bai tap mach dien tu

8/2/2019 Bai Tap Mach Dien Tu

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Khoa ien ien T. Ky thuat mach ien T I

-1- Tom Tat Bai Giang. -1-

BAI GIANG TOM TAT MON:BAI GIANG TOM TAT MON:BAI GIANG TOM TAT MON:BAI GIANG TOM TAT MON: IEN T INgi soan: TS. Pham Hong Lien.

Giao trnh chnh: Mach ien T 1 Le Tien Thng, HBK Tp.HCM.

Chng 1: Diode ban dan.Chng 1: Diode ban dan.Chng 1: Diode ban dan.Chng 1: Diode ban dan.I.I.I.I. Diode chnh lu:Diode chnh lu:Diode chnh lu:Diode chnh lu:

1111---- Quan he gia ien ap va dong ien cua Diode (H2Quan he gia ien ap va dong ien cua Diode (H2Quan he gia ien ap va dong ien cua Diode (H2Quan he gia ien ap va dong ien cua Diode (H2----1):1):1):1):

= 1

nKTqVexpIi D0D (1-1)

iD : Dong ien trong Diode (A).

VD : Hieu ien the hai au Diode (V).I0 : Dong ien bao hoa ngc (A).

q : ien tch electron 1,6.10-19 J/V.

K : Hang so Bolzman 1,38.10-23 J/0K.

N : Hang so co gia tr trong khoang (12) phu thuoc vao loai ban dan.

Goi ien the nhiet:q

KTVT = (1-2)

T (1-1) ta co:

=

T

D0

T

D0D nV

VexpI1nVVexpIi (1-3)

nhiet o T=3000K, tng ng T=270C, ta co VT2526mV. Khi o ien trong cua Diode c tnh bi phng trnh:

)(i

nVIi

nVrD

T

CD

Td =

+= (1-4)

ac tuyen Volt-Ampere cua Diode tren (H2-2)Kieu mau mach tng ng cua Diode tren (H2-3a,b,c).

2222---- Phng trnh ng tai cua Diode (H2Phng trnh ng tai cua Diode (H2Phng trnh ng tai cua Diode (H2Phng trnh ng tai cua Diode (H2----5).5).5).5). Phng trnh ng tai mot chieu cua Diode (DCLL)

1DDS RIVV += (1-5)


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Phng trnh ng tai xoay chieu cua Diode (ACLL)

)R//R(ivv L1dds += (1-6)

T (1-5) va (1-6) tren he toa o tong quat ta co:

DQdDDQdD Iii&Vvv +=+= (1-7)

Vi:

VD va iD la thanh phan tc thi cua ien ap va dong ien. VDQ va IDQ la cac gia tr mot chieu cua ien ap va dong ien. vd va id la cac gia tr xoay chieu cua ien ap va dong ien.

Vay phng trnh ng tai xoay chieu ACLL trong he toa o tong quat se

la:

sDQDL1DQD v)Ii)(R//R(Vv += (1-8)

3333---- Chnh lu ien ap xoay chieu:Chnh lu ien ap xoay chieu:Chnh lu ien ap xoay chieu:Chnh lu ien ap xoay chieu:a- Chnh lu ban song: (H2-6)

ien ap au vao: tsinvv maxs =

ien ap trung bnh DC tren tai:

=

+= maxL

LS

LmaxDC

VRR

RVV (1-9)

b- Chnh lu toan song: (H2-8a,b,c)

ien ap trung bnh Dc tren tai:

= maxLDCV2V (1-10)

4444---- Mach loc:Mach loc:Mach loc:Mach loc: (H2(H2(H2(H2----9a,b)9a,b)9a,b)9a,b)

Khi co tu C mac song song vi RL trong cac mach chnh lu ta co quan hegia ien ap trung bnh tren tai vi bien o ien ap au vao va ien tr R L va tuien C nh sau:

maxL

LDCmaxDC V1CfR4

CfR4fC4

IVV

+== (1-11)

5555---- Mach nhan oi ien ap:Mach nhan oi ien ap:Mach nhan oi ien ap:Mach nhan oi ien ap: (H2(H2(H2(H2----11a,b)11a,b)11a,b)11a,b)


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ien ap ra gan gap oi ien ap vao.

II.II.II.II. Diode on ap Zener:Diode on ap Zener:Diode on ap Zener:Diode on ap Zener:1111---- Cac tham so c ban cua diode ZenerCac tham so c ban cua diode ZenerCac tham so c ban cua diode ZenerCac tham so c ban cua diode Zener:::: (H3(H3(H3(H3----1)1)1)1)

ien ap on nh VZ khi dong ien qua zener thay oi trong khoang I zminIzmax. Thc te maxzminz I10

1I . (1-12)

ien tr ong tai iem lam viec.

dIdVr Zd = (1-13)

Diode Zener ly tng c coi co rd 0.

ien tr tnh:Z

Zt IVR = (1-14)

He so on nh:

d

t

Z

Z

Z

Z

ZZ

ZZ

rR

IV

dVdI

V/dVI/dIS === (1-15)

2222---- Mach on ap dung Diode Zener:Mach on ap dung Diode Zener:Mach on ap dung Diode Zener:Mach on ap dung Diode Zener: (H3(H3(H3(H3----2)2)2)2)

Mach tren hnh 3-2 luon thoa man he phng trnh:

+=

+=

ZiRS

LZR

VRIVIII

(1-16, 1-17)

Trong o ch co VZ const, con cac ai lng khac co the bien oi nhngphai thoa man ieu kien:

IZmin khi ILmax va VSmin IZmax khi ILmin va VSmax

T (1-16) va (1-17) tuy tng trng hp cu the ma ta co the suy ra cac hephng trnh khac nhau.

V du neu Ri = const th ta co he phng trnh:

(VSmin VZ)(ILmin + IZmax) = (VSmax VZ)(ILmax IZmin) (1-18)

V du neu Ri =const va RL = const ngha la IL = const th ta co he phngtrnh:


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VSmin = (IZmin + IL)Ri + VZ = IminRi + VZ (1-19)

VSmax = (IZmax + IL)Ri + VZ = ImaxRi + VZ (1-20)

Chu y v VL = VZ const nen khi IL thay oi ta co:

maxL

ZminL IVR

= (1-21)

minL

ZmaxL I

VR = (1-22)


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Chng II: Transistor hai lp tiep giap (BJT)Chng II: Transistor hai lp tiep giap (BJT)Chng II: Transistor hai lp tiep giap (BJT)Chng II: Transistor hai lp tiep giap (BJT)

che o tn hieu ln. che o tn hieu ln. che o tn hieu ln. che o tn hieu ln.

I.I.I.I. Cac tham so c ban cua Transistor.Cac tham so c ban cua Transistor.Cac tham so c ban cua Transistor.Cac tham so c ban cua Transistor. (H2(H2(H2(H2----1)1)1)1)+ He so truyen at dong ien phat khi mac Base chung

Thong thng = 0,95 0,99, ly tng = 1.

+ He so truyen at dong ien khi mac Emitter chung:

=

1(vai chuc vai tram lan).

+ Dong ien ra cc Collector:IC = IE + ICBO (2-1)

Trong o ICBO la dong ien phan cc ngc hay con goi la dong nhiet,thng rat nho.

+IE = IC + IB (2-2)suy ra IB = (1-)IE ICBO (2-3)

=

= CCBO

CCBOCB

IIIII1I (2-4)

tan so thap (H2-1) ta co: hfe = = hFE (2-5)

II.II.II.II. Mach phan cc cho Transistor:Mach phan cc cho Transistor:Mach phan cc cho Transistor:Mach phan cc cho Transistor:1111---- Mach phan cc Collector:Mach phan cc Collector:Mach phan cc Collector:Mach phan cc Collector:

Ta co phng trnh tai mot chieu:

VCC = VCEQ + ICQRC + IEQRE

VCEQ + ICQ(RC + RE) (2-6)

So lng ien t ti c Collector

So lng ien t phat i t cc Emitter =

VCEQ+

-

RE

ICQ

+VCCRC


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CQ

CEQCCEC I

VVRR

=+ (2-7)

RE thng c tnh theo cong thc thc nghiem:

CQCC

EQREE I

V)3,01,0(IV

R

= (2-8)

Thay vao (2-7) de dang tnh c RC.

Neu RE = 0 t (2-7) ta co:CQ

CEQCCC I

VVR

= (2-7)

2222---- Mach phan cc Base:Mach phan cc Base:Mach phan cc Base:Mach phan cc Base:a- Mach nh dong Base:

Ta co: RbIBQ + VBE + IEQRE = VCC (2-8)

VBE la ien ap m cua Transistor, con ky hieu la Vnh H2-2 chng 1. VBESi 0,7v va VBEGe 0,2v. Ngaynay chu yeu dung Transistor Silic nen t (2-8) ta co :

CCEEQEQ

b VRI7,01I

R =+++

Suy ra:

1RR

V

1RR

7,0VIb

E

CC

bE

CCEQ

++

++

= v VCC >>0,7v (2-9)

Phng phap nay t c dung do dong IBQ phu thuoc nhieu vao nhiet o.Phng phap nay ch c dung oi vi mach mac Collector chung e nang cao trkhang vao.

b- Mach nh ap Base: (H2-3)

IBQ

RbICQ

+VCC

RC

RE

iiRb

IBQ

IEQ VCC

RL

REVBB

R2

IBQI1

+VCCRC

RE

I2

R1


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Ta co: CC21

1BB VRR

RV

+= (2-10)

21

21b RR

RRR+

= (2-11)

BBCC

CCb

CC

BBb1 VV

VR

VV1

1RR

=

= (2-12)

BB

CCb2 V

VRR = (2-13)

Phng trnh tai DC: VCC = VCEQ + ICQ(RC + RE) (2-14)

Ap dung nh luat KII ta co: Vkn = 0, suy ra:

BBEEQBEBQb VRIVIR =++ (2-15)

++

=

1RR

7,0VIIb

E

BBEQCQ coi VBE = 0,7v (2-16)

Thay vao (2-14) ta tnh c VCEQ.

Thong thng khi thiet ke ta thng chon RE >> (1-)Rb e on nh dong IEQ.V vay neu cha biet Rb ta thng chon:

EEb R101R)1(

101R += (2-17)

Phng phap phan cc Base nay hay c dung nhat.

c- Mach nh dong Emitter:Ap dung nh luat KII Vkn = 0 ta co:

EEEEQBEBQb VRIVIR =++ (2-18)

Suy ra:

1RR

7,0VIb

E

EEEQ

++

= vi VBE = 0,7v (2-19)

Rb

IBQ

+VCC

RC

RE

-VEE

IEQ


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Phng trnh tai DC trong trng hp nay se la:

VCC + VEE = VCEQ + ICQ(RC + RE) (2-20)

Phng phap phan cc Base nay ch c dung khi mach yeu cau chat lngcao nh mach khuech ai vi sai, mach khuech ai thuat toan (KTT) v no phai ton

them mot nguon cung cap.

III.III.III.III. Giai tch mach TransistGiai tch mach TransistGiai tch mach TransistGiai tch mach Transistor bang o th:or bang o th:or bang o th:or bang o th:1111---- Bo khuech ai mac Emitter chung:Bo khuech ai mac Emitter chung:Bo khuech ai mac Emitter chung:Bo khuech ai mac Emitter chung:

Ta co the chia thanh 4 loai mach c ban nh sau:

a- Khong co CE, khong co CC: (H2-3)Bo khuech ai co the c thiet ke che o toi u (song ra tot nhat) hoac

che o bat ky.Che o toi u:Che o toi u:Che o toi u:Che o toi u: Thiet ke sao cho song ra ln nhat va khong b meo (I cmmax

hoac VLmax), thng cha biet cac ien tr phan cc R1, R2.

T o th (H3-2), ta thay song ra se ln nhat khi:

ACDC

CCCQTmaxcm RR

VII+

== (2-21)

ACTCQTCEmaxcm RIVV == (2-22)

Vi s o (H3-1) ta co: RAC = RDC = RC + RE nen t (2-21) va (2-22) ta suy ra:

)RR(2V

IEC

CCTCQ

+= (2-23)

2VV CCTCEQ = (2-24)

Che obat ky:Che obat ky:Che obat ky:Che obat ky: Thng cho trc R1, R2 hoac VCEQ hoac ICQ. Ap dung caccong thc (2-10, 11, 14, 16) se xac nh c (ICQ, VCEQ)

Neu ICQ < ICQTth Icm = ICQ. Neu ICQ > ICQTth Icm = iCQmax ICQ.


