Do an OCL don

8/2/2019 Do an OCL don

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n k thut mch

SVTH: M.TunV. Quyn T.Vin Trang- 1 -

LI M UNgnh in t Vin thng l mt trong nhng ngnh quan trng v mang

tnh quyt nh cho s pht trin ca mt quc gia. S pht trin nhanh chngca Khoa hc Cng ngh lm cho ngnh in t Vin thng ngy cng phttrin v t c nhiu thnh tu mi. Nhu cu con ngi ngy cng cao l iukin thun li cho ngnh in t Vin thng phi khng ngng pht minh ra ccsn phm mi c tnh ng dng cao, sn phm a tnh nngNhng mt iucn bn l cc sn phm u bt ngun t nhng linh kin: R, L, C, Diode,BJTm nn tng l mn Cu kin in t.

Hin nay, nc ta c rt nhiu loi my khuyt i m thanh trn thtrng. m tng khuyt i cng sut c thit k t cc mch nh: mchkhuyt i OTL, mch khuyt i OCLNhng ph bin nht l loi mchkhuyt i OCL. Bi v dng mch ny c u im v: hiu sut, h s s dngBJT cng sut, li bng thng, bin tn hiu raChnh v th m chngem chn mch khuyt i cng sut dng OCL lm n mn hc.

Qua n lc nghin cu, tm hiu ca bn thn cng vi s hng dn tntnh ca thy gio m em hon thnh n ny.

Vi khong thi gian c hn cng nh trnh kin thc ca em cn hnch nn em tin chc rng h thng hot ng cha c ti u v khng trnhkhi nhng thiu st. Em knh mong thy thng cm gip v ch bo thmcho em nhng kinh nghim. Em xin chn thnh cm n.

Nng, ngy 1 thng 5 nm 2010


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MC LC

Phn A: L ThuytChng I i cng v BJT ................................................ 3Chng II Cc tng khuych i tn hiu nh ..................... 11

Chng III Diode bn dn.................................................... 16Chng IV Hi tip...............................................................20Chng V Khuych i cng sut ........................................25Phn B: Tnh tonPhn I Tnh ton ngun..................................................38Phn II Tnh ton tng cng sut .....................................39Phn III Tnh tng li........................................................41Phn IV Tnh ton v chn t C2, C3, C4, C5 .....................44Phn V Tnh h s khuyt i ton mch.........................45Phn VI Tnh mch bo v ..............................................46

Phn VII Kim tra mo phi tuyn..................................48


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C

2. Cc ch lm vic:

Tu theo yu cu k thut m ta chn ch lm vic thch hp cho Transistor

2.1. Ch khuch i.

BJT lm vic ch khuch i th JE phi c phn cc thun cn Jc

c phn cc nghch cho c hai loi BJT npn v pnp.

H thc lin h gia cc dng in:

CBEIII (1)

H s truyn t dng base-chung thun:

E

C

I

I (2)

H s khuch i dng Emitter-chung thun:

B

C

I

I (3)

T (1),(2) v (3) ta c:

1hay

1

Ch ny c s dng nhiu nht trong k thut mch tng t.

2.2 Ch bo ho.

Cc tip xc cc pht (JE) v tip xc cc thu (JC) u c phn cc thun.

Dng ng ra ca BJT ch bo ho gim hn so vi ch khuch i v dng

ra bo ho tho mn ICBbh

< IE

v ICEbh

< IB

cho cch mc EC v BC.

2.3 Ch ngng dn.

Cc tip xc JE v JC u phn cc nghch, dng qua ch c dng ngc qua

tip xc cc thu. Do phn cc nghch c hai tip gip nn in tr ca BJT ch

ngng dn tng ng k v in p hot ng ln.

UrB

C

UrUvB

E CUr

Uv

B

E

E

B

VB

VCC

I

E

IC

I

n

p

n

CCEC BC

Uv


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3. Cc c tuyn Volt- Ampe.

