56

Phan 2 : MOT SO BAI TAP MAU

Chng 1: AP NG TAN SO THAP

1.1 a) S o mach va s o tn hieu be :

RERb CE

RL

i

>

iL

5uF

i4k

1k

100

iL

RLfe

E

h

CRb

i

>ic

.Re

=1K

5k .ib

ib

i(1+hfe)

1/

>

1k

sCc2

Re 5kRb

hie

1khfe

ri

ieu kien tnh :

==

=

V

BB

kkk

b

V

R

81,1110

10.20

10100//10

62

K

ie

e

fe

b

BBCQ hmA

Rh

R

VI 34,0

7,3

25.50)(7,3

7,0===

+

=

at

-3

5

1

110

10.51]

1//)[1(

+=+=ssCc

RhZ efe

-ssCc

RZ L

43

2

2

10.510

1+=+=

Ham truyen :

12

..

.ZhR

R

ZR

Rh

i

i

i

i

i

iA

ieb

b

c

efe

i

b

b

L

i

Li +++

===

Xay dng Ai bang cach khac :

iem zero : sradCcRe

/10001

;01

===

iem cc

srad

h

hRRCc

fe

iebe

a /1485

)]1

//([

1

1

=

+

+=

25][

1

2

=+

=Lc

bRRCc

o li tang gia

24. =++

==

feb

b

Lc

cfe

i

Lim

hR

R

RR

Rh

i

iA

)1485)(25(

)1000(.

21

500

+++

=ss

ssAi

63

0)( =++ BEEECQefe

b VVIRh

R

Suy ra

)(2,4

150

1

7,05mAI

KK

VV

CQ =+

=

nen : Kieh 3,02,4

2550 == Zin

=

=

20

1

fe

EQ

h

mAI

A i

24

16

0 1485103

25

suy ra :

1.7 T s o ta co :

ri

Rb

RE

Vcc

=

-5V

+

-

Vi

5V

Cc2

RL 10k1k

1k

Cc1

10

Ta co

KKKK

efeiebin RhhRZ 1)503,0//(1])1([' +=++=

nen :

FZ

CcZCcZrCc ininini

20010.5

111

)(

131

11

===+

=

1.9

R c

R bR E

=

=

20

1

fe

EQ

h

mAI

B B

iRL=

>iL

1 k

C c2

C e1 k

V cc

1 k

+V

1 ki

)1485)(25(

)1000(24

+++

=ss

ssAi

64

K

ieh 5,01

25.20 ==

>

1/

Lhfe/

Li

i

>ib

Vb

-

hie

Rbi sCehfeRe R

sCc1

Rchfeib

Ta co :

-e

fe

bibee

eC

h

RhRC

.75

1

)]//([

1

+=

-2

3

2

210.2

1

][

1

CcRRCc Lc=

+=

Xay ra iem cc kep tai srad /10= th srade /102 == nen

e tan so 3 db la 10 rad/s vi Ce=1333uF th neu khong ke en s gay meo cua Cc2 th sradeL /10== . Do o ta chon

sradL /12 =

65 hfe=100 hie = 1

K

12

..ZhR

R

ZR

Rh

i

i

i

i

i

iA

ieb

b

c

c

fe

i

b

b

L

i

Li +++

===

0)1)((

)1(.

1)(

..

2

2

+++

+

++=

sRChRR

CRR

sCcRR

sCcRhA

eeiebe

eeb

Lc

c

fei

]1)10(][)([

)1(..10

3

2

2

3

++++++

+=

sRCcRhRhsrRhC

ssRCRhCcA

Lbieefeebiee

eebfe

i

Thay so

)75,10)(1(

)75,0(67,6

+++

ss

ss

Cach khac

-iem zero sradCR ee

/75,010.1333.10

11,0

63===

-iem cc :- srad

h

RhRC

fe

bibee

/75,1010.1333.76,69

1

)]//([

161

+=

67,6. =++

=ieb

b

Lc

Lfe

imhR

R

RR

RhA

Vay : 1.11

)100093)(1(43

)10001333(20

+++

=ss

ssAi

)1)(7,10(

)75,0(.67,6

+++

=ss

ssAi

R1 R2

Ri

C

RC

ii 100k

10k

Vcc

1k

10k

10uF

66

ib

100K1K

i L1/

L'

L>hie nen co the bo qua ri o li vong T

''

0

'.

L

b

efe

L

b

efe

iL

L

v

i

i

iRh

v

iRh

v

vT

i

====

vi :

1)(

1

1

1

++=

++=

CshRRh

sC

sC

i

i

ieie

b

ieie

ie

ieL

hRRCshRCs

Cshr

hRsC

Rv

i

+++

++=

++=

2)(

1)(

)//(1

1'

nen

C iC

iL

Li

>i

b

R

hfei

b

R2R1

hierii

67

21011

10

]2)([

110

33

+

=+++

=shRshRCR

hTieie

fe

o li khong hoi tiep :

21011

)20(10

1//

1//

.3

0, +

+=

++

+===

=s

s

hsC

RR

sCRR

hi

i

i

i

i

iA

ie

fe

i

b

b

L

vi

Li

L

suy ra Tan so 3db )/(110 sradL = Gian o Bode

1-13

1,19

191

+=

sT

)110(11

)20(1000

1 ++

=

=s

s

T

AA iif

0

dbAif ,

24

39

20 L)(log scale

68

21 ,CcCc

Do Rg.>>ri nen : vg~vi

RgRs1

Rs2

Rd

RLri

L

VDD

VCc2

100k

100uF250

250100k

Cc1

+

-

Vi

5k

5k

S o tn hieu be :

ri

Rg 5K100k

250

250

D

gs

+

-

Vi1

5k

100k+

-

uv

rds

Rs1

RdRL

Rs2 100uF

Dung tng ng Thevenin nh tren ta co :

Lvg~vi

g~uvi

(u+1)

//Rl

V

Rd

Z

+

-

Vi1

5k

100k+

-

uv

rds

vi : sC

RRZ ss1

//21 +=+

25. ==+ mds gr

69

a) Xac nh ham truyen :

dsLd

Ld

i

Lv

rZRR

RR

v

vA

+++

==

)1(//

//

Do Rd

70

575,7)1(1

=+++

==

dsds

d

i

Lvm

RRr

R

v

vA

Do o ta cung co bieu thc giong tren:

