Dien Tu Can Ban Chuong4 5495

Chng 4: Mch phn cc transistor.

CHNG 4: MCH PHN CC TRANSISTORI. GII THIU: Vic phn tch hay thit k mt mch khuch i i hi s hiu bit v p ng dc v ac ca h thng. Ngi ta thng nhm ln rng transistor l mt linh kin khuch i tn hiu m khng cn ngun nng lng cung cp. Thc ra vic khuch i tn hiu ac l t qu trnh chuyn i nng lng t ngun cung cp dc. Do vic phn tch hay thit k bt k 1 mch khuch i in t u cha ng 2 phn: phn dc v phn ac. Cc mc hot ng dc ca 1 transistor c iu khin bi 1 s cc thng s bao gm 1 dy cc im lm vic c th c trn cc ng c tnh ca transistor. Cc dng in dc v cc mc in p dc phi c xc nh, mt mch in phi c xy dng m n s thit lp im lm vic mong mun cc mch in ny s c phn tch trong chng ny.

Hnh 4.1: c tuyn ng ra ca BJT. Mt h s phn cc khc rt quan trng cn ch n: s la chn v phn cc cho transistor ti im lm vic mong mun phi tnh n nh hng ca nhit . Nhit lm thay i cc h s nh ac v dng in ICEO . Nhit cng tng th dng in ICEO tng lm thay i im lm vic Q. Do cc mch in phi thit k c s n nh nhit khi c s thay i nhit th s thay i ca im lm vic l nh nht. S n nh ca im lm vic c ch nh bi h s n nh S xc nh mc thay i im lm vic ph thuc vo s thay i ca nhit . i vi BJT vic phn cc hot ng trong vng tuyn tnh cn phi ch : 1. Mi ni B-E phi phn cc thun vi in p phn cc vo khong 0,6 n 0,7V. 2. Mi ni B-C phi phn cc ngc vi in p phn cc nm trong gii hn cc i ca transistor. II. Mch phn cc cho BJT 1. Cc dng mch phn cc: a. Mch phn cc c nh: Mch phn cc c nh nh hnh 4-2 s dng transistor npn.

61


Hnh 4-2. Mch phn cc c nh. S mch hnh 4-2 c th chia ngun cung cp dc Vcc thnh 2 ngun nh hnh 4-3.

Hnh 4.3 Xt mch vng BE: Xt mch vng phn cc mi ni B-E ca hnh 4-4.

Hnh 4.4. p dng nh lut Kirchhoff:

+ VCC I B R B V BE = 0Suy ra dng in IB:

IB =

VCC V EE RB

Xt mch vng CE: Mch vng phn cc mi ni C-E c v li nh hnh 4-5.

62


Hnh 4.5 Mi lin h gia dng IB v IC:

I C = I BAp dng nh lut Kirchhoff:

VCE + I C RC VCC = 0Hay

VCE = VCC I C RCPhng trnh in p VCE:

VCE = VC V ETrong VCE l in p ca 2 im C v E cn in p V C v VE l in p im C v E so vi mass. Trong mch in trn th in p im E bng khng (VE=0V) nn phng trnh c vit li:

VCE = VCPhng trnh in p VBE:

(4-3) (4-4) (4-5)

VCE = VC V EVi VE = 0 nn:

V BE = V B

Hot ng bo ha ca transistor: i vi transistor hot ng vng bo ha th dng in c gi tr cc i ca mt mch c thit k c bit. Thay i thit k th mc bo ha tng ng c th tng hoc gim v d nhin gi tr bo ha ln nht xc nh bi dng in Ic cc i c cung cp trong bng thng s ca transistor. Cc iu kin bo ha thng nn trnh bi v mi ni CB khng cn phn cc ngc dn n tn hiu ng ra mch khuch i b mo dng. im hot ng trong vng bo ha trnh by hnh 4-7a. ch rng trong vng bo ha th cc ng cong c tnh c ni li vi nhau v in p C E nm ti mc hoc thp hn mc in p VCesat. Dng in Ic c gi tr tng i ln trn ng cong c tnh.

Hnh 4.7. Nu ta dng cc ng cong xp x nh hnh 4-7b xc nh nhanh cc mc gi tr trong vng bo ha. Trong hnh 4-7b, dng in tng i cao cn in p VCE xem nh bng 0V. p dng nh lut Ohm tnh in tr mi ni CE:

63


p dng kt qu trn ta v li mch c cu hnh nh hnh 4-8. i vi mch phn cc c nh khi transistor ch bo ha th s mch nh hnh 4-9, in p ri trn R chnh bng Vcc v dng in Ic bo ha c gi tr:

Hnh 4-8. Xc nh ICsat.

Hnh 4-9. Xc nh ICsat ca mch phn cc.

** Mch phn cc c nh n nh cc emitter: Mch phn cc dc hnh 4-10 c thm 1 in tr ti cc Emitter ci thin mc n nh ca cu hnh mch phn cc c nh.

