the signal which "reduces bias" will cause conduction. For audio circuits, it takes a

special arrangement of two tubes (or transistors) in a "push-pull" circuit to amplify without distortion. However, the class B amplifier has higher efficiency than the Class A type . . . because with no signal, there is very little conduction, thus little power required from the power supply. For RF circuitry, class B operation can be

used where there is a tuned circuit to "fill in the other half cycle" ... even with one tube or transistor.

Class C operation is the most efficient . . . but is only used in RF

circuits where the "tank circuit flywheel effect" can complete the

output waveform and prevent distortion of the input signal.

While on the subject of amplifiers, we'd better mention one "special" type that has an application in electronic circuitry. The cathode follower is a circuit often used for "impedance matching". The input impedance of this amplifier is high ... the output impedance is low. So, when it is desired to transfer a signal from a high impedance circuit, such as the conventional amplifier stage, to a low impedance load (speaker, cable, etc.,) this circuit is useful. You won't be required to recognize or draw this stage ... but should be

familiar with the name cathode follower, and its principle characteristic of high input impedance, and low output impedance.


Key things to remember here are: 1. Class A = conduction for 100% of input

cycle 2. Class A = least efficient 3. Class B = conduction for 50% of input

cycle 4. Class B = more efficient than class A 5. Class C = most efficient-conducts less

than 50% of time 6. Class C = used in RF circuits with tuned

circuits (to prevent distortion)

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INFORMATION

If we distilled the theory down into the barest minimum . . . we

could say that an oscillator is a "hot -rod" amplifier. All we mean

here is that we take a little chunk of the output signal ... and pur-

posely feed it back to the input circuit ... in proper phase (positive or regenerative feedback) to sustain oscillations.

To make an oscillator work there are four basic requirements: 1. DC power supply or source. 2. Amplification 3. Frequency determining system 4. Feedback in proper phase (to overcome circuit losses)

The important things to watch for ín our discussions of several

types of oscillators is "what determines frequency" and "how is

feedback obtained". Watch for these as we point out the features

of the common types of oscillators and you'll probably be able to

recognize the various types.

Two very popular oscillator circuits are the Hartley oscillator and

the Colpitts oscillator ... so, we'll discuss them first.

Look at the vacuum -tube and transistor versions of the Hartley.

The key identifying feature of the. Hartley is the tapped coil in the

frequency -determining "tank" circuit. The DC voltage and current

for this circuit are provided by the B+ supply. The frequency -

determining device is the tank circuit composed of I., and C. Feedback is accomplished by the tapped coil in the tank circuit .. and of course amplification is produced by the tube or transistor. We have the four ingredients for an oscillator . . . and it works!

The Colpitts oscillator is quite similar in circuit configuration ... but instead of a tapped coil to control feedback . . . the Co!piIts uses a capacitive voltage divider (2 capacitors in series). Look at

the diagrams of the standard Colpitts oscillator and look for the

capacitor voltage divider in the frequency -determining tank circuit and feedback path.

Again, we meet the four requirements for oscillation in the Colpitis circuit. DC comes from B+ or Vcc source. The l.0 "tank" circuit (L, and C1 & n) determines frequency. Relative values of C, and C:

determine the amount of feedback . . . and amplification is pro-

vided by the tube or transistor.

Two other types of oscillators that might appear on the FCC test

are the Pierce crystal oscillator and the Electron -coupled oscillator (generally an adaptation of the Hartley or Colpills . . . but using a

pentode tube).

Lr

REFERENCE & DATA B+

SERIES HARTLEY

VCC

B+

COLPITTS OSCILLATOR

SHUNT HARTLEY

CHECKPOINTS

Emitter and Collector Circuit of Transistor Hartley

Oscillator NOTE: Tapped coil L, in each case.

Vcc

Colpitts Oscillator Emitter and

Collector Circuit NOTE: Capacitive voltage divider C, -C2 in

tank circuit (coil NOTE tapped).

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INFORMATION

THE SYSTEM APPROACH

We've spent a great deal of time so far teaching you about single components, such as: resistors, capacitors, inductors, tubes and transistors. We have brought some of these components together to

form basic stages or circuits, such as amplifiers, and oscillators. Now it's time we give you an overview of some systems that combine the components and stages into a total functional system. The two most important systems to radio communications, of course, are the transmitter and receiver.

As you already know, the transmitter has the job of producing a

radio wave, which is in turn radiated into space by an antenna system . . . and then it's the job of the receiving antenna to pick up this transmission and feed it to a receiver ... which processes the incoming signal and retrieves the "intelligence" contained in it

(code, music, speech or teletype).

Every system ... your phonograph, an automobile, a transmitter or a

receiver, etc. . . . has three basic elements . . . at least. These elements are: 1. An input 2. A processor 3. An output.

In the case of the phonograph . . . the input might be the grooves in the record ... the processor includes the pickup, the audio amplifier and the speaker system . . . the output is the sound waves, or disturbances of the air in front of the speaker/s.

The automobile takes fuel as its input .. . processes it through its engine and power train . . . and creates the output of motion of the wheels. Are you beginning to get the idea of what is meant by a "system"?


