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BI 2: MCH HI TIPTRANSISTOR FEEDBACK CIRCUITS

I. Dng c th nghim:- F.A.C.E.T base unit- Board mch TRANSISTOR POWER AMPLIFIERS- 1 VOM- 1 My o dng sng- 1 My to sng

II. Ni dung th nghim:Hon thnh 5 ch

Ch 1: Lm Quen Vi Board MchGii thiu:

Bo mch gm 4 mch sau:+ Hi tip ni tip hi tip song song+ Mch hi tip song song ni tip nhiu tng.+ Mch hi tip ni tip song song nhiu tng.+ Mch khuych i vi saiMch ATTENATOR lm gim bin tn hiu vo

Sau y l mch n gin ca khuch i mt tng dng hi tip:

Mch multistate shunt-series feedback v multistate series-shunt feedback lnhng mch nhiu tng gm 2 tng khuch i v l s kt hp hi tip ni tip vhi tip song song.

y l s n gin ca mch hi tip nhiu tng

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- Khi mch th t l bo DIFFERENTAL AMPLIFIERS. N minh ha ngdng ca transistor trong vic khuch i tn hiu nh:

- Tn hiu vo c th t vo mt u, cn mt u ni t, hoc c th tn hiuvo l vi sai vi 2 tn hiu vo 2 u. tng t, tn hiu ra cng c th ly 1trong hai hoc c hai u ra.

Ch 2: Hi Tip Ni Tipa. Gii thiu:

- Hi tip m xut hin khi mt phn ca tn hiu ra chuyn li v nh hngi khng ln tn hiu vo.

- Hi tip m lm gim li v tng bng thng, v cn nh hng n trkhng vo v tr khng ra ca mch.

- Hi tip dng l dng hi tip t l vi dng ra.- Dng ra i qua in tr hi tip RE. to nn p hi tip Vf- Vf i khng vi tn hiu vo Va.

- T s hi tip l t s ca tn hiu hi tip vi tn hiu ra

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- i vi mch trn, Vo ly trn RE do h s hi tip bng 1.- Mi quan h gia li khi khng c hi tip(A) vi khi c hi tip(Af) l:

Af = .1 AA

- hi tip n bng thng. Bng thng l di tn s trong h s khuch i

khng i.y l p ng tn s

- Tn s ct trn v tn s ct di l tn s m ti tn hiu ra gim i 3 dB.- Hi tip m l gim bin tn hiu ra, v m rng gii hn tn s ct trn v

tn s ct di, nn lm tng bng thng.

b. Tin hnh th nghim:kho st nh hng ca hi tip ni tip n liMc mch nh hnh v:

- iu chnh in p cung cp l 10 VDC, iu chnh R3 in p trn ccCollector l 5.5 VDC.

- t knh 1 hin th in p trn cc B ca Q1. t knh 2 hin th in p ra.- iu chnh tn hiu vo c tn hiu ra l sng sin 3 Vp-p tn s 10 kHz. o

c: Vb= 20 mVp-p- li khi khng c hi tip A =Vo/ Vb= 150.- Ngt kt ni t C2, iu chnh tn hiu vo l sng sin 3 Vp-p tn s 10 kHz.

o c: Vb= 0.32 m Vp-p- li khi c hi tip Af = Vo/ Vb= 9,375

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- Tnh Af theo cng thc: A f = RCL/RE= 3.9k /390 = 10.- Kt qu o Af xp x tnh bng cng thc.

Kt lun:- Hi tip ni tip m lm gim li ca mch khuch i.- Ta c th tnh li ca mch c hi tip t in tr RE v in tr RCL

Kho st nh hng ca hi tip ni tip n bng thngMc mch nh hnh v

- iu chnh in p cung cp l 10 VDC, iu chnh R3 in p trn cccollector l 5.5 VDC.

- Kt ni knh 1 ca my hin sng o ng ra v cho tn hiu vo c sngvung 2 Vp-p, 25kHz. Thit lp my hin sng hin th 5s/DIV.

