Chng 5 1

Chng 4: MACH TRANSISTOR GHEP LIEN TANG 4.1 Gii thieu 4.2 Ghep Cascade cac mach khuech ai 4.3 Mach khuech ai vi sai (difference amplifier) 4.4 Cau hnh Darlington 4.5 Mach khuech ai ghep Cascode

Chng 5 2

4.1 Gii thieu Yeu cau thiet ke: o li (gain), cong suat ra, ap ng tan so, S dung nhieu hn mot TST. Mach khuech ai DC: Ghep trc tiep (direct coupling) Mach khuech ai AC: Ghep ien dung (capacitive coupling)

4.2 Ghep Cascade cac mach khuech ai Ghep Cascade: Ngo ra cua tang 1 la ngo vao cua tang 2, Mach co the gom nhieu cau hnh ghep cascade (vd: CE-CC, CE-CE, ) Phan tch: Xac nh tnh iem Phan tch mach tng ng tn hieu nho Xet mach ghep AC (ac-coupling) sau:

Phan tch DC, xac nh tnh iem: Hai tang oc lap (do ghep AC): Chng 2, 3. Phan tch AC (tn hieu nho): Mach tng ng: Chng 4

Chng 5 3

o li dong:

+

+

+===

1'1

'1

2'2

'21

2

221

1

2

2 ieb

b

ieb

bfe

LC

Cfe

i

b

b

b

b

L

i

Li

hRR

hR

RhRRRh

ii

ii

ii

ii

A

Gia s: hie1

Chng 5 4

Tang 1: Rb1 = R1 // R2 = 9.09K; VBB = VCCR1/(R1 + R2) = 1.82V mAhRRVV

Ifebe

BEQBBCQ 3.1/ 111

1 +=

Mach tng ng tn hieu nho: = 19201

11CQ

Tfeie I

Vhh ; = 260

222

CQ

Tfeie I

Vhh

o li ap:

( )[ ]

++

+=== 9.1)9//1()9//1(

11)50//(100//1100

5050

22

1

1

2

2 KKhKK

hKK

vi

iv

vv

vv

A ieiei

b

b

b

b

L

i

Lv -32

Bien o dao ong cc ai ien ap ngo ra:

MaxSwing = min(MaxSwing2 , Av2 MaxSwing1) vi Av2 : o li ien ap tang 2.

Chng 5 5

9 Tang 2:

RDC1 = 1K; Rac1 = 1K // 1K = 0.5K T DCLL va ACLL cua tang 2 MaxSwing2 = 5V

9 Tang 1:

Tang 2 mac CC: Zin2 = Rb2 // [hie2 + (hfe2 + 1)(Re2//RL) 33K RDC1 = 1K + 1K = 2K; Rac1 = 1K // 33K 0.97K T DCLL va ACLL cua tang 1 MaxSwing1 = 2.6V

Tang 2 mac CC: Av2 = 222

22

12

2

)//)(1()//)(1(

ieLefe

Lefe

outin

in

hRRhRRh

ZZZ

+++

+ 1 vi Zout1 = RC1 = 1K 9 Suy ra: MaxSwing = 2.6V

Chng 5 6

V du 2: Cho mach khuech ai ghep trc tiep sau. Xac nh tnh iem, o li ap, maxswing ngo ra

Xac nh tnh iem: e n gian, xem IB = 0 trong cac tnh toan tnh iem. VBE1 = 0.7V I3 = 0.7/600 = 1.17 mA IC2 = IE2 = I3 = 1.17 mA VCE2 = 9 (1.17mA)(1.3K + 1.8K + 0.6K) = 4.7V VE2 = (1.17mA)(1.8K + 0.6K) = 2.8V VC1 = VB2 = VBE + VE2 = 0.7 + 2.8 = 3.5V = VCE1 IC1 = (9 3.5)/2.2K = 2.5 mA

Xac nh MaxSwing: V tang 2 mac CE (Av thng >> 1) MaxSwing = MaxSwing2. Xet tang 2:

RDC = 1.3K + 1.8K + 0.6K = 3.7K Rac = 1.3K T DCLL va ACLL cua tang 2 MaxSwing = MaxSwing2 = 1.5V