INPUT

INPUT- RECORD

OUTPUT PROCESSOR

A SYSTEM

PROCESSOR (PICKUP, AMP, SPEAKER)

SOUND WAVES) )

A PHONOGRAPH "SYSTEM"

FUEL IN ENGINE AND POWER TRAIN

MOTION OF WHEELS

AN AUTOMOBILE "SYSTEM"

CHECKPOINT: Every system has at least 3 impertant parts. These are: 1. The 2. The 3. And the

ANSWERS: 1. Input 2. Processor 3. Output

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INFORMATION

Now, to add audio intelligence directly, via microphone, tape deck or whatever ... a couple more blocks are added to our basic transmitter block diagram.

Note in the diagram that we have added a speech amplifier and a

modulator block. The speech amplifier obviously amplifies or builds up the level of the audio signal. The modulator stage builds the audio signal up further, to a level that can he fed to the modulated RF stage. The modulated RF stage, in addition to building up the RF signal level, as in the CW transmitter . . . also, combines the audio information with the RF to produce a modulated RF output signal . . . which is fed to the antenna system to be transmitted. The inputs to this system are: the power line AC plus the audio speech, music, or whatever. The transmitter processes this and generates an output of modulated RF wave or signal.

There are several systems of generating an F.M. (frequency modulated signal). The most direct method: the transmitter's oscillator is directly caused to change frequency by the audio signal. This method is called the "Direct" method of F.M. A simplified diagram is shown in column 2.

A quick "tour" through the transmitter yields the following: The oscillator generates an RF signal. The frequency multiplier stage amplifies and multiplies the frequency of the oscillator signal. The final RF power amplifier (P.A.) builds up the signal to an appropriate level for transmission by the antenna. The modulator takes the audio signal as input and processes it so that the output of the modulator changes the frequency of the oscillator in accordance with the audio intelligence . . . thus directly an F.M. signal. The power supply does its usual job of providing appropriate DC

voltages and currents for all the stages ... having processed the AC

input from the power mains.

Now, our special type of A.M. transmitter . . . the SSB (single- sideband) transmitter. Look at its functional diagram.

Incidentally, this is known as the filter -type sideband transmitter. There is another way of creating a single sideband signal . . . the phasing method . . . but we won't discuss it here. as the more popular method by far is the filter method.

"Walking our thoughts" through this transmitter, we see that there are several new blocks not shown in our other A.M. transmitter. DON'T GET WORRIED . . . YOU DON'T HAVE TO MEMORIZE THIS! Just get the concepts in mind ... and you've done well.


OSC --*BUFFER --« RF AMP

vANTENNA

f

I

MICROPHONE

POWER SUPPLY

SPEECH AMP MODULATOR

A.M. (amplitude modulated) TRANSMITTER

OSC. -- -

MO

FRED. MULT.

RF P. A

I

POWER SUPPLY

1

ANTENNA

AC POWER IN

DULATOR AF INPUT

SIMPLIFIED FM TRANSMITTER

OSC BUFFER

OUTPUT BALANCED BAND BAND LINEAR

MODULATOR PASS FILTER

MIXER PASS FILTER RF AMP

1

AF

I ) SPEECH

MICROPHONE AMP

SIMPLIFIED SSB TRANSMITTER

OSC.

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INFORMATION REFERENCE & DATA

How did you do on that? Do you know how to find the upper and lower sideband frequencies now? I hope so! Since we are on the subject of the modulated wave and transmitters . . . here's something that might be on the test you're preparing for. For an A.M. signal (single -frequency modulated), what are the relative powers of the carrier, sidebands, etc?

Look at the illustrations in column 2 . . . then come back to this column and we'll discuss the important facts.

Without modulation, the power out of the transmitter is the carrier power ... whatever that might be. When we 100% modulate the RF with a sine -wave, single -frequency audio signal . . . the amplitude of the RF wave changes at the audio rate. As we said before ... we create sidebands. Note from the sketch that the peak amplitude of the modulated wave is 2 times the amplitude of the unmodulated carrier. Assuming that the RF is fed to a constant impedance load (50 ohms for example) ... if we double the E across the load, the current through it must double. Since E is double and I is double, and power equals E x I . .. then we have 2E x 21 ... thus 4 times the power . . . at the "instant" of peak amplitude of the modulated wave. This business of calculating power at a given instant of time gives us an instantaneous power answer. If you were to find the average power in a 100% sine -wave modulated RF wave you would find that there is the carrier power + the power in the side - bands coming from the audio signal . . . the result is 1.5 times the power of the unmodulated carrier. So, briefly listing the important numbers for you to remember ... 1. Average power of unmodulated RF = carrier power 2. Average power of 100% modulated RF = 1.5 times carrier power. 3. Instantaneous peak power for 100% modulation = 4 times carrier

power.

By the way ... it is possible to see this "graphical" representation of an A.M. signal by looking at the output of an A.M. transmitter using an oscilloscope. The oscilloscope is an instrument that you will probably want to put high on your list of test equipment you should obtain.