3.1. Cch mc EC.

a. c tuyn ng vo :

3.2. Cch mc BC.

a. c tuyn ng vo :

3.3. Cch mc CC.

S BJT mch EC

UBE(V)

UCE=0 UCE


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t2=500C t1=25

0C

IC0

I

UBEIC

UBEUBE00

a. c tuyn ng vo:

4. Dng bo ho ngc v n nh nhit im cng tc ca Transistor.

4.1. Dng bo ho ngc.

Do tip gip Jc phn cc ngc nn tn ti dng bo

ho ngc ICB0, dng ICB0 ph thuc nhiu vonhit .

Ic = IE + ICB0 = ( IB + IC ) + ICB0

(1 - ) IC = IB + ICB0 IC = IB + ( 1 + ) ICB04.2. n nh nhit im cng tc ca

Transistor.

S bin thin nhit nh hng n cc

tham s ca Transistor. Chu nh hng nhiu nht l

in p Emitter Base v dng ngc ICB0.

Khi ICB0 tng th IC tng, mt cc ht dn qua chuyn tip Colectter tng, dn

n s v chm gia cc ht vi mng tinh th tng lm cho nhit tng v ICB0 tng.

Nh vy, chu k c lp li lm cho dng IC v nhit Transistor tng mi. Hin

tng ny gi l hin tng hiu ng qu nhit. Hin tng ny lm thay i im

cng tc tinh, nukhng c bin php khc phc c th lm hng Transistor.Khi nhit thay i, UBE cng thay i dn n dng IC thay i lm thay i

im cng tc tnh.Tuy nhin, iu kin thng, ICB0nh hng nhiu hn so vi

UBE. Nh vy, khi ni n nh hng ca nhit , ta thng quan tm n dng ICB0.

UBC(V)

IB(A

S BJT mc CC

UBC(V)


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+Ub

Rc

Vcc

Rb IbIb

i vi Transistor Silc, dng ICB0 tng nhanh hn theo nhit nhng gi tr

tuyt i li nh hn so vi Transistor Gecmani cng nhit . Do vi Transistor

Si c th b qua ICB0.V th m bo cho mch in n nh, c bit nhit

cao, hay dng Transistor Si, lc ny cn quan tm n nh

hng ca (VBE).in p tri (VBE )c th do s bin i ca nhit

gy ra, cng c th do tp tn ca tham s Transistor

gy ra.

Nu IE = const v nhit mt ghp thay i lng

t th c tuyn IC = f(VBE) ca Transistor Si v Ge s tnh

tin song song vi trc tung mt lng l 2,5(t (mV/0C).

H s n nh nhit ca Transistor:

S cng nh mch cng n nh

Ta c: IC = IB + ( 1 + )ICB0 IC = IB + (1 + ) ICB0

C

CB

C

B

I

I

I

I

011

C

BC

CB

I

II

IS

1

10

5. Phn cc cho Transistor.

5.1. Phn cc Transistor bng dng c nh.

Ta c:

B

BEB

BCR

UUII

UCE = UCC - ICRC

11

10

C

BC

B

C

B

I

IS

dI

dI

I

I

Nh vy, h s n nh nhit ph thuc vo h s khuch i in p ca

Transistor thng ln nn h s S ca mch ny kh ln do n nh nhit km.