Hnh 4-10. Mch phn cc BJT c thm in tr cc E. Hnh 4-11. Xt mch vng BE: Hnh 4-10 c th v li nh hnh 4-11, p dng nh lut Kirchhoff c phng trnh:

+ VCC I B R B V BE I E R E = 0Phng trnh dng in chng 3:

(4-7) (4-8)

I E = ( + 1) I B

Th vo pt (4-15) c:

+ VCC I B R B V BE ( + 1) I B R E = 0Rt gn v suy ra dng in IB:

IB =

VCC V BE R B + ( + 1) R E

(4-9)

Ch rng ch c 1 s khc nhau trong phng trnh dng in IB so vi mch phn cc c nh l thnh phn (+1)RE. T phng trnh (4-17) ta c mch in tng ng nh hnh 4-12. Nu nhn t pha in p V BE th in tr RE phn hi tr li dng in ng vo IB bi h s (+1). Mc khc in tr cc E l mt phn ca mch vng cc CE c gi tr (+1)RE i vi mch vng BE. Do thng c gi tr vo

Ri = ( + 1) R E

(4-10)

64


khong 50 hoc cao hn nn in tr cc E s tr thnh mt in tr tht ln i vi ng vo cc B E nh hnh 4-13. in tr ng vo:

Hnh 4-12. Hnh 4-13. Hnh 4-14. Xt mch vng CE: Mch vng phn cc mi ni C-E c v li nh hnh 4-14. p dng nh lut Kirchoff v p:

+ I E R E + VCE + I C RC VCC = 0Thay th IE IC v suy ra in p VCE:

VCE = VCC I C ( RC + R E )in p cc Emitter VE:

(4-11)

VE = I E REin p cc Collector VC:

(4-12) (4-13) (4-14) (4-15) (4-16)

VC = VCE + V EHay

VC = VCC I C RCin p cc Base VB:

V B = VCC I B R BHay

V B = V BE + V E

b. Mch phn cc bng cu phn p: Mch phn cc bng cu phn p dc hnh 4-15. C 2 phng php phn tch mch: phn tch chnh xc v phn tch gn ng.

Hnh 4-15. Phn tch chnh xc:

65


Mch in ng vo c th v li nh hnh 4-18. Dng mch tng ng Thevenin v cn xc nh in tr tng ng Thevenin v in p Thevenin: Xc nh in tr Thevenin RTh: ngn mch ngun in p nh hnh 4-16.

RTh = R1 R 2Xc nh in p Thevenin Eth: mch in nh hnh 4-17.

(4-17)

E Th = V R 2 =

R 2VCC R1 + R 2

(4-18)

Mch ng tng Thevenin nh hnh 4-18 v dng in I B c th xc nh bng nh lut Kirchhoff:

E Th I B RTh V BE I E R E = 0Thay th dng IE = ( + 1) IB vo suy ra dng IB:

IB =

E Th V BE RTh + ( + 1) R E

(4-19)

Hnh 4-16. Xc nh RTh. Hnh 4-17. Xc nh ETH. ng. Phng trnh mch vng CE khng c g thay i kt qu c:

Hnh 4-18. Mch tng

VCE = VCC I C ( RC + R E )

(4-20)

Phn tch gn ng: Cu hnh mch phn p ng vo c th thay th bng mch in hnh 4.19. in tr Ri l in tr tng ng gia cc B v mass xc nh phn trc bng ( + 1) RE . Nu Ri ln hn nhiu so vi R2 th dng IB nh hn dng qua R2 rt nhiu c ngha l dng I1 xp x bng dng I2. Khi d in p trn R2 bng in p VB xc nh bi phng trnh:

VB =

R 2VCC R1 + R 2

(4-21)

C th xem Ri = ( + 1) RE RE th iu kin tha mn php tnh gn ng l

R E 10R 2Xc nh in p VE:

V E = V B V BEXc nh dng in IE:

IE =V c th xem:

VE RE

I CQ I Ein p VCEQ:

VCEQ = VCC I C ( RC + R E ) 66


im tnh Q khng ph thuc vo h s .

VCE = VCC I C ( RC + R E )

Hnh 4-19. c. Mch phn cc hi tip t collectror : Mch phn cc hi tip in p nh hnh 4-20.

Hnh 4.20 Hnh 4.21. Xt mch vng BE: Hnh 4-20 c th v li nh hnh 4-21, p dng nh lut Kirchhoff c phng trnh:' + VCC I C RC I B R B V BE I E R E = 0

C th xem' I C I C = I B

I E IC

Th vo phng trnh trn c:

+ VCC V BE I B ( RC + R E ) R B I B = 0Suy ra dng in IB:

IB =

VCC V BE R B + ( RC + R E ) V' R B + R '

(4-23)

Tng qut, phng trnh dng IB c dng nh sau:

IB =Do IC = IB nn:

I CQ =

V ' R B + R ' V ' V ' V ' = R B + R ' R ' R'

Do R ln hn RB rt nhiu nn c th xem:

I CQ =

67


Cho thy ICQ khng ph thuc vo h s . Xt mch vng CE: Hnh 4-20 c th v li nh hnh 4-21.