"ENVELOPE"

Eo

t UNMODULATED RF (CARRIER)

MODULATED RF

AVERAGE POWER HERE= POWER OF CARRIER

PEAK OF ENVELOPE INSTANTANEOUS PEAK ENVELOPE POWER

BECAUSE EP .2 x Eo AND IPK 2 x 10 P=E x I, THUS PEAK POWER= 4 x CARRIER POWER

AVERAGE POWER OVER COMPLETE °CYCLE'' 1.5 x CARRIER POWER DUE TO ADDITION OF SIDEBANDS

CHECKPOINTS

CHECKPOINT: 1. If a transmitter's carrier frequency is 3.5

MHz, at what frequencies do the upper and lower sidebands appear if the modulated RF stage is being modulated by a 10 kHz. sine -wave audio signal?

USB freq. = MHz. LSB freq. = MHz.

ANSWER: USB freq. = 3.51 MHz. LSB freq. = 3.49 MHz.

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INFORMATION Another area of information we want to mention before leaving the broad subject of transmitters . . . and that is the topic of neutralization of RF amplifiers.

If you will think back a little, you will remember that we said an oscillator was simply a "hot rod amplifier". As you can recall .. .

it really boils down to the fact that any amplifier that has some feedback "in proper phase" can become an oscillator.

It's undesirable for an amplifier to function as an oscillator when the circuit is to be used as an amplifier. In a transmitter ... if an amplifier goes "into oscillation" ... it will generate spurious radia- tions that can interfere on the radio bands . . . not to mention it causes loss of efficiency and other unwanted side effects.

The method that is used to minimize the effects of unwanted feedback in an RF amplifier is called neutralization.

Generally, this unwanted feedback from output circuit back into input circuit is due to the "interelectrode capacitance" between plate and grid of a tube.

To neutralize the EFFECTS of this Cgp . . . there are several methods of "balancing out" this undesired feedback. The main thought here is that a signal path must be provided which will ALSO feedback a signal from output to input ... but will cancel the effects of the undesired feedback signal. The most common way of achieving this is with a NEUTRALIZING CAPACITOR. The two basic methods . . . from which other methods are derived, are shown.

Plate (sometimes called Hazeltine) neutralization is characterized by the Cn being connected from the "bottom" of the plate tank (LC) circuit and the grid. Grid (sometimes called Rice) neutralization is identified by the C being connected directly at the plate ... and going to the bottom of -thegrid inductor.

You are not required to be able to "draw these circuits from scratch". You should be able to recognize them ... or be able to draw in the neutralizing capacitor between the two correct points if you are provided a drawing with everything but Cn.

REFERENCE & DATA

PLATE (Hazeltine) NEUTRALIZATION (CN = from bottom of plate tank directly to grid)

GRIDE (Rice) NEUTRALIZATION (CN = directly from plate to bottom of grid circuit inductor)

CHECKPOINTS

CHECKPOINT: 1. For plate neutralization of an RF ampli-

fier, the neutralizing capacitor is connected between

2. For grid neutralization of an RF amplifier, the neutralizing capacitor is connected between the

ANSWERS: 1. Between "bottom" of the plate tank and

the grid of the tube.

2. Between the plate of the tube and the bottom of the grid inductor.

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INFORMATION The superheterodyne receiver takes the incoming signal ... mixes it with a signal developed by an oscillator stage in the receiver .. .

and produces new frequencies . . . from which ONE band of frequencies is selected by tuned circuits to be passed along to the rest of the receiver circuits for further "processing".

This "new" frequency (really band of frequencies in which there is the "carrier" and the "sidebands" in which the intelligence is contained) is generally chosen to be much lower in frequency than the transmitted signal . . . but much higher than the audio range ... or "super" audible . .. hence the term "superheterodyne" receiver. The reason for Mixing and producing this new frequency (called the Intermediate Frequency) is it is much easier to build stable, high -gain amplifiers at lower frequencies. Also, if we design our receiver oscillator so that we tune (change) its frequency at the same time we are tuning to different stations ... we can make it so that the Difference Frequency between the local oscillator signal and the incoming RF signal is always the same. This means our intermediate frequency (I.F.) amplifier stage(s) can be "fixed tuned" . . . and we don't have to "track" several stages' tuning mechanisms to make the receiver work. Do the Checkpoint.

ANTENNA

All right! turn on your "biological computer" (your brain) and let's take a mental walk through a typical superheterodyne receiver and follow the signal as it travels from the receiver antenna . . .

through the various stages of "processing" . . . on out to the speaker and of course, you know what happens from there on.

Let's think our way through the "block diagram" of a receiver at the right. We'll try to mention the key job (or processing) each block performs in this functional "system" ... the receiver.

First, the RF amplifier does just what its name implies ... it builds up the RF signal coming from the antenna.

132


RF AMP

"CONVERTER" r 1

MIXER

LOCAL OSC.

I.F. AMP.

2ND DETECTOR

f

B.F.o.

BLOCK DIAGRAM OF SUPERHETERODYNE RECEIVER

WEAK MODULATED RF SIGNAL

AMPLIFIED FF SIGNAL

RF AMPLIFIER -BUILDS UP WERK SIGNALS

1.

CHECKPOINT: The term heterodyne means

a. To mix two signals together to produce new signals.

b. To separate two signals in order to isolate them.

c. Neither of the above.