- iu chnh my o c hin th nh hnh v:

- o c: T= 4s.- Tn s ct trn khi khng c hi tip f2 = 1/2T= 0.159/T= 40 kHz.- Ngt kt ni t C2, cho tn hiu vo l sng vung 2 Vp-p, 100kHz. Thit lp

my hin sng hin th 1s/DIV.- o cT= 0.18s.- Tn s ct trn khi c hi tip f2 = 1/2T= 0.159/T= 883 kHz.

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- Kt ni t C2, t sng vo l sng vung tn s 50Hz. Chnh bin sng vo sng ra c bin 2Vp.

- iu chnh my o c hin th nh hnh v:

- o c: T= 0.3 ms.- Tn s ct di khi khng c hi tip f1 = 1/2T= 0.159/T= 530 Hz.- Ngt kt ni t C2, t sng vo l sng vung tn s 50Hz. Chnh bin

sng vo sng ra c bin 2Vp.- o c: T= 1.8ms.- Tn s ct di khi c hi tip f1 = 1/2T= 0.159/T= 90Hz.

Kt lun:- Tn s ct di f1 khi c hi tip thp hn f1 khi khng c hi tip.- Tn s ct trn f2 khi c hi tip cao hn f2 khi khng c hi tip.- Hi tip m ni tip lm tng bng thng.

Kho st nh hng ca hi tip ni tip n tr khng voMc mch nh trn, c kt ni t C2

- iu chnh in p cung cp l 10 VDC, iu chnh R3 in p trn cccollector l 5.5 VDC.

- t knh 1 o tn hiu vo. t tn hiu vo l sng sin 80mVp-p, 10kHz.- t knh 2 ca my hin sng o in p trn cc B. t my hin sng

hin th ch o ADD- INVERT, 50 mV/DIV.o c: VR1= 50mVp-pDng vo Ii= VR1/R1= 5ATr khng vo Zi= 16k

- Ngt kt ni t C2. o tng t nh trn:VR1= 40mVp-pDng vo Ii= VR1/R1= 4ATr khng vo Zi= 20k

Kt lun: Zi tng khi c hi tip m ni tip.

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Ch 3: Hi Tip Song Songa. Gii thiu:

- Hi tip song song a mt phn tn hiu t ng ra v song song vi tn hiung vo.

- H s khuch i ca mch hi tip Af = Rf/Ri nn hon ton c lp vi ccthng s ca BJT hay ca mch

- H s hi tip = Ri/Rf.- Hi tip m song song l gim li, ta c th tnh li t gi tr in tr hi

tip v in tr vo.- Hi tip m song song lm gim li nhng tng bng thng.- Mch hi tip m song song lm gim tr khng vo v tr khng ra.

b. Tin hnh th nghim

kho st nh hng ca hi tip song song n liMc mch nh hnh v:

- iu chnh in p cung cp l 10 VDC, iu chnh R3 in p trn ccCollector l 5.5 VDC.

- t knh 2 o sng ra. Cho tn hiu vo c tn hiu ra l sng sin 3 Vp-ptn s 10 kHz.

- t knh 1 o sng Vb: Vb = 25 mVp-p

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- li khi khng c hi tip: A = Vo/ Vb= 120- Ni in tr R2 vo mch. iu chnh R3 in p trn cc Collector l 5.5

VDC. iu chnh li tn hiu vo tn hiu ra l sng sin 3 Vp-p.- t knh 1 o sng Vi: Vi = 0.68 Vp-p- li khi c hi tip : Af = Vo/ Vi= 4.4

Kt lun:- li in p gim khi c hi tip.- Af c th tnh theo cng thc: Af = R2/R1= 4.7

kho st nh hng ca hi tip song song n bng thngMc mch nh hnh v:

- iu chnh in p cung cp l 10 VDC, iu chnh R3 in p trn ccCollector l 5.5 VDC.