5.2. Phn cc cho Transistor bng in p phn hi.

Ta c: UCC = (IC + IB ) RC + IBRB + UBE

UCC= IB [(1 + ) RC + RB] + UBE

IC


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Rc

Vcc

Rb

IbIb

Vcc

R2

R1 Rc

Re

Ib

Re

+Ub

Rc

Vcc

Rb

Ie

Ib

=> BC

BECCB

RR

UUI

1

IC = IB =

BC

BECC

RR

UU

1

UCE = UCC - (IB + IC) RCH s n nh nhit :

Ta c: UCC= (IB +IB )RC + IBRB + UBE

UCC = (RC+ RB)IB + ICRC + UBE

=> IB =BEC

BECC

C

BC

C

RR

UUI

RR

R

=>`BC

C

C

B

C

B

RR

R

dI

dI

I

I

Nu RC >> RB th S 1

Vy : in p phn hi qua in tr RB trong mch phn cc lm tng n

nh nhit, ng thi lm gim h s khuch i tn hiu xoay chiu. Ta thy rng,

khng th nng n nh nhit ln cao v im cng tc tnh v n nh nhit

ca mch ph thuc ln nhau.

5.3. Phn cc Transistor bng dng Emiter.

p dng nh l Thevenin v Norton ta bin i mch in Hnh a thnh mch

in Hnh b.Trong , RB v UB c xc nh nh sau:

21

2.

RR

RVU CC

B

21

21.

RR

RRR

B

Ta c: UB = Ib RC + IERE + UBE

=>

BE

BEBB

RR

UUI

1

Hnh a

IC

Hnh b

Ic+IB


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IC = IB= BE

BEB

RR

UU

1

UCC = IC RC + IERE + UCE

Coi IC IE :

=> UCC= IC (RC + RE )+ UCE=> UCE= Ucc - IC(RC + RE)

H s n nh nhit:

Ta c: UB = IB RB + IERE + UBE

UB = IB(RB+RE) + UBE + ICRE

=> IB =EC

BECC

C

BE

E

RR

UUI

RR

R

=>BE

E

C

B

C

B

RR

R

dI

dI

I

I

=>

BE

E

RR

RS

1

1

Nu RE >> RB th S 1

Nh vy, mch n nh phi thit k sao cho RE cng ln cng tt.Nhng

nu RE qu ln s lm tng phn hi m, do lm gim tn hiu xoay chiu ca

mch. khc phc, ta mc song song RE vi mt t in CE c tr s ln d saocho i vi tn hiu xoay chiu th tr khng ca n gn nh bng 0, cn i vi tn

hiu xoay chiu th xem nh h mch.

u im ca mch phn cc bng dng Emiter l h s n nh nhit khng

ph thuc vo in tr RC, ngha l khng ph thuc im cng tc.


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Vcc

Ur

+

-Un

Rn

C1

R2

R1

Rt

C2

CeRe

Rc

B

Rc

rcrb Ur

Rtre

E

+

-En

Rn

ChngII: CC TNG KHUYCH I TNHIU NH

1. Mch khuch i EC.

R1,R2 : in tr phn cc cho BJT.

RC : in tr ti cc C ca BJT.

Re : in tr n nh nhit.

Rt : in tr ti.

Rn : Ni tr ngun tn hiu.

Un : Ngun tn hiu.

Ce : T thot xoay chiu.

C1 : T lin lc ng vo.

C2 : T lin lc ng ra.

1.1. Tr khng vo ca Transistor (rv) v mch EC (Rv).

RV = R1 // R2 // rV.

Ta c: U1 = ibrb + iere = ib[ rb + (1 + ) re]

=> rv = rb + (1+ ) re

1.2. H s khuch i dng in ca mch (Ki).

citi

bici

vibi

viti

iK ..

Ta c: iV.RV= iV.rV

ic=.ib => bi

ci

i

S mch EC

S tng ng

U1

ib

Rv rv

iv i

icR1//R2


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C2

Ucc

RcR1

Ur

R31k

CbR2

+

-Un

Rn

C1

Re

ReRc

rcre Ur

Rtrb

B

+

-En

Rn

B

S mch BC

iT.RT=iC(RC//RT) =>tR

tRcR

citi //

1.3. H s khuch i in p (Ku).

vnt

ivnv

tt

n

r

u RR

R

KRRi

Ri

U

U

K

.