Chng 5 7

Mach tng ng tn hieu nho: KIV

hhCQ

Tfeie 1

111 = ; KI

Vhh

CQ

Tfeie 14.2

222 =

Suy ra: [ ]

++

+===

1)8.0//6.0()8.0//6.0(

)//8.1//6.0(2.01

2.22.21003.1100

12

1

1

2

2 KhKhK

vi

ii

iv

vv

Aieiei

b

b

b

b

L

i

Lv

Av = 4000 ( 72dB) On nh phan cc: Mach khuech ai AC: Cac tang oc lap DC: Chng 3 Mach khuech ai DC: Big problem !!! V du 3: Xac nh thay oi cua dong tnh gay ra do anh hng cua nhiet o len VBE trong v du 2. Hoi tiep: Xac nh o on nh: IC1 /T va IC2 /T:

VB2 = 9V 2.2K(IC1 + IB2) = 9 2.2K(IC1 + IC2 / hfe2) VE2 = VB2 VBE2 = 9 2.2K(IC1 + IC2 / hfe2) VBE2 Mat khac: VE2 = 1.8KIE2 + VBE1 1.8KIC2 + VBE1 IC2(1.8K + 2.2K / hfe2) = 9 2.2KIC1 VBE1 VBE2 Tai B1: IC2 IE2 = IB1 + VBE1 / 0.6K IC1 / hfe1 + VBE1 / 0.6K

Chng 5 8

IC1(2.2K + 1.8K / hfe1 + 2.2K / (hfe1hfe2)) = 9 - VBE1(1 + 1.8K / 0.6K + 2.2K / (hfe20.6K)) - VBE2 IC1 K

VV BEBE2.2

49 21

Khi nhiet o thay oi: VBE / T = -k = -2.5 mV/0C KCmV

TIC

2.2/5.25 01 =

= 5.7 A/0C

Tai B1:IC2 = IC1 / hfe1 + VBE1 / 0.6K KCmV

TV

KTI

hTI BEC

fe

C

6.0/5.2

6.011 011

1

2 +

=

= -4.2 A/0C

4.3 Mach khuech ai vi sai (difference amplifier)

S dung: Phan tch: Gia s mach oi xng, cac TST giong nhau, mach cc B giong nhau

Chng 5 9

Phan tch tnh iem: VE1 = VE2 = (IE1 + IE2)Re VEE = 2IE1Re VEE = 2IE2Re VEE Do tnh oi xng, tach thanh 2 mach (Re 2Re):

IEQ1 = IEQ2 = febe

EE

hRRV

/27.0

+

VCEQ1 = VCEQ2 = VCC + VEE ICQ(Rc + 2Re)

V du 4: Trong mach tren, cho VCC = VEE = 10V; Rb = 0.2K; Re = 0.9K; Rc = 0.2K; RL = 10. Tnh dao ong cc ai dong tai. Xem 2Re >> Rb / hfe Theo phan tch tnh iem: ICQ = (10 0.7) / (20.9) = 5.17 mA VCEQ = 10 + 10 5.17(0.2 + 20.9) = 9.66V

DCLL: RDC = Rc + 2Re = 2K ACLL: Rac = Rc // RL 10 (???) Da vao o th: IC2max = 5.17 mA ILmax 5.17 mA

Chng 5 10

Phan tch tn hieu nho: Phan anh mach cc B (nguon i1 va i2) ve cc E:

at i0 = (i1 + i2)/2 va i = i2 i1 i1 = i0 (i/2) va i2 = i0 + (i/2)

Dung phng phap chong trap cho mach tng ng tn hieu nho, tach thanh 2 mode: Mode chung (common mode): i1 = i2 = i0