2. The new frequency (or band of frequencies) produced by mixing action in a superheterodyne receiver is called the

frequency.

3. This new frequency is usually (lower? higher?) than the "incoming" RF from the radio station ... but is higher than the

range.

ANSWERS: 1. A 2. Intermediate frequency 3. Lower than RF incoming

. higher than audio

AUDIO AMP

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INFORMATION

Out of the 2nd detector circuit comes our long-awaited audio signal. This signal is then amplified by the audio amplifier stage(s) and built up to drive the speaker.

There ... that trip through the receiver wasn't so bad after all, was it? Try the checkpoint to see if you picked up the most important facts.

How did you make out? Learning the block diagram of ANY system will help you understand it. It sort of gives you "pegs to hang your thoughts on". Makes it easier for you to "think" your way through a system. It gives you what someone has jokingly called the "GO- ZINTA-GOZOUTA" theory. Meaning: it helps you to learn what signal or input "goes into" a stage, or functional block . . . and what signal "goes out of" that stage or block. Once you have learned a system's block diagram and what type signal should go into each block and come out of each block ... you've gone a long way toward understanding how the system works. It also helps in practical troubleshooting of electronic systems. For example, if you know that all the appropriate signals are arriving at the input/s of a stage ... but the output is missing or abnormal ... you are pretty sure that the trouble is in that stage ...

REFERENCE & DATA

2ND DET.

nnAF

AF AMP

AF

2.

3.

4.

5.

CHECKPOINTS

CHECKPOINT: 1. What stage in a receiver demodulates the

modulated I.F. signal?

What stage(s) in a receiver changes the incoming signal(s) to a new frequency range?

Which stage or stages in a receiver build up the I.F. signals to a higher level?

Which stage in a communications receiver is needed in order to produce an audio tone for C.W. reception?

Which stage in a receiver builds up the incoming RF signal to a larger amplitude signal?

6. What is the job of the audio amplifier?

7. What stage(s) in a superheterodyne receiver is sometimes referred to as "fixed - tuned" stages?

ANSWERS: 1. 2nd Detector (or detector) 2. Converter (or mixer & local oscillator) 3. I.F. amplifier 4. B.F.O. (beat -frequency -oscillator) 5. R.F. amplifier 6. Builds up audio to level needed to drive

speaker. 7. I.F. amplifier stage(s)

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INFORMATION

IF you'll take your "memory banks" back quite a few pages ... into the Novice chapter, you will remember we talked about radio energy and electrical energy traveling at the speed of light . . .

which is approximately 186,000 miles per second ... or, 300,000,000 meters per second. This is how fast radio energy travels in "free space". We also mentioned that a wavelength is the distance that a given frequency of energy would travel during the time of one cycle (1 Hertz) of that frequency. You will recall that the formula for wavelength (symbolized by the greek letter lambda A was the velocity of the energy divided by the frequency.

Try the checkpoint!

I'll bet you are ahead of me already regarding why the half -wave and quarter -wave antennas are so -named. Yep! It's because their physical length is about 1/2, or 1/4 wavelength, respectively.

Because antennas are normally reasonably close to the ground, surrounding buildings, trees and objects . . . there is a "fudge - factor" involved when coming up with a useable and practical formula for determining the length of these antennas. It has been found that this "fudge -factor" is about 5% .. . that is, the antenna will end up being about 5% shorter than it would be if we used "free space" numbers for determining its length.

For example: the "free -space" formula for a Hertz (1/2 wave) antenna is:

The number that we'll call the "magic number" for a practical situation is:

In words, that says that the length of a half -wave Hertz antenna equals the "magic number" 468, divided by the frequency in MHz. The answer will be in FEET. See, we did all the converting from meters to feet for you. MEMORIZE THIS FORMULA! Try the problem in the checkpoint.

Naturally, the Marconi 1/4th wavelength antenna would be approximately half the length of a Hertz.


Á = 300 - 10 meters ;

30 MHz.

HERTZ ANTENNA LENGTH IN FEET

(t- 468/F(MHZ)+(

Antenna Leng h = 492

F (Mhz)

Hertz antenna length = 468

F (Mhz)

CHECKPOINT: 1. By what means does radio energy "get

transferred" from a transmitter into space?

2. What is the wavelength in meters of a frequency of 15 MHz.?

ñ = meters

3. The higher the frequency ... the (shorter? longer?) the wavelength?

ANSWERS: 1. antenna & its transmission line 2. 20 meters 3. shorter

CHECKPOINT: What is the proper length, in feet, for a Hertz antenna which is to operated at approximately 14.3 MHz.

FEET.

ANSWER: Approximately 32.72 feet.

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INFORMATION Each of the above -mentioned types have certain characteristics ... but let's just mention the main ones to remember.

Each has its own impedance characterisitc. This is called its characteristic impedance and is symbolized by: Zo. This is also sometimes referred to as its surge impedance. Don't get frightened by all these terms ... just be able to recognize them if you see them again. The things that effect the Zo of a transmission line really boil down to just a couple of items. Namely, the spacing between conductors . . . and the characteristic of the insulating or dielectric spacing material.