)7,55(

)40(575,7

++

=s

sAv

b)Neu ghep tu 2 au Rs=Rs1 + Rs2 th :

dsd

d

vrZR

RA

+++

=

')1(

vi Z'=(Rs1+Rs2)//1/sC

Thay so ta co :

46

)20(5,12

++

=s

sAv

->Bang thong giam va o li tang 1-12

R1

R2

R31K

RE

=10

V

E1k

1k

5k

5k

Vcc

V

Cc

+

-

Vi

ri 1K

1/

hfeRe

100k

hie

0.6k

Rb

3.5k

sCc

+

-

Vi1

ri

ieu kien tnh

KKKK

b

V

BB RV 5,35//51;51055

5=+==

+=

suy ra :

kieK

K

VV

CQ hmAo

I 6,015.4

2510015,4

1100

5,3

7,5===

+

=

71

Che o xoay chieu : do hfeRE = 100K>>hie->vE~VB

Ta co :

sCcrRhhR

RhhR

v

v

v

vA

iefeieb

efeieb

i

b

i

Ev 1

)//(

)//(

+++

++==

Thay so ta co ;

521.10.83,22

.10.62,175

5

+=

sCc

sCcAv

Tan so 3db :

CcCcL

83,22

521.10

10.83,22

521 5

5

==

Theo gia thuyet pipi 102 == LL f nen

)(26,710.83,22

521.10 5FCc

pi==

72

i

B=B1

b//rb'e b'e mv

c

iL

bi gC

RR

rbb'

bi b'e b'e M b'vm

LiL

LRg

V

CC1i r1R2

R41k

R31k

Vcc

+ Vbb

IiR

g

V

CC1i r1R2

R41k

R31k

Vcc

+ Vbb

Ii

Vi gia thiet Cb'c = 0 th CM = 0; rbb' = 0 : ngan mach B - B'.

S o ch con lai nh sau :

Theo gia thiet ta co :

ICQ = 2 mA suy ra :

hie = 2

25.feh = 12,5.hfe

ma : hie = rbb' + rb'e = rb'e (rbb' = 0) Do vay :

12,5.hfe = rb'e (1) gm = 5,12

1

'

=eb

fe

r

h= 0.08 mho

Tan so cao 3 dB :

fh = ebebMebeb CRCCR '''' 2

1

)(2

1

pipi=

+

vi : Rb'e = (ri // Rb + rbb') // rb'e = Rb // rb'e (rbb' = 0; ri = ) nen :

2-1

73

fh = ebebb CrR '' .//2

1

pi (2)

o li dong tan gia :

Aim = i

b

b

L

i

L

i

v

v

i

i

i '

'

.= = -gm.Rb // rb'e

Theo gia thiet : imA = 32 dB = 40

Suy ra :

Rb // rb'e = == 50008,0

40

m

im

g

A

vi :

Rb = 103 rb'e = =

50010

10.5003

3

103

T (2) ta co :

Cb'e = 500.10.800.2

1

//.2

1

3' pipi

=ebbh rRf

= 400 pF

T (1) suy ra :

hfe = 5,12

10

5,12

3

' =ebr

= 80

Vay : hfe = 80; rb'e = 1K; Cb'e = 400 pF 2-2 Cho s o mach nh sau :

Cac thong so : - r = 10

9 rad/s - hfe = 100 - Cb'c = 5pF - rbb' = 0 - IEQ = 10 mA i

CC

L

iLQ1

NPN

+V

i20uF

20uF

20uF

R

1k

1001k10k

1k10k

74

==

==

==

pFgC

mhoIg

KKKR

Tmeb

EQm

b

400/

4,040

9,010//1

'

i

Li

L

i

c1

b

b'e

b'e

ee

c2

mvb'e

+CM

c

B'

E

g

R

C

R

r

i C

C

C

RRr

a) Tnh o li tan gia Aim : cho ngan mach cac tu Cc1, Cc2, Ce va bo qua cac phan t Cb'e, Cb'c (cho h mach hai au cac phan t ay). Khi o :

Aim = i

eb

eb

L

i

L

i

v

v

i

i

i '

'

.=

vi : * Lc

c

m

eb

L

RR

Rg

v

i

+= .

'

* ebbi

bi

eb

i

eb

rRr

Rrr

i

v

'

'

'

//

//.

+=

Vay :

38)//(

).//(..

'

' =++

=ebbi

ebbi

Lc

c

mimrRr

rRr

RR

RgA

b) Tm tan so 3 dB fh : Ta co :

Cb'e = 40010

4,09==

T

mg

pF

75

Xet tan so cao ta se thay rang cac tu ghep ngoai Cc1, Cc2, Ce co tr khang rat be do rat ln ngan mach cac tu ghep ngoai. S o ch con :

bi b'eb'er

i//r

B' Li

cv

m+CM Lb'c

C

RR

gCRi

Va : CM = [1 + gm.(Rc//RL)]Cb'c CM = (1 + 0,4.500).5 pF CM = 1000 (pF) Tan so cao 3 dB :

h = )(

1

'' Mebeb CCR + (Rb'e = ri // Rb // rb'e = 196 )

h = 1210).1000400.(196

1+

= 3,64 (Mrad/s)

76

R = rbe + hfeRe, C = ''

1 em

eb

Rg

C

+

* o li tan gia : h mach Cb'c, C' :

Avm = ebiLefe

Lefe

rrRRh

RRh

')//(

)//(

++ = 0,9

* Tan so cao 3 dB : R' = rb'e + hfeRe' >> Re' : bo qua Re'. Do o tan so xay ra iem cc :

1 = )')((

1

)'('

1

'

'

'' CCRhrCCR cbefeebcb ++=

+

Thay so :

R' = hfe.CQI

25 + hfe.RE//RL = 10,5 K

C' = ''

'

1

/

1 em

Tm

em

eb

Rg

g

Rg

C

+=

+

= 2 pF

Suy ra : 1 = 12 Mrad/s : cho Zi e tnh tan so h ta xet hai trng hp sau :

= 12'' 10.40.500

1

.