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b- Co CE, khong co CC (Tu Bypass Emitter) (H2-5)Che o tChe o tChe o tChe o toi u:oi u:oi u:oi u:

RDC = RC + RE va RAC = RC thay vao (2-21) ta c:

EC

CC

ACDC

CCTCQmaxcm RR2

VRR

VII+

=+

== (2-25)

C

E

CC

EC

CCCACTCQTCEQmaxcm

RR2

VRR2

RVRIVV+

=+

=== (2-26)

Che o bat ky:Che o bat ky:Che o bat ky:Che o bat ky: c tnh toan theo cac cong thc (2-10, 11, 14, 16) va actuyen tai AC c ve nh sau:

( )CEQCEAC

CQC VvR1Ii = (2-27)

Cho VCEQ = 0 ACCEQ

CQmaxC R

V

Ii += (2-28)

Cho iC = 0 vCEmax = VCEQ + ICQRAC (2-29)

Phng trnh (2-28) va (2-29) e xac nh iCmax va vCEmax trong cac trng hpiem tnh Q bat ky

Q2

VCE(V)

iC(mA)iCmax1

ACDC

CC

RRV+

ACT R

1ACLL

0

ICQT

DCR1DCLL

Q1

QT

iCmax2

VCET

VCEmax1

VCC

VCEmax2

2ICQT

2VCET


10/74



c- Khong co CE, co CC:Che o toi u:Che o toi u:Che o toi u:Che o toi u:

ECDC RRR +=

LC

LCEAC RR

RRRR+

+=

Thay vao (2-21) ta c:

LC

LCEC

CCTCQmaxCm

RR

RR

R2R

VII

+++

== (2-30)

ACTCQTCEmaxCE RIVV == (2-31)

LC

LCEC

CC

LC

CmaxCm

LC

CmaxLm

RRRRR2R

VRR

RIRR

RI

++++

=+

= (2-32)

LC

LCEC

CC

LC

LCLmaxLmmaxLm

RR

RRR2R

VRR

RRRIV

+

+++== (2-33)

Che o bat ky nh tren nhng chu y:

ICQ < ICQT : ICm = ICQ. ICQ > ICQT : ICm = iCmax ICQ. Cm

LC

CLm IRR

RI+

= (2-34)

VLm = ILmRL. (2-35)

d- Co CE, co Cc: (tu ghep vo han) (H2-6)ECDC RRR +=

LC

LCAC RR

RRR+

=

CC

RLii

iL

VBB

Rb

+VCC

RC

RE


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thay vao (2-21) ta c:

LC

LCEC

CCTCQmaxCm

RRRRRR

VII

+++

== (2-36)

LC

LC

LC

LCEC

CCTCEQmaxCm RR

RR

RRRRRR

VVV+

+++

== (2-37)

LC

LCEC

CC

LC

CmaxCm

LC

CmaxLm

RRRRRR

VRR

RIRR

RI

+++

+

=+

= (2-38)

LC

LCEC

CC

LC

LCLmaxLmmaxLm

RR

RRRR

VRR

RRIV

+

+++== (2-39)

Che o bat ky xac nh nh tren.

So sanh 4 trng hp tren ta nhan thay tac dung cua cac tu C E va CC la lamtang bien o dong ien ra va ien ap ra (so sanh cac cong thc (2-21), (2-25), (2-30)va (2-36)).

e- Tnh toan cong suat: Cong suat nguon cung cap:

PCC = VCCICQ (2-40)

Cong suat trung bnh tieu tan tren tai:

L

2Lm

L2LmL R

V21PI

21P == (2-41)

che o toi u: ICmmax = ICQT nen oi vi trng hp a va b taco:

C

2

TCQC

2

maxCmmaxL RI2

1

RI2

1

P == v RL = RC (2-42)

Con oi vi trng hp c va d th do RC RL nen ta co:

L2

maxLmmaxL RI21P = (2-43)

Cong suat tieu tan tren Collector:


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2IRI)RR(PP

2Cm

AC2CQECCCC += (2-44)

Hieu suat:CC

L

PP

= (2-45)

CC

maxLmax P

P= (2-46)

che o lp A hieu suat cc ai %25max =

He so pham chat: 2PP

maxL

maxC = (2-47)

2222---- BoBoBoBo khuech ai mac Collector chung:khuech ai mac Collector chung:khuech ai mac Collector chung:khuech ai mac Collector chung:

Che o toi u:Che o toi u:Che o toi u:Che o toi u: Trong ca 3 hnh neu khong co CC ta co:

ECDC RRR +=

LE

LECAC RR

RRRR+

+=

thay vao cong thc (2-21), (2-22) ta se co:

LE

LE

EC

CCTCQmaxEm

RR

RRRR2

VII

+++

= (2-48)

++

+++

==LE

LEC

LE

LEEC

CCACTCQTCEQ RR

RRR

RRRRRR2

VRIV (2-49)

VL

R2 CC

CE

RL

+VCC

RC

RER1

-VEE

VLRb

CC

CE

RL

+VCC

RC

REVL

Rb CC

CE

RL

+VCC

RC

RE


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Trong ca 3 hnh neu co CC ta co:ECDC RRR +=

LE

LEAC

RR

RRR+

=

thay vao cong thc (2-21), (2-22) ta se co:

LE

LEEC

CCTCQmaxEm

RRRRRR

VII

+++

= (2-50)

+

+++

==LE

LE

LE

LEEC

CCACTCQTCEQ RR

RR

RRRRRR

VRIV (2-51)

Ta luon co:

maxCmLE

EmaxLm IRR

RI+

= (2-52)

LE

LEmaxCmLmaxLmmaxLm RR

RRIRIV+

== (2-53)

Che o bat ky:Che o bat ky:Che o bat ky:Che o bat ky: Tnh theo cac cong thc c xay dng trong phan machphan cc cho Transistor.

ac tuyen tai mot chieu DCLL va ac tuyen tai xoay chieu ACLLc ve tng t nh trong mach khuech ai Emitter Common.

3333---- Bo khuech ai mac Base chung:Bo khuech ai mac Base chung:Bo khuech ai mac Base chung:Bo khuech ai mac Base chung:

Che o toi u:CDC RR =

LC

LCAC RR

RRR+

=

Thay vao (2-21), (2-22) ta c:

Vi+

-

Ri

VL

R1

CC

RL

+VCC

RC

R2

Cb


14/74



LC

LCC

CCTCQmaxCm

RRRRR

VII

++

== (2-54)

+

++

===LC

LC

LC

LCC

CCACTCQTCEQmaxCm

RR

RR

RR RRR

VRIVV (2-55)

maxCmLC

CmaxLm IRR

RI

+= (2-56)

LmaxLmmaxLm RIV = (2-57)

Che o bat ky: c tnh trc tiep t mach


15/74



Chng III:Chng III:Chng III:Chng III: On nh phan cc cho Transistor BJTOn nh phan cc cho Transistor BJTOn nh phan cc cho Transistor BJTOn nh phan cc cho Transistor BJT

Chng nay nham nghien cu s dch chuyen iem Q theo ICBO, VBE khi thayoi nhiet o va theo khi b lao hoa. Coi gan ung cac ai lng VCC, VBB khongthay oi.

Neu s thay oi ICBO, VBE va la nho th bien xet ICQ se la ham tuyen tnhtheo cac bien khac.

ICQ = ICQ(ICBO, VBE, ) (3-1)

Tha so on nh dong ien:

E

b

CBO

CQI R

R1II

S +

= (3-2)

Tha so on nh ien ap:EBE

CQV R

1VI

S

= (3-3)

Tha so on nh he so khuech ai:

( )

++

+

=

=

E2b

Eb

1

1CQCQ

R1RRRII

S (3-4)

Vi 1CQ2CQCQ12 III& == (3-5)

Khi o s thay oi cua ICQ se c tnh bang cong thc:

++= SVSISI EVCBOICQ

( )

++

+

+

+=

E2b

Eb

1

1CQBE

ECBO

E

b

R1RRRIV

R1I

RR1 (3-6)

Chu y VBE thng co gia tr am.


16/74



Chng IV. Thiet ke va phan tch tn hieu nho tan so thap.Chng IV. Thiet ke va phan tch tn hieu nho tan so thap.Chng IV. Thiet ke va phan tch tn hieu nho tan so thap.Chng IV. Thiet ke va phan tch tn hieu nho tan so thap.

I. Cac thong so Hybrid:Cac thong so Hybrid:Cac thong so Hybrid:Cac thong so Hybrid: Tr khang vao khi ngan mach tai:

0vi

vh21

1i

=

=

o li ien ap ngc khi h mach nguon:0iv

vh12

1r

==

o li dong ien thuan khi ngan mach tai:0vi

ih21

2f

==

Tong dan ngo ra khi h mach nguon:0iv

ih12

2o

==

ng vi cac cach mac khac nhau EC, BC hay CC ma ch th hai c chnh. V du: hie, hib, hic, ...

II. Cach mac Emitter chung:Cach mac Emitter chung:Cach mac Emitter chung:Cach mac Emitter chung:Tr khang vao khi ngan mach tai:

EQ

Tfeie I

Vmhh = (4-1)

Trong o: - VT =25mV 3000K (270C) (4-2)

- m = 1 2 phu thuoc vao chat ban dan. V du BJT Silic co m = 1,4khi o:

EQ

Tfeie I

Vh4,1h = (4-3)

* oi vi H4-1a, co mach tng ng rut gon H4-4, ta co:

He so khuech ai dong ien:( )

b

ie

fe

ieb

bfe

i

b

b

L

i

Li

Rh1

hhR

Rhii

ii

iiA

+

=+

=== (4-4)

Neu hie


17/74



* oi vi H4-5, co mach tng ng H4-6 ta co:


( )

bi

ieLC

Cfe

iebi

bife

LC

C

i

b

b

L

i

Li

R//rh

1

1RR

RhhR//r

R//rhRR

Rii

ii

iiA

++

=+

+=== (4-8)

Neu RC >> RL & Rb//ri >> hie ta co: Aimax = -hfe (4-9)

Tr khang vao: iebii h//R//rZ = (4-10) Tr khang ra: CC

oeo RR//h

1Z

= (4-11)

* oi vi H4-17 ta co:


( ) ( ) Efeiebi

bife

RC

C

i

b

b

L

i

Li Rh1hR//r

R//rhR

Rii

ii

iiA

L+++

===

+

(4-12)

Tr khang vao: ( )[ ]Efeiebii Rh1h//R//rZ ++= (4-13) Tr khang ra: Co RZ = (4-14)S o EC hay c dung nhat do co A i, Av ln

III.Cach mac Base chung:Cach mac Base chung:Cach mac Base chung:Cach mac Base chung:T mach H4-9, H4-10, va H4-11 cac tham so cua cach mac Base chung (BC)

co the a ve cac tham so cua cach mac Emitter chung (EC) nh sau:

Tr khang vao khi ngan mach tai:

fe

ieib h1

hh+

= (4-15)

o li dong ien thuan khi ngan mach tai:fe

fefb h1

hh+

= (4-16)

Tong dan ra khi h mach nguon:


18/74



fe

oeob h1

hh+

= (4-17)

Nh vay e tnh cac tham so cua s o B.C ch can biet cac tham so cuas o E.C. V hfb 1 nen s o B.C t c dung pham vi tan so thap,

nhng c dung rat nhieu pham vi tan so cao e giam anh hng cuacac ien dung ky sinh.

IV.Cach mac Collector chung:Cach mac Collector chung:Cach mac Collector chung:Cach mac Collector chung:T H4-14 va H4-15 ta co:

Vb = Vbe + iERE (4-18)

Vbe = ibhie (4-19)

VE = iERE = ib(1 + hfe)RE (4-20)

Suy ra:

Vb = ibhie + ib(1 + hfe)RE = ib[hie + (1 + hfe)RE] (4-21)

i

b

b

b

b

E

i

Ev V

VVi

iV

VVA == (4-22)

Efeb

E R)h1(iV

+= (4-23)

Efeieb

bR)h1(h 1Vi ++

= (4-24)

Goi: [ ]Efeieb'b R)h1(h//RR ++= (4-25)

'bi

'b

'bi

i'b

ii

b

RrR

RrVR

V1

VV

+=

+= (4-26)

Thay (4-23, 24, 26) vao (2-22) ta c:

'bi

'b

Efeie

Efe

i

Ev Rr

R

R)h1(h

R)h1(

V

V

A ++++

== (4-27)

vi Rb theo (4-25).