Do oi xng: ie1c = ie2c iRe = 2ie1c = 2ie2c ve = (2Re)ie2c Tach oi: Re 2Re ie2c =

febibe

b

hRhRiR

/20

++

Chng 5 11

Mode vi sai (differential mode): i2 = - i1 = i/2

ie1d = - ie2d iRe = 0 ve = 0. Ngan mach Re ie2d = )/(2 febib

b

hRhiR

+

Chong trap (superposition): ie2 = ie2c + ie2d = ihRh

Ri

hRhRR

febib

b

febibe

b ++++ )/(2/2 0

iL = 2ieLc

c iRR

R+

= Aci0 + Adid

trong o: Ac = febibe

b

Lc

c

hRhRR

RRR

/2 +++

: o li dong mode chung

Ad = )/(2 febibb

Lc

c

hRhR

RRR

++

: o li dong mode vi sai

4.3.1 Ty so triet tn hieu ong pha CMRR (Common Mode Rejection Ratio): Mach khuech ai vi sai ly tng: Ac = 0: iL = Adi Mach thc te: nh ngha: CMRR =

c

d

AA

CMRR = febib

e

febib

febibe

hRhR

hRhhRhR

/)/(2/2

++++

(Gia s Re >> hib + Rb/hfe)

Chng 5 12

V du 5: Cho mach trong v du 4. Tnh CMRR. Gia s i0 = 1A, xac nh gia tr tn hieu ngo vao mode vi sai e ngo ra mode vi sai toi thieu ln hn 100 lan ngo ra mode chung. Ac - 0.1 Ad - 14 iL = -0.1i0 14i CMRR = Ad / Ac = 140 (43dB) e ngo ra mode vi sai 100ngo ra mode chung: 14i 100(0.1i0) i 100(i0 / CMRR) = 0.7 A 4.3.2 Nguon dong cc phat (Emitter) e tang CMRR: Tang Re : S dung nguon dong tai cc E.

Dung TST T3 tai cc E:

iC3 = e

BBEE

RVV 7.0

= const.

Xem T3 la nguon dong

Phan tch tnh iem: ICQ3 =

e

BBEE

RVV 7.0

Chng 5 13

Do tnh oi xng: ICQ1 = ICQ2 = e

BBEECQ

RVVI2

7.02

3 =

VC3 = VE1 = VE2 = - RbIb VBE = fe

CQb

hIR 17.0

VCEQ1 = VCEQ2 = VC1 VE1 = (VCC RcICQ1) (fe

CQb

hIR 17.0 )

VCEQ3 = VC3 VE3 = (fe

CQb

hIR 17.0 ) - (-VEE + ReICQ3)

Phan tch tn hieu nho: Tng t phan tren, thay Re bang 1/hoe. Nhan xet: 1/hoe rat ln: CMRR c tang ang ke. Chnh can bang: (Balance control) Thc te: T1 va T2 khac nhau Dung bien tr Rv gia E1 va E2 e chnh can bang.

Chng 5 14

ieu kien can bang: ICQ1 = ICQ2 KVL: RbIB1 + VBE1 + R1IEQ1 = RbIB2 + VBE2 + R2IEQ2 (Rb / hfe1 + R1)IEQ1 + VBE1 = (Rb / hfe2 + R2)IEQ2 + VBE2 Gia s VBE1 = VBE2, can bang Rb / hfe1 + R1 = Rb / hfe2 + R2 Mat khac: R1 + R2 = Rv

R1 =

21

1122 fefebv

hhRR

va R2 =

+

21

1122 fefebv

hhRR

Phan tch tn hieu nho: Khi can bang: ICQ1 = ICQ2 hib1 = hib2 = hib Mach tng ng tn hieu nho: Giong trng hp oi xng, trong o

hib1 + Rb1 / hfe1 hib1 + Rb1 / hfe1 + R1 =

+++

21

1122 fefebv

ib hhRR

h

hib2 + Rb2 / hfe2 hib2 + Rb2 / hfe2 + R2 =

+++

21

1122 fefebv

ib hhRR

h

Ad = )]/1/1)(2/(2/[2 21 fefebvibb

Lc

c

hhRRhR

RRR

++++

: Giam so vi trng hp khong dung Rv

Ac = )]/1/1)(2/(2/[)/1(2 213 fefebviboeb

Lc

c

hhRRhhR

RRR

+++++

Chng 5 15

V du 6: Thiet ke mach sau e co CMRR = 100 (40dB). Tai 1K ghep AC. TST co hfe = 100.

Theo hnh ve: R1 = R2 = 50 ; hfe1 = hfe2 = 100.