1

=ebeb Cr

= 50 (Mrad/s)

a) Neu < : bo qua C'. Khi o :

Av = i

b

b

e

i

e

v

v

v

v

v

v '

'.=

Trong o :

* 1'

b

e

v

v

* 1/

1'

+=

ii

b

sv

v

vi i = '

'

1

icb RC ='iR ri + rbb'

77

h = 230 Mrad/s Aim 1

Do ta ang xet < ma > . Vay : co the lay xap x h = 1' 2-6 Cho s o mach nh sau :

- T = 109 rad/s

- Cb'c = 6p

78

- rbb' = 0 - IEQ = 1 mA - hfe = 20

Transistor Q2 c ghep dang B chung vi Q1 nham muc ch lam giam ien dung Miller CM . Tan so 3 dB ngan mach B chung f fT. Do o ta ch xet Q2 tan so thap va Q1 tan so cao nen Re1 b ngan mach.

S o thay the :

b ngan mach (noi mass). ay ta a bo qua mot tu Cbe mac song song Rc2 do qua nho. Trong o :

CM = (1 + gm.Rc1 // 1

'

+fe

eb

h

r).Cbc = 11,85 pF 12 pF

a) Tnh Aim : ngan mach cac tu ghep ngoai, h mach cac tu ghep trong ta se co s o sau e tnh Aim va ngan mach cac ien tr trong tr ien tr rbe. Ta co :

Aim = i

eb

eb

e

e

L

i

L

i

v

v

i

i

i

i

i1'1

1'1

..=

Aim = -hfb. )//.(.1

1

2

2ieb

ibc

cm

Lc

c hRhR

Rg

RR

R

++

Aim = ))((

)//(..

1

12

ibcLs

iebcmc

hRRR

hRRgR

++

Thay so :

Aim )2510)(1010(

)500//10.(10.04,0.10333

333

++

= -6,5.

Vay : Aim = -6,5 b) Tm tan so cao 3 dB h :

79

h = 12'' 10).1240.(33,333

1

))(//(

1+

=+ Mebebb CCrR

= 57,7 Mrad/s

Vay : h = 58 Mrad/s

* Ta thay tu CM co gia tr rat nho la do Rc1 // 1

'

+fe

eb

h

rma

1

'

+fe

eb

h

r= hib

co gia tr nho. Neu khong mac them Q2 vao th : CM = (1 + gm.Rc1 // RL).Cbc =120 pF Khi o :

h = 1210).12040(33,333

1+

= 18,75 Mrad/s : nho hn trng

hp tren rat nhieu. Do o ghep them Q2 lam tang bang thong cua mach. 2-9

i

iL

Z

Z

i

o

Q22N4223

100k

Vcc20V

20uF

1k4k

r6 20uF

+

-

v1

rds = 5K, Cgs = 6p, Cgd = 2p, gm = 0,003 mho

gmrds = =15 Gia s FET c phan cc che o tnh vi cac thong so nh tren. * Xet s o mach tan so thap :

a) Tr khang vao :

80

=+= igc

i ZRsC

Z1

' 1

'iZ khi 0

Neu mach hoat ong tan so > 2

1

cLCR= 50 Hz th

1//

//

++

=

ds

Ls

Ls

g

s

rRR

RR

v

v= 0,72

81

5,01

1

+=

++=

s

s

sCrR

R

v

v

cig

g

i

g

Vay :

Av = )5,0)(50(

72,0

1)(

..72,0.

1.

.

1

1

2

2

++=

+++ sssCrR

sCR

sCR

sCR

cig

cg

cL

cL

* Tan so gia : a) Tr khang vao : Zi = Rg =100K

b) Tr khang ra : Zo = Rs // 1+

dsr = 0,29K

c) o li ap : Avm =

1//

//

++

==

ds

Ls

Ls

g

L

i

L

rRR

RR

v

v

v

v= 0,72

Tan so thap 3 dB : L = 50 Hz (Tren thc te ham truyen Av co 3 iem cc song co 2 iem cc xap x bang nhau). * Tan so cao : ngan mach cac tu ghep ngoai :

1) Neu

82

Z = ri // 1

1

+=

sCr

r

C gdi

i

gd

Z1 = gsC

1

Suy ra :

+=

++=

sg

osgms

vZZ

Zv

ZZivvgv

1

1)]()([

T o suy ra :

Zo = 1

1

1 Zg

ZZ

i

v

mo

s

+

+=

hay :

Zo = )1)(/1(

)(1[.

1

gdimgs

igd

m CrjgCj

rCgsCs

g ++

++

Z0 = 28

811

)10.5(

)10.25,1(10.67,6

+

+

s

s

o li ap : a) < gm/Cgs : bo qua R. Khi o :

1210.95,3.1

1

]).1(

[1

1+

=

+++=

sRg

CCs

v

vA

m

gsgd

i

g

v

Tan so cao 3 dB : h = 2531.10

2 Mrad/s > gm/Cgs = 500 Mrad/s : loai b) >> gm/Cgs : bo qua C.

Av = 8

8

10.24,121

10.5

]).//(1[

//

+=

+=

gdii

i

i

g

CrRsr

rR

v

v

h = 12,24.108 rad/s = 1224 Mrad/s >> gm/Cgs

Gian o Bode :

83

1 = RC '.1

2 = RCC gd1.1

'1

+

Zi =

+'

1//

1

sCR

sCgd

2-10 S o mach : Cac thong so : + ri = 1K, Rg = 100K + Rs1 = 1,5K; Rs2 = 2,5K + RL = 1K + Cgd = 2p, Cgs = 6p + gm = 0,003 mho, Rds = 5K = 15 Ta xet cac khoang tan so sau :

a) Tan so thap : S o tng ng nh sau : Trong o :

+ Zi =

K

K

K

RR

R

ss

g

4

5,2.

16

151

100

]/[1

1 2 =

+

= 241K : rat ln so vi

ri.