Tr khang vao cua s o H3-7 : Zi = hie + (1 + hfe)RE (4-28)

T s o H3-8 ta xac nh c tr khang ra:


19/74



fe

'bi

ibiE

Eo h1

R//rh0Vi

VZ+

+==

= (4-29)

Nh vay e tnh cac tham so cua mach C.C ta cung ch can biet tham so cuamach E.C. S o C.C co Av


20/74



Chng V:Chng V:Chng V:Chng V: Transistor hieu ng trng.Transistor hieu ng trng.Transistor hieu ng trng.Transistor hieu ng trng.

I. Ly thuyet hoat ong cua JFET:Ly thuyet hoat ong cua JFET:Ly thuyet hoat ong cua JFET:Ly thuyet hoat ong cua JFET:VDS: ien ap gia cc mang va cc nguon.

VDS (tai iem nghen) = Vp = Vpo + VGS (5-1)

Vpo: ien the nghen c tra tren o th. VGS: ien ap gia cc cong va cc nguon.

ien the anh thung Breakdown la mot ham cua ien ap GS:GSDSSDSX VBVBV += (5-2)

Trong o BVDSS la ien the Breakdown ng vi VGS = 0.

Tai vung bao hoa, dong dien mang c tnh gan ung:

++=

23

po

GS

po

GSpoD V

V2VV31II vi VGS < 0 (5-3)

Ipo: dong ien nghen c tra tren o th.

T (5-3) ta thay:

Khi VGS = 0 ta co: ID = Ipo (5-4) Khi VGS = -Vpo ta co: ID = 0 (5-5)

II. Ly thuyet hoat ong cua IGFET:Ly thuyet hoat ong cua IGFET:Ly thuyet hoat ong cua IGFET:Ly thuyet hoat ong cua IGFET:ien ap gia cc mang va cc nguon:

VDS(tai iem nghen) = Vp = Vpo + VGS (5-6)

BVDSX = BVDSS + VGS (5-7)

2

po

DSpoD VV1II

+= (5-8)

Khi VGS = 0 ta co: ID = Ipo (5-9) Khi VGS = -Vpo ta co ID = 0 (5-10)


21/74



III. Giai tch o th va phan cc:Giai tch o th va phan cc:Giai tch o th va phan cc:Giai tch o th va phan cc:1111---- Phan cc JFET:Phan cc JFET:Phan cc JFET:Phan cc JFET:

Phng trnh tai DC (DCLL): VDD = VDS + ID(Rd + RS) (5-11)

Do IG 0 nen mach t phan cc: VGS = -IDRS (5-12)

2222---- Phan cc JGFET:Phan cc JGFET:Phan cc JGFET:Phan cc JGFET: H5-14Phng trnh tai DC (DCLL): VDD = VDS + ID(Rd + RS) (5-13)

nh ngha nguon ap cung cap cho cc cong la:

DD21

1GG VRR

RV+

= (5-14)

Theo nh luat KII Vkn = 0 ta co:

IG(Rb + R3) + VGS + IDRS = VGG (5-15)

Do IG 0 nen ta co:

SDDD21

1SDGGGS RIVRR

RRIVV +

== (5-16)

e ID t thay oi theo nhiet o ta phai chon:

2 VVVVGGpo

GGGSQ += (5-17)

Suy ra: poGSQGG VV2V += (5-18)

DQ

poGSQS I

VVR

+= (5-19)

Cac gia tr cho bi phng trnh (5-17, 18, 19) se xac nh iem tnh Q va cctieu hoa s phu thuoc vao nhiet o cua tnh iem.

IV. Giai tch tn hieu ln, s sai dang:Giai tch tn hieu ln, s sai dang:Giai tch tn hieu ln, s sai dang:Giai tch tn hieu ln, s sai dang:oi vi IGJET:

2

po

GSpoD V

V1II

+= (5-20)

a tn hieu AC vao cc cong:


22/74



tcosVVvVV 0imGSQiGSQGS +=+= (5-21)

Thay (5-21) vao (5-20) ta c:

44 344 2144444 344444 214444 34444 21

haibachaiphanThanh

0po

impo

nhatbachaiphanThanh

0po

im

po

GSQ

po

DCbnhtrungphanThanh

2

po

im

2

po

GSQ

Dt2cos

V

V

2

Itcos

V

V

V

V1I2

V

V

2

1

V

V1I +

++

+

+= (5-22)

Khi Vim = 0 ta co:

2

po

GSQpoDQ V

V1II

+= (5-23)

Khi Vim


23/74



1)1)1)1) Bo khuech ai cc nguon chung:Bo khuech ai cc nguon chung:Bo khuech ai cc nguon chung:Bo khuech ai cc nguon chung: H5-17Tr khang vao nhn t nguon: Zi = R3 + (R1//R2) (5-30)

Tr khang ra nhn t tai: Zo = Rd//rds (5-31)

( )

( )

Lm

213

ioLm

i

gs

gs

L

i

LV Rg

R//RRr1

1Z//RgVV

VV

VVA

++

=== (5-32)

vi ri


24/74



Tr khang vao:1Rr

iV

R ddsi

sgsg

+

+== (5-42)

He so khuech ai:

1Rrr

RVV

A ddsi

didV

+

++

== (5-43)


25/74



Chng VI: Mach Transisrtor ghep lien tangChng VI: Mach Transisrtor ghep lien tangChng VI: Mach Transisrtor ghep lien tangChng VI: Mach Transisrtor ghep lien tang

1)1)1)1) Transistor ghep Cascading:Transistor ghep Cascading:Transistor ghep Cascading:Transistor ghep Cascading:a) Tang E.C-E.C: H6-1

o li dong ien:

+

+

+

===

1e'

1b

'1b

2ie'

2b

'2b1fe

L2C

2C2fe

i

1b

1b

2b

2b

L

i

Li hR

RhRRh

RRRh

ii

ii

ii

iiA (6-1)

vi Rb1 = ri//Rb1 va Rb2 = Rc1//Rb2

Tr khang vao: Zi = ri//Rb1//hie1 (6-2)

Tr khang ra: Zo = Rc2 (6-3)

Neu trong H1-1 khong co CE1 va CE2 th:

++

++

+

=

1E1fe1e'

1b

'1b

2E2fe2e'

2b

'2b1fe

L2C

2C2fei RhhR

RRhhR

RhRR

RhA (6-4)

Zi = ri//Rb1//[hie1 + (1 + hfe1)RE1] (6-5)

Zo = Rc2

b) Tang E.C-C.C hnh bai tap 6-3

o li dong ien:

( ) ( )

+

++

+=

==

1e1b

1b

L2E2fe2ie2b1C

2b1C

L2E

2E2fe

i

1b

1b

L

L

L

i

Li

hRR

R//RhhR//RR//R

RRRh

i

i

i

'i

'i

i

i

iA

(6-6)

Zi = Rb1//hie1 (6-7)

ZoZi

hie1

iC1

hfe1ib1

RE2hfe2

RLhfe2

hie2

ib1

ii

iL

Rb2 VLRb1 RC1

ib2


26/74



+=

2fe

2b1C2ib2Eo h

R//Rh//RZ (6-8)

* Neu khong co CE1 khi o:

( ) ( )

++

++

+=

==

1E1fe1e1b

1b

L2E2fe2ie2b1C

2b1C

L2E

2E2fe

i1b

1bL

LL

iLi

RhhRR

R//RhhR//RR//R

RRRh

iii'i'iiiiA(6-9)

Zi = Rb1//[hie1 + hfe1RE1] (6-10)

+=

2fe

2b1C2ib2Eo h

R//Rh//RZ (6-11)

2)2)2)2) Mach khuech ai vi sai:Mach khuech ai vi sai:Mach khuech ai vi sai:Mach khuech ai vi sai:T H6-4 ta co:

fe

bE

EE2EQ1EQ

hRR2

7,0VII+

== (6-12)

VCEQ1 = VCEQ2 VCC + VEE ICQ(RC + 2RE) (6-13)

Cac dong ien co the c phan thanh hai thanh phan:

+ Mot chung2

iii 210+

= (6-14)

+ Mot vi sai 12 iii = (6-15)o li dong ien mot chung:

++

+=

fe

bibE

1b

LC

CC

hRhR2

RRR

RA (6-16)

o li dong ien mot vi sai:

+

+=

fe

bib

b

LC

Cd

hRh2

RRR

RA (6-17)


27/74



v vay iL = ACi0 + Adi (6-18)

Ty so nen tn hieu ong pha:

fe

b

ib

E

h

R

h

RCMRR

+

(6-19)

e tang CMRR ta dung nguon dong cc phat.

3)3)3)3) Cach ghep Darlington:Cach ghep Darlington:Cach ghep Darlington:Cach ghep Darlington:T H6-10, ap dung nh luat K II. Vkin = 0 ta co:

IB1Rb + VBE1 + VBE2 + ICQ2RE = VBB1 (6-20)

1BBE2CQb2fe1fe

2CQ VRI7,07,0Rhh

I=+++ (6-21)

Tong quat dong ien ra ICQ2 c tnh:

2fe1fe

bE

1BB2CQ

hhRR

4,1VI+

= (6-22)

Neu2fe1fe

bE hh

RR >> th (6-22) tr thanh:

E

1BB2CQ R

4,1VI (6-23)

Trong o

CC21

11BB VRR

RV+

= (6-24)

21

21b RR

RRR+

= (6-25)

Phng trnh tai mot chieu au ra:

( )EC2CQ2CEE2C2fe

2CQ2CC2CECC RRIVRIh

IIRVV +++

++= (6-26)

VCE1 = VCE2 VBE2 = VCE2 0,7 (6-27)


28/74



2fe

2CQ1CQ h

II = (6-28)

T H3-2 ta co:

b

1ieLC

C2fe1fe

2ie1ib1fe

b

b

LC

C2fe

i2b

2bL

iLi

Rh21

1RR

Rhh

hhhR

RRR

Rh

ii

ii

ii

A

++=

++

+=

==

(6-29)

Chu y:

1ib1fe

1ie

2fe1EQ

3

2fe1EQ

3

2fe2ie hhh

hI10.25h4,1

I10.25h4,1h ====

(6-30)

Tr khang vao: Zi = hie1 + hfe1hie2 2hie1 (6-31)

Tr khang ra: Zo = RC

* Mach Darlington cung co the lam viec che o toi u khi o:

RDC = RC + RE (6-32)

RAC = RC//RL (6-33)

Dong ien ra che o toi u:

LC

LCEC

CC

ACDC

CCCQmaxcm

RRRRRR

VRR

VIITU22

+++

=+

== (6-34)

ACCEQCEQmaxcm R.IVV TU2TU22 == (6-35)

maxcmLC

CmaxLm 2

IRR

RI+

= (6-36)

LmaxLmmaxLm RIV = (6-37)

* Mach Darlington cung co the mac theo kieu C.C.


29/74



He so khuech ai:

( ) ( ) 2fe1feLE1iebi

bi

LE

E2fe1fe

i

1b

1b

L

L

L

i

Li

hhR//Rh2R//RR//R

RRRhh

ii

i'i

'ii

iiA

+++=

==

(6-38)

( ) ( ) 2fe1feLEiebibi

LE

EL2fe1fe

i

LL

i

LT hhR//Rh2R//R

R//RRR

RRhhiiR

iVA

1+++

=== (6-39)

ii

L

ii

LL

i

LV AR

RiRiR

VVA === (6-40)

Tr khang vao:

( )[ ]2fe1feLE1iebii hhR//Rh2//R//RZ += (6-41)

Tr khang ra:

+=

2fe1fe

bi2ibEo hh

R//Rh2//RZ (6-42)

VL

+VCC

T1

Ri

C1

RL

ZoZi

IC1

IE1=IB2ii

R2

RE

RC

R1 C2

T2

hie1

RLhfe1hfe2

ib1

iL

VLRi

Zo

hfe1hie2

ii REhfe1hfe2Rb


30/74



4)4)4)4) Mach khuech ai Cascode:Mach khuech ai Cascode:Mach khuech ai Cascode:Mach khuech ai Cascode: H6H6H6H6----13131313T1 mac theo kieu EC, T2 mac theo kieu BC. Do o ay la bo khuech ai lien

tang E.C-B.C.

