Giao Trinh DTTT

PHAN I

LY THUYET

Chng 1: MACH LOC TCH CC

1-1 Ham truyen co ap ng phang toi a:

Con goi la ham Butterworth. Khi bac cua bo loc tang len, tan so cat khong thay oi, nhng o doc cua bo loc tang dan en ly tng. Khi thiet ke cac bo loc bac cao: 3, 4, 5 ta da vao bang cac ham Butterworth a chuan hoa.

1-2 Mach loc tch cc bac nhat

a- Mach loc thong thap bac nhat: LTT1

Ham truyen: SCR1

ASVSVSH

11

0V

1

2

(1)

3

20V R

R1A (2)

12C CR

1 (3)

C1 V2(S)

R1 +-

R2 R3

Bo khuech ai ao

C 0

AV0

H(S)

C1

R1

R2

+ -

Bo khuech ai khong ao

Ham truyen: SCR1

ASVSVSH

12

0V

1

2

(1)

1

20V R

RA (2)

12C CR

1 (3)

b- Mach loc thong cao bac nhat

Ham truyen:

SRC11

ASVSV

SH

11

0V

1

2

(1)

1

20V R

RA (2)

C1

V2(S)

R1

R2

+ -

111 RC

1

0

AV0

H(S)

11t CR

1 (3)

1-3 Mach loc tch cc bac hai

a- Mach LTT2

AV0 = 1 (1)

212

20 CCR

1 (2)

10C22R

(3)

122

02 CR

1C

(4)

1

20V R

RA (1)

3221

20 RRCC

1 (2)

C1

V2(S)

R + - V1(S)

R

R R

C2

C2

Mach hoi tiep am mot vong

C1

V2(S)

R1 V1(S) R3

R2

C2 + -

Mach hoi tiep am 2 vong

Neu chon: 21

0V2

1

2

bA1b4

CC

(3)

10

12 Cf4

bR

(4)

0V

21 A

RR (5)

22120

22

3 RCCf4bR

(6)

Trng hp 1: AV0 = 1 (R3 = 0).

Neu chon 21

2

1

2

bb4

CC

(1)

Th 10

121 Cf4

bRR

(2)

2121

20 CCRR

1 (3)

Trng hp 2: R1 = R2 = R; C1 = C2 = C; AV0 1.

RC1

0 (1)

4

30V R

R123A (2)

C1 V2(S) R1

R4

V1(S) R3

R2

C2 + -

Mach LTT2 dung hoi tiep dng

59,022RR

4

3 (3)

b- Mach LTC2

Trng hp 1: AV0 = 1 va C1 = C2 = C.

212

20 RRC

1 (1)

C2R

01 (2)

2R

R 12 (3)

Trng hp 2: C1 = C2 = C; R1 = R2 = R;

RC1

0 (1)

23RR

1A4

30V (2)

59,022RR

4

3 (3)

c- Mach LTD2:

C1 V2(S) R1

R4

V1(S) R3

R2

C2 + -

Bo LTC dung hoi tiep dng

C1

V2(S) R1 V1(S)

R3

R2

C2

+ -

Bo LTD hoi tiep am 2 vong

Xet trng hp C1 = C2 = C ta co:

3321

110 R'RC2

1RRRRR

C21f

(1)

1

3

010V R2

RCRQA

(2)

21

21330 RR

RRR21CR

21Q

(3)

CR1

Qf

D3

0

(4)

21

21

RRRR'R

(5)

ieu kien: A0L > 2Q2.

min max

D

Q

H

C1 R1

R2

C2

+ -

LTD bac 2

Ham truyen: 2211

21

1

2

RSC1RSC1RSC

ZZSH

(1)

111 CR2

1f

(2)

222 CR2

1f

(3)

213 RC2

1f

(4)

3

10V f

flg10dBA

f1 f 0

A

H(dB)

f2 f3

-40dB

Chng 2: KHUECH AI CONG SUAT CAO TAN

(KCSCT)

2-1 Goc cat cua bo KCSCT:

Goc cat tnh theo o: T

tT180 000

(1)

Cac thanh phan dong ien ra c tnh da theo he so phan giai xung dong ien ra cua Transistor:

- Thanh phan trung bnh mot chieu:

0m0max0 I.II

- Thanh phan hai bac nhat:

1m1max1 I.II

- Thanh phan hai bac n:

nmnmaxn I.II

2-2 Cac mode hoat ong cua bo KCSCT lp C dung Transistor

Vm

Vm

Vmin C A 0

ib

V B

D

Vmax

Vb

iC

Imax

T

t0 t Imax

Hnh 2-1 Dang ac tuyen ong va gian o thi gian cua dong ien che o C

hfe

fcao

0 = hFE 0,7070

f

1

0 0,3f 3f f fT ftrungbnh fthap

Dai tan so lam viec cua Transistor c chia lam 3 oan:

- f0 f: tan so thap, cac tham so c coi la khong thay oi; hfe = 0;

- 0,3f f0 3 f: tan so trung bnh, cac tham so cua Transistor thay oi va xuat hien ien tr ky sinh (rbb), ien dung ky sinh (Cbe, Cbc)

2

0

fe

2

0

0*

ff

1

h

1

(3)

- f0 3 f: tan so cao, cac tham so cua Transistor thay oi, xuat hien rbb, Cbe, Cbc va cac iem cam ky sinh Lks.

00

00 f

fjj

(4)

Trong giao trnh ien t thong tin chu yeu chung ta se nghien cu bo KCSCT tan so thap va tan so trung bnh va ch xet che o kem ap. (Transistor nh mo nguon dong)

2-3 Bo KCSCT dung Transistor

1. Bo KCSCT dung Transistor che o kem ap mac Emitter chung.

CC LC +

VBB -

Lch

Cng

Cng

Lch

Cng

RE en

Rb

Rn

Cng

Cac bc thiet ke bo KCSCT khi cha ke en anh hng cua mach ghep au vao va au ra (Chu y: cac bc thiet ke khong nhat thiet theo trnh t a ra)

0- Xac nh pham vi lam viec cua Transistor theo (2-2) e ve s o tng ng tn hieu nho cho ung.

1- VCC = (0,5 0,8)VCEmax cho phep

2- Chon goc cat: = 600 900

3- Chon he so li dung ien ap: 1 = 0,85 0,95 = VCm1/VCC.

4- Xac nh bien o hai bac nhat tren Collector: VCm1 = 1VCC.

5- Xac nh cac dong ien:

*

1

1Cmn

I'I

; 1t1c'bTnn RC1'II

BOnB Itcos'Ii ; bmn I'I ; *COBOI

I

; 1Cm1

0CO II

6- *M* e'b*b CC'C ;

1

e'b*e'b

CC ;

1

11tc'be'bT*M

RCCC

1

1tc'bTe'b*b

RC1C'C

*'b

iEC Cj1Z

;

1

*

c'b*

e'b 1CC

Neu ke ca rbe ta co: ZiEC = rbe//ZiEC

Neu rbe >> ZiEC ta co ZiEC ZiEC

LC

|hfe|ib CC

Cbe rbe C*M

Rt1 In

R1 C*be

Neu rbe so sanh c vi ZiEC ta co:

20

e'b

20e'be'b

e'biEC

1

r

Cr1

rZ

7- Bien o ien ap kch thch vao: Vbm1 = In|ZiEC|

8- Cong suat vao cua nguon kch thch:

iEC2ni Z.'I2

1P

9- Xac nh tr khang nguon tng ng

e'bnn CR

T

fee'be'b

h1Cr

e dong ien au vao khong b meo th: n

e'be'bT

fen C

1C

hR

2

0

n*

n

1

RZ

10- Thien ap Base

)(Z'I7,0

)(Z'IVV

0

*

nn

0

*

nnBEB

11- ien tr tai tng ng: 1Cm

1Cmt

*

L IV

RZ

12- Cong suat nguon cung cap: PCC = ICOVCC.

13- Cong suat hu ch tren tai

t

21Cm

t2

1Cm1Cm1CmL RV

21RI

21IV

21P

14- Cong suat tieu tan tren Collector: PC = PCC PL.

15- Hieu suat cua mach: )()(

21

PP

0

1

CC

L

Trong thc te thng cong suat ra tren tai c biet trc nen ta co the tnh cac bc 0 4, 13, 11, 5, . . .

2. Bo KCSCT dung Transistor che o kem ap mac Base chung.

2

0

0*

1

; e'b* e'b

CC ; *

e'biEC Cj

1Z

Cac bc thiet ke tng t nh tren.

16. CC

0 CL1f vi C* c'bC 'CCC

Rt1 = 0Q0LC 00

1tC Q

RL

vi Q0 = 50 100

C20

2C Lf41C

c'bCC CC'C

Neu au vao bo KCSCT co mach cong hng Lb, Cb th ta cung xac nh tng t nh tren vi:

bb0 CL

1f ; Rt1 = 0Q0LC; b

20

2b Lf41C

b*

'bb 'CCC vi C*b tnh theo bc 6 tren.

2-4 Bo nhan tan dung Transistor

ie LC |h*fb|ie

CC Cbc

fe

e'b

hr

Rt1 In

Rn C*

be

CC LC + VBB -

Lch

Cng

Cng

Lch

Cng

RE Lb Rb Cb

Cng

Muc ch cua bo nhan tan:

- Nang cao tan so song mang

- M rong thang tan so lam viec

- Nang cao ch so ieu che trong may phat FM

- Nang cao o on nh tan so v khong co hien tng hoi tiep ky sinh qua Cbc do tan so hoat ong au vao va au ra khac nhau.

