Mch khuch i bn dn

Mch khuch i bn dn
Bi:
L Sc

Chng ny trnh by v cc mch khuch i dng tranzito. c th khuch i c

th cc tranzito hot ng ch tch cc, v do phi c cc mch in cung cp in
p phn cc cho tranzito.

C 3 cch mc mch khuch i l emit chung (EC), baz chung (BC) v colect
chung (CC). Mi cch mc u c u im v nhc im chung, nhng mch EC c
s dng rng ri nht v c h s khuch i in p v dng in ln.

Tng ng cc mch khuch i dng trazito lng cc, cng c cc mch khuch i
tng ng dng tranzito trng l SS, GS v DC. Cc mch khuch i dng FET c h
s khuch i thp nhng li c n nh v trnh nhiu tt hn so vi BJT.

Mt trong cc khi mch quan trng trong cc thit b in t l khi mch khuch i
cng sut. y thng l khi mch cui cng , c nhim v khuch i tn hiu ln
cng sut a ra ti. Cc tranzito dng trong cc mch khuch i cng sut thng
l cc tranzito chu c dng ln. Ty vo cng sut yu cu m c cc loi mch
khuch i cng sut khc nhau, hot ng ch khc nhau. Cc ch lm vic ca
tng khuch i cng sut bao gm ch A, AB, v B.

NH NGHA, CC CH TIU V THAM S C BN CA MCH

KHUCH I

nh ngha mch khuch i

Thc cht khuch i l mt qu trnh bin i nng lng c iu khin, nng

lng mt chiu ca ngun cung cp, khng cha thng tin, c bin i thnh nng
lng xoay chiu theo tn hiu iu khin u vo, cha ng thng tin, lm cho tn hiu
ra ln ln nhiu ln v khng mo. Phn t iu khin l tranzito. S tng qut
ca mch khuch i nh hnh sau

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Mch khuch i bn dn

Phn t c bn l phn t iu khin tranzito c in tr thay i theo s iu khin ca

in p hay dng in t ti cc iu khin (cc B) ca n, qua iu khin quy lut
bin i dng in ca mch ra bao gm tranzito v in tr Rc. Ti li ra gia cc C
v cc pht, ngi ta nhn c mt in p bin thin cng quy lut vi tn hiu vo
nhng ln c tng ln nhiu ln. n gin, gi thit in p t vo cc gc c
dng hnh sin.

T s hnh 2-2 ta thy rng dng in v in p xoay chiu mch ra (t l vi

dng in v in p tn hiu vo) cn phi coi l tng cc thnh phn xoay chiu dng
in v in p trn nn ca thnh phn mt chiu I0 v U0. Phi m bo sao cho bin
thnh phn xoay chiu khng vt qu thnh phn mt chiu, ngha l Io m v U0
m .Nu iu kin khng c tho mn th dng in v in p mch ra trong
tng khong thi gian nht nh s bng khng v s lm mo dng tn hiu.

Nh vy m bo cng tc cho tng khuch i (khi tn hiu vo l xoay chiu) th

mch ra ca n phi to nn thnh phn dng mt chiu I0 v in p mt chiu U0.
Chnh v vy, mch vo ca tng, ngoi ngun tn hiu cn khuch i, ngi ta cng
phi t thm in p mt chiu UV0 (hay dng in mt chiu IV0). Cc thnh phn
dng in v in p mt chiu xc nh ch lm vic tnh ca tng khuch i.
Tham s ca ch tnh theo mch vo (IV0, UV0) v theo mch ra (I0, U0) c trng
cho trng thi ban u ca s khi cha c tn hiu vo.

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Mch khuch i bn dn

Cc ch tiu v tham s c bn ca mt tng khuch i

H s khuch i K size 12{ widevec {K} } {}

Ni chung v tng khuch i c cha cc phn t in khng nn K l mt s phc.

Tr khng u vo v u ra

Tr khng u vo v tr khng u ra ca tng khuch i c nh ngha

Z = U ,Z = U
V r

I I
V r

V r

Mo tn s

Mo tn s l mo khi khuch i ca mch khuch i b gim vng tn s thp

v vng tn s cao.

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Mch khuch i bn dn

Mo phi tuyn.