Che o DC:Che o DC:Che o DC:Che o DC: T hnh ve neu I2 >> IB1 va I2 >> IB2 ta co:CC

321

212BB VRRR

RRV++

+= (6-43)

2BB21

11BB VRR

RV+

= (6-44)

1fe

21E

1BB1E1C

hR//RR

7,0VII+

= (6-45)

VCE1 = VBB2 VBE2 RCIC1 REIC1 = VBB2 0,7 IC1(RC + RE) (6-46)

VCE1 = VCC IC1RL VBB2 + 0,7 (6-47)

Che o AC:Che o AC:Che o AC:Che o AC:

( )( )( )

+=

==

1ie21

211fe2fb

i

1b

1b

2e

2e

L

i

Li

hR//RR//Rhh

ii

ii

ii

iiA

(6-48)

iLi

LL

i

LT ARi

Rii

VA === (6-49)

Tr khang vao: Zi = R1//R2//hie1 (6-50)

Tr khang ra: Zo = (6-51)


31/74



Chng VIIChng VIIChng VIIChng VII: Mach khuech ai hoi tiep.: Mach khuech ai hoi tiep.: Mach khuech ai hoi tiep.: Mach khuech ai hoi tiep.

1. Khai niem c ban ve mach hoi tiep:Khai niem c ban ve mach hoi tiep:Khai niem c ban ve mach hoi tiep:Khai niem c ban ve mach hoi tiep:

Hoi tiep la hien tng a tn hieu t ngo ra cua bo khuech ai ngc tr vengo vao. Hoi tiep gom 3 loai:

+ Hoi tiep noi bo sinh ra do tnh chat vat ly cua Transistor. V du: hoi tiep quaCbc khi mac E.C.+ Hoi tiep k sinh: sinh ra do cac phan t ghep mach: ien cam, bien ap, tcam ...

Hai loai hoi tiep tren la khong mong muon v no lam xau i cac ch tieuky thuat cua bo khuech ai.

+ Hoi tiep ben ngoai: do ta mac vao e cai thien cac ch tieu ky thuat cua bokhuech ai. Khi cac phan t hoi tiep la thuan tr ta co hoi tiep khong phu thuoc

tan so va khong ao pha, khi hoi tiep bao gom cac khau RC, L, bien ap,Transistor, ... th hoi tiep phu thuoc tan so va ao pha.

Neu hoi tiep a ve ong pha vi tn hieu vao, ta goi la hoi tiep dng va chdung trong cac bo dao ong. Neu hoi tiep a ve ngc pha vi tn hieu vao ta goi lahoi tiep am va chung c s dung rat nhieu trong cac bo khuech ai e cai thiencac ch tieu ky thuat. Trong chng nay ta ch xet hoi tiep am.

Phan loai hoi tiep:Phan loai hoi tiep:Phan loai hoi tiep:Phan loai hoi tiep:

+ Tuy theo ien ap hoi tiep (V f) ty le vi ien ap ra (Vo), dong ien ra(Io) hay ty le vi ca hai ma hoi tiep thuoc loai hoi tiep ien ap, hoitiep dong ien hay hoi tiep hon hp. e phan biet ba loai hoi tiep nayta dung phep th:

Ngan mach tai ma mat hoi tiep (Vf= 0) th o la hoi tiep ien ap. H mach tai ma mat hoi tiep (Vf= 0) th o la hoi tiep dong ien. Ca khi ngan mach va h mach tai ma van con hoi tiep (V f 0) tho la hoi tiep hon hp.

Bo khuech ai

AV

Mach hoi tiep

TaiNguon


32/74



+ Tuy theo ien ap hoi tiep a ve ngo vao mac noi tiep hoac mac songsong vi nguon tn hieu vao ma ta co hoi tiep noi tiep (sai lech ap)hoac hoi tiep song song (sai lech dong).

+ Nh vay ta co bon loai hoi tiep hay dung Hoi tiep ap, sai lech ap. Hoi tiep ap, sai lech dong. Hoi tiep dong, sai lech ap. Hoi tiep dong, sai lech dong.

2. Anh hng cua hoi tiep am en cac tham so cua bo khuech ai:Anh hng cua hoi tiep am en cac tham so cua bo khuech ai:Anh hng cua hoi tiep am en cac tham so cua bo khuech ai:Anh hng cua hoi tiep am en cac tham so cua bo khuech ai:a) oi vi o li ap khi co hoi tiep am AVf:

V

VVf A1

A

A += (7-1)

+ AV: o li ap khi cha co hoi tiep.+ : he so hoi tiep.

b) oi vi s mat on nh cua o li ap:+ He so bat on nh cua AV:

V

V

AdAq = (7-2)

+ He so bat on nh cua AV khi co hoi tiep am:V

f A1 qq += (7-3)

c) Anh hng en meo tan so va meo pha:+ He so meo tan so: 1

AAM

f

0 = (7-4)

A0: o li day gia. Af: o li tai f.

+ He so meo tan so khi co hoi tiep am f la:V

f A11M1M

+

= (7-5)

V

f A1MM+

= (7-5)

+ o dch pha khi cha co hoi tiep la , o dch pha khi co hoi tiep am f:


33/74



Vf A1 +

= (7-6)

d) Anh hng en he so meo phi tuyen va tap am He so meo phi tuyen:

m1

2nm

2m3

2m2

m1

2nm

2m3

2m2

IIII

VVVV +++

=+++

=LL

(7-7)

Trong o Vnm (Inm) la thanh phan hai bac th n (n = 1, 2, 3, ...)

He so meo phi tuyen khi co hoi tiep am:V

f A1 +

= (7-8)

Tap am khi co hoi tiep am:V

TATAf

A1

VV

+

= (7-9)

e) Anh hng en tr khang vao: Hoi tiep noi tiep: ( )Viif A1ZZ += (7-10) Hoi tiep song song:

V

iif A1

ZZ+

= (7-11)

f) Anh hng en tr khang ra:

Hoi tiep dong ien: ( )Voof A1ZZ += (7-12)

Hoi tiep ien ap:V

oof A1

ZZ+

= (7-13)

3. Hoi tiep ien ap, sai lech dong ien:Hoi tiep ien ap, sai lech dong ien:Hoi tiep ien ap, sai lech dong ien:Hoi tiep ien ap, sai lech dong ien:T H7-2 ta co o li dong thuan:

'i

iifi A0G

AA ==

= (7-14)

o li dong khi co hoi tiep am:

T1A

RGA1AA i

Lii

iif

=

+= (7-15)

o li vong c nh ngha:


34/74



Liii

'L

L RGA0iV

VT ==

= (7-16)

Neu T >> 1 ta co:L

f

i

Lif R

RiiA == (7-17)

i

f

i

Lvf r

RVVA == (7-18)

Tr khang vao khi co hoi tiep:T1

ZZ iif

= (7-19)

Tr khang ra khi co hoi tiep:T1

ZZ oof

= (7-20)

4. Hoi tiep ien ap, sai lech ien ap:Hoi tiep ien ap, sai lech ien ap:Hoi tiep ien ap, sai lech ien ap:Hoi tiep ien ap, sai lech ien ap: o li ien ap khi khong co hoi tiep:

'v

vvfV A0K

AA ==

= (7-21)

o li ien ap khi co hoi tiep:

T1A

AK1A

VVA V

VV

V

i

Lvf

=

+== (7-22)

o li vong T:VV

'VV

i'L

L AKAK0VV

VT ===

= (7-23)

Tr khang vao khi co hoi tiep:)T1(ZZ iif = (7-24)

Tr khang ra khi co hoi tiep:

T1ZZ oof

= (7-25)


35/74

Khoa ien ien t Ky thuat mach ien T I

- 35 - Mot so bai tap mau - 35 -

MOT SO BAI TAP MAU CHO QUYENMOT SO BAI TAP MAU CHO QUYENMOT SO BAI TAP MAU CHO QUYENMOT SO BAI TAP MAU CHO QUYENGiao trnh mach ien t IGiao trnh mach ien t IGiao trnh mach ien t IGiao trnh mach ien t I

Chng I: DIODE BAN DAN.Chng I: DIODE BAN DAN.Chng I: DIODE BAN DAN.Chng I: DIODE BAN DAN.

I.I.I.I. Diode ban dan thong thng:Diode ban dan thong thng:Diode ban dan thong thng:Diode ban dan thong thng:1) Ve dang song chnh luVe dang song chnh luVe dang song chnh luVe dang song chnh lu: (Bai 1-1 trang 29)

Cong thc tong quat tnh VL:

LLi

DSL RRR

VVV+=

VD = 0,7V (Si) va VD = 0,2V (Ge)

aaaa---- Ve VVe VVe VVe VLLLL(t) vi V(t) vi V(t) vi V(t) vi VSSSS(t) dang song vuong co bien o 10 va 1V(t) dang song vuong co bien o 10 va 1V(t) dang song vuong co bien o 10 va 1V(t) dang song vuong co bien o 10 va 1V

Ket qua vi gia thiet: Ri = 1, RL = 9, VD = 0,7V.V Diode chnh lu ch dan ien theo mot chieu nen:

Trong 0T21 > , Diode dan iD 0 iL 0 VL 0.

V37,8991

7,010V 1L =+

= va V27,09

917,01V 2L =

+

=

Trong 0T21

< , Diode tat iD = 0 iL = 0 VL = 0.

bbbb---- Ve VVe VVe VVe VLLLL(t) vi V(t) vi V(t) vi V(t) vi VSSSS(t) dang song sin co bien o 10 va 1V(t) dang song sin co bien o 10 va 1V(t) dang song sin co bien o 10 va 1V(t) dang song sin co bien o 10 va 1V....

iL

iD

RL

RiVLVs

-

-+VD

10

-10

0 1 - -

++

VS

2 3 4 t(ms)

1

-1

0 1 - -

++

VS

2 3 4 t(ms)

8,37

0 1

VL1

2 3 4 t(ms)0,27

0 1

VL2

2 3 4 t(ms)


36/74



Khi VS = 10sinot ngha la VSm = 10V >> VD =0,7V ta co:99

9110R

RRVV L

Li

Sm1L =

+

+

tsin9V 01L

(Ta giai thch theo 0T21

> va 0T21

< ) Khi VS = 1sin0tngha la VSm = 1V so sanh c vi 0,7V:

+ VS > 0,7V, Diode dan, iD 0, iL 0, VL 0.6,0tsin9,09

917,0tsin1

V 00

2L =+

=

Tai sin0t = 1, |VL2| = 0,27V.+ VS < 0,7V, Diode tat, iD = 0, iL = 0, VL = 0.

Vi dang song tam giac ta co ket qua tng t nh song sin.