Tan so cong hng au vao:

bb0V CL

1 vi * e'bbb C'CC

vi e'b* e'b

CC ; b0101t LQR

Tan so cong hng au ra:

CC0ra CL

1k ; C020tn LQkR

Goc cat toi u cua bo nhan tan dung Transistor

k180

T ; k: he so nhan tan cua bo nhan

Cac bc thiet ke cua bo nhan tan:

0- Xac nh pham vi lam viec cua Transistor theo (2-2)

1- VCC = (0,5 0,8)VCEmax cho phep

ib LC |h*fe|ib

CC Cb Lb Rt2 Rt1 rbe C*be

2- Chon goc cat toi u: k

180T

3- Chon he so li dung ien ap:

k = 1 =0,85 0,95 = VCm1/VCC = VCmk/VCC

VCmk = kVCC

4- Xac nh xung dong hai bac k

1bm*

km

*

kCmk I.'I.I

5- Xac nh cong suat hu ch tren tai ng vi hai bac k

6- ien tr cong hng tng ng cua mach ra ng vi hai bac

k: 1tk

1

1Cm1

k

1Cm

Cmk

Cmktk R

I

VIV

R

7- Hieu suat cua bo nhan tan:

CC

Lkk P

P ; vi PCC = ICO.VCC

8- Do khong co hien tng hoi tiep qua Cbc nen

*

k

Cmk*

1

1Cmbnnn

III'II

iB = Incost IBO vi Cmkk

0**

COBO I.

1II

9- Tr khang vao cua tang

*'b

iEC Cj1Z

; vi

1

*'b

e'CbC

Neu ke ca rbe ta co ZiEC = rbe//ZiEC (tnh nh tren)

1 1 1 11 1 1

1 1

2 2k k k

Lk Cmk Cmk Cm Cm L LP I V I V P P

10- Bien o ien ap kch thch vao: Vbm1 = In|ZiEC|

11- Cong suat cua nguon kch thch: iEC2ni ZI21P

12- CBn P,V,Z tnh nh bo KCSCT

13- Tnh mach cong hng vao:

bb0 CL

1 ; vi

010

1tbb

*e'bb Q

RL;'CCC

14- Tnh mach cong hng ra

CC0 CL

1k ; vi 020

tnC Qk

RL

Chng 3: CAC MACH TAO DAO ONG

3-1 Cac van e chung ve mach tao dao ong

- Bo tao dao ong tan so thap, trung bnh: dung bo khuech ai thuat toan + RC hoac dung Transistor + RC.

- Bo tao dao ong tan so cao: 0,3f f0 3f dung Transistor + LC hoac dung Transistor + thach anh

- Bo tao dao ong tan so sieu cao: dung Diode Tunel, Diode Gunn.

- Cac tham so c ban cua mach dao ong: tan so dao ong, bien o ien ap ra, o on nh tan so, cong suat ra, hieu suat.

- Trong chng 3 ta ch xet mach dao ong LC, dao ong thach anh va ch xet ieu kien dao ong cua mach

- He so khuech ai cua bo khuech ai

*1

*

2A

**

V

VjexpAA

+ Modul he so khuech ai: *1

*

2*

V

VA

+ A goc di pha cua bo khuech ai.

- He so truyen at cua bo hoi tiep

ht**

jexp

Bo khuech ai A

Bo hoi tiep

V1 V2

+ Modul he so hoi tiep: *2

*

1*

V

V

+ B goc di pha cua bo hoi tiep

- ieu kien pha e mach dao ong: = B + B = 0,2

- ieu kien bien bo e mach dao ong: 1.A**

3-2 Bo dao ong LC dung Transistor

a- Mach tao dao ong 3 iem C kieu Colpits mac EC

Rb = R1//R2; Rb = Rb//rbe rbe (neu Rb >> rbe)

Cac bc thiet bo tao dao ong 3 iem C:

1- Xac nh pham vi tan so lam viec cua mach

2- Xac nh ieu kien pha

0C1XX

2BE1

; 0C1XX

1CE2

;

X3 = XCB = L > 0

Lch

R1

+VCC

Cng

L

R2

RE Cbe

Cng C1

C2 Vk

C E

B

L C1

C2 Vk

C

E

B

Cbe

3- Xac nh he so hoi tiep

nCC

VV

2

1

CE

BE (1)

- Ta thng biet f0, L t o suy ra:

21

2120

2t CCCC

Lf41C

(2)

- n co the tnh theo cong thc (3-45) nhng nhieu khi khong u d lieu e tnh

- Neu mach lam viec tan so thap ta co the chon

n = 0,01 0,05, t o tnh C1, C2.

- Neu mach lam viec tan so trung bnh, e mach hoat ong o nh ta chon: C2 = 10Cbe C2 = 11Cbe roi t (2) tnh C1, thay vao (1) tnh n

4- He so khuech ai cua s o mac EC

211

K2

e11

e21C n

h//RphhSZA (3)

- tan so thap: h21e = hfe, h11e = hie = rbe

- tan so trung bnh: 2

0

fe*

e21

1

h|h|

; e'b

e'b C.f21r

- Rk = 0LQ0 thng biet trc 0, L, Q0 (4)

- p: he so ghep au ra cua Transistor vi khung cong hng

n11

CCC

CCC

CC

VV

p21

2

1

21

21

K

CE

(5)

5- ieu kien bien o e mach dao ong: 1.A**

(6)

b- Mach tao dao ong 3 iem C kieu Colpits mac BC

Gia thiet RE >> hib

- Bc 1 va 2 lam nh tren, thng mach mac BC lam viec tan so thap.

- Bc 3: He so hoi tiep: 21

1

2

21

21

K

CE

CCC

CCC

CC

VV

p

Neu mach lam viec tan so thap ta co the chon n = 0,1 0,5; t o tnh C1, C2 v Ct thng tnh c.

Neu mach lam viec tan so trung bnh, tnh nh tren

- Bc 4:

2

11K

2

b11

b21C n

h//Rp

hh

SZA

tan so thap: h21b 1, h11b = hie/hfe

tan so trung bnh:

L C1

C2 Vk

Cbe

+VCC R1 R2

RE Cng

L C1

C2 Vk

Cbe

2

T

0

*

b21

1

1|h|

; e'bT

*

b11 Cf21h

Rk tnh nh tren.

He so ghep au ra cua Transistor vi khung cong hng:

1VV

VV

pBC

BC

K

BC

- Bc 5: ieu kien bien o e mach dao ong: 1.A**

c- Mach tao dao ong 3 iem C kieu Clapp mac EC

- Cac bc thiet ke tng t nh mach dao ong 3 iem C kieu Colpits mac EC, ch khac ve Ct va he so ghep p cua Transistor vi khung cong hng.

t0 LC2

1f

vi 021t C

1C1

C1

C1

Lch

R1

+VCC

Cng

L

R2

RE Cbe

Cng C1

C2

Vk

C

E

B

C0

L C1

C2

Vk

C

E

B

C0

Neu ta chon C1, C2 >> C0 th Ct Co khi o nhanh cong hng noi tiep L, C0 se quyet nh tan so cong hng cua mach va mach se on nh tan so hn

- He so ghep p: 1

0t

K

CE

CC

1CC

VV

p

d- Mach tao dao ong 3 iem C kieu Clapp mac BC

Cac bc thiet ke tng t nh mach dao ong 3 iem C kieu Colpits mac BC, ch khac ve Ct va he so ghep p.

- Khi biet f0, L ta tnh c Ct, ta se chon C0 ln hn Ct mot chut v du: Ct = 25pF th ta chon C0 = 30pF.

- He so ghep p: 21

21t

21

21

t

K

BC

CCCC

C

CCCC

CVV

p

3-3 Cac mach dao ong dung thach anh

a- S o tng ng cua thach anh

- Lq, Cq, rq la L, C, r cua thach anh (rq = 0)

- Cp: ien dung gia (Cp = 10 100pF) (Cq = 0,01 0,1pF)

Lch

R1

+VCC

L R2

RE Cng

C1

C2

Vk

C

E

B

C0

L

RE

C1

C2

Vk

C

E

B

C0

Tan so cong hng noi tiep: qq

q CL1

(1)

- Tan so cong hng song song:

p

qq

p

qq

pq

p

q

pq

pqq

p C2C

1CC

1

CCC

CCCC

L

1

- Tr khang tng cua thach anh:

Zq = Xq = j0Lt (3)

Vi

p

2

q

0qp

20

2

q

0

t

CCC

1

L (4)

e thay oi tan so cong hng rieng cua thach anh ta mac CS noi tiep vi thach anh:

pqq

20qp

Spqq20Spq

St CCLCC

CCCLCCCCj1Z

(5)

khi o tan so cong hng noi tiep cua mach se la

Sp

qqq CC

C1f'f

(6)

rq

Lq Cp

Cq

CS TA

sp

q

q CCC

21

ff

e giam anh hng cua Cp ngi ta mac tu C0 song song vi

Cq p0qq0

qqp CCneufCC

C1ff

b- Mach tao dao ong dung thach anh vi tan so cong hng song song

e mach dao ong theo kieu 3 iem C kieu Colpits, thach anh phai tng ng nh cuon cam, ngha la: q < 0 < p

Thc te 0 p nhng e tnh toan n gian do p q ta coi

2qp

0

(1)

p

qqp C2

C1 (2)

Lch

+VCC

Cng R2

C1

C2 LTA

Cng

Cng R1 RE

C1

C2 LtTA

jXq

0 q p

0

Lt

Ct Ct

Biet 0 , Cq, Cp ta tnh c q

- ien cam rieng cua thach anh: q

2q

q C.1L

(3)

- ien cam tng ng cua thach anh:

pqq20qp20

2

q

0

t CCLCC

1

L

(4)

t0tTA LjZ (5)

- 21

21

t20

t CCCC

L1C

(6)

Cac phan con lai tnh toan tng t nh mach dao ong 3 iem C kieu Colpits mac EC.