- t knh 2 o sng ra. Cho tn hiu vo c tn hiu ra l sng sin 3 Vp-ptn s 10 kHz.

- Kt ni knh 1 ca my hin sng o ng ra v cho tn hiu vo c sngvung 2 Vp-p, 25kHz. Thit lp my hin sng hin th 5s/DIV.

- iu chnh my o c hin th nh hnh v:

- o c : T= 2s.- Tn s ct trn khi khng c hi tip f2 = 1/2T= 0.159/T= 78kHz.- Ni in tr R2, cho tn hiu vo l sng vung 2 Vp-p, 100kHz. Thit lp my

hin sng hin th 1s/DIV.- o c T= 0,12s.- Tn s ct trn khi c hi tip f2 = 1/2T= 0.159/T= 132,5 kHz.

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- B ni R2, cho tn hiu vo l sng vung 50 kHz. iu chnh tn hiu vo tn hiu ra c bin 2Vp.

- iu chnh my o c hin th nh hnh v:

- o c: T= 0.4ms.- Tn s ct di khi khng c hi tip f1= 1/2T= 0.159/T= 390 Hz.- Ni in tr R2, cho tn hiu vo l sng vung 2 Vp-p, 50 Hz. t my hin

sng 1ms/DIV.- o c: T= 0.8ms.- Tn s ct di khi c hi tip f1 = 1/2T= 0.159/T= 198 kHz.

Kt lun:- Tn s ct di f1 khi c hi tip thp hn f1 khi khng c hi tip.- Tn s ct trn f2 khi c hi tip cao hn f2 khi khng c hi tip.

Kho st nh hng ca hi tip ni tip n tr khng vo raMc mch nh trn, b kt ni R2

- iu chnh in p cung cp l 10 VDC, iu chnh R3 in p trn ccCollector l 5.5 VDC.

- t sng vo l sng sin 80mVp-p, 10kHz. t my hin sng hin th ch o ADD- INVERT, 50 mV/DIV.o c: VR1= 90mVp-pDng vo Ii = VR1/R1= 9ATr khng vo Zi = 8,8 k

- Ni in tr R2 vo mch.o c: VR1 = 100 mVp-pDng vo Ii = VR1/R1= 7ATr khng vo Zi = 8 k

- Ngt kt ni R2, iu chnh R3 in p trn cc collector l 5.5 VDC.iuchnh tn hiu vo tn hiu ra l sng sin 3 Vp-p, 10 kHz.

- Tng in tr R8 ln gi tr cao nht. Ni R8 vo mch, iu chnh R8 sao choVo gim xung 1.5 Vp-p. B ni R8, o in tr tng ca R8 v R9, ta c Zo( khi khng c hi tip): Zo = 3k

- Ni in tr R2 vo mch, iu chnh R3 in p trn cc C l 5.5 V.iuchnh tn hiu vo tn hiu ra l sng sin 0.8 Vp-p, 10 kHz.

- Ni R8 vo mch, iu chnh R8 sao cho Vo gim xung 0.4 Vp-p. B ni R8,o in tr tng ca R8 v R9, ta c Zo ( khi c hi tip): Zo = 300

Kt lun: hi tip m song song lm gim tr khng vo, ra ca mch.

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Ch 4: Khuch i Nhiu Tnga. Gii thiu:

- C hai dng c bn ca khuch i hi tip nhiu tn l ni tip song song,v song song ni tip.

- y l dng mch hi tip song song ni tip

- Trong khuch i a tng vi hi tip song song ni tip t s ca Rsf/Refquyt nh li dng Ai.