Mo phi tuyn l do tnh cht phi tuyn ca cc phn t bn dn nh tranzito gy ra. Khi
uv ch c thnh phn tn s th ur ni chung c cc thnh phn tn hiu vi tn s l bi
ca tc l n.(vi n = 1, 2...) vi cc bin cc i tng ng lmax . H s mo
phi tuyn do tng khuch i gy ra c nh gi l:

Hiu sut ca tng khuch i

Hiu sut ca mt tng khuch i l i lng c tnh bng t s gia cng sut tn
hiu xoay chiu a ra ti Prvi cng sut tiu th ngun cung cp mt chiu: P0.

=
P r

P 0

HI TIP TRONG CC TNG KHUCH I

Hi tip l vic thc hin truyn mt phn tn hiu t u ra tr v u vo b khuch

i. Thc hin hi tip trong b khuch i s ci thin hu ht cc ch tiu cht lng
ca n v lm cho b khuch i c mt s tnh cht c bit.

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Mch khuch i bn dn

C th phn chia hi tip thnh cc kiu nh: Hi tip ni tip hoc song song (khi in
p hi tip v mc ni tip hoc song song vi in p vo). Hi tip in p hoc dng
in (khi in p hi tip v t l vi dng in/ in p ra).

Nu in p hi tip v ngc pha vi in p vo (khi n s lm gim tn hiu vo)

th l hi tip m. Hi tip m lm gim h s khuch i ca mch nhng b li
n li lm tng tnh n nh ca mch v tng di tn lm vic. Do trong cc mch
khuch i ngi ta thng s dng hi tip m.

Ngoi ra hi tip m cn c tc dng tng n nh ca h s khuch i, v n c

dng rng ri ci thin c tuyn bin , tn s (hnh 2-5) ca b khuch i nhiu
tng ghp in dung. V min tn s thp v cao h s khuch i b gim. Tc dng
hi tip m min tn s k trn s yu v h s khuch i K nh v s dn n tng
khuch i di bin tn v m rng di thng f ca b khuch i.

Nu in p hi tip v cng pha vi tn hiu vo (n s lm tng bin tn hiu vo

mch khuch i), th gi l hi tip dng. Hi tip dng lm tng h s khuch i
nhng lm mch khng n nh thm ch nu hi tip nhiu s lm mch xy ra hin
tng t kch v mch s dao ng, do hi tip dng thng dng trong cc mch
to dao ng.

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Mch khuch i bn dn

Hi tip m cng lm gim mo phi tuyn ca tn hiu ra v gim nhiu (tp m) trong
b khuch i.

Nhng quy lut chung nh hng ca hi tip m n ch tiu b khuch i l:

Mi loi hi tip m u lm gim tn hiu trn u vo b khuch i v do lm

gim h s khuch i Kht, lm tng n nh ca h s khuch i ca b khuch i.

Ngoi ra hi tip m ni tip lm tng in tr vo.

Hi tip in p ni tip lm n nh in p ra, gim in tr ra Rrht. Cn hi tip dng

in ni tip lm n nh dng in ra It, tng in tr ra Rrht.

Hi tip m song song lm tng dng in vo, lm gim in tr vo Rvht, cng nh

in tr ra Rrht.

Cn ni thm l hi tip dng thng khng dng trong b khuch i nhng n c

th xut hin ngoi mun do ghp v in bn trong hay bn ngoi gi l hi tip k
sinh, c th xut hin qua ngun cung cp chung, qua in cm hoc in dung k sinh
gia mch ra v mch vo ca b khuch i.

Hi tip k sinh lm thay i c tuyn bin - tn s ca b khuch i do lm tng

h s khuch i cc on ring bit ca di tn hoc thm ch c th lm cho b
khuch i b t kch ngha l xut hin dao ng mt tn s xc nh.

loi b hin tng trn c th dng cc b lc thot, b tr mch in v cc linh kin

hp l.

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Mch khuch i bn dn

CC S KHUCH I C BN DNG TRANZITO LNG CC

Tng khuch i Emit chung (EC)

S d ngi ta gi l tng emit chung l v nu xt v mt xoay chiu th tn hiu u

vo v u ra u c chung mt cht t l cc E ca tranzito.