+ Z16

5

1

' Krdso =+=

= 0,32K

+ A +

=1

'

v 1

* Tan so cat thap :

1 = 361 10].1241[10.20

1

][

1

+=

+ iic rZC= 0,2 (rad/s)

84

2 = ]320//10.410[10.20

1

]//[

1336'

2 +=

+ osLc ZRRC= 39 (rad/s)

Suy ra : L 38 (rad/s) * Dan nap ngo ra :

Yo = 33' 10.32,0

1

10.4

111+=+

os ZR= 3,38(kmho)

* o li ap :

Av = i

g

g

s

s

L

i

L

v

v

v

v

v

v

v

v..=

Trong o :

501.10.20.10

.10.20.10

1.

..

1 63

632

2

+=

+=

+=

+=

s

s

s

s

sCR

CsR

sCR

R

v

v

sL

cL

cL

L

s

L

39

)50(714,0

1//

1//

'

2

2

++

=+

+

+

=s

s

ZsC

RR

sCRR

v

v

oc

Ls

cLs

g

s

2,0.10.20.10.2421

.10.20.10.241

1 123

123

1

+=

+=

++=

s

s

s

s

sCrZ

Z

v

v

cii

i

i

g

Suy ra :

Av = )2,0)(39(

714,0 2

++ sss

Cung co the tnh bang cach khac : - Da vao s o mach ta thay Av co iem zero kep tai = 0. - iem cc :

+ 1 = 1)(

1

cii CrZ += 0,2 (rad/s)

85

+2 = 6332

' 10.20).320//10.410(

1

]//[

1+

=+ cosL CZRR

=39 (rad/s)

+ Aim = g

L

i

g

g

L

i

L

v

v

v

v

v

v

v

v=

.

(do Zi >> ri) = '//

//

oLs

Ls

ZRR

RR

+= 0,714

Vay :

Av = Aim.)39)(2,0(

714,0

))(( 21

2

++=

++ sssss

b) Tan so cao : hay rut gon con : Trong o : bo qua Rg do Rg >> ri .

+ C= 670.10.31

10.6

1 3

12

+=

+ Rg

C

m

gs= 2p

+ R = rds//Rs//RL = 0,67K

* Neu

86

- o li :

Av = ]10.2).10//670(1[10

10//670

]).//(1[

//1233

3

+=

+ sCrRsr

rR

gdii

i

Av = )1(

4,0

h

s+

- Dan nap ngo vao :

Yo = ssCrRr

gd

isds

2

33310.2

10

1

10.4//10.5

11

//

1 ++=++

Yo = 0,00145 + 2.10-12s, iem zero : 725 Mrad/s

Gian o Bode : 2-11 Cac thong so mach : - T = 10

9 rad/s, hfe = 100, Cbc = 5p, Cbe = gm/T = 40ICQ/T - ICQ = 10 mA Cbe = 400 pF, R1 = 10K, R2 = 1K Rb 1K - rbb = 0, ri = 10K, RL = 1K

- Cc1 = Cc2 = Cc3 = 20 F; rbe = hie = 100.10

25= 0,25K

S o tng ng tong quat : Trong o : + Rb1 = Rb2 = 1K; rbe1 = rbe2 = 0,25K + Rc1 = Rc2 = Rc =1K at R1 = ri // Rb1 //rbe1 =10K // 1K // 0,25K = 0,2K R2 = Rc1 // Rb2 //rbe2 =1K // 1K // 0,25K = 0,17K R3 = Rc2 // RL = 1K // 1K = 0,5K S o thay the bi tu Miller : Trong o : CM = (1 + gm. Rc2 // RL).Cbc2 = (1 + gm.R3).Cbc2 = (1 + 0,4.500).5p = 1000p CM _ phan anh tr khang tang sau ve tang trc. + C2 = Cbe2 + CM = 1400p

87

+ C = R2gmCbc1 = 0,17.103.0,4.5p = 340 pF

+ R = 1'

2

cbmCg

C= 1400p/(0,4.5p) = 700 = 0,7K

* Tan so 3 dB tren : Ta co :

= 1

''

C

CC cbeb +

C1 = Cb'e + (1 + gmR2)Cb'c C1 = 400p + (1 + 0,4.170).5p = 745p

nen :

= p

pp

745

5100 + = 0,544

1 = 200.10.745

1112

11

=

CR = 6,71 (Mrad/s)

2 = 170.10.1400

1112

22

=

CR= 4,2 (Mrad/s)

nen :

h =

+

+++

++ 2

2

2

1

1

2

2

1

1

2

2

21 4)1(2)1(22

= 2544,0.2

2,4.71,6

+

++ )544,01(2

2,4

71,6

71,6

2,4

+

+

++ 2

2

544,0.4)544,01(22,4

71,6

71,6

2,4

= 47,61[-(3,13) + 18,1)13,3( 2 + ]

h = 2,95 (Mrad/s) * o li tan gia : tan so cao ta co :

88

Av =

.

111

1.

)(

).(

21

2

21

2

2122

ss

RR

RRRg

Lc

cm

+

++

+

Suy ra :

Aim = Lc

cmv

s RR

RRRgA

+=

2

2122 ..

lim

Thay so :

Aim = 33

32

1010

190.200.10.4,0

+ = 2720

* Neu bo qua anh hng cua tu Miller, dung cong thc ghep n tang (n = 2).

Aim = (-gmRbe)n/2 = (-gmRbe)

2/2 vi : Rbe = Rc // Rb // rbe = 1K // 1K // 0,25K = 0,17K Suy ra :

Aim = (-0,4.170)2/2 = 2312 (

Lc

c

RR

R

+=

2

1)

Tan so 3 dB tren cua mot tang :

1 = 12'' 10.400.170

11

=ebeb CR

= 14,7 (Mrad/s)

Suy ra tan so 3 dB tren khi ghep 2 tang :

h = 0,64.1 = 9,4 (Mrad/s) (0,64 = 122/1 )

Ta thay ro rang hai ket qua qua chenh lech nhau. ieu nay ung bi le trong cong thc ghep n tang, ngi ta xem Cbc = 0 trong khi theo gia thiet Cbc = 5p. 2-12 S o mach : + ri = 1K, Rg = 1M + Rd = 10K, RL = 10K

89

+ Cgs = 6p, Cgd = 2p + gm = 0,03 mho, rds = 5K S o thay the : at : R1 = Rg // ri 1K R2 = Rd // rds // Rg Rd // rds = 3,33K R3 = Rd // rds // RL = 2,5K Tng t nh BJT : bien oi s o thanh : Trong o :