1.4. H s khuch i cng sut (Kp).

KP = Ku.KI

1.5. Tr khng ra ca mch khuch i (Zr).

Khi h mch Rt, , Zr = rce // Rc do rce >> Rc => Zr = RC.

1.6. Quan h gia tn hiu vo v tn hiu ra.

bn k dng (+) ca tn hiu vo lm ib tng -> ic tng -> UC gim -> tn

hiu ra gim.

bn k m (-) ca tn hiu vo lm ib gim -> ic gim ->UC tng -> tn hiu ra

tng.

Vy, vi mch EC th tn hiu vo v tn hiu ra nghch pha nhau.

2. Mch khuch i BC.

2.1. Tr khng vo ca Transistor (rV) v mch khuch i (RV)


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Vcc

Ur

+

-Un

Rn

C1

R2

R1

Rt

C2

Re

Rv = Re // rv.

Ta c: U1= iere + ibrb=ie(re +1

br

)

rv=re+1

br

2.2. H s khuch i dng in (Ki).

t

tc

v

v

i

R

RR

r

RK

//.. , Ki < 1.

2.3. H s khuch i in p (Ku).

vn

tc

v

v

u

RR

RR

r

RK

//..

2.4. H s khuch i cng sut.Kp = Ki.Ku.

2.5. Tr khng ra ca mch khuch i (Zr).

Khi khng c ti Rt th Zr = rr // Rc ( Rc vi >> Rc.

2.6. Quan h gia tnh hiu vo v tn hiu ra.

bn k dng ca tn hiu vo lm ie gim -> ic gim -> Uc tng -> tn hiu ra

tng.

bn k m ca tn hiu vo lm ie tng -> ic tng -> Uc gim -> tn hiu ragim .

Vy, vi mch BC th tn hiu vo v tn hiu ra ng pha nhau.

3. Mch khuch i CC.

iv S mch CC


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Re

rcrb

Ur

Rt

re

+

-En

Rn

rb

3.1. Tr khng vo ca Transistor (rV) v mch khuch i (RV)

Rv = R1//R2//rv

Ta c:

teebb

teteebb

RRrriURRiririU

//1//

1

1

rv = rb + (1+)(re+Re//Rt)3.2. H s khuch i dng in ca mch (Ki).

t

tc

v

v

i

R

RR

r

RK

//.1.

3.2. H s khuch i in p ca mch (Ku).

vn

te

v

v

u

RRRR

rRK

//.1. , Ku < 1.

3.4. H s khuch i cng sut ca mch (Kp).

KP= Ki.Ku

3.5. Quan h gia tn hiu vo v tn hiu ra.

bn k dng (+) ca tn hiu v lm dng ib tng -> ie tng -> Ue tng -> tn

hiu ra tng.

bn k m (-) ca tn hiu v lm dng ib gim -> ie gim -> Ue gim -> tnhiu ra gim.

Vy, tn hiu vo v tn hiu ra ng pha nhau.

4. Nhn xt.

Tham s BC EC CC

Ki Nh(0.98) Ln(-47.2) Ln(48.2)

ib

ib

U1

rvRv

R1//R2

S tng ng


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Ku Ln(72) Ln(-72) Nh(0.99)

Zin Nh(20.4) Trung bnh (986) Ln(73K)

Zout Ln(1.03M) Trung bnh(52.6K) Nh(32)

ZL//Zout 1.5K 1.46K 31

Mch EC c Ku, Ki ln nn Kp ln, do c dng trong cc mch khuch

i cng sut .Tr khng vo v tr khng ra trung bnh nn tin li cho vic ghp vi

ti v ngun tn hiu .

Mch CC c tr khng vo ln nn thng dng lm mch phi hp tr

khng.

Mch BC v EC c hi tip m qua in tr Re nn thng c dng lm

ngun dng, cn mch CC thng c dng lm ngun p.

tn s cao th mch BC c nhiu u im hn so vi mch EC v CC.