Trong s ny Cp1, Cp2 l cc t ni tng, n ngn cch in p mt chiu trnh nh

hng ln nhau, R1, R2, RC xc nh ch tnh ca tng khuch i.

RE in tr hi tip m dng in mt chiu c tc dng n nh nhit, CE t thot thnh

phn xoay chiu xung t ngn hi tip m xoay chiu.

c im ca tng khuch i EC l tng khuch i o pha, tn hiu ra ngc pha vi

tn hiu vo.

Nguyn l lm vic ca tng EC nh sau: khi a in p xoay chiu ti u vo xut

hin dng xoay chiu cc B ca tranzito v do xut hin dng xoay chiu cc C
mch ra ca tng. Dng ny gy st p xoay chiu trn in tr RC. in p qua t
CP2 a n u ra ca tng tc l ti Rt. C th thc hin bng hai phng php c
bn l phng php th i vi ch mt chiu v phng php gii tch dng s
tng ng i vi ch xoay chiu tn hiu nh.

Phng php th da vo c tuyn vo v ra ca tranzito c u im l d dng tm

c mi quan h gia cc gi tr bin ca thnh phn xoay chiu (in p ra m v
dng in ra r) v l s liu ban u tnh ton. Trn c tuyn hnh (2-7a), v ng
ti mt chiu (a-b). S ph thuc UCE0 = f(IC0) c th tm c t phng trnh cn

bng in p mch ra ca tng: UCE0=EC-IC0.RC-IE0RE=EC-IC0RC-

I C0
- RE

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Mch khuch i bn dn

V h s gn ng 1, nn c th vit UCE0 = EC - IC0 (RC+RE)

Da vo c tuyn vo IB= f (UBE) ta chn dng cc gc tnh cn thit IB0, chnh l xc

nh c to im P l giao im ca ng IB = IB0 vi ng ti mt chiu trn
c tuyn ra hnh 2-7b.

xc nh thnh phn xoay chiu ca in p ra v dng ra cc C ca tranzito phi

dng ng ti xoay chiu ca tng. Ch rng in tr xoay chiu trong mch cc E
ca tranzito bng khng (v c t CE mc song song vi in tr RE) cn ti Rt c
mc vo mch cc C, v in tr xoay chiu ca t C2 rt nh.

Nu coi in tr xoay chiu ca ngun cung cp EC bng khng, th in tr xoay chiu

ca tng gm hai in tr RC v Rt mc song song, ngha l Rt~ =Rt // RC. T thy r
in tr ti mt chiu ca tng Rt= = RC + RE ln hn in tr ti xoay chiu Rt~. Khi
c tn hiu vo, in p v dng in l tng ca thnh phn mt chiu v xoay chiu,
ng ti xoay chiu i qua im tnh P.

dc ca ng ti xoay chiu ln hn dc ng ti mt chiu. Xy dng ng

ti xoay chiu theo t s s gia ca in p v dng inUCE =IC.(RC // Rt). Khi
cung cp in p vo ti u vo ca tng th trong mch cc gc xut hin thnh phn
dng xoay chiu ib lin quan n in vo uv theo c tuyn vo ca tranzito.V dng
IC=IB nn trn mch cc C cng c thnh phn dng xoay chiu iC v in p xoay
chiu ura lin h vi iC bng ng ti xoay chiu. Khi ng ti xoay chiu c

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Mch khuch i bn dn

trng cho s thay i gi tr tc thi dng cc C iC v in p trn tranzito uc hay ngi

ta ni l s dch chuyn im lm vic. im lm vic dch t P i ln ng vi 1/
2 chu k dng v dch chuyn i xung ng vi 1/2chu k m ca tn hiu vo. Nu
chn tr s tn hiu vo thch hp v ch tnh ng th tn hiu ra ca tng khuch i
khng b mo dng. Vic chn im lm vic tnh v tnh ton s c thc hin theo
mt tng khuch i c th. Nhng tham s ban u tnh ton l bin in p ra r
v dng in ti t , cng sut ti Pt v in tr ti Rt. Gia nhng tham s ny c quan
h cht ch vi nhau, nn v nguyn tc ch cn bit hai trong nhng tham s l
tnh cc tham s cn li.

tnh ton theo phng php gii tch dng s tng ng i vi ch xoay
chiu tn hiu nh.