S dung cong thc tnh Ad va Ac phan chnh can bang, thay 1/hoe bang Re, suy ra:

CMRR = ib

e

ib

ibe

c

dh

Rh

hRAA

++++++=

60100/20001002100/1000502

Yeu cau: CMRR 100 Re 100(60 + hib) Gia s ICQ1 = ICQ2 = 1mA hib = 25 Re 8.5K. Chon Re = 10K. Tnh VEE: VEE = Rb1IB1 + VBE1 + R1IE1 + Re(2I1) = 20.8V

DCLL: RDC = Rc + R2 + 2Re ACLL: Rac = Rc // RL

Chng 5 16

4.4 Cau hnh Darlington

Phan tch tnh iem: 9 DCLL cho T2: VCC = VCE2 + Rc(IC1 + IC2) + ReIE2 Do IC1 IE1 = IB2 = IC2 / hfe2

Chng 5 17

Phan tch tn hieu nho: Phan anh mach cc B1 cua T1 Cc E1 va mach cc E2 cua T2 Cc B2

Tnh iem: ICQ2 = hfe2ICQ1 hie2 = 112

22

2CQ

T

CQfe

Tfe

CQ

Tfe I

VIh

Vh

IV

h == = hib1

Suy ra: Ai = 211

112

)/()/(

ieibfeb

febfe

Lc

cfe

i

LhhhR

hRhRRRh

ii

+++ =

11

12

2 ibfebbfe

Lc

Cfe

hhRRh

RRRh

++

Ai = 1

21 2)(

ieb

b

Lc

cfefe hR

RRR

Rhh ++

Xem 2TST ghep Darlington 1 TST co: he = 2hie1 va hfe = hfe1hfe2

Chng 5 18

V du 7: Xac nh tnh iem cua mach sau. Gia s hfe = 100.

Nguon dong T5: VB5 = 3.19.29.26 + = -4.14V

Gia s IB5

Chng 5 19

4.5 Mach khuech ai ghep Cascode MACH 1:

Cau hnh: CE CB: Thng dung trong cac mach tan so cao. Phan tch tn hieu nho:

o li truyen at (Transfer gain): 121

2112

1

1

2

2 ////

)(ie

feLfbi

b

b

e

e

L

i

LT hRR

RRhRh

ii

ii

iv

iv

A +===

Chng 5 20

Phan tch DC: Gia s bo qua IB1 va IB2: VB1 = VCCR1 / (R1 + R2 + R3) VE1 = VB1 - VBE1 = VB1 0.7 IC1 =

e

BR

V 7.01 = IC2 VB2 = VCC(R1 + R2) / (R1 + R2 + R3) VE2 = VB2 - VBE2 = VB2 0.7 VCE1 = VC1 - VE1 = (VE2 RcIC1) VE1 VCE2 = VC2 VE2 = (VCC RLIC2) VE2 MACH 2:

T1: CE T2 va T3: Mach Cascode, dung e chuyen mc DC cua (vL) en 0 (level shifting) dung trong cac mach K ghep trc tiep (direct coupled amplifiers).

Chng 5 21

Phan tch tnh iem: Gia s bo qua IB2, IB3. VB3 = (-6)(2.3K) / (2.3K + 3.7K) = -2.3V VE3 = VB3 - VBE3 = -2.3 0.7 = -3V IC3 = 3V / 3K = 1 mA Can phai xac nh Rb sao cho VLDC = 0. IB1 = (12 0.7) / Rb VC1 = 12 8KIC1 = 12 8K[hfe1(12 0.7) / Rb] = VB2 VLDC = VE2 3.3KIC2 = (VB2 0.7) 3.3 = VB2 4 e VLDC = 0 VB2 = 12 8K[hfe1(12 0.7) / Rb] = 4 Rb = hfe1(12 0.7) / 1mA Phan tch tn hieu nho: T1: Mac CE. T2: Mac CC. T3: Mac CB.

Xac nh vL / vc1 : Mach tng ng tn hieu nho cua T2: Phan anh tr khang E2 B2 , trong o R03 = 1/hob3 la tong tr nhn vao cc C cua T3 (CB).

)/1(3.3)/1(

3222

32

1 obfefeie

obfe

C

LhhKhh

hhvv

++= 1 Mach Cascode (T2, T3) ch lam thay oi mc DC ngo ra ma khong thay oi o li ap cua mach K T1 (CE).