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KA

Chng III DIODE BN DN.

1. Cu to.

Diode bn dn l dng c bn dn c mt tip gip p-n v hai in cc a ra t

hai min.

in cc a ra t min bn dn p l in cc Anod.

in cc a ra t min bn dn n l in cc Katod.

K hiu:

2. Phn cc v c tuyn tnh ca Diode.

Quan h gia dng in i qua diode v in p t trn Anod v Katod c

tnh bng biu thc sau :

11 kT

qV

S

V

V

SeIeII T

Trong : I : dng qua diode

IS: dng bo ho ngc.

VAK: in p t gia hai cc diode.

VT= 26 mV : p nhit. K=1,381.1023J/K, q=1,6.10-19C

Phn dng thun: khi in p thun tng i nh,in trng trong vn ng

k so vi in trng ngoi, m in trng trong ngn trdng khuch tn nn dng

in thun rt nh.Khi in p thun vt qu gi tr V c gi l in p m, ph

thuc vo nhit v vt liu bn dn.

Bn dn Si : V( =0.7 V)

p n

I

UUAK

KA

Vdt

Dngnghch

Dng thun


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R

D

R

D

Hnh b

Vi

t

Bn dn Ge : V(=0.3 V)Th in trng trong b kh bi in trng ngoi, in tr ca diode rt nh,

dng in thun tng nhanh theo in p.

Phn dng nghch: Khi t in p nghch ln diode, dng in nghch rt b.

Dng ny tng theo nhit , trong mt gii hn nht nh ca in p th khng phthuc vo in p.

Hin tng nh thng: nh thng diode c hai dng c bn

nh thng in l s tng t ngt lng ht dn qua tip gip p-n dotc dng ca in trng mnh ln cc nguyn t ca tinh th.

nh thng nhit xy ra khi tip gip b t nng qu gii hn cho php. 3. c tnh ng m ca diode bn dn.

Khi in p vo Vi bin i t +V1 n -V1, nu diode l kho ng m l

tng, th dng sng dng in qua ti RL c dng Hnh b. Dng thun bng, dng

nghch gn bng 0.

Hnh c biu th dng in thc t. Dng thun bng, dng nghch c t bin ,

ch sau thi gian phc hi nghch tre th diode mi tin n trng thi ngt mch, dng

in xp x 0. V vy, nu tn s in p vo Vi rt cao sao cho na chu k m ca V i

b hn thi gian re thi diode khng cn tc dng dn in mt chiu na.

4. Cc tham s c bn ca diode bn dn.

-V2

Vi i

VD

Hnh a

C2C1

C5

trtS

tre

Hnh c

IRL

V2/RL

V1/RL


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4.1. Dng chnh lu trung bnh cc i IF.

l dng din thun trung bnh cc i cho php chy qua diode trong thi

gian s dng lu di, c xc nh bi in tch chuyn tip p-n v iu kin to

nhit. Ta cn ch iu kin to nhit v bo m dng in trung bnh nh hn IF

diode khi hng.4.2. in p nghch cc i VT.

Nu in p nghch t vo diode t n VT th dng in nghch tng nhanh,

tnh dn in mt chiu ca diode b ph hng, lm hng diode.

4.3. Dng in nghch IR.

l tr s dng in nghch cc i khi diode cn cha bi nh thng, IR cng

nh th tnh dn din mt chiu cng tt.IR ph thuc nhiu vo nhit .

4.4. Tn s cng tc.

in dung chuyn tip p-n v in dung khuch tn ca diode l yu t ch yu

gii hn tn s cng tc, khi tn s vt qua gii hn ny th diode khng th hin tnh

nng dn in mt chiu na.

4.5. Thi gian phc hi nghch trr.

Thi gian trr c o trong cc iu kin qui nh v ph ti, dng in thun,

dng in nghch tc thi cc i.

trr = tS +tr

Vi: tS : thi gian lu gi

tr : thi gian chuyn tip

4.6. in dung tip gip p-n.