- y l dng mch hi tip ni tip song song

- Trong khuch i a tng vi hi tip song song ni tip t s ca Rsf/Refquyt nh li p Av.

b. Tin hnh th nghimKho st li dng v tr khng ra mch hi tip song song ni tipMc mch nh hnh v

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- iu chnh in p a vo ng vo sao cho V0=100mVpk-pk, tn s 10kHzdng sng sinceI0 = V0/R11 = 100/100 = 1mApk-pk

- Dng knh 2 o R1VR1 = 67mVpk-pkIR1 = VR1/R1 = 67mV/10k = 0.0067mApk-pk

- Ai = I0/IR1 =1/0.0067 = 149.25- Tnh theo cng thc: Ai = R5/R7 = 33k/220 = 150- Av = V0/VR1 = 100/67 = 1.49- Z0 = RC0//Rc

Kt lun :- Mch ch yu dng khuch i dng v Ai ln hn nhiu so vi Av- Ai = in tr hi tip song song /in tr hi tip ni tip.- in tr ra cao xp x bng in tr ra trn collector ca Q2- Dng ra khng i khi in tr ti thay i v in tr ra cao ch ph thuc vo

in tr trn collectorKho st li p v tr khng ra ca mch hi tip ni tip song songMc mch nh hnh v

- Knh 1 ni cc base ca Q1- Knh 2 ni vi ng ra ca Q2- iu chnh sao cho ng ra 10kHz ,1.0Vpk-pk

o c Vi=18mVpk-pkAv=V0/Vi=1V/18mV=55

- tnh theo cng thc:Av=R8/R5=12k/220=54.5- tnh in tr ra ca mch c hi tip- Z0=(R8/)//R9=(15k/100)//5.6k=146

Kt lun :- in tr ra ca mch khuch i hi tip ni tip song song thp.- Tr khng ra t l vi gi tr in tr hi tip song song.- H s khch i p Av khng ph thuc vo in tr collector ca Q2 v

Av=R8/R5

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Ch 4: Khuch i Nhiu Tnga. Gii thiu:

- Mch khuch i vi sai c hai BJT c li bng nhau, b cho sai s cacc linh kin, R4 c gi tr bin thin c, iu chnh R4 c p ti thiugia hai collector gi l cn bng khuch i vi sai

- Mch khuch i vi sai c nhiu ch hot ng, ta c th ly tn hiu ra tmt u cc C cn u kia ni t (ng ra n), hoc c th ly t hai u(ng ra vi sai). Tng t, vi tn hiu vo ta cng c ng vo n hoc ngvo vi sai.

b. Tin hnh th nghimMc mch nh hnh v

- iu chnh R4 in p gia hai cc C ca hai BJT bng nhau.- Dng volmeter o din p trn R3 V R3 = 1.5 Vdc

IR3=VR3/R3=1.5V/3.3K=0.45mAIR5=2.IR3=2.0.45mA=0.9mA

Cho tn hiu xoay chiu vo

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- Cho tn hiu vo cc B ca Q1 c in p nh nh l 500mVpk-pk, c tn s1KHz sng sin

- Ta thy tn hiu ti cc C ca Q1 ngc pha vi tn hiu ti cc B ca Q1, tnhiu ti cc C ca Q2 cng pha vi tn hiu ti cc B ca Q1.

- S dng phng php ADD INVERT o p gia hai chn C ca hai BJTta thy tn hiu gp i tn hiu khi ch ly mt cc.

- in p ng ra Vc1=1.2Vpk-pk, Vc2=1.2Vpk-pk- li p ng ra n: Av1 = Vc1/Vi = 1.2/0.5 = 2.4

Av2 = Vc2/Vi = 1.2/0.5 = 2.4- Ng ra vi sai Vc3 = 2,4Vpk-pk- li p ng ra vi sai l AV = Vc3/Vi = 2,4/0.5 = 4.8

Kt lun:- li ng ra vi sai l tng ca 2 li ng ra n- li 2 ng ra n th bng nhau

Mc mch nh hnh v:

Ta c vic1=vic2Ng ra n ca Q1 v01=Avcm.vic1Ng ra n ca Q2 v02=Avcm.vic2Nn ng ra vi sai l vo=v01-v02=0Kt lun:tn hiu n pha khng c khuch i ng ra v vy mch khuch i vi sai ctc dng chng nhiu ng pha tt