2) Bai 1Bai 1Bai 1Bai 1----3:3:3:3: e co cac ket qua ro rang ta cho them cac gia tr ien tr: R1 =1K, Rb = 10K, RL = 9K.

aaaa---- Ve VVe VVe VVe VLLLL(t) vi dang song vuong co bien o 10V va 1 V.(t) vi dang song vuong co bien o 10V va 1 V.(t) vi dang song vuong co bien o 10V va 1 V.(t) vi dang song vuong co bien o 10V va 1 V. 0T21 > , Diode dan, RthD 0, dong iL chay qua Ri, D, RL nen ta co:

V37,810.9.10.9107,010R

RRVV

V 333LLi

DS1L =

+

=

+

=

V27,010.9.10.9107,01R

RRVV

V 333LLi

DS2L =

+

=

+

=

0T21

< , Diode tat, Rng = , dong iL chay qua Ri, Rb, RL nen ta co.

iLRL9K

Ri=1K

VLVs+

-

-+ VD

Rb=10K

10

0-10

9

- -+ +

12 3

4 t(ms)

VS

VL1

0 1 2 3 4 t(ms)

1

0-1

12 3

4 t(ms)

VS

VL2

0 1 2 3 4 t(ms)

0,7

0,27


37/74



V5,410.9.10.91010

10RRRR

VV 3343L

Lbi

S1L =

++=

++=

V45,010.9.10.91010

1RRRR

VV 3343L

Lbi

S1L =

++=

++=

bbbb---- Ve VVe VVe VVe VLLLL(t) vi dang song sin co bien o 10V va 1 V.(t) vi dang song sin co bien o 10V va 1 V.(t) vi dang song sin co bien o 10V va 1 V.(t) vi dang song sin co bien o 10V va 1 V. e n gian khi VSm = 10V (>>VD = 0,7V) ta bo qua VD. Khi o:

+ 0T21

> , Diode dan, RthD 0, dong iL chay qua Ri, D, RL nen ta

co:

)V(tsin910.9.10.910

tsin10R

RRV

V 03

330

LLi

S1L =

+

=

+=

+ 0T21


)V(tsin5,410.9.10.91010

tsin10R

RRRV

V 03

3430

LLbi

S1L =

++

=

++=

Khi VS = 1sin0t so sanh c vi VD ta se co:+ 0T

21

> , khi VSm 0,7, Diode dan, RthD 0, dong iL chay qua Ri,

D, RL nen ta co:

)V(63,0tsin9,010.9.10.910

7,0tsin1R

RR7,0tsin1

V 03

330

LLi

02L =

+

=

+

=

Tai 2t0 = , sin0t = 1, ta co VL2m = 0,9 - 0,63 = 0,27V

+ 0T21

> , khi VSm < 0,7, Diode tat, RngD = , dong iL chay qua Ri,

Rb, RL nen ta co:

tsin315,010.9.10.91010

tsin7,0R

RRRtsin7,0

V 03

3430

LLbi

02L =

++

=

++

=

10

-100 1 - -

++

VS

2 3 4 t(ms)

1

-10 1 - -

++

VS

2 3 4 t(ms)

8,370 1

VL1

2 3 4t(ms)

0,270 1

VL2

2 3 4 t(ms)-4,5

-0,45


38/74



+ 0T21


tsin45,010.9.10.91010

tsin1R

RRRtsin1

V 03

3430

LLbi

02L =

++

=

++

=

2)2)2)2) Dang mach Thevenin ap dunDang mach Thevenin ap dunDang mach Thevenin ap dunDang mach Thevenin ap dung nguyen ly chong chap:g nguyen ly chong chap:g nguyen ly chong chap:g nguyen ly chong chap:Bai 1-20 vi Vi(t) = 10sin0t

a- Ve mach Thevenin:Ap dung nguyen ly xep chong oi vi hai nguon ien ap VDC va Vi: Khi ch co VDC, con Vi = 0 th ien ap gia hai iem A-K:

V3

10.5,110

10.5,15

rR

rVV 33

3

ii

iDCAK =

+

=

+

=

Khi ch co Vi, con VDC = 0 th ien ap gia hai iem A-K la:)V(tsin4

10.5,11010tsin.10

rRRVV 033

3

0ii

iiAK =

+=

+=

Vay khi tac ong ong thi ca VDC va Vi th sc ien ong tngng Thevenin gia hai iem A-K la:

VL

+

-Vi

+

-

iD

RL1,4K

Ri=1K

VDC=5v

KA

ri=1,5K

RT id

VTK

A

RLRi//ri iL

VT

KA

10

0-10

9

- -

+ +t(ms)

VS

VL1

t(ms)

1

0-1

t(ms)

VS

VL2

t(ms)

0,7

0,315+ +

- --4,5 -4,5

0,585


39/74



)V(tsin43rR

RVrR

rVV 0ii

ii

ii

iDCT +=

++

+=

ien tr tng ng Thevenin chnh la ien tr tng ng cu aphan mach khi Diode h mach la:

=++

=++

= K210.4,110.5,110

10.5,1.10R

rR

r.RR 3

33

33

Lii

ii

T

b- Ve ng tai DC khi2

,3

,2

,3

,0t0

= .

Tai V3V0t T0 == Tai )V(46,6

2343V

3t T0 =+=

=

Tai )V(71.43V2

t T0 =+=

=

Tai )V(46,023

43V3t T0 ==

= Tai )V(11.43V

2t T0 ==

=

Theo nh luat Ohm cho toan mach ta co.

T

TD

TT

DT

RV

V.R1

RVV

i +=

=

Tai )mA(15,110.237,0.

10.21i0t 330 =+==

Tai )mA(88,210.246,67,0.

10.21i

3t 330 =+=

=

Tai )mA(15,310.277,0.

10.21i

2t 330 =+=

=

iD (mA)

3,152,88

1,15

3 6,46 7-1

VT

t


40/74



Tai )mA(58,010.246,07,0.

10.21i

3t 330 ==

=

Tai )mA(85,010.217,0.

10.21i

2t 330 ==

=

c- Ve

( )

( ) )V(tsin8,21,2tsin437,0V7,0

10.2V10.4,1Rr//R VRRV.Ri.R)t(V

00T

3T3Lii

TLTTLDLL

+=+==

=+

===

II.II.II.II. Diode Zenner:Diode Zenner:Diode Zenner:Diode Zenner:1) Dang dong IL = const (bai 1-40); 200mA IZ 2A, rZ = 0

a- Tm Ri e VL = 18V = const.Imin = IZmin + IL = 0,2 + 1 = 1,2 A.Imax = IZmax + IL = 1 + 2 = 3 A.Mat khac ta co: Vimin = 22V = IZmin.Ri + VZ.Suy ra:

==

=

= 3,32,1

42,11822

IVV

RminZ

Zminii

Vimax = 28V = IZmaxRi + VZSuy ra

==

=

= 3,33

103

1828I

VVR

maxZ

Zmaxii

Vay Ri = 3,3.b- Tm cong suat tieu thu ln nhat cua Diode Zenner:

PZmzx = IZmax.VZ = 2.18 = 36W.

2) Dang dong IL const: (bai 1-41), 10mA IL 85mA.IZmin = 15mA.

VL

0-0,7

2,1

4,9V

t

RLVZ=10v13v


41/74



a- Tnh gia tr ln nhat cua R imaxLminZ

Zii

minLmaxZ

Zi

IIVV

RIIVV

+

+

Khi VDC = 13V ta co=

+

30

085,0015,01013R maxi

Khi VDC = 16V ta co=

+

60

085,0015,01016R maxi

Vay ta lay Rimax = 30.b- Tm cong suat tieu thu ln nhat cua Diode Zenner.

PZmax = IZmax.VZ.Mat khac: Vimax = IZmaxRi + VZ

mA20030

1016R

VVIi

Zmaximax =

=

=

mA19019,001,02,0III minLmaxmaxz ==== W9,11019,0P maxz ==

3)

Dang IZ const; IL const (Bai 1-42)30 IL 50mA, IZmin = 10mA.rZ = 10 khi IZ = 30mA; Pzmax =800mW.

a- Tm Ri e Diode on nh lien tuc:mA80

108,0

VPI

Z

maxZmaxZ ===

Vay 10mA IZ 80mATa co: Imin = IZmin + ILmax = 60mAImax = IZmax + ILmin = 110mAMat khac: Vimin = Imin.Ri + VZ = 20V

RLVZ=10v20v


42/74



== 7,16606,0

1020R maxi

Vimax = Imax.Ri + VZ = 25V

== 36,13611,0

1025R mini

Suy ra: 136,4 Ri 166,7Vay ta chon Ri =150

b- Ve ac tuyen tai:Ta co: VZ + IZRi = VDC ILRi Vi VDC = 20V ta co:

==

===+

mA50IkhiV5,1215005,020mA30IkhiV5,1515003,020

150IVL

LZZ

Vi DC = 25V ta co:

==

===+

mA50IkhiV5,1715005,025

mA30IkhiV5,2015003,025150IV

L

LZZ

Tng ng ta tnh c cac dong IZ:

mA7,36150

105,15I 1Z =

= ; mA7,16150

105,12I 2Z =

=

mA70150

105,20I 3Z =

= ; mA50150

105,17I 4Z =

= ;

IZ(mA)

VZ

36,7

50

30

80

70

10

20,5 17,5

15,5

VZ =10V 0

rZ =10

16,7

12,5


43/74



Chng IIChng IIChng IIChng II: TRANSISTOR HAI LP TIEP GIAP: TRANSISTOR HAI LP TIEP GIAP: TRANSISTOR HAI LP TIEP GIAP: TRANSISTOR HAI LP TIEP GIAP

I.I.I.I. Bo khuech ai RBo khuech ai RBo khuech ai RBo khuech ai R----C khC khC khC khong co Cong co Cong co Cong co CCCCC va khong co Cva khong co Cva khong co Cva khong co CEEEE (E.C).(E.C).(E.C).(E.C).1) Bai 2-10: 20 60, suy ra ICQ khong thay oi qua 10%.

Phng trnh tai mot chieu:VCC = VCEQ + ICQ(RC + RE).

mA81010.5,1

525RRVV

I 33EC

CEQCCCQ =

+

=

+

=

Neu coi ay la dong ien ban au khi = 60 sao cho sau mot thi gian ch con = 20 th yeu cau ICQ 7,2mA.

Ta giai bai toan bai toan mot cach tong quat coi 1 = 20; 2 = 60.E22bbE11b R10

1RRR101R ==

==== K610.60.101RRK210.20.

101R 32bb31b

Vay 2K Rb 6K

Mat khac

+

=

bE

BBCQ R

R

7,0VI , neu coi VBB const th ta co:

9,0R

R

RR

II

1

bE

2

bE

2CQ

1CQ

+

+

= (1)

Co the tnh trc tiep t bat phng trnh (1):

+

+

+

12bE

1

bE

2

bE

9,01RR1,0R

R9,0R

R

==

+

=

+

K53,310.3,28

100

209,0

601

10.1,09,01

R1,0R 3

3

12

Eb

Chon Rb = 3,5K.

VCEQ = 5V+

-

+25V

R2

R1

RC=1,5K

RE=1K


44/74



Neu bo qua IBQ ta co VBB VBE + IEQRE = 0,7 + 8.10-3.103 = 8,7V. Suyra:

==

=

= K4,55368652,010.5,3

257,81

110.5,3

VV

1

1RR3

3

CC

BBb1

=== K06,10100577,8

2510.5,3VVRR 3

BB

CCb2

Ta co the tnh tong quat: Chon Rb = 4K thay vao (1):%9,88

12001067

2010.410

6010.410

II

33

33

2CQ

1CQ==

+

+

= , b loai do khong thoa man (1).

Chon Rb =3K thay vao (1): 91,011501050

20

10.310

6010.310

II

33

33

2CQ

1CQ==

+

+= thoa

man bat phng trnh (1), ta tnh tiep nh tren.

2) Bai 2-11: Vi hnh ve bai (2-10) tm gia tr cho R1, R2 sao cho dong iC xoaychieu co gia tr cc ai. iem Q toi u c xac nh nh sau:

ACCQTTCEQ

ACDC

CCTCQmaxCm

R.IVRR

VII

=

+==

T hnh ve: RDC = RC + RE = 1,5.103 + 103 = 2,5K.

RAC = RC + RE = 1,5.103 + 103 = 2,5K.Suy ra: mA510.5,210.5,2

25I 33TCQ =+=

VCEQT= 5.10-3.2,5.103 = 12,5V

Chon === K1010.100.101R

101R 3Eb (bo qua IBQ)

VBB VBE + ICQT.RE = 0,7 + 5.10-3.103 = 5,7V

VCE(V)

iC(mA)

VCEQT= 12,5 25

10RV

DC

CC =

( )5

RR2V

EC

CC =+

310.5,2

1ACLLDCLL

QT

0


45/74



==

=

= K13K95,12772,0

10

257,51

110.10

VV

1

1RR4

3

CC

BBb1

=== K44K85,437,5

2510VV

RR 4BB

CCb2

V RDC = RAC nen phng trng tai DC va AC trung nhau.

3) Bai 2-14: iem Qbat ky v biet VBB = 1,2V; = 20. Tm gia tr toi a cua daoong co the co c C va tnh .

Biet = 20, VBEQ = 0,7V.