Gi tr in dung ny bao gm in dung khuch tn v in dung chuyn tip.

in dung khuch tn (in dung tip gip khi diode phncc thun):

T

TD

T

TSD

D

DD

V

vi

V

Ii

dv

dQC

)(

Trong : T : gi l thi gian chuyn tip.VT : in p nhit

IS : dng bo ho ngc

in dung chuyn tip (in dung tip gip khi diode phn cc ngc):


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j

R

j

R

n

jv

C

dv

dQC

1

0

Vi : in dung tip gip khi phn cc 0 ca diode.

Trong : A : tit din ca tip gip p - n.

Chng IV HI TIP

Hi tip l ly mt phn tn hiu ra (in p hoc dng in) ca mng 4 cc

tch cc a tr v u vo thng qua mt mng 4 cc gi l mng hi tip. Ngi ta

chia hi tip thnh hai loi l hi tip m v hi tip dng. Hi tip ng vai tr rt

quan trng trong k thut mch tng t. Cho php thay i tnh cht ca b khuch

i, nng cao cht lng ca b khuch i .

Hi tip c hai loi:

+ Hi tip m c tn hiu hi tip ngc pha tn hiu vo nn lm gim tn

hiu vo.Hi tip m mt chiu c dng n nh ch cng tc, hi tip m

xoay chiu dng n nh cc tham s ca b khuch i.

+ Hi tip dng c tn hiu hi tip ng pha tn hiu vo nn lm mnh tn

hiu vo. Hi tip dng thng lm cho khuch i mt n nh nn thng c s

dng to dao ng.

Phn loi mch hi tip:

Hi tip ni tip in p: tn hiu hi tip a v u vo ni tip vingun tn hiu v t l in p u ra.

Hi tip song song in p: tn hiu hi tip a vo u vo song songvi ngun tn hiu ban u v t l in p ra.

Hi tip ni tip dng in: tn hiu hi tip a v u vo ni tipngun tn hiu v t l dng in ra.

Hi tip song song dng in: tn hiu hi tip a v u vo song songngun tn hiu v t l dng in ra.


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(-)

Xh

1. Hi tip m.

1.1. Cc phng trnh c bn ca mang 4 cc c hi tip m.

Ta c cc quan h sau:

Xr = K.Xh (1)

Xv= Kn.Xn (2)Xh = Xv - Xht (3)

Xht= Kht.Xr (4)

T (1),(2),(3) v (4) =>

Hm truyn t ton phn:

n

n

rtp KK

X

XK '.

su hi tip: g = 1 + K.Nu |g| > 1 th |K| < |K| => hi tip m.

Nu |g| < 1 th |K| > |K| => hi tip dng

1.2. nh hng ca hi tip m n cc tnh cht ca b khuch i.

1.2.1.nh hng n h s khuch i .

Ta c: K < K

Xv Xr

Xht

Xn K

Kht

Kn

S khi ton phn ca b khuch i c hi tip.


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hU

Iv1Iv

+

- Kht

hU

Vy, h s khuch i khi c hi tip m nh hn khi khng c hi tip m.

1.2.2. nh hng n tr khng vo.

Hi tip m lm thay i tr khng vo ca phn mch nm trong vng hi tip.S thay i ny ch ph thuc vo phng php mc mch hi tip v u vo (ni

tip hay song song), khng ph thuc phng php ly tn hiu u ra a vo

mch hi tip.

a. Tr khng vo ca b khuch i c hi tip m ni tip.