One formula that you should at least be able to recognize if you see it again relating to the impedance of an air -insulated parallel conductor transmission line. The formula is:

Without going into a deep discussion of this formula . . . you can see that if the spacing between conductors for the "open -wire" transmission line is increased . . . the Zo will increase. Also, the smaller the radius of the conductors ... the higher the Zo.

You, as a radio amateur, may want to construct an "open -wire" transmission line ... but chances are pretty good that most of the time you will buy ready-made transmission line in the form of "coaxial cable". Cables are designed with various values of Zo . .

so normally, you would simply buy coax (as it is called for short) that is approximately the Zo you need ... and which can handle the amount of power used.

While we are on the subject of impedances in transmission lines and antennas ... let's talk about a very important subject to radio amateurs. The subject is standing -wave ratio, abbrieviated SWR. If you listen on the amateur bands much at all . . . you will hear hams using this term frequently.

If we terminate a transmission line (no matter which type) whose characteristic impedance (Zo) is 73 ohms, with a dipole whose feed -point impedance is 73 ohms ... we would say that there is an impedance match between the two. In real life, if we did this .. .

virtually all the electrical (radio) energy traveling along the transmission line would be transferred to the "load" ... in this case the antenna. When the impedances are matched as in our example .. .

the SWR (standing -wave -ratio) would be considered "1". You'll see why in a moment.

When there is an impedance "mismatch" between "source and load" . . . we cause a situation where some portion of the energy is not absorbed by the load but is "reflected".


o SPACER

e -- END -VIEW

OPEN -WIRE LINE

Zo = 276 log b/a

where: Zo is the characteristic impedance b = the center -to -center distance

between conductors a = radius of conductor log = logarithm (base 10)

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INFORMATION

One other term we will mention so you won't be "shaken" if it appears on a test is the term feedline. This term is sometimes used instead of the word transmission line. It means the same thing. Its the line that "feeds" the energy to the antenna.

140


CHECKPOINT: 1. If an antenna feed -point impedance is

100 ohms and we are feeding the antenna with a transmission line whose Z0 = 25 ohms ... what is the SWR along the line?

2. Transmission lines may come in several forms. Among them are: the open - line twin- and

cable.

3. If we double the spacing between the conductors of an open -wire transmission line . . . what will happen to its characteristic impedance? . . . will it increase, decrease or remain the same?

4. One other name that is sometimes used for the term "characteristic impedance" is? " impedance". The symbol used to represent characteristic impedance is

5. If the minimum current value points along a transmission line where there is an SWR of 3, is 1 ampere ... what would be the maximum current value points along the line? A.

ANSWERS:

1. 4 2. open -wire; twin -lead; coxial cable 3. increase 4. surge -impedance; Zo 5. 3 amperes

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INFORMATION

SMORGASBORD!

O.K. Fasten your seathelts! We've come to the portion of the book where we will expose you to a big variety of facts ... some related ... some not . . . for you to glean for test preparation purposes. It

would be impossible to go into detail on all these ... but we want you to al least get a glimpse of these facts so you'll recognize them if they appear on the FCC exam.

Our first course in the Smorgasbord deals in the general area of significant terms and definitions.

The FCC has certain designations for the various types of emissions, or transmissions. Two broad categories of classification are . .

the type of modulation used (amplitude, frequency or phase and pulse), and the type of transmission (CW, phone. TV, etc.)

The type of modulation is indicated by the letter at the beginning of the FCC designation and the number denotes transmission mode ... for instance A3 transmission is amplitude modulated phone ... F3 is frequency modulated phone . . . and pulse modulated transmissions use the prefix P.

We don't expect you to memorize the whole chart at the right . . .

but you should look at it until you think you could recognize the definitions if you saw them. We'll point out in a couple of minutes which designations you will see most often as a radio amateur.

Wow! Looks like a lot of info! It is ... but the main ones to be aware of are A0, Al, A3, A3J and F3. AO is simply unmodulated and un - keyed carrier. Al = keying of an unmodulated carrier to send code. A3 = standard A.M. phone; A31 = single-sideband phone (which is

SSB-suppressed carrier AM). F3 is F.M. phone operation. That's not so bad, is it? You'll get a chance to see if you can recognize some of these designations when you take the sample "multiple- choice" FCC type test we'll give you at the end of this chapter.

Next on our "variety show" is the term TVI. This means television interference. This can be in many forms ... but basically is disrup- tion of the picture and/or sound on a TV receiver due to electrical radiation from outside sources. Some sources of TVI can be: machin- ery (such as diathermy machines), electric ovens, amateur transmitters and so on. If the transmitter has good shielding, and uses good engineering practice in design (by-passing, shielding and filtering), then the problem probably lies with the receiver. Two good methods of at least decreasing the possibility of TVI are: make sure the transmitter has a good "low-pass" filter in its output circuit (to eliminate frequencies which might appear in the TV channels ... and, on the receiver that is being troubled ... properly install a "high-pass" filter (which will reduce the strength of any signal from a nearby amateur transmitter).

Checkpoint.