R = mgd gC

C 1.2

C = R2.gm.Cgd1 C2 = Cgs2 + CM = Cgs2 + (1 + gmR3)Cdg2 * o li ap : Ta tnh cac tham so : C1 = Cgs + (1 + gmR2)Cgd C1 = 6p + (1 + 0,03.3,33.10

3).2p = 206p C2 = 6p + (1 + 0,03.2,5.10

3).2p =158p nen :

= p

pp

C

CC gdgs

206

62

1

+=

+ = 0,039

1 = 31211 10.10.206

11

=RC

= 4,85 (Mrad/s)

2 = 31222 10.33,3.10.158

11

=RC

= 1,9 (Mrad/s)

Suy ra :

+

+++

++= 2

2

1

2

2

1

1

2

2

1

2

212 4)1(2)1(22

h

Ta tnh :

90

a = 22

21

039,0.2

9,1.85,4

2=

= 3029 (Mrad/s)2

b = )039,01(285,4

9,1

9,1

85,4)1(2

1

2

2

1 ++=++

= 4,87

c = =+ 22 4a 22 039,0.487,4 + = 4,870624

Suy ra :

2h = a(-b + c)

nen :

h = 000624,0.3029 = 1,375 (Mrad/s)

* o li tang gia :

tan so cao :

Ai =

21

2

21

212

111

1.

])//[(

)//.()(

ss

RrR

RRrRg

Ldsd

dsdm

+

++

+

nen :

Aim = Ldsd

dsdmv

RrR

RRrRgA

+=

)//(

).//.(lim 21

2

0

Aim = 33

3332

10.1010).5//10(

10.33,3.10.10).5//10.()03,0(

+

Aim = 748 Avm = Aim.i

L

r

R

91

2-13 Cac s o mach nh sau : + rbe = 1K + Cbe = 1000p + Cbc = 20p + gm = 0,05 mho

hfe = 50. a) S o tan so thap : Tan so 3 dB thap :

fL = )]///([2

1

febibee hRhRC +pi

fL = )]20020//(10[10.20.2

136 +pi

Suy ra : fL = 44 Hz * o li tan gia :

Aim = ieb

bfe

i

bfe

i

L

hR

Rh

i

ih

i

i

+=

=

Aim = -50.KK

K

110

10

+ = -45

* tan so cao ta co s o nh sau : Trong o :

CM = (1 + gmRc).Cbc CM = (1 + 0,05.10

3).10p = 500p Tan so 3 dB tren :

fh = )1000//1000.(10).1000500(2

1

)//)((2

112

''+

=+ pipi ebbebM rRCC

fh = 116 KHz Suy ra :

GBW = )4410.21,0(25)( 6 = Lhim ffA = 5,249 (MHz)

92

Vay : GBW = 5,249 MHz

b) S o tan so thap : Tan so 3 dB thap :

fL = )]//([2

1

1 efeiebic RhhRrC ++pi

Thay so :

fL = )]10.5010//(1010[10.20.2

133536 ++pi

= 0,23 (Hz) = fL

* o li tan gia :

Aim = efeiebi

bii

i

L

RhhRr

Rrr

i

i

++=

//

)//.(

Aim = KKKK

KKK

501100//1

)100//1.(1

++= 0,02

* tan so cao : bo qua Rb do Rb >> ri , ta co s o : Vi : R 'i = ri + rbb ri

R = rbe +hfe 'eR = 1K +50.1K = 51K

'eR = Re = 1K

C = 3''

10.1.05,01

1000

1 +=

+

p

Rg

C

em

eb = 20p

Ta xet hai kha nang sau : Neu < th Av 1 GBW = : loai.

Neu >> = 123'' 10.1000.10

1

.

1

=ebeb Cr

= 1 Mrad/s :bo

qua R. Khi o :

Av = i

b

v

v ' (do ve < 'bv )

93

Av = )'...(].').[(1

'.1

''2

'''

'

cbeicbiie

e

CCRRsCRCRRs

CRs

++++

+

Av = )10.8,22/1)(10.2,2/1(

10.5/1

10.2.10.5.1

10.2.177

7

1628

8

ss

s

ss

s

++

+=

++

+

h = 22 (Mrad/s) (h >> : thoa) fh = 3,5 (MHz) Vay : GBW = imLh Aff ).( 3,5 (MHz)

S o ghep C chung co bang thong ln hn s o ghep E chung nhng o li lai nho hn.

2-14 Cbc = 10p, gm = 0,05 mho Cbe = 1000p; rbe = 1K * Cach 1 : bo qua Cbc, coi tan so fL rat nho. Ap dung cong thc ghep 2 tang giong nhau, ta co :

Aim = (-gm.Rbe)(-gm.Lc

c

RR

R

+).rbe

vi : Rbe = ri // Rb // rbe = 1K // 10K // 1K = 0,5K = 2

1rbe

nen : Aim = (-0,05.0,5.10

3)2 = 625

Tan so 3dB cao cua mot tang :

f1 = 123'' 10.1000.10.5,0.2

1

2

1

=pipi ebeb CR

= 0,318 (MHz)

* Cach 2 : tnh toan chnh xac bang s o tng ng : a) Xet tan so thap :

94

vi Z = hie + hfe.

e

esC

R1

//

95

Chng 3 : KHUECH AI CONG SUAT AM TAN ____________

3-16

V IC = 0,2A nen ay la iem Q

bat ky. a) o doc cua ng tai AC :

50

1

20.)58,1(

11122

===LAC RNR

= 0,02 b) Phng trnh ng tai AC

trong he toa o tong quat :

iC ICQ = -ACR

1(vCE VCEQ)

+ Cho vCE = 0 maxCi = ICQ +

50

202,0 +=

AC

CEQ

R

V= 0,6A

+ Cho iC = 0 maxCE

v = VCEQ + ICQ.RAC = 20 + 0,2.50 = 30V

c) Gia tr nh cc ai cua ien ap collector khi khong b sai dang :

maxCmV = min[VCEQ, ICQ.RAC] = min[20 , 0,2.50] = min[20,10] = 10V

iC(A)

vCE(V) 0

Q 0,2

iCmax= 0,6

VCC= 20 vCEmax=30

DCLL( )

ACLL(-0,02)

20b

+CC

=

1,58:1

L

V 20V

R

R

96

d) PL = 50

10.2

1.2

1.2

1 2

2

22

==L

Cm

L

Lm

RN

V

R

V = 1W

e) PCC = ICQ.VCC = 0,2.20 = 4W

= 4

1=

CC

L

P

P= 0,25 = 25%

97

= 24.14,3

4.2 = 61,15W

c) 2PC = PCC PL = 61,15 24 = 37,15W

PC = 2

15,37 = 18,575W

d) = 15,61

24=

CC

L

P

P = 39,24%

3-20

ien ap collector nh trong moi transistor la cc ai : === CCCmp VVV maxmax 24V; maxmax Cmp II = = 8A

(v dong nh tang cung ty le ln gap 2 lan so vi 3-19)

a) 2

1.