Ap dung cac cong thc (1) (6) tren va cac cong thc trong mach dao ong 3 iem C kieu Colpits mac BC.

Khi tu CS mac noi tiep vi thach anh no ong vai tro nh tu C0 trong mach dao ong 3 iem C kieu Clapp. Khi tnh Lt, Ct ta se chon CS ln hn Ct mot chut, roi tnh C1, C2 nh cac mach tren.

c- Mach tao dao ong dung thach anh vi tan so cong hng noi tiep

C1

C2 TA

RE Mach B.C

Trong loai mach nay thach anh ong vai tro mach hoi tiep. Ch ung tai tan so cong hng noi tiep cua thach anh th Zq 0 khi o B B va mach se hoat ong nh 3 iem C kieu Colpits hoac Clapp va cung co the mac EC hay BC

3-4 Mach tao dao ong RC

n gian va thong dung nhat la mach dao ong cau Wiew

- Tan so dao ong: RC1

- ieu kien dao ong ve bien o: 1.A.A***

Ma 31

nen 3A*

- Mat khac 212

1*

R2R3RR1A

B

R

B

Z3

Z1

Z2

D2

C

R2

Vin R

R1

+ -

R

D1

Vout

Chng 4: IEU CHE TNG T

4-1 ieu bien

a- Pho cua tn hieu ieu bien va quan he nang lng trong ieu bien.

V(t) = Vcost (1)

V0(t) = V0cos0t (2)

VAM(t) = V0(1+mcost)cos0t (3)

)1(VVm

0

(4)

V

0 t

Vo

0 t

VAM

0 t

VAM

0 t

V0

2mV0

0 - 0 + 0

tcos2

mVtcos

2mV

tcosVtV 00

00

00AM

- Cong suat tai tin:L

20

0 R2V

P (6)

- Cong suat hai bien tan: 2

mPP2

0bt (7)

- Cong suat ieu bien:

2

m1PPPP2

0bt0AM (8)

- He so li dung cong suat: AM

bt

PP

k (9)

- Cong suat ieu bien ln nhat: 20maxAM m1PP (10)

ay la ieu kien e chon Transistor sao cho

PAMmax < PCmax

b- ieu bien Collector

ien ap Collector bien oi theo ien ap am tan:

tcosVVV CC*

CC (11)

vi 1VV

CC

1Cm

e am bao Transistor khong b anh thung, phai thoa man ieu kien: CEOmaxCE0 BVVVV (12)

maxCECCCC000 VV2m1Vm1VmVV2 oi vi ieu bien th maxCECC V5,0V (13)

Neu au ra cua mach ieu bien la mach loc co hieu suat CH th ieu bien Collector co cong suat nh la:

phepchomaxC

CH

20

maxAM Pm1P

'P

(14)

ay la ieu kien e chon Transistor co PCmax cho phep

e thiet ke bo ieu bien Collector ta se tien hanh theo hai phan nh sau:

- Cho trc

2m1

PPPPP2

AM0

CH

AAM

mA khi a biet 0P ta

tien hanh cac bc thiet ke nh oi vi mach KCSCT (muc 2-3)

- Thiet ke phan ieu bien: 1Cm0 mVmVV

- Pho cua ieu bien tVAM (theo 3) va ve pho

- Tnh cong suat hai bien tan (theo 7).

- Tnh he so li dung cong suat k (theo 9).

- Kiem tra ieu kien ien ap (theo 12)

- Kiem tra ieu kien cong suat (theo 14).

4-2 ieu tan va ieu pha

a- Quan he gia ieu tan va ieu pha

Dao ong ieu hoa song mang:

tcosVtcosVtV 00000 (1)

Tn hieu ieu che am tan: tcosVtV (2)

Tn hieu ieu tan

FM:

000FM tsintcosVtV (3)

Trong o: tcost 0 (4)

Vi : lng di tan cc ai

Ch so ieu tan:

V

km f ; he so ty le (5)

Tn hieu ieu pha PM:

000PM tcostcosVtV (6)

Trong o: tcost 0 (7)

Vi : lng di pha cc ai

Ch so ieu pha: kVm p (8)

vi k: he so ty le

Quan he gia o di tan va o di pha:

tsin..dt

d

(9)

T 3, 6, 9 ta nhan thay ch can biet tn hieu ieu tan FM se tm c tn hieu ieu pha PM va ngc lai.

b- Pho cua tn hieu ieu tan va ieu pha

Khi ch tnh cac thanh phan Im(mf) 0,01I0(mf) th be rong dai tan cua tn hieu ieu tan chiem la:

maxffFM 1mm2D (1)

- Khi mf > 1 ta co bieu thc gan ung:

DFM 2mfmax 2 (2)

goi la ieu tan bang rong

- Khi mf < 1 DFM 2max (3)

0 + 2

In(m)

0 t 0 -

0 + 0

I0 I1

I1

I2

0 - 2

Goi la ieu tan bang hep

e mf const khi tan so thay oi pha phat phai co mach pre-emphasis va pha thu co mach de-emphasis.

c- ieu tan bang Varicap

n

inV eCC

(1)

Cin: ien dung ban au khi e = 0

: hieu ien the tiep xuc si 0,7V

n: he so phu thuoc loai varicap .2,1,21,

31n

eVe pc (2)

vi Vpc: ien ap phan cc ban au cho varicap

tcosVtcosVe 00 (3)

RD

CD

CC1

CC2 CC0

CV

0 Vpc

V

V

Trong thc te ta phai tm moi cach e giam anh hng cua ien ap cao tan tren varicap, khi o: tcosVe (4)

Goi ien ap AC tren varicap a chuan hoa: pcV

ex

(5)

n

pcin0V V

CC

(6)

0Vn

V Cx1C (7)

Tuy theo cach mac varicap vao khung cong hng ta co the tnh gan ung o di tan do varicap gay ra theo ien ap ieu che V.

pc0a V

Vnf5,0f (8)

30V

0V

pc0b CC

CV

Vnf5,0f (9)

40V

0V

pc0c CC

CV

Vnf5,0f (10)

Mac Varicap n:

a b c

L CV0 L L CV0 CV0 C3

C4

C

+VCC

R2 LK Ra

Cng CV0

R Lch

RE R1

Neu chon 2

VV CCCEQ th 2

VV CCRE tao phan cc ngc cho varicap.

ien tr R thng c chon vai tram k. Do dong tren R bang 0 nen VR = 0V. e thiet ke mach ieu tan varicap can tien hanh 2 phan

- Phan th nhat: thiet ke e mach thoa man ieu kien dao ong ve pha va bien o (giong phan 3-2-b).

- Phan th hai: thiet ke mach ieu tan varicap. Tuy theo cach mac varicap vao khung cong hng theo s o a, b, c ma chon cong thc (8) hoac (9) hoac (10) e tnh ffV .

Tnh f1 = f0 - f Lf41C 2

121td

CV1

Tnh f2 = f0 + f Lf41C 2

222td

CV2

Ve ac tuyen CV = f(V).

Mac varicap ay keo:

C

E LK

B

CV0

C

R

RE

C1

+VCC

V

VR L

E

Cng CV1

C2

R2

Lch

RE R1

Cng CV2 Lch

Cng

S o mac varicap ay keo triet tieu c hoan toan song cao tan tren varicap nen cac cong thc 8, 9, 10 tren c tnh

chnh xac hn. 2

C2

CCC

CCC 2V1V

2V1V

2V1V0V neu 2V1V CC . Ve ly thuyet

pF1001C 0V , tren thc te gia tr hay gap pF5010C 0V ; v du pF50CC 2V1V pF25C 0V .

Cac bc thiet ke c tien hanh nh 2 phan tren

d- On nh tan so trung tam cua tn hieu ieu tan

Cac bien phap on nh tan so trung tam f0 c xep t n gian en phc tap:

- ieu tan trc tiep bang thach anh: o di tan hep, ch dung trong phat thoai quoc te.

- S dung thach anh lam bo dao ong: o di tan hep.

- On nh nguon cung cap, s dung cac ien tr bu nhiet.

- Ha thap tan so trung gian cua bo ieu tan e nang cao o on nh tan so.

- S dung he thong t ong ieu chnh tan so AFC-F: ch ieu chnh tho.

- S dung he thong t ong ieu chnh tan so hon hp AFC-F va AFC-P: AFC-F ieu chnh tho, con AFC-P ieu chnh tinh a 0f .

C1 V

E

B

CV1

C2 Lch

RE

C

CV2

Chng 5: VONG GI PHA PLL

5-1 Nhng u, khuyet iem cua vong gi pha PLL

u iem:

- Kha nang lam viec tan so cao.

- S oc lap ve kha nang chon loc va ieu hng tan so trung tam.

- Nhng linh kien ben ngoai t.

- De dang trong viec ieu hng

Khuyet iem:

- S thieu thon thong tin ve bien o tn hieu.

- T ong ieu chnh he so khuech ai kho.

5-2 S o khoi va nguyen ly hoat ong cua PLL

Vong ieu khien pha co nhiem vu phat hien va ieu chnh nhng sai sot ve tan so gia tn hieu vao va tn hieu ra, ngha la PLL lam cho tan so ra 0 cua tn hieu song song bam theo tan so vao i cua tn hieu vao.

Bo so pha LTT

VCO

K

S o khoi cua PLL

Vp(t) Vd(t)

Vi(t) = Visinit

Vo(t) = Vocos(ot + )

Khi tn hieu vao a lot vao dai bat cua PLL, th tan so f0 cua VCO se bam theo tan so vao i .