142


Designations for Types of Transmissions

0 = absence of modulation (just RF carrier) 1 = Telegraphy without using a modulating tone (unmodulated C.W.) 2 = Telegraphy in which an audio tone is keyed on and off with the carrier (tone -modulated

C.W.) 3 = Telephony (including broadcasting) 4 = Facsimile (sending pictures, etc., by radio) 5 = Television (visual portion of transmission only) 6 = Four frequency diplex-telegraphy 7 = Multichannel voice -frequency telegraphy 9 = Cases not covered by 0-7

Other special designations regarding transmissions are:

Suffix A = single-sideband, reduced carrier H = single-sideband, full carrier J = single-sideband, suppressed carrier

B = two independent sidebands C = "vestigial" sideband (one sideband partially suppressed as in a TV transmission)

For Pulse type transmissions, the following:

D = amplitude modulated pulse transmission E = width (duration) modulated pulse transmission F = phase (or position) modulated pulse transmission G = code -modulated pulse transmission

Note: these designations are used as "suffixes", for example A3J = amplitude modulated phone using single-sideband, suppressed carrier transmission.

TRANSMITTER

ANTENNA ANTENNA

TV RECEIVER

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INFORMATION When dealing with radio frequencies, you sometimes hear someone talk about skin effect. This refers to the fact that high frequency electrical energy traveling along a wire or conductor tends to concentrate its current flow on the outside surface of the conductor . . . or on its "skin". This is due to the way the "induced counter- emf" is distributed throughout the cross section of the wire . . .

so there is less opposition to RF currents on the outside of the conductors. This phenomenon of the current traveling along the outside surface of the conductor is called skin effect.

Another rather important term we'll mention in our "smorgasborg" of terms is the decibel, abbreviated dB. This has to do with the comparative power levels of two signals. The mathematical formulas that you should recognize for this are:

O.K. Let's try a checkpoint on our variety of important terms.

144

REFERENCE & DATA

dB (gain, or loss in decibels) =

P2 10 10g10

P,

where: logro means logarithm to the base 10. P2/Pr means the ratio of the higher power level to the lower power level.

The dB gain or loss can also be found by comparing voltage levels. Here the formula is:

dB = 20 logia V2/V,

CHECKPOINTS

CHECKPOINT: 1. What is the third harmonic of a frequency

of 2.5 MHz? MHz.

2. Skin effect has to do with the way low frequency currents travel along a conductor. True or False?

3. When calculating for the number of decibels gain, or loss . . . the term 10 logo is used in conjunction with the power ratios . . . or, the voltage ratios of the signals being compared?

ANSWERS: 1. 7.5 MHz 2. False (high frequency currents) 3. Power ratios

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INFORMATION

RULES & REGULATIONS . . . OPERATING PROCEDURES .

& MISCELLANEOUS INFORMATION

Key fact to learn! What is the maximum permissible plate power input to the final stage of a transmitter as stated by the FCC? The basic answer to this is 1000 watts. This means that the DC plate voltage times the DC plate current must not exceed 1000 watts. There is one exception to this rule ... and that is on the 1.8 - 2.0 MHz. band the power is restricted to lower levels ... and is dictated by where you live. If your transmitter does not have accurate meters for measuring these parameters ... the maximum permissible power is 900 W. The FCC likes amateur stations to use the MINIMUM power necessary to carry out the desired communications . . . so there's less possiblity of interference with other operators. One other time we mentioned the regulation that the FCC says that the maximum permissible percentage of modulation is 100% ... but it bears repeating here. Remember that fact! It should be mentioned too, that you cannot modulate even 100% if your transmitter is not capable of producing a "clean" signal with 100% modulation. If a

transmitter is well -designed, however, the standard is 100%.

You may be asked why crystal control of transmitters is sometimes used. The basic answer to this is that crystals (used as frequency - determining devices) are capable of much higher Q's than coils and capacitors ... and hence are more accurate and stable. The FCC may ask you to know what precautions can be taken to avoid danger from shock from high -voltage electrical circuits. Sev- eral of the preventative measures include:

Don't allow high -voltage (including AC line voltage) circuits to be exposed where people can touch them.

Never work on a "live" electronic piece of equipment. This means be sure power is removed when working on equipment which has power supplies, etc., that contain high voltage circuits.

BLEED THE CHARGE OFF all power supply capacitors by shorting from the B+ side to chassis or ground with a "shorting bar". The shorting bar consists of a conductor connected to ground .. with an insulated handle that you can use to position the other

end of the conductor on the "hot" side of B+ ... thus, discharging any charge on the power supply capacitor(s).

All metal cabinets, chassis, etc., should be permanently connected to a good ground in addition to the normal AC power mains ground. Time for a Checkup!


CHECKPOINT: 1. What is the maximum permissible DC

plate power input to the final stage of an amateur radio transmitter? watts.

2. What is the maximum percentage of modulation that can be used with a well - designed amateur transmitter?

3. Which type tuned circuit gives better frequency stability and accuracy ... an LC tank circuit, or a piezoelectric crystal?

4. Name three precautions that can be used (on transmitters for example) to avoid shock

ANSWERS: 1. 1000 Watts 2. 100% 3. crystal 4. Be sure things are enclosed and no high -

voltage circuitry is exposed so someone could touch it. Discharge capacitors by means of a shorting bar. Be sure all equipment is well-grounded.