Cc tham s ca mch EC tnh gn ng nh sau:

+ in tr vo ca tng: RV=R1// R2 // rV ; rV= rB + (1+).rE.

+ H s khuch i dng in: K = R R C

// t

R
i

Nh vy tng EC c h s khuch i dng tng i ln, v nu nh RC>> Rt th n

gn bng h s khuch i ca tranzito.

+ H s khuch i in p: K =
R R
C
// t
(du tr th hin s o pha)
R R
u

n
+ V

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Mch khuch i bn dn

+ H s khuch i cng sut K = P = K .K ; rt ln khong t (0,2 ? 5).103 ln

r

P
P u i

+ in tr ra ca tng. Rr=RC // rC (E); V rC(E) >> RC nn Rr = RC.

Tng EC c h s khuch i in p v dng in ln nn thng c s dng nhiu.

Tng khuch i Colect chung (CC)

in tr RE trong s ng vai tr nh RC trong mch EC, ngha l to nn mt in

p bin i u ra trn n. T C c nhim v a tn hiu ra ti Rt. in tr R1, R2
l b phn p cp in mt chiu cho cc B, xc nh ch tnh ca tng. tng
in tr vo thng ngi ta khng mc in tr R2. Tnh ton ch mt chiu tng
t nh tnh ton tng EC. kho st cc tham s ca tng theo dng xoay chiu, cn
chuyn sang s tng ng xoay chiu.

Cc tham s:

+ in tr vo ca tng: RV R1 // R2 // (1+).(RE // Rt)

Nu chn b phn p u vo R1, R2 ln th in tr vo s ln. Tuy nhin khi khng

th b qua in tr rC(E) mc song song vi mch vo, nn in tr vo phi tnh:

RV = R1 // R2 // [(1+).(RE // Rt) ]//rE(E)

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Mch khuch i bn dn

in tr vo ln l mt trong nhng u im quan trng ca tng C chung, dng lm

tng phi hp vi ngun tn hiu c in tr trong ln.

+ H s khuch i dng in:

K = (1 + ). R . R R
V E
// t

r R
i

V t

+ H s khuch i in p:

K = (1 + ).
R R E
// t

R R
u

n
+ V

Khi RV >> Rn v gn ng RV (1+).(RE + Rt) th Ku 1. Nh vy tng khuch i C

chung khuch i cng sut tn hiu trong khi gi nguyn tr s in p ca n.

V Ku = 1 nn h s khuch i Kp xp x bng Ki v tr s.

+ in tr ra ca tng: R = R //(r + r R R R ) = R //r

r E E
B
+ n
//
1+
1
// 2
E E

in tr ra ca tng nh c (1?50)?. N c dng phi hp mch ra ca tng khuch

i vi ti c in tr nh, khi tng C chung dng lm tng ra ca b khuch i c
vai tr nh mt tng khuch i cng sut n ch A khng c bin p ra.

Tng khuch i cc B chung (BC)

Cc phn t R1, R2, RE dng xc nh ch tnh IE. Cc phn t cn li cng c

chc nng ging s mch EC.

+ in tr vo: RV = RE //[rE + ( 1 )rB]

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Mch khuch i bn dn

in tr vo ca tng c xc nh ch yu bng in tr rE vo khong (10?50)?.

in tr vo nh l nhc im c bn ca tng BC v tng s l ti ln i vi ngun
tn hi vo.

+ H s khuch i dng ca tng: K = . R R C

// t

R
i

+ H s khuch i in p: K = .
R R C
// t

R R
u

n
+ V

+ in tr ra ca tng: Rr = RC // rC(E) RC .

Cn ch rng c tuyn tnh ca tranzito mc BC c tuyn tnh ln nn tranzito c

th dng vi in p cc C ln hn s EC. Chnh v vy tng khuch i BC c
dng khi cn c in p u ra ln.

TNG KHUCH I O PHA

Tng o pha dng khuch i tn hiu v cho ra hai tn hiu c bin bng nhau
nhng pha lch nhau 1800 (hay ngc pha nhau).

S tng khuch i o pha chia ti v hnh 2-11a. Tn hiu ly ra t cc E v cc

C ca tranzito. Tn hiu ra ur2 ly t cc E ng pha vi tn hiu vo uv cn tn hiu ra
ur1 ly t cc C ngc pha vi tn hiu vo. Dng tn hiu v trn hnh 2-11b,c,d.