5-3 Mot so ng dung cua PLL.

- Tach song tn hieu ieu tan.

- Tach song tn hieu ieu bien.

- Tong hp tan so.

- Nhan tan so bang khoa hai PLL.

- ieu che tan so (FSK) va ieu che pha (PSK).

- ong bo tan so.

- Bo loc bam theo thong dai hoac loc chan.

Chng 6: MAY PHAT

6-1 nh ngha va phan loai may phat

Mot so ch tieu ky thuat c ban cua may phat:

- Cong suat ra cua may phat.

- o on nh tan so: 730

1010ff

- Ch so ieu che AM (m), ch so ieu tan FM (mf)

- Dai tan so ieu che.

6-2 S o khoi tong quat cua cac loai may phat

- S o khoi tong quat cua may phat ieu bien (AM).

- S o khoi tong quat cua may phat n bien (SSB).

- S o khoi tong quat cua may phat ieu tan (FM).

- S o khoi tong quat cua may phat FM stereo.

6-3 Cac mach ghep trong may phat

Yeu cau chung oi vi cac mach ghep.

- Phoi hp tr khang.

- am bao dai thong D.

- am bao he so loc hai cao.

- ieu chnh mach ghep.

Cac loai mach ghep c ban.

- Ghep bien ap.

- Ghep ho cam.

- Ghep hai mach cong hng.

6-4 Cac mach loc c ban trong may phat

a- Mach loc n.

He so pham chat cua mach vao:

i

0

i

Li R

LRXQ

(1)

He so pham chat cua mach ra:

L0C

L0 CRX

RQ (2)

He so pham chat tng ng cua mach:

0i

0it QQ

xQQQ

(3)

e truyen at cong suat ln nhat va ap tuyen tan so

bang phang nhat ta co 2

QQQQ it0i (4) vi tan so loc

cua mach: LC1

0 (5)

b- Mach loc n.

Vi C

Ri

RL

L

C1 Vi

C2

Ri

RL

L

Khi mach oi xng C1 = C2 =C.

C

i

C

L0i X

RXRQQ (1)

vi

2CL

10 (2)

Khi mach bat oi xng: C1 C2 ta co.

10t

i1C C

1Q

RRX

(1)

20t

L2C C

1Q

RRX

(2)

2C1CL XXX (3)

vi LiRRR (4)

0it QQQ (5)

c- Mach loc oi.

1QRQR

X 2t

ti1C

(1)

RXX

X 3C1C2C (2)

1QRQR

X 2t

tL3C

9 (3)

LiRRR (4)

C1 Vi

C2

Ri

RL

L1 L2

C3

2C1C1L XXX (5)

3C2C2L XXX (6)

2QQ

Q 0it

(7)

Chng 7: MAY THU

7-1 nh ngha va phan loai may thu:

Mot so ch tieu ky thuat c ban cua may thu:

o nhay: bieu th kha nang thu tn hieu yeu cua may thu ma van am bao:

- Cong suat ra danh nh PL.

- Ty so tn hieu tren nhieu (S/N).

o chon loc: la kha nang chen ep cac dang nhieu khong

phai la tn hieu can thu: 1AA

Sf

0E

Chat lng lap lai tin tc

7-2 S o khoi tong quat cua may thu:

S o khoi tong quat mach khuech ai trc tiep.

S o khoi tong quat mach oi tan AM va FM.

S o khoi tong quat mach n bien (SSB).

7-3 Mach vao may thu:

Mot so ch tieu ky thuat cua mach vao:

He so truyen at: A

0MV E

VA (1)

o chon loc f

0E A

AS (2)

Dai thong D (BW)

Tan oan lam viec.

a- Anh hng cua Anten hoac tang au en mach vao

Tan so cong hng cua mach vao: LC1

0 (1)

ien dan tng ng luc cong hng:

Ld

CdgR1

0

0000

t (2)

vi 0

0 Q1d he so ton hao (3)

Khi co anh hng cua ien dan anten:

A0a CjgY (4)

hoac ien dan cua mach vao tang khuech ai au:

A0i CjgY (5)

ien dan tong ng cua mach cong hng tr thanh:

i2

0t gmgg (6)

i2

t CmCC (7)

He so pham chat cua mach cong hng se giam t Q0 xuong Qi theo quan he:

0

i2

0

t0

0 ggm

1gg

QQ

dd

(8)

Neu i2CmC (9) kha nho th o sai lech tan so cong hng

se la: CCm

21

CC

21

ff i

2

0

0

Nguyen tac xac nh he so mac mach (m)

- Mc tang ton hao khong vt qua gii han cho phep.

m2Ci m2gi C g0 L

S bien thien cua tham so anten va dan nap vao tang au khong gay anh hng en ch tieu cua mach vao.

Cac mach loc nhieu lot thang (nhieu tan so trung gian)

b- Cac mach ghep anten vi mach cong hng vao

f < 30MHz CA = 50 250pF; rA = 20 60.

f > 30MHz CA = 10 20pF; rA = 10.

7-4 Bo tron tan

Thc chat la mot tang khuech ai cao tang PF co hai tan so vao khac nhau va au ra bo oi tan ta co vo so tan so thns nfmf . V au ra bo oi tan ta at mot mach cong hng tai tan so trung gian: thnstg fff

bang song trung va ngan: KHz455ftg

bang song FM: MHz7,10ftg

7-5 Tach song

a- Nhng ch tieu ky thuat c ban cua bo tach song bien o.

He so truyen at: 0iTS

TS0TS mV

VVV

A (1)

Tr khang vao cua bo tach song: iTS

iTSiTS I

VZ (2)

Meo phi tuyen: %100.I

...II

m1

2m3

2m2 (3)

b- Tach song Diode

Tach song noi tiep: Da vao qua trnh phong nap cua tu ta se co dang ien ap am tan a tach song tren ien tr tai RL.

CK R LK C

ieu kien e tu C loc tan so trung gian au vao:

R1

C

(1)

ieu kien e RC khong gay meo tn hieu la:

mm1RC

2 (2)

V vay trong thc te 1m 8,07,0m

ien tr vao tach song: 2RR i (3)

e tnh R ta da vao bang sau:

SR 20 50 100 200 1000

ATS 0,75 0,84 0,89 0,93 0,98

Tach song song song t c dung do 3RR i

c- Tach song tan so

Yeu cau cua bo tach song tan so.

He so truyen at cao Smax ln: fddV

S TS0f

ac tuyen truyen at phang trong mot pham vi tan so rong.

ien ap vao khong can ln.

e ffV TS0 ma khong phu thuoc bien o ien ap vao nen trc bo tach song tan so can co bo han che bien o.

ac tuyen truyen at phai oi xng qua goc toa o ffff

Co 3 nguyen tac e thc hien tach song tan so.

- Bien tn hieu vao FM thanh tn hieu AM, roi dung tach song bien o e tach song thng dung IC chuyen dung nh IC e tach song.

- Bien tn hieu FM thanh tn hieu ieu che o rong xung roi thc hien tach song tn hieu o rong xung nh mot mach tch phan.

- Lam cho tan so cua tn hieu FM bam theo tan so VCO cua PLL, ien ap sai so chnh la ien ap can tach song.

PHAN II

BAI TAP

Chng 1: MACH LOC TCH CC

Bai 1.3 Thiet ke bo LLT bac 2 co tan so cat tren fc = 1Khz, C1 = 0,1F, |Av0| = 1 trong hai trng hp:

1. Bo LLT co hoi tiep dng.

2. Bo LLT co hoi tiep am nhieu vong.

Bo LLT co hoi tiep dng.

1b

22121

20

2b

0211

21

SRRCCSRRC11)S(H

Chon: 221.4

bb4

CC

21

2

1

2

C2 = 2C1 = 2x0,1F = 0,2F

11231010.14,3.4

2Cf4

bRR 73

10

121

Bo LLT hoi tiep am hai vong.

10

121

1

2v Cf4

bRR1RRA

0

R2 +-V1(S)

C2

C1 R1 V2(S)

V2(S) V1(S) C2

C1

R1

R2

+-R3

6,11221010.14,3.4

2RR 7321

4

22.1.4

b

A1b4

CC

21

v2

1

2 0

C2 = 4C1 = 4x0,1F = 0,4F

5576,1122.10.4.1010.10.4

1RCCf4

bR 776221

20

22

3

Bai 1.6 Thiet ke bo LLT bac 3 co tan so cat tren fc = 1Khz, C1 = 0,16F, Av0 = 3.

V cho moi C bang nhau nen ch co the dung bo LLT2 hoi tiep dng 1 vong.

Theo bang Butterworth ta co: B3(S) = (S+1)(S2 + S + 1), (b22 = 1; b12 = 1).

LLT1:

K110.16,0.10.28,6

1C.

1R 63C

1

K310x3R3R3RRA 312

1

2v0

LLT2: Chon

K110.16,0.10.28,6

1C

1RR 630

43

59,0RR

59,123RR

1A6

5

6

5v0

Chon R6 = 1K R5 = 590.

R4 V2(S) V1(S)

C

R1

R2

R3 + - +

- . V2(S) R5

R6

C

C

Bai 1.8 Thiet ke bo LLT bac 4 co tan so cat tren fc = 1Khz, tat ca cac C = 0,16F.

V tat ca cac tu ien bang nhau nen chon s o hoi tiep dng mot vong la n gian nhat. Khi o ta cung chon cac R bang nhau.