146

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INFORMATION

In an SSB transmitter, if the final amplifier stages are improperly loaded and/or there is excessive driving signal voltage (input signal), the stage(s) can be driven into a condition of "saturation". This simply means that further increases in driving voltage do not cause appropriate increased power output from the stage . . . in other words . . . the stage is not operating "linearly" and produces distorted output signals. If you were to look at the output of the SSB transmitter on an oscilloscope . this condition would be seen as flat -topping of the waveform. This shows that the peaks of the signal are being "clipped" due to the non-linear operation of the amplifier. We've talked about the Class A amplifier and the "linear" amplifier(s) used in SSB transmitter output stages. Now let's get in one last "tidbit" on amplifiers and mention something about the Class C amplifier. It is possible to have too much plate current flowing in a Class C amplifier if it is operated incorrectly. Some of the main causes for excessive plate current: incorrect load (too heavy a current drain from the circuit); too little negative bias (improper biasing); the plate circuit tank circuit not tuned to resonance (plate current dip . . . or minimum plate current is present when tuned to resonace); too small an input signal (driving power too small) ... particularly if grid -leak bias is being used to develop some of the operating bias.

Let's have another little checkup on these miscellaneous facts just before we move on to give you the final bits of information we want to convey before you try your SAMPLE FCC TEST at the end of this chapter.

As a radio amateur, you must always be alert to the "edges of the band" you are working, or transmitting on at the time. YOU ARE RESPONSIBLE to make sure that NONE of your signal "slops" over the edge of the band. One thing that is very important to keep in mind is that all the equipment you operate has "tolerance" ratings. For example, if you are using crystal control of your transmitter frequency . . . the crystal has a tolerance rating . . . and you must be sure not to get closer to the edge of the band than the amount of frequency which the crystal can he "off" (and still be within its tolerance rating). This is true of variable frequency oscillators, frequency meters for measuring frequency, etc. NEVER operate closer to the band edge than the tolerance of your equipment would dictate as being safe!


CHECKPOINT: 1. Which type transmission takes up the least

bandwidth in the radio spectrum . . .

Al, or A3?

2. Which occupies less spectrum or bandwidth ... A3 or A3)? What is the approximate comparison in bandwidths of these two types of transmissions?

3. To help prevent interference on the bands ... a "dummy load", or "dummy antenna" can be used during what part of the operating procedure?

4. Should a class A amplifier stage normally draw "grid current"?

5. When observing the output of a SSB transmitter with an oscilloscope . . . is "flat - topping" a sign of good operation?

6. Excessive plate current in a class C RF amplifier can be caused by too much bias. True or False?

7. When a class C RF amplifier plate tank circuit is tuned to resonance ... is plate current minimum or maximum? -

ANSWERS: 1.'A1. '

2. A3J"(about 1/2 as much as A3) 3. Tune up of 'the transmitter "

4. No 5. No ,

6: False -

7. Minimum

148

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1. The maximum input power to the final an amateur transmitter should not exceed:

a. 5 watts b. 10 watts. c. 100 watts d. 1KW e. 1000 KW

2. A series resonant circuit is characterized at resonance as having:

a. b. c. d. e.

Minimum current Maximum impedance High resistance Maximum current None of the above

3. The main characteristic of an Electrolytic capacitor is:

a. b. c. d. e.

A high capacitance -to -size ratio Not polarity conscious A low opposition to DC current Economical None of the above

4. The inductance of two 5 Henry inductors in series is:

a. 5 Henries b. 2.5 Henries c. 10 Henries d. 25 Henries e. None of the above

5. A 1 of capacitor is connected in series with a 2 of capacitor. The total capacitance will be:

a. 0.66 of b. 3 of c. 3/4 of d. 1/2 of e. None of the above

6. The formula for finding the resonant frequency of an LC circuit is:

a. fr = 2 ft. b. f r = XL + Xc c. fr = 0.159/ CT or d. fr=ZxQ e. None of the above

1

2 A CLC

SAMPLE FCC -TYPE GENERAL CLASS LICENSE EXAM

stage of 7. When XL = Xc in a parallel LC circuit . the circuit: a. current will be maximum b. impedance will be minimum c. phase angle will equal 90 degrees d. will not be at resonance e. current will be minimum

8. If a series resistive circuit has a 10K and a 20K resistor in series ... what is its total resistance?

a. 10K b. 20K c. 15K d. 30K e. 6.66K

9. What is the inductive reactance of a 10 Henry inductor at a frequency of 1000 Hz?

a. 628 ohms b. 6280 ohms c. 62.8K d. 10K e. None of the above

10. Using the same transformer in each case ... what would be the average DC output of a bridge rectifier (unfiltered) compared to a conventional FW rectifier (unfiltered)?

a. The same output voltage b. Twice the output voltage c. one half the output voltage d. three times the output voltage e. None of the above

11. Which of the circuits below represent a "capacitor - input" pi type filter?

A.

B.

C.