Ta co: mA3,3501007,02,1

RR

VVI

bE

BEQBBCQ =

+

=

+

=

e tm gia tr toi a cua dao ong co the co c C ta phai vephng trnh tai DC, AC

VCEQ = VCC ICQ(RC + RE) = 6 3,3.10-3.1,1.103 = 2,37V Vay gia tr toi a cua dao ong la:

ICmmax = iCmax ICQ = 5,45 3,3 = 2,15mASuy ra VLmax = ICmmax.RC = 2,15.103.10-3 = 2,15V

PCC = ICQ.VCC = 3,3.10-3.6 = 19,8mW( ) ( ) mW31,210.10.15,2

21R.I

21P 3

23C

2maxCmL ===

+6V

Rb = 1K

RC = 1K

RE = 100VBB = 1,2V

45,5RV

DC

CC =

ICQ = 3,3

iC (mA)

VCE(V)2,37 3 60

2,725 QTQbk

=

11001ACLLDCLL


46/74



Hieu suat: %7,1110.8,1910.31,2

PP

3

3

CC

L ===

II.II.II.II. Bo KRC khong co CBo KRC khong co CBo KRC khong co CBo KRC khong co CCCCC, C, C, C, CEEEE (tu bypass Emitter) (EC)(tu bypass Emitter) (EC)(tu bypass Emitter) (EC)(tu bypass Emitter) (EC)1) Bai 2-15: iem Q bat ky.

a- Tm R1, R2 e ICQ = 01mA (Rb


47/74



T hnh ve ta nhan thay e ICm ln nhat va khong b meo th ICmmax =10mA.Ta co the tm iCmax va VCemax theo phng trnh

( )CEQCEC

CQC VVR1Ii =

Cho VCE = 0 mA60150

5,710R

VIi 2C

CEQCQmaxC =+=+=

Cho iC = 0 V95,7150.10VR.IV 1CEQCCQmaxCE =+=+=

2) Bai 2-16: iem Q toi u (hnh ve nh hnh 2-15).e co dao ong Collector cc ai ta co:

ACDC

CCCQTmaxCm RR

VII

+== (1)

VCEQT = RAC.ICQT (2)RDC = RC + RE = 150 + 100 = 250RAC = RC = 150Thay vao (1) ta c: mA25

15025010I CQT =+

=

V75,310.25.150V 3CEQT ==

VBB 0,7 + ICQT.RE = 3,2V.=== K1100.100.

101R

101R Eb

=

=

= K47,168,0

10

102,31

10

VV

1

1RR33

CC

BBb1

=== K1,331252,3

1010VV

RR 3BB

CCb2

VCE(V)

iC(mA)

VCEQT= 3,75

2ICQT= 50

40RR

V

EC

CC

=+

150

1ACLL

2VCEQT=7

10

ICQT= 25

2501DCLL


48/74



e ve ACLL, rat n gian ta ch can xac nh:iCmax = 2ICQTva VCemax = 2VCEQT.

III.III.III.III. Bo K RBo K RBo K RBo K R----C co CC co CC co CC co CCCCC va Cva Cva Cva CEEEE (E.C).(E.C).(E.C).(E.C).1) Bai 2-20: iem Q toi u

RDC = RC + RE = 900 + 100 =1K=

+=

+= 450

900900900.900

RRRR

RLC

LCAC

mA9,6RR

VII

DCAC

CCCQTmaxCm

+==

VCEQT = ICQT.RAC = 6,9.10-3.450 = 3,1VVBB = 0,7 + RE.ICQT= 0,7 + 100.6,9.10-3 = 1,4V

=== K1100.100.10

1

R10

1

R Eb

=

=

= 116386,0

10

104,11

10

VV1

1RR33

CC

BBb1

CE

Vcc=10V

R2

R1

RC=900

RE100

CC

RL=900K

VCE(V)

iC(mA)

VCEQT= 3,1

2ICQT = 13,8

10RR

V

EC

CC =+

4501ACLL

6,2 100

ICT= 6,9

10001DCLL


49/74



=== 71434,1

1010VV

RR 3BB

CCb2

Ta co dong xoay chieu:

V1,3V

mA45,39,6900900

900I.RR

RI

Lm

CmLC

CLm

=

=+

=+

=

2) Van bai 2-20 neu ta bo tu CE th ta se co bo khuech ai R.C co CC ma khongco CE. Khi o ket qua tnh toan se khac rat t v RE


50/74



(V

>> bER

R nen co the tnh gan ung theo cong thcE

BBCQ R

7,0VI

= )

VCEQ = VCC ICQ(RC + RE) = 25 2,1.10-3.3.103 = 18,7V

T hnh ve ta thay: ICQ < ICQT nen ICm = ICQ = 2,1mA

mA05,110.1,2.10.210.2

10.2IRR

RI 333

3

CmLE

LLm =

+=

+=

VLmmax = RL.ILm = 2.103.1,05.10-3 = 2,1V* Cach ve DCLL va ACLL cua bo K R.C mac C.C tng t nh cach mac E.C

( )CEQCEAC

CQC VVR1Ii =

vi =+

+= k2RR

RRRR

LE

LE

CAC

Cho VCE = 0 suy ra mA45,1110.2

7,1810.1,2RV

Ii 33

AC

CEQCQC =+=+=

iC = 0 suy ra V9,2210.1,2.10.27,18IRVV 33CQACCEQmaxCEQ =+=+=

* Vi bai toan tren neu cha biet R1 va R2 ta co the thiet ke e dong ien raln nhat: RDC = RC + RE = 103 + 2.103 = 3K.

Q

VCE(V)

iC(mA)

VCEQ

= 18,7

ICmax = 11,45

3,8RV

DC

CC =

310.2

1ACLL

100

ICQT= 5

310.3

1DCLL

22,9 25

ICQ = 2,1


51/74



Ta co: mA510.210.3

25RR

VI 33

ACDC

CCCQT =

+=

+=

VCEQT = ICQT.RAC = 10V.

2) Bai 2-24: Mach c nh dong Emitter.Theo nh luat K.II: Vkn = 0 ta coRbIBQ + VBEQ + RE.IEQ VEE = 0

Suy ra mA93100

7,010R

R

7,0VI

bE

BBEQ =

+

=

VCEQ = VCC + VEE ICQ(RC + RE)= 10 + 10 93.10-3.150 = 6,05V

mA5,4610.93.100100

100IRR

RI 3Em

LE

ELm =

+=

+=

VLm = ILmRL = 46,5.10-3.102 = 4,65V ay la iem Q bat ky nen ta co:

( )CEQCEAC

CQC VVR1Ii =

+ Cho VCE = 0 suy ra mA214RV

IiAC

CEQCQmaxC =+=

+Cho iC = 0 suy ra V675,1050.10.9305,6RIVV 3ACCQCEQCE =+=+=

iL

I.

CC

VL

CE

VEE=-10v

Rb


52/74



Neu bai nay c tnh che o toi u th:RDC = RC + RE = 150

=+

= 50RR

RRR

LE

LEAC khi o

mA100A1,05015020

RRV

I DCACCCCQT ==+=+=

VCEQT= ICQT.RAC = 5V


53/74



Chng IVChng IVChng IVChng IV:THIET KE VA PHAN TCH TN HIEU NHO TAN SO THAP.THIET KE VA PHAN TCH TN HIEU NHO TAN SO THAP.THIET KE VA PHAN TCH TN HIEU NHO TAN SO THAP.THIET KE VA PHAN TCH TN HIEU NHO TAN SO THAP.

I.I.I.I. S o mac Emitter chung E.C:S o mac Emitter chung E.C:S o mac Emitter chung E.C:S o mac Emitter chung E.C:1) Bai 4-7: Q bat ky.

a-

Che o DC

K3205,3

20.5,3RR

RRR21

21b

+=

+=

V320.205,3

5,3VRR

RV CC21

1BB

+=

+=

mA6,4

10010.3500

7,03I3CQ

+

=

VCEQ = VCC ICQ(RC + RE) = 20 4,6.10-3.2.103 = 10,8V

==

76010.6,410.25h.4,1h 3

3

feie

b- Che o AC:

i

b

b

L

i

Li i

iii

ii

A == (1)Zo

iC

Zi

Ri2K

ibRb3Kii

RC1,5K

iL

RL=1,5Khie100ib

1,2K

RL=1,5Kii

RC=1,5K

CC2-+

+VCC=20V

CE+

-

R13,5K

iL

R2=20K

Ri=2K RE1,5K

CC1-+


54/74



50100.10.5,110.5,1

10.5,1h.RR

Rii

ii

ii

33

3

feLC

C

b

C

C

L

b

L =+

=+

==

( )61,0

76010.2,110.2,1

hR//RR//R

ii

3

3

iebi

bi

i

b =+

=+

=

Thay vao (1) ta co: Ai = -50.0,61 = -30,6Zi = Ri//Rb//hie = 1200//760 = 465Zo = RC = 1,5K.

2) Bai 4-11: Q bat ky va hfe thay oi.a- Che o DC:

100R5010.50.101R

101R bE11b ==== , bo qua IBQ.

mA83

5010010

7,07,1RR

7,0VI

1

bE

BB1EQ =

+

=

+

=

mA10010

7,07,1R

7,0VIE

BB2EQ =

=

=

2110.8310.25.50.4,1h 3

3

1ie

==

5,5210.10010.25.150.4,1h 3

3

2ie

suy ra 21 hie 52,5b- Che o AC:

RL=100Rb=100

VBB=1,7vii

RC=100

CC-+

+VCC=20v

CE+-

iL

RE10


55/74



ieb

bfe

LC

C

i

b

b

L

i

Li hR

R.h.

RRR

ii

ii

ii

A++

===

66,2021100

100.50.100100

100A 1i =++

=

1,495,52100

100.150.100100

100A 2i =++

=

Zi = Rb//hie suy ra Zi1 = 100//21 = 17,36Zi2 = 100//52,5 = 34,43

Vay 20,66 Ai 49,1817,36 Zi 34,43

3) Bai 4-12: Dang khong co tu CEa- Che o DC:

mA5,4

1001010

7,07,5

hRR

7,0VI 43

fe

bE

BBCQ =

+

=

+

=

(co the tnh ICQ = 5 mA)VCEQ = VCC ICQ(RC + RE) = 20 4,5.10-3.(3.103) = 6,5V

==

77810.5,4

10.25

.100.4,1h 3

3

ie

b- Che o AC:

Rb=10K

VBB=5,7Vii

RC=2K

CC-+

+VCC=20V

iL

RE=1K

RL=100

RL=100

ib

Rb10KiiRC2K

iLhfeRE

100ib

hie=778

iC

ibRb100

ii

RC100

iLhie hfeib RL = 100


56/74



i

b

b

L

i

Li i

iii

ii

A == (1)

24,95h.RR

Rii

ii

ii

feLC

C

b

C

C

L

b

L =+

==

09,01077810

10RhhR

Rii

54

4

Efeieb

b

i

b =++

=++

=

Thay vao (1) ta c Ai = -95,24.0,09 = -8,6

[ ] =+= K1,910//10Rhh//RZ 54Efeiebi

II. S o mac B.C:S o mac B.C:S o mac B.C:S o mac B.C: Bai 4-21, hoe = 4101) Che o DC:

91,01110

h1hh

fe

fefb ==

+=

==+

=

3210

10.25.10.4,1.111

h1hh 3

3

fe

ieib

54

fe

oe

ob

1011

10

h1

hh

==+

=

2) Che o AC:

VCC

R2

Vi+

- R1 Cb

ri=50 RL=10K

RL10K

iL1/hob

105

iC

hfbie0,91ib

hib

32

ieRi 50

Vi

+

-


57/74



i

e

e

L

i

LV V

iiV

VV

A == (1)

82791,0.1010

10.10h.

h1R

h1R

ii

.iRi

iV

54

54

fb

obL

obL

e

C

C

LL

e

L =+

=

+

==

012,03250

1hR

1hR

V.V1

Vi

ibiibi

i

ii

e =+

=+

=+

=

Thay vao (1) ta c AV = (-827).(-0,012) = 10,085 10

III. S o mac C.C:S o mac C.C:S o mac C.C:S o mac C.C: Bai 4-231) Che o DC

VCC = IBQRb + VBEQ + REIEQ

mA65,4

1001010

7,010

RR

7,0VI53b

E

CC

EQ=

+

=

+

=

VCEQ = VCC REIEQ = 10 4,65.10-3.103 = 5,35 V

2) Che o AC

RL1K

Vi+

-RE

1K

ZoZi

Cc2ri 500

100KRb

Cc1

+VCC =10V

iLri 500

Vi+

-

hie 753ib

Rb100K

Re.hfe100K

RL.hfe100K

Vb VL


58/74



=

75310.65,4

10.25h4,1h 33

feie

ib

bL

iLv V

VVV

VV

A == (1)( )

( )[ ]

985,0000.50753

500.100R//Rhhi

R//Rh.iVV

LEfeieb

LEfeb

b

L

=+

=

+=

(2)

Rb = Rb//[hie + hfe(RE//RL)] = 33,3

994,010.3,33500

K3,33Rr

RRr

V.R.