B4(S) = (S2 + 0,765S + 1)(S2 + 1,848S + 1)

Mat loc 1:

b21 = 1; b11 = 0,765

C1 = C2 = C3 = C4 = 0,16F

K110.16,0.10.28,6

1C

1RRRR 630

6521

235,1RR

RR

1765,03A4

3

4

3v01

Chon R4 = 1K R3 = 1235.

Mat loc 2:

152,0RR

RR

1848,13A8

7

8

7v02

Chon R8 = 1K R7 = 0,152. R8 = 152.

Bai 1.5 Thiet ke bo loc thong dai LTD bac 2 co tan so cong hng f0 = 10Khz; D = 2000Hz; Av0 = 10; C1 = C2 = 0,1F.

V1(S) R1 R2

R3

R4

R5 C1 +-

. +-

R6

R7

R8

C2

C3

C4 V2(S)

V2(S) V1(S) R1

R2

R3 C1 C2

+-

510.2

10Df

Q 34

00

K6,110.10.2.14,3

1DC1R 733

802010.6,1

A2R

R3

0V

31

21

21143822

320

'

RRRR16

4,6100

10.10.6,1.10.41

C.R.1R

R2 = 20.

Bai 1.11 Thiet ke bo LTD bac 2 co F = 1000 3000Hz; A0(dB) = 10dB; C1 = 0,1F f1 = 1000Hz; f2 = 3000Hz.

K6,110.1016,0

Cf21R 73

111

V2(S) V1(S) R1

R2

C1

C2

+-

1000 f 3000 0

1

H(dB)

10ffdB10

fflg10A

3

1

3

10V

Hz10010

100010f

f 13

K1610.10.28,6

1Cf2

1R7213

2

nF3,310.16.3

16,010.16.10.3.28,6

1Rf2

1C 63322

2

Chng 2:

KHUECH AI CONG SUAT CAO TAN

I. Khuech ai cong suat cao tan: (KCSCT)

Bai 2.2

PL = 0,1W; f0 = 10MHz; Q0 = 50;

f1 = 3500MHz; hfe = 100; Cbe = 10-9F;

Cbc = 1pF; PCmax = 2W; VCEmax = 40v;

maxiC = 1A; = 900; Zn = 1K.

1. MHz5,10f.3,0MHz10fMHz5,3100

10.3500hf

f 06

fe

T suy ra s o

lam viec tan so thap, ta co s o tng ng sau:

2. Chon VCC = 0,5VCEmax = 0,5x40 = 20V

3. Chon = 0,9.

4. Vcm1 = .VCC = 0,9x20 = 18V.

5. A1imaxmA11A011,018

1,0x2V

P2I C1Cm

L1Cm

6. ien tr cong hng tng ng:

LC CC Lch

Cng Rn

Rb en

Cng Lch

+ VBB -

+ VCC - Cng

LC CC 100ib

Rn

en hie Rt

ib

1636011,018

IV

R1Cm

1Cmt

7. H521,050.10.28,6

10.636,1Qf2

RL 7

3

00

tC

8. pF6,48655,20

1010.521,0.10.86,9x4

1Lf4

1C8

61420

2C

9. A10.22100.5,0

011,0).(

III 5

1

1Cmbmn

10. VBB = VBEO In.Zn.0( - ) = 0,7 22.10-5.103.0,3 = 0,634V

11. ICO = 0()..Ibm = 0,3x100x22.10-5 = 6,6mA

12.

53010.6,610.25.100.4,1

I10.25h4,1h 3

3

CQ

3

feie

13. Neu Rn = 1K th Zi = Rn//hie = 103//530 = 346

14. Bien o ien ap kch thch:

Vbm = Ibm.Zi = 22.10-5x346 = 76mV

15. PCC = ICO.VCC = 6,6.10-3x20 = 0,132W.

16. Hieu suat: %75132,0

1,0PP

CC

L

Bai 2.1: Gia thiet giong bai 2.2, ch khac fT = 350MHz

1. MHz5,10f3MHz10f;MHz5,3100

10.3500hf

f 06

fe

T

suy ra mach hoat ong tan so trung bnh co cac CKS. Cac bc 2 8 giong het nh tren

9. He so khuech ai dong ien dai tan so trung bnh:

Cbe Rt LC CM |hfe|ib

Rn

en rbe .

ib . -

CC

33027,3

100

10.5,3.210.21

100

1

hh

2

6

720

fe.

fe

.

10.

mA67,033x5,0

011,0.

II

1

1Cm'n

11. e dong vao Base khong b meo phai thoa man ieu kien:

e'bnn C.R1

5,4598,21

1010.10.5,3.28,6

1C.f2

1R3

96c'b

n

15027,3

5,45

10.5,3.210.21

5,45

1

RZ

2

6

720

n.

n

12. In = In[1 + TCbc.1().ZL] = 0,67.10-3[1 + 6,28.108.3,5.10-12.0,5.1,64.103] = 0,67.10-3[1 + 1,8] = 1,88mA

13. mV2,1810.88,1.15I.ZV 3n.

nnm

14. V7,05,0.15.10.63,07,0)(ZIVV 30.

n'nEO'BBB

15.

86,2j8,2.10.10.28,6.j

5,0)(ZC1Cj

)(Z 971Lc'bTe'b

1iEC

Mat khac:

86,2j'Cj

1Z'b

iEC

nF55,586,2.10.28,6

1'C 7'b

Chu y: Cb = Cbe + CM (khac vi ien t 2 la co them goc cat)

16. 51271.

c'bOEC 10.110j5,0.33110.10.28,6.j)(.1CjY

33

OECOEC 10.j10.j

1Y

1Z

17. mA7A10.710.67x33x32,0'I..I 35n.

0CO

18. PCC = ICO.VCC = 7.10-3.20 = 134mW = 0,134W

19. %6,74134,0

1,0PP

CC

L

Bai 2.8: Giong bai 2.1, ch khac au vao khung cong hng ch c ghep mot phan vao au vao e giam anh hng cua ien dung ky sinh va giam anh hng cua rbe.

Cac bc t 1 10 giong bai 2.2, nhng cach tnh mach vao

khac v au vao co khung cong hng. Cac bc t 15 19 giong bai 2.2.

Nh tren ta co: Cb = 5,55nF, khi o ien dung au vao: Cb = Cb + a2Cb = Cb + (1/25)5,55.10-9 = Cb + 222.10-12.

Coi he so pham chat cua khung cong hng au vao va au ra giong nhau: Q01 = Q02 = 50.

H517,04,3110.1624

50.10.21624

QR

L8

7010

1tb

pF51039,20

1010.517,0.10.86,9.4

1Lf4

1C8

614b

20

2b

Khi o ien dung can mac au vao: Cb = Cb a2Cb = 510pF 222pF = 288pF.

Tr khang vao:

e'b2e'b

tbi r25//1636ar

//RZ

16510.710.25..33.4,1

10.710.25h.4,1r 3

3

3

3.

fee'b

Suy ra: Zi = 1636//4125 = 1171,4 He so pham chat cua mach au vao:

Qi = 0RiCb = 6,28.107.1171,4.510.10-12 = 37,5 Neu au vao mac trc tiep vao Base, khi o ien dung ky

sinh rat ln (Cb = 5,55nF) >> Cb, nen mach lam viec khong on nh va khi o Zi rat nho (Zi = Rtb//rbe = 150) nen Qi = 0RiCb = 6,28.107.150.510.10-12 = 4,8

ngha la Lb, Cb cua tang KCSCT chnh la LC, CC cua tang tien KCSCT trc o.

II. Mach nhan tan

Bai 2.5

6

0

3500.1035 ; 10 0,3 10,5

100T

fe

ff MHz f MHz f MHz

h Suy ra mach hoat

ong tan so thap:

1. Chon goc cat toi u: 00

902

180

2. Chon nguon cung cap: VCC = 0,5VCEmax = 0,5x40 = 20V 3. Chon 1 = 2 = 0,9

Suy ra VCm1 = VCm2 = xVCC = 0,9x20 = 18V. 4. Cong suat ra cua bai 2.1 la cong suat hai bac nhat. Do o

cong suat ra do hai bac 2 gay ra se la:

W042,01,0x5,0

212,0P.)()(P.

)()(P 1L

1

21L

1

22L

5. Dong ien hai bac hai:

mA7,4180424,0x2

VP2I

2Cm

2L2Cm

6. ien tr tng ng cua mach cong hng ra:

383010.7,4

18IV

R 32Cm

2Cm2t

7. Neu coi gia thiet cua bai 2.1 trong o Q0 la he so cham chat cua khung cong hng au ra: Q02 =50 ta co:

LC CC Lch

Lb Rb Cb

Lch

+ VBB -

+ VCC - Cng

Rt2 LC Cb gmVbe

CC Lb Rt1 rbe

H61,08,62

10.38350.10.28,6.2

3830Q2

RL

7

7020

2tC

8.

pF10423,96

1010.61,0.10.86,9.4.4

1L.2

1C8

614C

20

C

9. Hieu suat bo nhan: %32318,075,0.5,0

212,0. 11

22

10. tan so thap: 30 22

( ) 0,318. 4,7.10 7,05

( ) 0,212CO CmI I x mA

11.

50049610.05,7

10.25.100.4,1I10.25.h.4,1hr 3

3

CO

3

feiee'b

12. tan so thap: A10.22100.212,010.7,4

h.90I

I'I 53

fe0

2

2Cm1bmn

13. Bien o ien ap kch thch au vao:

mV84V08425,0500//163610.22r//RIZ.IV 5e'b1t1bmi1bm1bm

14.

16243830x

5,0212,0R.

)()(R 2t

1

21t

15. Coi mach cong hng au vao va au ra co he so pham chat rieng bang nhau: Q01 = Q02 = 50. Thc te khi co rbe th he so pham chat Q01 < Q01.