Hnh 4.21 p dng nh lut Kirchhoff c phng trnh:

I E R E + VCE + I ' C RC VCC = 0Do IC IC v IE IC nn:

I C ( R E + RC ) + VCE VCC = 0hay

VCE = VCC I C ( R E + RC )2. Phn tch ng ti: Xt mch in nh hnh 4.22

(4-24)

nh:

Hnh 4.22 ng ti dc (DCLL DC load line) Mch in hnh 4-22 thit lp mt phng trnh ng ra din t mi lin h gia 2 bin IC v VCE

VCE = VCC I C RCHay:

IC =

V 1 VCE + CC RC RC

(4.25)

Phng trnh 4.25 chnh l phng trnh ng ti dc ca mch hnh 4.22. th ng ti DCLL ca mch hnh 4.22 trn ng c tuyn ng ra ca transistor trn hnh 4.23

68


Hnh 4.23: ng ti DCLL. Nu dng in IB thay i bi cc gi tr khc nhau ca R B th im tnh Q s di chuyn ln hoc di chuyn xung nh hnh 4-24. Nu in p Vcc v IB gi c nh v in tr Rc thay i th ng ti s dch chuyn nh hnh 4-25.

Hnh 4-24. Nu RC c nh v Vcc thay i th ng ti dch chuyn nh hnh 4-26.

Hnh 4-25

Hnh 4.26 ng ti ac (ACLL AC load line): Ta c phng trnh ng ra ch vi tn hiu ac:

ic ( RC \ \ R L ) + vce = 0 ic = 1 vce ( RC \ \ R L )

hay:

m v

iC = ic + I CQ hay ic = iC I CQ 69


vCE = vce + VCEQ hay v ce = vCE VCEQVy phng ng ra khi c ngun tn hiu ac l:

1 (vCE VCEQ ) ( RC \ \ R L ) VCEQ 1 iC = vCE + + I CQ ( RC \ \ R L ) ( RC \ \ R L ) (iC I CQ ) =

(4.26)

Phng trnh 4.26 chnh l phng trnh ng ti ac ca mch hnh 4.22. vy th ca phng trnh ng ti ac chnh l ng thng ACLL trn hnh 4.27. Nhn xt: ACLL v DCLL lun lun gia o nhau ti im lm vic tnh Q.

Hnh 4.27: ng ti ACLL , DCLL v dng sng tn hiu vo ra. Tm dao ng cc i ca tn hiu ng ra (maxswing): Maxsing [vce(p-p)] = 2 x vce (p) =2 x min[VCEQ , ICQRac] Trong Rac = RC||RL i vi mch hnh 4.22. 3. H s n nh nhit Cc yu t gy bt n nh im lm vic l: in p ngun cung cp, nhit . y ta ch xt n yu t nhit v n lin quan n vn cch phn cc cho transistor. Khi nhit thay i s nh hng n cc thng s ca transistor, th hin bi cc tham s sau: - Dng r:

I CO (T2 ) = I CO (T1 )210 .oC

T2 T1 T *

(4.27)

Trong T * l bin thin nhit lm dng in bo ha ngc tng gp i thng bng - H s truyn t dng in , :

T2 = T1 (1 +- in p VBE ng vi IB = const:

T2 T1 ) 75

(4.28)

(4.29) Vy khi nhit lm vic ca transistor b thay i lm cc thng s trn ca transistor thay i theo kt qu l im lm vic Q b dch chuyn trn c tuyn ng ra hnh 4.28.

V BE = (2 2,5)mV / o C

70


Hnh 4.28: im lm vic Q b dch chuyn khi nhit thay i a. 25oC, b. 1000C. Tiu chun nh gi s bt n nh ca mch theo nhit l S, cc h s bt n nh l:

I C I CO I C S (V BE ) = V BE I S ( ) = C S ( I CO ) =

(4.30) (4.31) (4.32)

V d 4.1: Xt s bt n nh nhit ca mch hnh 4.29a:

(a) Hnh 4.29. Gii: Xt mch vng hnh 4.29b: ta c: hay

(b)

VCC = I B R B + VBE + I E R E I C = I E + I CO IE = I C + I CO

(4.33)

(4.34) :

V

71


I B = I E IC =Thay cng thc 3.34 v 3.35 vo 4.33:

I C I CO IC

(4.35)

IC = Thay =

vo biu thc trn: +1

(VCC V BE ) RE + RB + I CO R E + R B (1 ) R E + R B (1 )

IC = T cng thc 4.35, ta c:

(VCC VBE ) RE + RB + ( + 1) I CO RB + ( + 1) R E R B + ( + 1) R E RE + R B R B + ( + 1) R E

(4.35)

S ( I CO ) = ( + 1)

(4.36)

Hnh 4.30: th ca S(ICO) theo RB/RE Nu

RB

RE

( + 1) , ta c S ( I CO ) ( + 1)

Nu 1

Documents

Dien Tu Can Ban Chuong4 5495