Ta s kho st ch tiu ca tng tnh tng t nh tng CC.

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Mch khuch i bn dn

hoc tnh gn ng:

H s khuch i in p u ra 1 xc nh tng t nh s EC, cn u ra 2 xc

nh tng t nh s CC.

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Mch khuch i bn dn

Nu chn RC= RE v c Rt1=Rt2 th gi tr h s khuch i Ku1 gn ng bng Ku2 v

s ny cn gi l mch o pha chia ti.

Tng o pha cng c th dng bin p, s nguyn l nh hnh 2-12.

Hai tn hiu ly ra t hai na cun th cp c gc pha lch nhau 1800 so vi im 0.

Khi hai na cun th cp c s vng bng nhau th hai in p ra s bng nhau. Mch
ny c h s khuch i ln, d dng thay i cc tnh ca in p ra v cn c tc dng
phi hp tr khng nhng cng knh, nng n v mo ln nn hin nay t c dng.

CC S C BN DNG TRANZITO TRNG (FET)

Nguyn l xy dng tng khuch i dng tranzito trng cng ging nh tng dng
tranzito lng cc. im khc nhau l tranzito trng iu khin bng in p. Khi chn
ch tnh ca tng dng tranzito trng cn a ti u vo (cc ca G) mt in p
mt chiu c gi tr v cc tnh cn thit. Cc s ngun chung (SC), cc mng chung
(DC) v cc cng chung (GC) v nguyn l mch cng tng t.

Cn ch thm mt s c im ca mch khuch i dng tranzito trng l cc mch

ny thng c h s khuch i nh hn so vi tranzito lng cc, tuy nhin n nh
v trnh nhiu li tt hn.

PHNG PHP GHP CC TNG KHUCH I

Trn thc t khi khuch i 1 tn hiu nh ln n mt cng sut ln theo yu cu

th mt tng khuch i cha th p ng c m ngi ta thng phi s dng nhiu
tng khuch i. Khi ghp ni cc tng khuch i thnh mt b khuch i th ta mc

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Mch khuch i bn dn

u ra ca tng ng trc vo u vo ca tng sau. in tr vo v ra ca b khuch

i s c tnh theo tng u v tng cui.

H s khuch i ca b khuch i nhiu tng bng tch h s khuch i ca mi tng

(tnh theo n v s ln) hay bng tng ca chng (tnh theo n v dB)

Vic ghp gia cc tng c th dng t in, bin p hay ghp trc tip.

Ghp tng bng t in

u im ca phng php ny l mch n gin, cch ly c thnh phn mt chiu

gia cc tng, thun li cho vic tnh ton cc ch mt chiu.

Nhc im l lm gim h s khuch i min tn s thp, Ku 0 khi f 0.

Ngoi ra vi tn s thp th mch lm tng mc hi tip m dng xoay chiu trn cc

in tr RE v do lm gim h s khuch i.

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Mch khuch i bn dn

Ghp bng bin p

Trong s cun s cp W1 mc vo cc C ca T1, cun th cp W2 mc vo cc B

ca T2 qua t CP2. Ghp tng bng bin p cch ly in p mt chiu gia cc tng m
cn lm tng h s khuch i chung v in p hay dng in tu thuc vo bin p
tng hay gim p.

u im ca mch ny l in p ngun cung cp cho cc C ca tranzito ln v in p

mt chiu cun dy b, do cho php ngun c in p thp. Ngoi ra tng ghp bin
p d dng thc hin phi hp tr khng v thay i cc tnh in p tn hiu trn cc
cun dy. Tuy nhin n c nhc im l c tuyn tn s khng bng phng trong di
tn. Kt cu mch nng n, cng knh, h hng sa cha thay th phc tp.

Mch ghp trc tip

Trong mch ny cc C ca tranzito trc u trc tip vo cc B ca tranzito sau. Cch

trc tip ny lm gim mo tn s thp trong b khuch i, c dng trong b khuch
i tn hiu c thnh phn mt chiu (tn hiu bin thin chm).

Nhc im ca mch l khng tn dng c khuch i ca tranzito do ch cp

in mt chiu.