Khi khng c hi tip (Kht.Xr=0):

=>rhth

v

h

v

v

vrr

I

UU

I

UZ

'

Khi c hi tip:

=>

rhth

v

hth

v

v

vrrg

I

UKKU

I

UZ

.'.1

'

Nu rrht Zv = g.Zv

b. Tr khng vo ca b khuch i c hi tip m song songKhi khng c hi tip:

rhthv

vrrZ

Y111

'

Khi c hi tip:

rhthv

hrht

v

vv

rr

g

U

IIXK

U

I

ZY

1'

'

1'

Nu reht >> rh th Zv= Zv /g

hi tip song songhi tip ni tip

rh

rrht

rh

rrhtU

KhtXr


21/49

n k thut mch


+

-

Vy: hi tip m ni tip lm tng tr khng vo phn mch nm trong vng

hi tip ln g ln v hi tip m song song lm gim tr khng vo g ln.

1.2.3.nh hng n tr khng ra.

S thay i tr khng ra khi c hi tip khng ph thuc vo phng php ly tnhiu v m ph thuc vo phng php ni u ra b khuch i vo mch hi tip.

a. Tr khng ra ca b khuch i c hi tip m in p.

Khi khng c hi tip: Zra = rr // rvht ( rr (v rrg

r

g

r

KK

r

I

UZ rr

hth

r

rng

rh

r

11

' , rr g

ZZ r

r'

b. Tr khng ra ca b khuch i c hi tip m dng in.

Khi khng c hi tip: Zra = rr + rvht ( rr , rr >> rvht )

Khi c hi tip:

rvngrhngrh

htng

vng

vngrng

rXKrXKU

KK

XKXKI

1.

=> rnghtngrng

rara rgKKr

I

UZ

)1('

=> Zra= g.Zra

KhXh

raU Ura

rht

r

R

t

R

r

rvh

t

hi tip m in p hi tip m dng in

KngXh

Ira


22/49

n k thut mch


Vy, hi tip m in p lm gim tr khng ra g ln, cn hi tip m dng in

lm tng tr khng ra g ln.

1.2.4.nh hng n nhiu v tp m.

Khi c tn hiu t u vo b khuch i th u ra ngoi tn hiu c

khuch i cn c tn hiu nhiu v tp m (domch sinh ra).

[(Xv - Xht)K1 + Xta].K2 = Xr

K1K2Xv-K1K2Xht-K2Xta= Xr

Thay Xht =Kht.Xr : K1K2Xv + K2Xta = Xr(1+K1K2Kht)

=>r

ht

ta

ht

vX

KKK

KX

KKK

KKX

212

21

21

11

=>r

ht

ta

ht

v XKK

X

K

X

1

Nhn xt:Hi tip m lm gim tn hiu Kht ln nhng lm gim tp m hn i K1Kht ln.

1.2.5. nh hng n mo phi tuyn v di ng.

Xh = Xv - KhtXr = Xv - K.KhtXh

=> Xh =g

X

KK

Xv

ht

v .1

Nhn xt:

i lng in gim g ln nn mo phi tuyn sinh ra do on cong vnhu c tuyn vo cng gim g ln.

Khi i lng t trc tip vo b khuch i gim g ln th di ngtng g ln.

2. Hi tip dng

Gi s khi khuch i v khi hi tip c e Khtj

htht KK.

Xr

Xta

Xht

XVK1 K2

Kht

Xh


23/49

n k thut mch


mch to ra dao ng th: 1..)(

e KhtKj

htKK

=>,...2,1,0,2

1.

nn

KK

KhtK

ht

)2(

)1(

(1) l iu kin cn bng v bin cho bit mch ch dao ng khi h s

khuch i ca b khuch i b dc tn hao do mch hi tip gy ra.

(2) l iu kin cn bng v pha cho bit dao ng ch c th pht sinh khi tn

hiu hi tip ng pha tn hiu vo.

K : moun h s khuch i .


24/49

n k thut mch


Chng V KHUYCH I CNG SUT.

1. Ch cng tc v nh im lm vic cho tng khuch i cng

sut.

Tu thuc vo ch cng tc ca Transistor ngi ta phn bit thnh cc

ch A, AB, B v C.