V1

VV

3'bi

'b

'bi

i'b

ii

b =+

=

+=

+= (3)

Thay (2), (3) vao (1) ta co: AV = 0,985.0,994 = 0,979 0,98

[ ] +

+= 37,12553,7//10

hR//rh//RZ 3

fe

biibEo

( )[ ] ==+= K3,33RR//Rhh//RZ 'bLEfeiebi

hie/hfe 7,53ie

RE1K

ri/hfe5

Rb/hfe1K

Zo


59/74



Chng VI:Chng VI:Chng VI:Chng VI: MACH TRANSISTOR GHEP LIEN TANG.MACH TRANSISTOR GHEP LIEN TANG.MACH TRANSISTOR GHEP LIEN TANG.MACH TRANSISTOR GHEP LIEN TANG.

I. Transistor ghep Cascading:I. Transistor ghep Cascading:I. Transistor ghep Cascading:I. Transistor ghep Cascading:1) E.C1) E.C1) E.C1) E.C C.EC.EC.EC.E

Bai 6Bai 6Bai 6Bai 6----1111: iem Q bat ky, 2 tang hoan toan oc lap vi nhau.aaaa ---- Che o DCChe o DCChe o DCChe o DC

==>=+

=+

= 500R.h.101RK1,2

10.710.310.7.10.3

RRR.RR Efeb33

33

2111

21111b

suy ra, khong c bo qua IBQ1;

V310.10.710.3

10.3V.RR

RV 33

3

CC2111

111BB =

+=

+=

mA2,16

502100100

7,03

hR

R

7,0VI

1fe

bE

1BB1EQ

1

1

=

+

=

+

=

VCEQ1 = VCC IEQ1(RC1 + RE1) = 10 16,2.10-3.300 = 5,14V

===

10810.2,16

10.25.50.4,1I

10.25.h4,1h 33

1EQ

3

1fe1ie

==


60/74



50h.1ii.

ii

ii

2fe2b

2C

2C

L

2b

L === (2)

( )

06,550.1458164

164

h.hR//R

R//Rii.

ii

ii

1fe2ie2b1C

2b1C

1b

1C

1C

2b

1b

2b

+

+

==

(3)

951,01082100

2100hR

Rii

1ieb

b

i

1b =+

=+

= (4)

Thay (2), (3), (4) vao (1) ta coAi = (-50).(5,06).(0,951) 241

Zi = Rb//hie1 = 2,1.103//108 103Zo =

e tm bien o nh oi xng cc ai ta ve DCLL va ACLL.

T ( )CEQCEAC

CQC VvR1Ii =

vCE = 0 suy ra, ICmax = ICQ +VCEQ/RAC= 1,2.10-3 + 7,3/2.103 = 4,85mA

iC = 0 suy ra, vCemax = VCEQ.RAC= 7,3 + 1,2.10-3.2.103 = 9,7V

T ac tuyen DCLL va ACLL ta co ICmmax = 1,2mA

Bai 6Bai 6Bai 6Bai 6----2222: iem Q toi u nen phai tnh tang th hai trc, tang 1 sau.

aaaa---- Che o DC:Che o DC:Che o DC:Che o DC:RDC2 = RC2 + RE2 = 2250; RAC2 = RC = 2K.

mA35,220002250

10RR

VI

2AC2DC

CCT2CQ =

+=

+=

VCEQ2T = ICQ2T.RAC2 = 2,35010-3.2.103 = 4,7V

Q

VCE(V)

iC(mA)

7,3

iCmax = 4,85

4,4RV

DC

CC =

310.2

1ACLL

0

2501DCLL

9,7

10

ICQ = 1,2ICmmax


61/74



===

74510.35,2

10.25.50.4,1I

10.25.h4,1h 33

2EQ

3

2fe2ie

RDC1 = RC1 + RE1 = 200 + 100 = 300;RAC1 = RC1//Rb2//hie2 = 200//900//745 134,4

mA234,134300

10

RR

V

I 1AC1DCCC

T1CQ =+=+=

===

7610.2310.25.50.4,1

I10.25.h4,1h 3

3

1EQ

3

1fe1ie

b- Che o AC:Che o AC:Che o AC:Che o AC: S o tng ng tn hieu nho nh tren ch co hie1 va hie2 cogia tr khac. Ta ap dung luon cong thc (1) tren:

( )434

7621002100.

745164164.2500

hRR

.hR//R

R//Rh.hA

1ieb

b

2ie2b1C

2b1C2fe1fei

++=

++=

2) E.C2) E.C2) E.C2) E.C C.C:C.C:C.C:C.C:

Bai 6Bai 6Bai 6Bai 6----3333 iem Q toi uaaaa---- Che o DC:Che o DC:Che o DC:Che o DC:

Tang 2:RDC2 = RE2 = 1K; RAC2 = RE2//RL = 500.

mA7,650010

10RR

VI 3

2AC2DC

CCT2CQ =

+=

+=

VCEQ2T= ICQ2T.RAC2 = 6,7.10-3.500 = 3,35V

===

52210.7,610.25.100.4,1

I10.25.h4,1h 3

3

2EQ

3

2fe2ie

==== K101010.100.101R.h101R432E2fe2b

VBB = 0,7 + ICQ2T.RE = 0,7 + 6,7.10-3.103 = 7,4V

==

=

= K46,3826,0

10

104,71

10

VV

1

RR

44

CC

2BB

b12

=== K5,134,7

1010VV

RR 42BB

CCb22

Tang 1:

RDC1 = RC1 + RE1 = 400 + 100 = 500;RAC1 = RC1//Rb2//[hie2 + hfe(RL//RE)]= 400//104//[261 + 100.500] = 400//8333 382.

mA34,11382500

10RR

VI

1AC1DC

CCT1CQ =

+=

+=

VCEQ1T= ICQ1T.RAC1 = 11,34.10-3.382 = 4,33V


62/74



===

30934,11

10.25.100.4,1I

10.25.h4,1h3

2EQ

3

1fe1ie

=== K1100.100.101R.h

101R 1E1fe1b

VBB1 = 0,7 + ICQ1T.RE1 = 0,7 + 11,34.10-3.100 = 1,834V

==

=

= K25,128166,010

10834,11

10

VV

1RR 33

CC

1BB

1b11

=== K45,5834,11010

VV

RR 31BB

CC1b21

bbbb---- Che o AC:Che o AC:Che o AC:Che o AC:

i

1b

1b

2b

2b

LT i

i.

ii

.iV

A = (1)

( )( ) 3LEfe2b

L 10.5,50R//Rh1iV

=+= (2)

( ) ( )

75,051407

10.385100.50500522385

385

h.R//Rh1hR//R

R//Rii

.ii

ii

2

1feLEfe2ie2b1C

2b1C

1b

1C

1C

2b

1b

2b

=++

=

+++

==

(3)

764,030910

10hR

Rii 3

1ie1b

1b

i

1b =+

=+

= (4)

Thay (2), (3), (4) vao (1) ta coAT = (50,5.103).(-0,75).(0,764) -29000AT = -29000V/A = -29V/mA.

Zi = Rb//hie1 = 103//309 = 236

[ ] =+=

+= 907,9//1085,322,5//10

hR//R

h//RZ 332fe

2b1C2ibEo

iL

VL

ZoZi

Rb12,1Kiihie1309 100ib R

C1400 Rb210K

hie2 522

(1+hfe)RL101K

ib1 ib2iC1

(1+hfe)RE101K

RE1K

Zo

hib2ie2

Rc1//Rb2hfe2


63/74



3)3)3)3) Dang bai hon hp E.CDang bai hon hp E.CDang bai hon hp E.CDang bai hon hp E.C C.C:C.C:C.C:C.C: Bai 6-4

Tm R ei

02

i

01

iV

iV

= (1)

i

1b

1b

2b

2b

01

i

01

ii

.ii

.iV

iV

= (2)

i

1b

1b

2b2E2fe

i

02

ii

.ii

.R)h1(i

V+= (3)

suy ra2E2fe2b

01 R).h1(i

V+= (4)

2b

2C

2C

3b

3b

01

2b

01

ii

.ii

.iV

iV

= (5)

T (5) suy ra

2E2fe2fe3E3fe3ie2C

2C3E3fe

2b

01 R)h1(hR)h1(hRR

R.R)h1(

iV

+=++++

+=

5050100.505010R10

10.5050 333

=+++

(6)

105 = 7050 + RR = 100K - 7,05K 93K

Tmi

01

iV

T (2) ta co:

505050501010.9310

10.10.5050

R)h1(hRRh.R.R)h1(

ii

ii

iV

iV

333

23

3E3fe3ie2C

2fe2C3E3fe

2b

2C

2C

3b

3b

01

2b

01

+++

=

++++

+==

(7)

( )

( )63,7

655010.5

50501000500100.500

h.

R)h1(hR//R

R//R

i

i.

i

i

i

i

4

1fe

2E2fe2ie2b1C

2b1C

1b

1C

1C

2b

1b

2b

==++

=

+++

==

(8)

( )

( )5,0

101010

hR//RR//R

ii

33

3

1ie1bi

1bi

i

1b =+

=+

= (9)

Thay (7), (8), (9) vao (2) ta c:

iC2

hie11K

Ri

100Kii

hie2 1K

100ib1

RC1

1K

Rb2

1Khfe2RE2

5050

ib2ib1 iC1

Rb1

1K

Vo2100ib2

hie3 1K

RC2

1Khfe3RE3

5050

ib3


64/74



( ) ( )mAV27,195,0.63,7.5050

iV

i

01 =

Zi = Ri//Rb1//hie1 500

[ ] ++=

++= 5,471093010//50

hR

hRh//RZ

3fe

2C

3fe3ib3Eo

II.II.II.II. Transistor mac vi sai va DarlingtnTransistor mac vi sai va DarlingtnTransistor mac vi sai va DarlingtnTransistor mac vi sai va Darlingtn1) Bai 61) Bai 61) Bai 61) Bai 6----23: E.C23: E.C23: E.C23: E.C E.C.E.C.E.C.E.C.

aaaa---- Che o DCChe o DCChe o DCChe o DC

V25,29.10.310

10V.RR

RVV 33

3

CC2111

112BB1BB =

+=

+==

mA55,142,7107,025,2

hR

R2

7,0VII 3

fe

bE

1BB2EQ1EQ =

+

=

+

==

IE = 2IE1 = 3,1mAVCEQ1 = VCC 2RE.ICQ1 = 9 2.500.1,55.10-3 = 7,45VVCEQ2 = VCC 2RE.ICQ1 RC2.ICQ2

= 9 103.1,55.10-3 2,5.103.1,55.10-3 = 3,575V

RE3

50

Zo

hib3 10R/hfe3 930

Rc2

hfe3 =10

+VCC =9V

ii

R213K

iL

ib3

R11K

RE1500

T1 T2

RC22,5K

R223K

R121K

T3

T4

RC4=RL60

RE460

hfe=100

iC4hie1

hfe2ib2100ib2

RL(1+hfe)RE1

(1+hfe)ib4ii

hie3

RC22,5K

Rb2750

h2feRE4

ib1 ib2 hfe3hie4

Rb1750

hie2


65/74



V725,34,1875,39VVI.RVV 4BE3BE2CQ2CCCR 4E ===

mA6260725,3

RV

II4E

R4EQ4CQ

4E ===

mA62,010

10.62h

IIII 2

3

fe

4CQ4BQ3EQ3CQ ===

VCEQ4 = VCC ICQ4(RC4 + RE4) = 9 62.10-3.120 = 9 7,44 = 1,56VVCEQ3 = VCEQ4 VBE4 = 1,56 0,7 = 0,86V

===

225810.55,1

10.25.100.4,1I

10.25.h4,1h 33

1EQ

3

1fe1ie

===

564510.62,0

10.25.100.4,1

I10.25

.h4,1h 33

1EQ

3

3fe3ie

===

45,5610.6210.25.100.4,1

I10.25.h4,1h 3

3

1EQ

3

4fe4ie

bbbb---- Che o ACChe o ACChe o ACChe o AC

( ) 50500101.500h1RRfe1E

'

E==+=

( )[ ]( ) [ ] =+=+++= K76,617101.606045,56h1Rh1hR 3fe4E4fe4ie' 4E

i

1b

1b

2c

2c

3b

3b

L

i

Li i

i.

ii

.ii

.ii

ii

A == (1)

( )( ) 10201101.101h1h1ii

4fe3fe3b

L ==++= (2)

( )

3

33

3

'4E4iefe3ie2C

2C

2c

3b

10,4605,631 5,2

10.76,6175700564510.5,210.5,2

Rhh1hRR

ii

=

+++

=

++++

=

(3)