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Mch khuch i bn dn

MT S MCH KHUCH I KHC

Mch khuch i Darlington

Khi cn tr khng vo tng khuch i ln dng vo nh, h s khuch i ln ta ni

mch khuch i theo Darlington. Mch in gm hai tranzito T1 v T2 u nh hnh
2-17.

H s khuch i dng ton mch l = 1.2

Mch Kaskode

Mch gm hai tranzito ghp vi nhau, T1 mc E chung cn T2 mc B chung.

Khi tn hiu vo T1 khuch i t tn hiu ra Ura1 ln cc E gc T2 iu khin tip T2

khuch i cho Ura2

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Mch khuch i bn dn

Ta chng minh c h s khuch i in p ca T1: Ku1 = -1

ca T2: K R 2
. C
nn h s khuch i chung:
r
u2

V2

trong rV2 l in tr vo ca tranzito T2.

u im c bn ca mch ny l ngn cch nh hng ca mch ra n mch vo ca

tng khuch i, c bit tn s cao.

TNG KHUCH I CNG SUT

c im chung v yu cu ca tng khuch i cng sut

Tng khuch i cng sut l tng khuch i cui cng ca b khuch i, c tn hiu
vo ln. N c nhim v khuch i tn hiu a ra ti mt cng sut ln nht c th
c. Vi mo cho php vo bo m hiu sut cao.

Do khuch i tn hiu ln, tranzito lm vic trong min khng tuyn tnh nn khng th
dng s tng ng tn hiu nh nghin cu m phi dng phng php th.

Cc tham s c bn ca tng khuch i cng sut l:

- H s khuch i cng sut Kp l t s gia cng sut ra v cng sut vo :

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Mch khuch i bn dn

- Hiu sut l t s cng sut ra v cng sut cung cp mt chiu P0:

Hiu sut cng ln th cng sut tn hao trn cc C ca tranzito cng nh.

Tng khuch i cng sut c th lm vic cc ch A, AB, B v C tu thuc vo

ch cng tc ca tranzito.

Ch A l ch tng khuch i c 2 na chu k (+) v (-) ca tn hiu vo. ch

ny gc ct =1800, dng tnh lun ln hn bin dng in ra nn mo nh nhng
hiu sut rt thp, ch dng khi yu cu cng sut ra nh.

Ch AB tng khuch i hn na chu k (+) ca tn hiu vo, gc ct 900< <1800.

Lc ny dng tnh b hn ch A nn hiu sut cao hn. im lm vic ca ch
AB gn khu vc tt ca tranzito.

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Mch khuch i bn dn

Ch B tng khuch i na tn hiu hnh sin vo, c gc ct = 900. ch ny

dng tnh bng khng nn hiu sut cao.

Ch AB v B c hiu sut cao nhng gy mo ln. gim mo phi dng mch

khuch i kiu y ko.

Ch C tng khuch i tn hiu ra b hn na hnh sin, gc ct <900. N c dng

trong cc mch khuch i cao tn c ti l khung cng hng chn lc sng hi
mong mun v c hiu sut cao.

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Mch khuch i bn dn

Tng khuch i cng sut ch A

Trong tng khuch i ch A, im lm vic thay i i xng xung quanh im lm

vic tnh. So vi tng khuch i tn hiu nh, n ch khc l tn hiu vo ln nn IC0
phi ln theo. Xt tng khuch i n mc EC v c h s khuch i ln v mo nh.

Cng sut ra ca tng:

Cng sut tiu th ca ngun cung cp:

Hiu sut ca mch cc C:

T hnh v ta thy khi C=UC0=EC/2 v C = IC0 th c ?max=25%.

Nu dng mch ra ghp bin p, thc hin c phi hp tr khng v tn dng c

ngun nui EC, hiu sut cc i c th t 50%.

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Cng sut tiu hao trn mt ghp gp:

Theo cng thc ny ta thy cng sut PC ph thuc vo tn hiu ra. Khi khng c tn
hiu PC = P0 nn ch nhit ca tranzito phi tnh theo cng sut P0.

Tng khuch i cng sut y ko ch B v AB c bin p.