2

1maxmaxmax

== CmCmL IVP .24.8 = 96W

b) PCC = ITB.VCC = CCC

VI

.2

max

pi =

14,3

8.2.24 = 122,3W

c) = 3,122

96=

CC

L

P

P = 78,5%

e PC max ta co :

ICm = pi

28.

2max pi

=CmI = 5,1A

2PC = PCC - CmCmCmCCL VIIVP .2

1.

2 2max

=pi

= pi

2.24.5,1 -

2

1.5,1.24

77,96 61,2 = 16,76W PC = 8,38W

98

3-21 He so ghep bien ap :

N = 40

160=

s

p

N

N= 4

ILm = 8

40.22=

L

L

R

P= 3,16A VLm = LLRP2

VLm = 8.40.2 = 25,3V VCm = VLm.N = 101,2V

Vay minCC

V = 101,2V

3-23 Goi sai dang khi cha co hoi tiep la D = 10%. Goi sai dang khi co hoi tiep la Df = 1%.

Ta co :

Df = A

D

+1 = 10% 1 + A = 10 A = 9 = T

A = 02,0

9=

T = 450

3-24

Avf = v

v

A

A

+1 = 20 1 + Av =

20

80=

vf

v

A

A= 4

Mat khac :

Df = + vA

D

1 D = Df(1 + Av) = 0,1.4 = 0,4 = 40%

3-26

99

a) LL

CCLLL R

R

VRIV ..

maxmax== = 20V

V mach khuech ai mac collector chung nen gia tr nh ngo vao ln nhat la : Vin = VL = 20V (Av 1)

b) 12

20.2

1.2

1 22

max

max==

L

Lm

LR

VP = 16,67W

c) Vdt = 2VCC = 2.20 = 40V

3-28

L

c1in

+

c2 L

E1

E2

B

B

1

2

30V

R680

R

680

+

C

500uF

R

8

R

0.5

R

0.5

+

Cv

T1

T2

a) Ipa = 680.2

4,130

2

2 =

R

VV DCC = 21mA

-

L

CC

CC

=20V

=-20V

c1 Y h

+

C vin

T2

V

+V

R 12

T1

100

b) 1B

V = VCC Ipa.R = 30 21.10-3.680 = 15,7V

2B

V = 1B

V 2VD = 15,7 1,4 = 14,3V

c) PL = 8

10.2

1.2

1 22

=L

p

R

V= 6,25W

PCC = ITB.VCC = 69,26

300

)85,0(

30.10

)(

.. =

+=

+=

pipipi LE

CCp

CCCm

RR

VVV

I

= 11,24W

= 24,11

25,6=

CC

L

P

P = 55,6%

d) fL = ==+ 7,266

10

10.5.5,8.28,6

1

)(2

1 4

4CEL CRRpi

37,5 Hz

3-29

a) RL + RE = =cLCfpi2

1

471

10

15.10.5.28,6

1 4

4=

= 21,23

'LR = 21,23 0,5 = 20,73

b) +

=+

=)73,205,0(2

30

)(2 '1

max

LE

CCLm

RR

VI 0,7A (= 0,7065A)

73,20.)7,0.(2

1.)(

2

1 2'2maxmax

== LLL RIP = 5,174W

PCC = ITB.VCC = =2

.2

max CCL VI

pi30.

14,3

7065,0= 6,75W

= 75,6

174,5max =CC

L

P

P = 76,65% ( =

EL

L

RR

R

+'

'

.4

pi = 76,65%)

Bai mau :

101

Cho mach ien nh hnh ve : Q3, Q4 co hfe = 20; rbe = 10; Q1, Q2 co hfe = 50; rbe = 100 a) Tm

maxmaxmax,, CCCL PPP

tren moi BJT, bo qua tieu tan cua thanh phan phan cc. b) Hay ve o th bieu dien PCC, PC, PL, theo ICm cua Q3, Q4. Bai giai : V Q1 va Q2 giong nhau, Q3 va Q4 giong nhau nen ta

ch can tnh cho Q1 va Q3. a) e PL max ta co :

41

15maxmax3max4 +

=+

===LE

CCLmCmCm

RR

VIII = 3A

20

3

3

3

3==

fe

cm

bmh

II = 0,15A

max 3,4 max

2 21 1. .3 .42 2

L Lm LP I R= = = 18W

3maxC

P tai ICm = pimax

2 CI= 1,91A

2 4.)91,1.(2

115.91,1.

2 2max

=pi

CP = 18,256 7,296 = 10,96W

maxC

P = 5,5W

333,4

22 . .

Cm

CC TB CC CC

IP I V V

pi

= = = 28,66W

c1

Yh

+

B

B

1

2

in

c2

VCC=

CCV- =

LR

4

-15V

R7

1

R6

1

Q4

Q3

v

C

C

R2

15V

R3

R1

R5

100

R4

100

Q1

Q2

102

28,66

P

Pcc

PL

PC

1,91 3

Icm 5,33 5,5

18

0

= max 3,4

3,4

18

28,66

L

CC

P

P= = 62,8%

6' .. 3334 RIrIV CmebbmR += = 0,15.10 + 3,1 = 4,5V

100

5,4

4

4

4==

R

VI

R

R = 0,045A = 45 mA

341 bmRCm

III += = 45 +150 =195 mA 0,2A

50

10.195 3

1

1

1

==fe

cm

bmh

II = 3,9 mA

+=+= 23]20.110//[100]//[ 36341 feieAC hRhRR

WRIP ACcmL 46,023.)2,0.(2

1..