( ) 1001.hii

ii

ii

2fe1b

2b

2b

2c

1b

2c === (4)

(V RE rat ln nen coi ib2 ib1)3

''E1ieb

b

i

1b 10.4,12828322258750

750RhR

Rii =

++=

++= (5)

iC2hie1 2258

100ib2

RC60

RE50,5K

Zo

(1+hfe)2ib310201ib3

ii

hie35645

RC22,5K

Rb750

RE4618K

ib1 ib2hie4(1+hfe)5700

Rb1750

hie2 2258 ib3 iL

Zo


66/74



vi [ ] =+= 28323008//5050Rh//RR b2ie'E

''E

Thay (2), (3), (4), (5) vao (1) ta co:Ai = 10201.(-4.10-3).(-100).128,4.10-3 = 522,3 (lan)

Zi = Rb1//(hie1 +RE) 750//(2258 + 2832) = 654Zo = Zo = Zo//RC = RC = 60

2) Bai 62) Bai 62) Bai 62) Bai 6----24:24:24:24: E.C C.C

aaaa---- Che o DCChe o DCChe o DCChe o DCAp dung nh luat K.II Vkn = 0 cho vong 2 ta co:

VBE3 + IEQ3RE3 VEE = 0 (1)

mA3,210

7,03R

VVI 3

3E

3BEEE3EQ =

=

=

mA15,12

III 3EQ2EQ1EQ ===

VCE1 = VCE2 = VCC RC1ICQ1 VE1 (2)Mat khac ap dung nh luat K.II Vkn = 0 cho vong 1 ta co:

-VBB1 + RbIBQ1 + VBE1 + VE1 = 0 (3) VE = VBB1 RbIBQ1 VBE1 = 1 104.1,15.10-5 0,7 =0,185V

Thay vao (2) ta c:VCE1 = VCE2 = 6 103.1,15. 10-3 0,185 = 4,665V 4,67V

Ta co VE1 = VCE3 + RE3.IEQ3 - VEE (4)Suy ra VCE3 = VEE + VE1 RE3IEQ3

= 3 + 0,185 103

.2,3.10-3

= 0,885VVRE6 = VCC RC2ICQ2 VBE4 - VBE5 - VBE6= 6 103.1,15.10-3 2,1 = 2,75V

mA2751075,2

RV

I6E

6RE6EQ ===

VCE6 = VCC VRE6 = 6 2,75 = 3,25VVCE5 = VCE6 VBE6 = 3,25 0,7 = 2,55VVCE4 = VCE5 VBE5 = 2,55 0,7 = 1,85V

VL

ZoZi

ii

-3V

Rb110K

RC11K

Rb210K

RE1K

VBB2

1V

T4T5

RC21K

VBB11V

T3 RE6

10

T6


67/74



===

304310.15,110.25.100.4,1

10.15,110.25.h4,1h 3

3

3

3

1fe1ie

mA75,2hI

I6fe

6EQ5EQ == ; A10.75,2h

II 5

5fe

5EQ4EQ

==

===

72,1210.27510.25

.100.4,1I10.25

.h4,1h 33

6EQ

3

6fe6ie hie5 = 1272; hie4 = 127.200

bbbb---- Che o ACChe o ACChe o ACChe o AC

( ) += 766E3

fe6E'

6E 1010.Rh1RR

i

2b

2b

4b

4b

L

i

LT i

i.

ii

.iV

iV

A == (1)

= 7' 6E4b

L 10RiV

(2)

4

4

2

733

23

'6E4ie2C

2fe2C

2b

2C

2C

4b

2b

4b

10.3,96106,3811

10

1010.6,38110

10.10

Rh3Rh.R

ii

.ii

ii

=++

=++

=

++==

(3)

485,01010.086,610

10

Rh2RR

ii

ii

434

4

2b1ie1b

1b

i

1b

i

2b

=+

=

++==

(4)

Thay (2), (3), (4) vao (1) ta coAT = 107.(-96,3).10-4.(-0,485) = 46728V/A = 46,7V/mA

Zi = Rb1//[2hie1 + Rb2] 6,15K

( ) =

+= 37,0382,0//10

hh3

hR

//RZ 3fe

4ie3fe

2C6Eo

Chng VII:Chng VII:Chng VII:Chng VII: MACH KHUECH AI HOI TIEP.MACH KHUECH AI HOI TIEP.MACH KHUECH AI HOI TIEP.MACH KHUECH AI HOI TIEP.

iC2hie1 3043

VLhfe2ib2100ib2

ZoZi

ii

hie4127,2K

RC21K

Rb210K

RE6

ib1

Rb110K

hie2 3043 ib4

ib2

hie5.hfe4127,2K

hie5.hfe4.hfe5127,2K

Zo

RE63hie4h3fe

=0,381Rc2

h3fe 10


68/74



I.I.I.I. Hoi tiep ap, sai lech dong.Hoi tiep ap, sai lech dong.Hoi tiep ap, sai lech dong.Hoi tiep ap, sai lech dong.1)1)1)1) Bai 7Bai 7Bai 7Bai 7----4444

=

=

40hK1h

GTfe

ie ;

==

=

?ii

A?T

KLi

Li

ay la dang hoi tiep ap, sai lech dong.

a- Tnh o li dong T: cho ii = 0'

1

1b

1b

2b

2b

1

i'

1

1

Vi

.ii

.iV

0iVV

T ==

= (1)

( ) 32fe22E2b

1 10.41h1RiV

=+= (2)

( ) ( )

941,08580

10.4110.411010.240.10.2

h1Rh1RhRh.R

ii.

ii

ii

3333

3

2fe22E2fe21E2ie1C

1fe1C

1b

1C

1C

2b

1b

2b

==+++

=

+++++==

(3)

634

1ief1ief

'1

'1

'1

1b 10.911010

1hR

1hR

V.

V1

Vi =

+=

+=

+= (4)

Thay (2), (3), (4) vao (1) ta co:

V1

iiRE221K

Rf10K

RC12K

RC22K

iL

RL100

+VCC

RE211K

T1VL

iC2

hie1 hfe1ib140ib1

RE21(hfe2+1)41K

ZoZi

iiRC12K

RC22K

ib1 ic1

40ib2Rf V1

hie2 1Kib2

V1+

-

RE22(hfe2+1)41K RL

100


69/74



T = 41.103.(-0,941).91.10-6 = -3,51

b- Tnhi

L

ii

iA = cho V1 = 0

i

1b

1b

2b

2b

L

i

Li

i

i.

i

i.

i

i

i

iA == (1)

6,3940.1010.2

10.2h.

RRR

ii

.ii

ii

3

3

2feL2C

2C

2b

2C

2C

L

2b

L =+

=+

== (2)

941,0ii

1b

2b = (nh (3) phan tren) (3)

234

4

1ief

f

i

1b 10.911010

10hR

Rii =

+=

+= (4)

Thay (2), (3), (4) vao (1) ta co:Ai = (-39,6).(-0,941).91.10-2 = 33,9 34

c- Tnh Aif, Zif, Zof.54,7

51,3134

T1A

A iif =+

=

=

Zi = Rf//hie1 = 104//103 910

=+

=

= 20251,31

910T1

ZZ iif

Zo = RC2 = 2K

=+

=

= 44351,31

10.2T1

ZZ

3o

of

2)2)2)2) Bai 7Bai 7Bai 7Bai 7----11111111

=

=

C10h100h

GT ibfe

;

=

==

?TZ;Z

?ii

A

KL oii

Li

hie = hib.hfe = 10.100 = 1K

ii

Rf= Rb =10KRC 2K

C iL

RL

+VCC

RE100

C


70/74



a- o li vong T: cho ii = 0'L

b

b

L

i'L

L

Vi

.iV

0iVV

T ==

= (1)

433

233

feLC

CL

b

C

C

L

b

L 10.5101010.10.10h.

RRR

.Rii

.iV

iV

=+

=+

== (2)

634

iefief

'L

'L

'L

b 10.911010

1hR

1hR

V.

V1

Vi =

+=

+=

+= (3)

Thay (2), (3) vao (1) ta co:T =(-5.104).91.10-6 = -4,55

b- Tnh Ai, Zi, Zo.i

b

b

L'Li

Li i

i.

ii

0Vii

A ==

= (1)

50100.1010

10h.

RRR

ii

.ii

ii

3

3

feLC

C

b

C

C

L

b

L =+

=+

== (2)

234

4

ief

f

i

b 10.91

1010

10

hR

R

i

i =+

=

+

= (3)

Thay (2), (3) vao (1) ta co:Ai = (-50).91.10-2 = -45,45Zi = Rf//hie = 104//103 = 910Zo = RC = 103 = 1K

c- Tnh Aif, Zif, Zof.2,8

55,4150

T1A

A iif =+

=

=

=

+

=

= 164

55,41

910

T1

ZZ iif

=+

=

= 18055,41

10T1

ZZ

4o

of

II.II.II.II. Hoi tiep ap, sai leHoi tiep ap, sai leHoi tiep ap, sai leHoi tiep ap, sai lech ap:ch ap:ch ap:ch ap:

hie1iL

VLhfeib100ib

Zo

iiRC1K

ib ic

Rf Rf10KVL

+

-

RL


71/74



1)1)1)1) Bai 7Bai 7Bai 7Bai 7----10101010

=

=

C

50h50h

GT ibfe

;

=

==

?T

Z;Z

?ii

A

KL oii

Li

a- Tnh o li vong T (cho ii = 0)'L

1b

1b

2b

2b

L

i'L

L

Vi

.ii

.iV

0iVV

T ==

= (1)

3334

34

2fe2Cf

2Cf

2b

2C

2C

L

2b

L

10.5,8350.10.67,150.10.210

10.2.10

h.RR

R.Rii

.iV

iV

==+

=

+==

(2)

( )

46,1210.5,210.83,0 50.10.83,0

hhR//R

R//Rii

.ii

ii

33

3

1fe2ie2b1C

2b1C

1b

1C

1C

2b

1b

2b

=+

=

+==

(3)

63533

3

1fef1E1fe1ie1b

1fe1E'L

1b

10.2,110.210.5

1.10.510.5,2890

10.5

'Rh.R1.

R.hhRh.R

Vi

=+++

=

+++

=

(4)

Thay (2), (3), (4) vao (1) ta co:

VL

ii

Rf=10K

C

+VCC

R218K

CR111K

RC11K

RE1100

R2210K

R1210K

RC22K

RE21K

iC1

hfe1ib150ib1

hie1 2,5K

hie22,5K

RC22K

Zo

iiRC11K

Rb25K

Rf10K

ib2ib1 iC2

hfe2ib250ib2

Rf.hfe15.105

RE1hfe15K

VL+

-

Rb1890


72/74



T = (-83,5.103).(-12,46).(1,2.10-6)= -1,25

b- Tnh AT, Zi, Zo.i

1b

1b

2b

2b

L

i

LT i

i.

ii

.iV

iV

A == (1)

2b

L

ii va

1b

2b

ii tnh nh tren theo cong thc (2), (3)

( )107,0

10.95,410.5,2890890

hRRhRR

ii

331fef1E1ie1b

1b

i

1b =++

=+++

= (4)

Thay (2), (3), (4) vao (1) ta co:Ai = (-83,5.103).(-12,46).(0,107) = 111.103V/A = 111V/mAZi = Rb1//[hie1 + (RE11//Rf)(1 + hfe)] = 890 //[2500 + 4950] = 795Zo = Rf= 10K

c- Tnh AVf, Zof, Zif.mAV49

AV10.49

25,1110.111

T1AA 33TTf ==

+=

=

( ) =+== 1788)25,11(795T1ZZ iif

=+

=

= 444425,11

10T1

ZZ

4o

of

2)2)2)2) Bai 7Bai 7Bai 7Bai 7----12121212

=

=

C50h20h

GT ibfe

;

=

=

?TZ;Z

?AKL oi

V

VL

Vi+

-

ri 1K

Rf=1K

C

+VCC

R2110K

CR111K

RC1500

RE1282

R2210K

R121K

RC2500

RE2282

C

RE1122

C

RE2122

hfe1ib1

20ib1

Rf(hfe1+1)21K hie2 1050

Vi+

-Rb1910 R

Cb2910

ib1 iC2iC1

VL+

-

ri 1K

RC1500

RC2 500ib2ii hie1 1050

20ib2


73/

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