16. F517,04,3110.1624

50.10.21624

QR

L8

7010

1tb

17. pF51039,20

1010.517,0.10.86,9x4

1Lf4

1C8

614b

20

2b

Bai 2.4

1. 6

0

350.103,5 ; 10 3 10,5

100T

fe

ff MHz f MHz f MHz

h Suy ra mach hoat

ong tan so trung bnh: Cac bc 2 en 10 giong nh bai 2.5.

12. tan so trung bnh:

. Rt2 LC Cb gmVbe

CC Lb Rt1 rbe

Cb

*

2 27

06

100 10033

3,0272 .1011

2 .3,5.10

fefe

hh

13.

1648,16310.05,7

10.25.33.4,1I10.25.h.4,1r 3

3

CO

3*

fee'b

14.

mA03,07,155

10.7,433.212,0

10.7,4

h90

II

33

*

fe0

2

2Cm'n

15. Vbm1 = Ibm1.Zi = Ibm1 = (Rt1//rbe) = 3.10-5(1636//164) = 4,47mV. Cac bc 15, 16, 17, 18 nh tren ch chu y: Cb = Cbmac + Cb*

20. Trong mach nhan tan do tan so cong hng au vao (0) va au ra (20) khac nhau, nen khong co anh hng cua CM*.

Khi o ta co: F10.25,0

10)(

CC 9

9

1

e'b*'b

Ta thay Cb* = 2000p >> Cb = 510pF. Do o mach nay lam viec khong on nh, ta phai cai tien mach nh bai 2.9

Bai 2.9:

Cac bc t 1 den 14 giong het nh bai 2.4 15. Vbm1 = Ibm1.Zi = Ibm1 = (Rt1//(rbe/a2) = 3.10-5(1636//4100) =

0,035mV = 35mV. Cac bc 16, 17, 18 giong bai 2.5

20.

F10.25,0

10CC 99

1

e'b*'b

suy ra pF802510.2C.a

9*

'b2

21. Khi o ien dung Cb = Cb Cb* = 510pF 80pF = 430pF.

Rt2 LC Cb gmVbe CC Lb Rt1

. Cb.a2 rbe

a2

VL

Chng 3: BO DAO ONG

I. Dao ong ba iem ien dung

Bai 3.2: Mach dao ong fthap. f0 = 10MHz, Q0 = 100, L = 1H, rbe = 500, fT = 3500MHz, hfe = 100, Cbe = 1000pF, PCmax = 4W, VCE0max = 40V, maxiC = 1A, CCE = 5pF.

Cac mach dao ong lam viec che o lp A nen tnh toan mach phan cc giong nh trong mon ien T I. Trong thc te: IEQ = 15mA va RE = 102103 (sinh vien t chon). V du: Chon IEQ = 5mA; RE = 103; VCC = 10V.

VRE = RE.IEQ = 103.5.10-3 = 5V Suy ra VCEQ = VCC VRE = 5V VBB 0,7 + RE.ICQ = 5,7V; Rb = (1/10).hfe.RE = 10K;

K26,237,510

1010VV

VRR 4BBCC

CCb1

K8,407,5

1010VVRR 4

BB

CCb2

50010.510.25100

I10.25hh 3

3

EQ

3

feie

1. ay la mach dao ong 3 iem ien dung nen thoa man ieu kien can bang pha v:

21 C

1X

; 1

2 C1X

; X3 = L.

R1

Cb

+VCC Lch

VK

R2

RE CE

C1

C2 E L

2. MHz35100

10.3500hf

f6

fe

T ;

MHz5,10f.3,0MHz10f0

S o lam viec tan so thap

3. He so hoi tiep: nCC

VV

2

1

CE

BE

n co the chon theo kinh nghiem: n = 0,010,05 (E.C) va n = 0,10,5 (B.C) hoac tnh theo cong thc. Chon n = 0,01 C2 = 100C1.

4. ien dung tnd ng cua khung cong hng:

pF5,25344,39

1010.10.86,9.4

1Lf4

1C8

61420

2t

pF5,253C100C

C100.CCC

CCC11

11

21

21t

21 CnF35,25C100 pF5,253C1

5. He so khuech ai cua s o mac E.C:

2

11K

2

e11

e21C n

h//Rp

hh

SZA

h21e = hfe = 100; h11e = rbe = 500; M510500

nh

Z 4211

ifa

6. RK = 0LQ0 = 6,28.107.10-6.100 = 6,28K. 7. He so ghep Transistor vi khung cong hng:

Cbe

C

Rb=R1//R2

B

CCE C1

C2 L

99,001,01

1n1

1CC

CC

CCCC

VV

p21

2

1

21

21

K

CE

8. 33632 10.23,110.155,6.5110.5//10.28,6.99,0

500100A

9. ieu kien bien o e mach t kch:

13,1201,0.10.23,1n.|A|||.|A| 3***

; Suy ra mach dao ong

Bai 3.1: Mach dao ong tan so trung bnh 1. Bc 1: nh tren.

2. MHz5,3100

10.350hf

f6

fe

T

MHz5,10f.3MHz10f0

Suy ra s o lam viec tan so trung bnh (co cac ien dung ky sinh)

3. He so hoi tiep co the chon nh tren, khi o C1, C2 giong het bai 3.2. Nhng e mach hoat ong on nh ta co the chon C2 = 10Cbe = 10.10-9 = 10-8F = 10nF, suy ra C2 = 9nF.

4. Nh tren ta co: Ct = 253,5pF

pF5,25310C

10.CCC

CCC 8

1

81

21

21t

20121

81 10.5,25310.5,253.C10.C

20418 10.5,25310.4,2531C10

121 10.5,253C.97465,0

pF255CpF260C '11

Chu y: C1 = C1 + CCE va C2 = C2 + Cbe trong o C1 va C2 mac ben ngoai, con CCE va Cbe la cac tu ien ky sinh.

Khi o: 026,010

10.260CC

n 812

2

1

5.

2

*

11K

2*

11

*

21C n

|h|//Rp

|h|

|h|SZA

tan so trung bnh ta co:

33

10.5,3.210.21

100

1

h|h|

2

6

722

fe*

21

5,4598,21

1010.10.5,3.28,6

1C.f2

1r|h|3

96e'b

e'b

*

11

K3,67

10.76,65,45

026,05,45

n|h|Z 422

*

11ifa

6. Nh tren RK = 6,28K.

7. 975,0026,01

1n1

1VV

pK

CE

8. 332 10.3,67//10.28,6.975,05,45

33A

397510.975,3

10.48,5.725,0K3,67//K97,5725,03

3


135,103026,0x3975n.|A|||.|A|***

Suy ra mach dao ong.

Dao ong 3 iem ien dung kieu Clapp Bai 3.3: Mach dao ong tan so trung bnh:

Bc 1, 2, 3 giong nh bai 3.1 tren 4. Nh bai tren ta co Ct = 253,5pF nhng

021t C1

C1

C1

C1

Chon C0 = 270pF e C0 quyet ng f0 trong mach

E

C0

VK

Cbe

C

B

CCE C1

C2

L

824

12

0

0

021

21

11

10.41,210.68445

10.5,16

.

11

.

11

t

t

t CC

CC

CCCC

CC

CC

18

188

18

1 C41,210C10.41,2.10.C10C

nF1,7C1

Khi o: 71,010.1010.1,7

CC

n 99

2

1

5. Giong bai 3.1 ch khac:

3,90

504,05,45

71,05,45

nhZ 22

11ifa

6. Nh tren: RK = 6,28K

7. 0357,01,7

2535,010.1,710.5,253

CC

VV

p 912

1

t

K

CE

8. 33,53,90//8725,03,90//10.28,6.0357,05,45

33A 32

9. ieu kien bien e mach t kch:

178,371,0x33,5n.|A|||.|A|***

Suy ra mach dao ong

10. Chu y: C2 = C2 Cbe = 10-8 10-9 = 9nF; C1 = C1 CCE = 7,095nF

Bai 3.4: Mach dao ong tan so cao. 1. Mach thoa ieu kien pha ve dao ong nh tren

2. MHz5,3100

10.350hf

f6

fe

T

f0 = 100MHz > 3f = 10,5MHz Vay mach hoat ong tan so cao. Ta khong the tnh cac Lk sinh, v vay phai dung s o dao ong mac Base chung. f = fT = 350MHz f0 = 100MHz < 0,3f = 105MHz Suy ra s o hoat ong tan so thap (khong can tnh Ck sinh)

Lch

RE

B R1

E

C0

VK

+VCC C1

Cb

L

R2

C2

+VCC

3. V s o lam viec tan so thap nen he so hoi tiep co the chon nh bai 3.2. ay la s o B.C nen chon n = 0,1.

1,0CC

CC

CCC.C

VV

21

1

2

21

21

BC

BE

4. Giong bai 3.3 ta co Ct = 253,5pF; chon C0 = 270pF va ta se

co: nF4,41CC1,01

C.n110.41,2

C.CCC

222

8

21

21

C1 = 0,1C1 + 0,1C2 = 0,1C1 + 4,14.10-9 0,9C1 = 4,14nF C1 = 4,6nF.