2

1 21

2

1max 1==

WVI

P cccm

CC 91,115.14,3

2,0.22 11max ===

pi

%;2491,1

46,01max == Pcmax1 tai mA

II cmcm 2,124

2 max11 ==

pi

WP

P

C

c

5,0

23.)10.2,124(2

115.10.2,124.

22

1max

233

max1

=

=

pi

b) Neu Vi = 10V hay tnh cac bc nh tren . Gia thiet R1 , R2 , R3 rat ln nen bo qua:

103

at R=R4hfe1//[rb'e3 + R6hfe1 +R6hfe1hfe3] =0,5//[0,5K +1k] = 225

' ' .hfe1

hfe3

hfe3R1hfe14K

R6hfe11KR4hfe1500

rbe3

500

rbe1

100

+

-

Vi

925,04225,1,0

4

311

31

'

=++

=++

==KKOKhhRRr

hhR

V

VA

K

fefeLeb

fefeL

i

Lv

AR

VII

VVAV

L

LcmLm

iVL

31,24

25,9

)(25,910.925,0.

3 ====

===

Ah

II

fe

cm

bm 1155,020

31,233 ===

WR

VRIP

L

Lm

LLmL 7,104

)25,9(

2

1

2

1.

2

1 2223max ===

WVI

VccIPcc cccm

TB 2215.14,3

31,2.2.2. 333 ===

pi

AIIP

Pcmcm

CC

L 47,12

%;6,4822

7,1033

'

3

=====pi

103

fo=30Mhz rb

,e=1

K rbb'=0 hfe=100 fT=500Mhz Cb'c=2p

Teb

fe

eb r

hC

1

'

' =

Cb'e=31,8pF

4_3 Cho mach khuech ai cong hng nh sau :

R2

R1

100

L

->oo

->oo

->ooi

Vcc

i L Ce

Cc1

RFC Cc2

R

1kRe1k

10k

S o thay the Rb=1

K//10K~1K,qm=hfe/rb'e= 1,0 mho

1K1K 1k

'B

b'e'b L

L

m 'bg v

ii L Rb r

C R

V

Ta co R=Rb//rb'e=0,5K Cb' = Cb'e + CM = 31,8p + (1+gmRL)Cb'c = 31,8p + (1+0,1.103)2p = 234pF Cong hng tai fo = 30 Mhz th

)(302

1

'

MhzLC

fb

o ==pi

suy ra : HfC

Lob

pipi

12,0)10.30(4.10.234

1

)2(

1262122

'

===

Bang thong :

)(36,110.234.500.2

1

2

112

'

MhzRC

BWb

===pipi

o li dong :

104

)(1

. '

'

oo

i

m

i

b

b

L

i

Li

jQ

Rg

i

v

v

i

i

iA

+===

Vi : 2210.234.10.5.10.3.28,6 1227' == boi RCQ

nen : Aim = -gmR = -0,1.500 = -50

)10.3

10.3(221

150

7

7

+

=

j

Ai

4_4 Thiet ke mach cong hng n co : Aim = 10db = 3,16 ri = 1

K fo = 40 Mhz RL = 1

K BW = 1Mhz Vcc = 10 V Cuon day co Q = 50; S o can thiet ke co dang nh sau :

iC

iL

1k 1ke'b L

L

m 'bg v

i irLi r

Rp C R

V

ay ta chon transistor co : gm=0,01 mho ; rbb'=0 ; hfe = 100

Cb'c = 10p, Cb'e=1000p 2

2

'10

10

==m

fe

ebg

hr

Ta co :

6102

1==

RCBW

pi

vi R = ri //Rp//rb'e C = Cb'e + (1+gmRL)Cb'c + C' : C' la tu ghep ngoai

105

va c

o

c

o

ebi Q

C

Q

C

rRprR

+=++=++= 3

'

10.4,12500

1

1000

11111

vi c

cr

LQ

= : he so ton hao cua cuon day

Do o :

)50

)'1110(10.4,1.(

)'1110(2

110 36

CpF

CpFBW o

++

+==

pi

hay

)50

)'1110(10.4,1[

10.2

1)'1110( 3

6

CpFCCpF o

++==+

pi

hay

6

3

6 10.2

10.4,1)

10.502

21(

pipipi

= of

C

suy ra :

111010.1,1

50

401

10.22,0 99

==

=

C pF

Ta nhan thay : C = 1110+C' = 1110pF -> transistor co cac thong so khong thoa. Chon lai transistor co Cb'c = 2p rb'e=2500 Cb'e = 32p hfe = 100

gm = 0,01 fT = 500Mhz Khi o : Cb'e + ( 1 + gmRL )Cb'c = 54 pF suy ra C' = 1100 -54 = 1046 pF = 0,001uF

S o nh sau : HC

Lo

0144,012

==

106

i i'

L

>i LL

V

RLi r

C

R = ri//Rp//rb'e Rp = Qc. Lo. = 1,81K

Aim = -gmR = -0,01.512 = -5,12 > 3,16 : cha toi u 4_5 :

RL

i i=1K

n:1 m:1

1kL2L1i r

Cac ieu kien : Aim cc ai tai fo = 30Mhz e n gian hoa ta gia s n=m S o can thiet ke :

RL

i i=1K

n:1 m:1

2N4223

Rg1k

L2L1i r

Cac thong so cua FET : rds = 5

K , Cgs = 6p , Cgd = 2p gm = 0,003 mho Phan khang tr khang t ai : R'L = n

2RL S o tng ng :

2N4223 FET

107

2i

a g vm gi 'L

a Cgd(1+gm(rds//R ))2 L'

>iL

RrdsCgsL1ri

at : Ci = a

2[Cgs + Cgd(1 + gm(rds//R'L))]

Gia s rds

108

Cho hfe = 50, hie = 1K , rbb' = 0

Cb'e = 10p , Cb'c = 1p RL= 100 , C" = 10p Mach cong hng tai fo = 10Mhz va BW = 1Mhz