INPUT OUTPUT

o T o

D. 0------1K o

INPUT OUTPUT

o o

E. None of these.

12. The main elements in a triode tube are:

a. emitter, base and collector b. cathode, control -grid, screen -grid, plate c. plate, control -grid, cathode d. heater, plate e. cathode, grid, screen supressor, plate

13. Between what two points is Cn appropriately connected to obtain the proper schematic for "plate neutralization".

a. A & B

b. B & D c. A & D d. A & E e. None of these

II

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14. If the maximum voltage along a transmission line is 4.5 volts, and the minimum is 2 volts ... what is the SWR of the line?

a. 2

b. 4.5 c. 2.25 d. 1

e. None of the above

15. Which of the trapezoidal waveforms shown below represents "overmodulation"

A B

E . None of these

C

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31. What is the power in the sidebands single -tone modulated A.M. signal if power is 200 watts?

a. 200 watts b. 65 watts c. 265 watts d. 100 watts e. 33.3 watts

32. SWR is caused by:

a. Too high an antenna b. An impedance mismatch c. Transmitter out of alignment d. A "matched" load e. None of the above

of a 100% the carrier

33. If a final RF amplifier tube in a transmitter has a maximum plate dissipation rating of 100 watts; could it safely be used if the DC input power were 200 watts and the plate efficiency 60%?

a. Yes b. No

34. Which circuit below represents a "constant K Pi section high-pass" filter?

A.

B.

INPUT OUTPUT

INPUT ¡ o

V--\----0 OUTPUT

o

35. What is the Beta of a transistor whose base current is 10 microamperes, emitter current is 1 mA, and collector current is 0.97 mA?

a. 1

b. 100 c. 97 d. 10 e. 9.7

36. Which type antenna is noted for having a

angle of radiation"?

a. The Hertz b. The Marconi c. None of these

37. Equalizing resistors are often used across series capacitors, or rectifier diodes to:

a. Equalize the voltage across each element b. Equalize the current through each element c. Equalize the power dissipated by the circuit d. To increase the circuit impedance e. None of the above

38. How much power is being dissipated by a 10K resistor which has 0.2 amperes passing through it?

a. 20 watts h. 40 watts c. 400 watts d. 4000 watts e. 200 watts

39. Is it safe to operate a crystal -controlled transmitter within 10 kHz of the 7.0 MHz "bandedge" if the crystal tolerance rating indicates a frequency accuracy of plus or minus 1%?

a. Yes b. No

40. Is a Zener diode "forward", or "reverse" biased for normal operation?

a. Forward b. Reverse

41. What frequency can a General Class licensee use for phone operation on the 40 meter band?

a. b. c. d. e.

7.2-7.3 MHz 7.225 - 7.3 MHz 7.025 - 7.150 MHz 7.150 - 7.225 MHz None of these

42. A class B amplifier conducts:

a. About 50% of the time during each input cycle. b. about 100% of the time during the input cycle c. less than 50% of the time of the input cucle d. 10% of the time during each input cycle e. None of these

43. Are there countries in the world with which the "low U.S. does NOT have "3rd party agreements" hut

yet amateur operators can talk to other amateur operators from that country?

a. Yes b. No

44. Which of the items listed below can help reduce interference to others on the band? a. Use maximum power for transmitting b. Use a directional antenna c. Use 100% modulation d. Tune up using your regular antenna e. None of the above

45. To help avoid shock from electrical equipment in your station, you should: a. Leave power supplies uncovered so you can

see them. b. Only work on equipment with power on, so

pilot lights illuminate your work area. c. Open all "bleeder" resistors d. Ground metal cabinet and enclosures. e. None of the above

46. Which symbol below represents an NPN transistor?

A. B.

47. Some causes of grid current flow in a Class A amplifier might be: a. Too much bias and too much input signal b. Too little bias and too little input signal c. Too much signal input, and too little bias d. Bad grid in tube e. None of the above

48. One cause of excess plate current in a Class C RF amplifier might be:

a. Too "light" loading of final amplifier b. Plate lank tuned to resonance c. Too little plate voltage d. Too little bias e. None of the above

49. Which circuit shown below represents the conventional FW rectifier system?

B.

RI_

50. What might be done to protect the amateur station equipment from lightning? a. Use a lower antenna b. Be sure equipment is properly plugged -in c. Use an antenna "grounding switch" when sta-

tion is not in use. d. Paint the antenna e. None of the above

152

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SAMPLE NOVICE EXAM ANSWERS SAMPLE GENERAL EXAM ANSWERS

1. C 2. D 3. C 4. E 5. C 6. B

7. C 8. B 9. A

10. E 11. D 12. B

13. C 14. B 15. A 16. B

17. E 18. D 19. B 20. D 21. E 22. C 23. B

24. B

25. C 26. B

27. B

28. C 29. B

30. B

31. C 32. D 33. C 34. B

35. D

1. 1)

2. I) 3. A 4. C 5. A 6. C 7. E 8. D 9. C

10. B

11. B

12. C 13. B

14. C 15. C 16. D 17. E

18. D 19. D 20. C 21. C 22. D 23. A 24. B 25. C 26. B 27. C 28. C 29. B 30. C 31. D 32. B

33. A 34. A

35. C 36. B

37. A 38. C 39. B

40. B 41. B 42. A 43. A 44. B

45. D 46. B 47. C 48. D 49. B

50. C

1 4

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