5. He so khuech ai cua s o mac B.C.

2

b11K

2

b11

b21C n

h//Rp

hh

SZA

h21b hfb 1;

5100500

hh

hfe

ieb11

500

1,05

nh

Z 22b11

ifa

6. Nh tren: RK = 0LQ0 = 6,28.108.10-6.100 = 62,8K

7. 3

9

12

8

12

21

21

10.6110.15,4

10.5,253

10.41,2

110.5,253

CC

CC

C

V

Vp

t

K

CE

8. 88,31500//2342,0500//10.8,62.10.372151A 36

9. ieu kien bien o e mach dao ong t kch:

1188,31,0x88,31n.|A|||.|A|***


III. Dao ong thach anh:

Bai 3.6: Mach dao ong tan so thap kieu Colpits Chu y gia thiet giong bai 3.2 nhng cha biet L. Lch

C

Cng

C1

Cb

+VCC

1. ieu kien pha: e mach dao ong theo kieu 3 iem C Colpits,

thach anh phai tng ng nh cuon cam, ngha la: fq > f0 > fp. e n gian ta coi f0 nam gia fq va fp: Mach Colpits E.C

q11

13

qp

qqp f005,110.2

101fC2

C1ff

Mat khac: qqqpq

0 f0025,12f005,1f

2ff

f

Suy ra: MHz975,90025,110

0025,1f

f7

0q

MHz025,10f005,1f qp

2. MHz35100

10.3500hf

f6

fe

T

f0 = 10MHz < 0,3f = 10,5MHz Suy ra s o lam viec tan so thap.

3. ien cam rieng cua thach anh:

mH55,2392410

10.10.5,99.86,9.41

C.f21L 1312

q2

qq

4. Tr khang tng tng cua thach anh tai tan so f0: Zt = j0Lt

nen: pqq20qp20qq

20

t CCLCC

1CLL

00572,110.55,2.10.10.86,9.4LCf4CL 31314qq20

2qq

20

fq f0 fp

f/2 f/2

H34H89,33

10.428,0.10.44,3910.572

10.00572,1101010.86,9.4100572,1L

1314

5

11131114t

5. pF5,710.34.10.86,9.4

1Lf4

1CC

CCC 614t

20

221

21t

6. Chon he so hoi tiep: n= 0,01.

nCC

VV

2

1

CE

BE

Suy ra C1 = 0,01C2 hoac C2 = 100C1.

Suy ra: pF5,757C100pF5,7C100C

C100.CC 111

11t

Vay C2 = 100C1 75,75nF va C1 7,75pF 7. He so khuech ai cua s o mac E.C

2

*

11K

2*

e11

*

e21C n

|h|//Rp

|h|

|h|SZA

h21e = hfe = 100; h11e = rbe = 500.

M510500

nh

Z 4211

ifa

8. RK = 0LtQ0 = 6,28.107.9,226.10-7.100 Suy ra RK = 57,94.100 = 5794

9. 99,001,01

1n1

1CC

CVV

p21

2

K

CE

10. 11355678.2,0M5//10.5794.99,0500100A 32


135,1101,0x1135n.|A|||.|A|***


Bai 3.5: Mach lam viec ftrung bnh mac kieu Clapp.

1. ieu kien pha: tnh nh tren, nhng lu y khong nhng thach anh phai tng nh cuon cam, ma thach anh cung CS cung phai tng ng nh cuon cam ngha la: q < q < 0 < p.

VK

CS=C0

E

C

C1

C2

B

fq = 9,975MHz va fp = 10,025MHz

2. MHz5,3100

10.350hf

f6

fe

T

f0 = 10MHz < 3f = 10,5MHz Suy ra s o lam viec tan so trung bnh phai ke ca CKS. Cac bc 3, 4, 5 nh tren: Lq = 2,55mH; Lt = 34H va Ct = 7,5pF = C1 noi tiep C2 noi tiep CS.

6. e CS quyet nh tan so cong hng trong mach ta chon CS

0833,010.03,3.10.275

C.CCCxC

CCCC

CVV

p

812

21

21t

21

21

t

K

CE

10. 1,2910.455//5794x0833,05,45

33A 32


191,21,0x1,29n.|A|||.|A|***


IV. Dao ong RC

Bai 3.8: Dao ong cau

e mach dao ong: 32

RRR

21

2

3R2 = 2R1 + 2R2 R2 = 2R1 Chon R1 = 1K, th R2 = 2K.

F16,010.16,010.10.28,6

1R

1C 6330

Bai 3.7: Dao ong cau Vien

VL

R1

C

+ -

R2

R

R

C

R1

-

R2

e mach dao ong: 2121

2 R2RRR

R31

Chon R2 = 1K R1 = 2K. F16,010.16,0

10.10.28,61

R1C 6330

Chng 4: IEU CHE TNG T

I. ieu bien Collector

Bai 4.4: PL = 100mW; f0 = 1MHz; f = 10KHz; Q0 = 50; m = 0,5; = 900. Tran sistor nh bai 2-1.

1. MHz5,3100

10.350hf

f6

fe

T

f0 = 1MHz < 0,3f = 1,05MHz Suy ra s o hoat ong tan so thap.

2. Chon nguon cung cap VCC = 0,5VCEmax = 0,5x40 = 20V. 3. Chon he so li dung ien ap = 0,9. 4. Bien o trung bnh phan hai bac nhat: VCm1 = V0 = VCC = 18V.

khi o V0 = 18cos(2.106t)

5. mA88,9

225,01

1,0x182

2m1

PxV

2V

P2I 2

AM

1Cm1Cm

01Cm

6. ien tr cong hng tng ng:

)(182210.88,9

18IV

R 31Cm

1Cmt

7. H8,510.5810.4,31

182250.10.28,6

10.822,1Qf2

RL 776

3

00

tC

8. nF37,475,228

1010.8,5.10.86,9.4

1Lf4

1C6

612C

20

2C

9. Bien o dong kch thch vao:

mA2,0A10.76,19100.5,010.88,9

h).(I

I 53

fe1

1Cmbm

10. mA610.2x100x3,0I.).(I 4bm0CO 11. Tr khang vao cua Transistor

LC CC Lch

V

Cng

+VCC

58310.610.25.100.4,1

I10.25h.4,1rZ 3

3

CQ

3

fee'bi

12. Bien o ien ap kch thch: Vbm = Zi.Ibm = 583.0,2.10-3 = 116mV

13. PCC = ICO.VCC = 6.10-3.20 = 120mW

14. %7012,0089,0

PP

CC

L

15. Bien o ien ap am tan: V = mV0 = 0,5x18 = 9V khi o: V(t) = 9cos(2.104t)

16. Pho cua tn hieu AM

t)1010(2cos5,4t)1010(2cos5,4)t10.2cos(18

t)cos(2

mVt)cos(

2mV

tcosVV

46466

00

00

00AM

17. mW8910.82,1.2

)18(R2V

P 32

L

20

0

18. mW125,112

)5,0(10.892

m.PP2

32

0bt

19. 1191

1m2

1PP

k

2AM

bt

20. Kiem tra: V0 + V = 18 + 9 = 27V < BVCEO = 40V PAMmax = P0(1 + m)2 = 89.10-3(1 + 0,5)2 = 200mW < PCmax.

Bai 4.5: Vtt = 10cos2.106t; V = 7cos2.104t; RL = 1K. 1. Giong nh bai 4-4: f0 = 1MHz < 0,3f = 1,05MHz Suy ra s o hoat ong tan so thap.

2. He so ieu che: 7,0107

VVm

0

3. Chon = 0,9 V11,119,0

10VV 1CmCC

106-104 106+104 106

VAM

4,5

18

0

4. Pho cua tn hieu ieu bien:

t)1010(2cos5,3t)1010(2cos5,3)t10.2cos(10

t)cos(2

mVt)cos(

2mV

tcosV)t(V

46466

00

00

00AM

5. mW500.2)10(

R2V

P 32

L

20

0

6. mW25,12249,010.5

2m.PP 2

2

0bt

7. Cong suat ieu bien: PAM = P0 + Pbt = 50 + 12,25 = 62,25mW

8. PAMmax = P0(1 + m)2 = 5.10-2(1 + 0,7)2 = 144,5mW < PCmax.

9. %201

5,02

1

1m2

1k

2

10. Kiem tra: V0 + V = 10 + 7 = 17V < BVCEO = 40V 11. Bien o hai bac nhat:

mA101010

RV

RV

I 3L

0

L

1Cm1Cm (chu y RL = Rt)

12. H18,3.4,31

1050.10.28,6

10Qf2

RL

4

6

3

00

L

13. nF97,74,125

1010.18,3.10.86,9.4

1Lf4

1C6

61220

2

14. Bien o dong kch vao:

mA2,0100.5,0

10h).(

II

2

fe1

1Cmbm

15. mA610.2x100x3,0I.).(I 4bm0CO

58310.610.25.100.4,1

I10.25h.4,1r 3

3

CQ

3

fee'b

16. V117,0583.10.2r.IV 4e'bbmbm

II. ieu tan Varicap

Bai 4.6: ieu tan dung Varicap n f0 = 100MHz; f = 75KHz; Q0 = 50; L = 0,1H;

106-104 106+104 106

VAM

3,5

10

0

Transistor nh bai 2-1

1. MHz5,3100

10.350hf

f6

fe

T

f0 = 100MHz > 3f = 10,5MHz Suy ra s o hoat ong tan so cao, nen phai chuyen sang dung mach dao ong B.C nh hnh ve

2. 21

219

71620

2t CCC.C

pF35,2544,39

1010.10.86,9.4

1Lf4

1C

Chon C1 = 30pF

pF5,16310)35,2530(10.35,25.10.3

CCxCC

C 121211

t1

t12

3. He so hoi tiep:

n155,010.5,16310.30

10.30CC

CC

CCCC

VV

1212

12

21

1

2

21

21

BC

BE

4. He so ghep Transistor vi khung cong hng:

1VV

VV

pCB

CB

k

CB

R1

L

C

Lch V

C

+VCC

R2

RE C2

C1

C

L

C

B

RE C2

C1

E

5. tan so thap nen: h21b 1; 1050500

hh

hfe

e11b11

6. Rt = 0LQ0 = 6,28.108.10-7.50 = 3140 7. He so khuech ai khi mac B.C:

8,36416//3140.1101

nh

//Rphh

SZA 22b11

K2

b11

b21C

8. ieu kien bien o e mach t kch: ongdaomach17,5155,0x37n.|A|||.|A|

9. Chon Vpc = 5V; n = ; = 0,7V; Suy ra ta co: 38

pc21

10 10.7557,0

V.5,16330

5,163.10.21.