2

310.5

10

50 ===ie

fe

mh

hg

)(63. MhzBWAGBW vm ==

4_7

100

n

L

L

L

"

n2 n1

nLV

R+

-

Vi

9K

C

Giai Mach dung cuon cam kep at a=n1/n2 S o thay the

9K

i'

B'

(n/n2)2C

)C"

+

- L_a_Vbe'

2a Cb m bg v

Vo1

Vo

00R2

R1

+V

R

R1

hay : ref1

2ref

1

2O V

)1(R

R]1

1

1[V

R

RV

+

=

+=

133

Gia thiet : RA = RB

va QC

C

R

R

2

1

2

1 ==

ay la OP-AMP khuech ai khong ao :

211R

R1k

A

B =+=+= (1)

7.17

V1

A

B

+VO/K Vo

C1

R1

+

-

Vi

R2

R

R

)1sCR(2

VV)2(

1sCR

V2V1sCR

1kV 22

o

1

22

11

22

O +=+=

+=

Tong dong tai nut 1:

0

sC

1

VV

R

Vk

V

R

VV

1

1O

2

1O

1

1i =

+

+

(3)

)4(0)1sCR(2

sCV

)1sCR(R2

V)1sCR(

R2

VsCV

R2

V

R

V

2211

22

2

O22

1

O1O

2

o

1

i

=+

++++

)5(02

sCVR

2

sRRCCV

R2

VR

2

sCVR

2

V

2

sCRVsRCV

R2

RVV

1O1

2

2121O

2

O1

2O1O22o

11O

2

1oi

=

++

)6(0sRRCCVQV

sCRVVsCRVsRCVQVV2

2

2121OO

21OO22O11Ooi

=

++

)7(1]CRCRRC[ssRRCC

2

V

VA

212211

2

2121i

O

v ++++==

134

a_Tm HoLp , o , Q

khi C1 = C2 =C; R1 = R2 = R3 = R4 = R AoL = -> v

+ = v-=0 -> V3 = 0 (1); Zi =

id = 0 ta co :

)8(RCQ

1

RCQ

1

QRRQCC

1

RRCC

122

2

O2

2

2

2

222122121

2

O ====

T (7) [ ]Q

sCRCRRCs

o=++ 212211

)9(1

22

11

21

21

11

22112122

2121

RC

RC

RC

CR

CR

CRRCCRCR

RRCCQ

+

=+

=

Thay vao v du 7 8 ta co : R1 = R2 = R; C1 = C2 = C ; fo = 1 khz

Vao 9 ta co oo fRC

Q pi 21

;1111

1===

+=

Chon C = 10nF kCf

Ro

1610.10.28,6

1

2

183===

pi

7_22

C1=C C2=C

=R2 =R =KR4

=R

V2 V3V1Vo

R3

KRR4R

+

-

Vi

R

22

4

3

4

32 kVVkR

V

R

V

kR

VV

R

VVO

oO ==

=

(2)

sC

V

R

VV

R

VV

R

Vv Oi

1

1

3

1

2

21

1

1 +

+

=

Do R nh nhau nen :

Vi = 3V1 -V2 -VO + sCRV1 = V1(3+sRC) + VO/k - VO (3)

135

Mat khac : 4

32

2

2

2

21

1 R

VV

sC

V

R

VV +=

Do V3 = 0; R2 = R4 = R C1 = C2 = C nen : V1 - V2 = sRCV2 + V2 -> V1 = V2 (2+sRC) = -(Vo/k)(2+sRC) (4)

Thay 4 vao 3 k

kVsRCsRC

k

VV o

Oi

+++=

1)3)(2(

222236)1()(

CRssRCsRCk

k

V

VsH

i

O

+++++

==

55

51

1

55)5( 222222

++

++

+=

+++

=

k

CRssRC

k

k

k

CRssRCk

k

RC

k

Q

sCRs

kk

kH O

o

OLP

5;

5

5;5

+==

++=

5

5

5

1

5

51

5

)5( +=

+

+=

+=

k

k

k

RC

kQ

O

136

C1 = C2 =C; HoLP > 1 V+ = V- = V2 (1)

AB

O

R

V

R

VV =

B

A

V1

V2Vo

R1

R

RR2

C2C1

+

-

Vi

22 )1()11

( kVR

RVV

RRV

R

V

A

BO

ABB

O =+=+= (2)

)1(1 2221

2

2

2

21 sCRVsCRVR

V

sC

VV+==

)1

1()1(

22

221

sCRk

V

sCR

CsRVV O +=

+= (3)

CsRVCsRVVsCRV

sC

VV

R

VV

sC

VVoi

Oi

1211121

1

11 )21(11

++=

+

=

CsRk

VsCR

sCR

sCR

k

VVCsRV oooi 11

2

21 )21(

)1(+

++=

21

22

12

2

21

2

)2()1()(

RRCsRRsCk

kCRRs

V

VsH

i

o

+++== (4)

Qk

RRC

RCR

CRR ool

1

1

)2()5(

11 12

2

2212

21

2 =

+==

7_20

137

kR

V

sRC

sCV O=+

1

1

)2(1

1

)1(1

+=

+=

sCRk

V

ksCR

sRCVV

o

O

)6()2(

1

12 RRC

kQ

o +

=

Thay 5 vao 6 ta c R2 can chng minh

7_23 . Cho C1 = C2 = C ; R1 = R2 =R

0=== + VVAOL (1); kR

V

sCR

V O=+

1

1

Vo

V+

V-V1

KR2=R=R==C

=C

+

-

Vi

R1C2

R2

C1

KR

SC

VV

SCR

V

R

VV oi11

111 ++

=

; oi SRCVSRCVSRC

SRCVVV +

+= 1

111

SRC

VRCSSCRVVCRSSCRVSRCVSRCVVV OOi +

+++=

1

222

1

222

1111

OOO V

kRCS

CRSkSRCSCRV

SCR

SCR

k

V

SRC

CRSSCR 222222 )1(11.

1

1 +++=

+

+

++=

222)1(1)(

CRSkSCR

kSCR

V

VsH

i

O

+++==

fo = 1khz ; Q = 10 ; k=1

138

k = 1 :

Qj

Q

j

CRSSRC

SRCsH

oO

o

+

=++

=2222

121

)(

QRC

RCCR OOO

1

;22

22

2 ===