S tng khuch i cng sut y ko gm hai tranzito T1, T2. Ti c mc vi tng

khuch i qua bin p ra BA2. Mch cc C ca mi tranzito mc vi na cun s cp
bin p. T s bin p l n2=W21/Wt=W22/Wt.

Hai tranzito T1 v T2 thay phin nhau hot ng hai na chu k ca tn hiu vo.T1
hot ng na chu k dng, T2 hot ng na chu k m, in p trn cun dy
W12 c o so vi tn hiu vo.

Bin p BA1 c h s bin p l n1=WV/W11=WV/W22 m bo cung cp tn hiu vo

cc B ca hai tranzito. Tng c th lm vic ch B hay AB. Trong ch AB thin
p ly trn cc B ca hai tranzito c ly t ngun nui EC bng b phn p R1, R2.
Trong ch B thin p ban u bng khng nn khng cn R1. Khi in tr R2
c dng m bo cng tc cho mch vo ca tranzito trong ch gn vi ch
ngun dng.

Cng sut ra ca tng tnh c theo din tch tam gic:

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Cng sut a ra ti c tnh n hiu sut ca bin p l:

Tr s trung bnh dng tiu th t ngun:

Cng sut tiu th t ngun cung cp:

Hiu sut ca mch cc C:

Hiu sut ca tng:

Nu ?b/a = 1 th ?= 78,5% khi C=EC.

Thc t C<EC. do vng cong ca c tuyn v ?b/a = 0,8 nn ? =0,6 ?0,17.

PCmax khi C=0,64.EC.

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trnh mo do tnh khng ng thng on u c tuyn vo tranzito khi dng cc

B b. Ngi ta cho tng lm vic ch AB. ch ny UBE0, IB0, IC0 b nn cc
cng thc dng cho ch B vn ng.

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Mch khuch i cng sut y ko khng bin p

MCH KHUCH I DNG IC

Ngy nay IC c dng ph bin trong cc thit b in t dn dng cng nh chuyn

dng do c rt nhiu u im. Ngoi tr IC khuch i thut ton c c im ring,
IC tuyn tnh dng khuch i c nhiu loi, do yu cu khc nhau ca cc nh ch to,
c cho sn trong s tay tra cu hay s mch. Thng thng cc IC ny c chia
ra nhiu vng (theo s khi) lm nhiu nhim v khc nhau. V d mt IC dng
cho my thu thanh thng c khi dao ng (OSC hoc VCO), trn tn (MIX), khuch
i trung tn (IF.AMP), tch sng iu bin (AM.DET), tch sng iu tn (FM.DET)
v c th c c mch khuch i m tn. Tuy nhin lm vic c cc chn tng
ng ca IC phi c ni vi cc linh kin mch ngoi ph hp v cp ngun nui.

Khi cn khuch i cng sut ra ln, c th dng IC khuch i cng sut ring. Mc
cng sut ra, mc ngun nui tu tng loi IC m ta la chn cho ph hp. Trong trng
hp cn mch khuch i hai ng cho my stereo ta c th dng IC kp hay hai IC

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n cng loi thc hin. Ngoi ra cn ch rng khi dng cc IC cng sut ln cn
c phin to nhit bo m nhit cho php khi lm vic.

BI TP

Cho mch in nh hnh 2-27

Hy tnh cc c tnh ca mch (Dng in v in p trong mch).

Gii:

Cho mch in nh hnh 2-28. Vi:

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Tnh cc c tnh ca mch?

Gii

Cho mch nh hnh 2-29. Vi:

Hy tnh R1, R2 mch n nh thin p vi UCE0 = 3V. Bit URE = 0,673V.

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Gii

Cho mch nh hnh 2-30.

Vi

Hy xc nh RB v UCE1 UCE2 = 6V

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Gii:

Cho mch in nh hnh 2-31. Vi:

1. Xc nh dng in v in p mt chiu trn cc cc.

2. Bit Rt=8k, xc nh ti mt chiu v ti xoay chiu ca tng khuch i. V
ng ti mt chiu ti v tr im lm vic Q u ra ca tng khuch i.
3. V dng tn hiu vo, ra ca mch.

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Gii:

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Cho mch in hnh 2-32.

Bit:

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Xem T1,T2 l l tng khuch i cho bin cc i. Iphn cc = 1mA.

Gii:

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