21

VV.

CCC.nf

21f

suy ra

38

10.75V8,4411

5,163.10f suy ra V = 20mV

10. f1 = f0 - f = 99,925MHz;

f2 = f0 + f = 100,075MHz;

pF39,259381,3

1010.10.9985.86,9.4

1Lf4

1C10

71221

21t

pF317,259499,3

1010.10.10015.86,9.4

1Lf4

1C10

71222

22t

pF06,3010.11,138

10.39,25.10.5,163CC

xCCC 12

1212

1t2

1t211

pF955,2910.183,138

10.317,25.10.5,163CC

xCCC 12

1212

2t2

2t212

30,06pF

29,96pF CV0 = 30pF = C1

C

V(

Bai 4.7: ieu tan dung Varicap ay keo 1. Bc 1 va 2 nh tren

2. pF35,25CC

xCCCC

xCCC

21

21

20v10v

20v10vt

Chon pF10CC

xCCCpF20CC

20v10v

20v10v0v20v10v

Suy ra pF35,15CC

CCC21

213

3. He so hoi tiep: v s o mac B.C nen ta chon n = 0,1

21212

21

2

1

21

21

BC

BE

C9CC1,0C1,0C

1,0nCC

CC

CCCC

VV

pF35,151C9C

C9CCC

CCC1

11

21

213

9C1 = 153,5pF = C2 C1 = 17pF

4. Cac bc 4, 5, 6 nh tren

7. 5,6001,0

10//3140.1101

nh

//Rphh

SZA 22b11

K2

b11

b21C

8. ieu kien bien o e mach dao ong t kch:

ongdaomach16,81,0x26,86||.|A| 9. Chon Vpc = 5V; n = ; = 0,7V

R1

CV1 Ra VFM

+VCC

R2

L

C

VR

RE

C2

C1

R

CV2 -

38

pc30v

0v0 10.7557,0

V.

35,151010.10.

21.

21

VV

.CC

C.nf

21f

suy ra

39

10.75V57810f suy ra V = 43,4mV

10. Bc 10 nh tren ta co: Ct1 = 25,39pF; Ct2 = 25,317pF Suy ra: Cv01 = Ct1 C3 = 10,04pF; Cv02 = Ct2 C3 = 9,967pF

Chng 6: MAY PHAT

I. KCSCT + mach ghep au vao va au ra

Thiet ke bo KCSCT nh hnh ve co f0 = 10MHz. Transistor co cac thong so fT = 350MHz; hfe = 100; Cbc = 1pF; Cbe = 1000pF; PCmax = 10W; VCEmax = 40V; max iC = 1A. Cac bc thiet ke:

1. MHz5,3100

10.350hf

f6

fe

T

f0 = 10MHz < 3f = 10,5MHz Suy ra s o hoat ong tan so trung bnh

2. Chon VCC = 0,5VCEmax = 0,5x40 =20V 3. Chon = 0,9 4. Bien o hai bac nhat: PCm1 = .VCC = 0,9x20 = 18V.

5. Cong suat hai bac nhat: W5,28,0

2PP2

A1L

6. Bien o hai bac nhat cua dong ien ra:

mA27818

5,2x2V

P2I1Cm

1L1Cm

A1imaxmA556278x2I2 C1Cm 7. ien tr cong hng tng ng mach ra

65278,018

IV

R1Cm

1Cm1t

Rt1 chnh la tr khang vao cua mach ghep au ra. 8. Mach ghep au ra:

8,69xRRxRRR AtLi

Lch

+ VBB -

+ VCC -

-

L2

Rb

L1

C2

Cng

C1

Ri = 50 Pi = ?

Lch

Cng

C3

PA = 2W RA = 75

1=0,7 Qt1=5 =900 1=0,7 Qt2=5

275

8,6965Q

RRXt

i2C

pF59027.10.28,6

1Xf2

1C 72C0

2

295

8,6975Q

RRXt

L3C

pF55029.10.28,6

1C 73

pF5721CCCCC 1*

c'b'2

*c'b

'2

'2

662927XXX 3C2C2L

H110.28,6

66f2

XL 70

2L

9. He so khuech ai tan so trung bnh

33027,3

100

10.5,3101

100

1

hh

2

6

72

0

fe.

fe

.

10. Thanh phan trung bnh DC cua dong ra

mA178278x5,0

32,0I.)()(

I 1Cm1

0CO

11. Cong suat nguon cung cap PCC = ICO.VCC = 0,178.20 = 3,56W

12. Hieu suat cua mach KCSCT

%22,7056,35,2

PP

CC

L

13. Tr khang vao

'b*12

12697

1Lc'bTe'b

1iEC

Cj18j87,7j

35,6j50

0114,0110.28,6j5,0

5,0.6510.10.3501.10.10.28,6.j5,0

)(ZC1Cj)(

Z

L2

C2

Ri = Rt = 65 C3

RA = RL = 75

)CCC(nF023,287,7.10.28,6

1C *M*

e'b'b*

7'b*

14. Tr khang vao cua Transistor tan so trung bnh

5,6178,010.25..33.4,1

I10.25h.4,1r

3

CO

3.

fee'b

15. Mach ghep au vao: e s truyen at cong suat ln nhat va ap tuyen tan so bang phang nhat ta thiet ke sao cho Qi = Q0 = 2Qt.

65,010

5,6Q2r

QRX

t

e'b

0

LC

nF5,2465,0.10.28,6

1Xf2

1C 7C0

'1

Mat khac: nF5,22023,25,24CCCCCC * 'b

'11

*b1

'1

50050x5x2RQ2RQX itii2L

H810.28,6

500f2

XL 70

L1

16. e'b

*'b0

e'be'biEC

'iEC r.Cj1

rr//ZZ

5

3,15,6

25,42.1010.092,4.86,9.41

5,6rC1

rZ

1418

2e'b

2*'b

20

e'b'iEC

17. Cong suat au ra cua Transistor

mW75,010.5,75.10.3.21

5.533,0278,0

21Z

I21ZI

21P

44

2

i

2

1

1Cmi

2'nb

18. Cong suat au vao cua tang KCSCT

mW07,17,010.75,0PP

3

1

bi

II. KCSCT + ieu bien

L1

C'1 Ri = 50 Qt1 = 5

rbe Cb

Thiet ke mach ieu bien Collector co gia thiet nh bai tren biet m = 0,7 va f0 = 10MHz; f = 10KHz. e n gian ta khong tnh mach vao

1 4 nh bai tren: VCm1 = 18V = V0 5. Cong suat ieu bien:

W5,28,0

2PP2

AAM

Cong suat hai bac nhat:

2

m1PP2

0AM

W2

249,01

5,2

2m1

PPP2

AM1L0

6. Bien o hai bac nhat cua dong ien ra

mA18718

2x2V

P2I1Cm

1L1Cm

7. )(25,96187,018

IV

R1Cm

1Cm1t

Rt1 chnh la tr khang vao cua mach loc au ra 8. Mach loc au ra

8575x25,96xRRR Li

2,365

8596Q

RRXt

i1C

nF44,02,36.10.28,6

1Xf2

1C 71C0

1

325

8575Q

RRXt

L2C

nF5,032.10.28,6

1C 72

+VCC

L1

+ VBB -

Rb KCSAT

Lb C1

Cng

Cng

Lch

C2

RA = 75 PA = 2W

pF5,175,0.331101CC 121*

c'b*

c'b

pF5,4225,17440C '1 2,68322,36XXX 2C1C1L

H1,1H10.86,1010.28,62,68

f2XL 77

0

1L1

9. Nh tren 33*

10. A12,0187,0x5,0

32,0I.)()(

I 1Cm1

0CO

11. PCC = ICO.VCC = 0,12.20 = 2,4W

12. %704,2

68,1PP

CC

1L

13. Nh tren )25,96Zvi(83,7jZ LiEC

14.

625,912,010.25..33.4,1

I10.25h.4,1r

3

CO

3.

fee'b

15. 2

e'b2*'b

20

e'b'iEC

rC1

rZ

6587,1625,9

64,92.10.16,4.10.86,9.41

625,91814

16. Cong suat au vao

mW4,06.33.5,0

187,021Z

I21ZI

21P

2

i

2

11

1Cmi

2'nb

17. Bien o ien ap am tan V = mV0 = 0,7x18 = 12,6V khi o: V(t) = 12,6cos(2.104t)

18. Pho cua tn hieu am tan

t)1010(2cos3,6t)1010(2cos3,6)t10.2cos(18

t)cos(2

mVt)cos(

2mV

tcosVV

47477

00

00

00AM

L1

C1

Rt1 = Ri = 96 C2

RA = RL = 75

107-104 f 107+104 107

VAM

6,3

18

0

19. W41,0249,068,1

2m.PP

2

0bt

20. %4,165,241,0

PP

kAM

bt

21. Kiem tra: V0 + V = 18 + 12,6 = 30,6V < BVCEO = 40V PAMmax = P0(1 + m)2 = 1,68(1 + 0,7)2 = 4,85W< PCmax = 10W (Chu y: cac bc 9 16 co the khong phai tnh khi thi).

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