1.1. Ch A.

Ch khuch i gn nh tuyn tnh, gc ct ( = T/2 =1800. Khi tn hiu vo

hnh sin th dng tnh lun lun ln hn bin dng in ra. V vy, hiu sut ca b

khuch i ch A rt thp (


25/49

n k thut mch


+

-

Un

Vcc

Cv

Rn

RL

T1

t

tn hiu ra

1.3. Ch B.

C gc = 900. im lm vic tnh c xc nh ti UBE = 0. Ch mt na chu

k m (hoc dng) ca in p c Transistor khuch i.

1.4. Ch C.

Gc ct 78%) nhng mo rt ln, nthng c dng trong cc b khuch i tn s cao v dng vi ti cng hng

c th lc ra c hi bc nht nh mong mun.

2. Khuch i cng sut hng A (Khuch i n).

Ta c Ucc=ICRL +UCE.

im lm vic tnh l trung im t Ucc n Ucc/RL:

2

2CC

CEQ

L

CC

CQ

UU

R

UI

Tn hiu xoay chiu v mt chiu u chy qua cng mt mch nn ta c ng ting trng vi ng ti tnh.

-Dng ra c bin : gi tr hiu dngL

CC

R

UI

220

tn hiu vo

tn hiu ra

tPDM

ICQ

UCC

t


26/49

n k thut mch


Cng sut tn hiu trao cho ti RL:L

CC

L

CCCC

LR

U

R

UUIUP

822.

22

2

00

+ Cng sut tiu tn ca in tr :L

CC

L

L

CC

LCQRR

UR

R

URIP

42

22

2

+ Cng sut ca ngun cung cp:

L

CC

L

CC

L

CC

RDCCR

U

R

U

R

UPPP

244

222

=> %25%100.2

.8 2

2

CC

L

L

CC

U

R

R

U

3. Tng khuch i y ko.

tng cng sut hiu sut v gim mo phi tuyn, ngi ta dng tng khuch

i y ko. Tng khuch i y ko l tng gm c hai phn t tch cc mc chung

ti.

Mt s c im:

im t ca mch song song l u m ca ngun mt chiu. im t ca

mch ni tip l im gia ca ngun mt chiu.

Cc mch y ko dng hai Transistor cng loi c kch thch bi cc tn

hiu ngc pha. to tn hiu ny c th dng tng khuch i o pha hoc dng

bin p m cun th cp ca n c im gia ni t v mt xoay chiu.

Cc mch y ko dng hai Transistor khc loi c kch thch bi cc tn

hiu ng pha.V vy, c th dng mt tn hiu kch thch cho c hai Transistor

Cc tng khuch i y ko c th lm vic ch A, AB, hoc B nhng

thng thng ngi ta thng dng ch B hoc AB.

3.1. Khuch i cng sut y ko hng B (y ko songsong).

RL: in tr phn nh t RL vo cun s cp

2

2

1'

N

N

R

R

L

L

Dng ra c bin l ICM, gi tr hiu dng l:


27/49

n k thut mch


RL

T2

UvUcc

T1

Trong ITB l dng trung bnh chy qua cc C ca BJT, tnh bi cng thc:

L

CCCM

CMTBR

UItdtII

'

'sin

2

1

0

Vy:

Khuch i y ko hng B c hiu sut cao nhng tn hiu ra b mo xuyn

tm. Ngy nay, b khuch i y ko song song ch cn c dng trong nhng

trng hp yu cu phi cch li in mt chiu i vi ti hoc yu cu yu cu hiu

sut cao trong khi ngun cung cp nh. Nhc im ca mch l kch thc ln, gi

thnh cao, di tn lm vic hp, khng th thc hin di dng mch tch hp.

3.2. Khuch i hng A-B.

khc phc hin tng mo xuyn tm khuch i cng sut hng B, ngi

ta ch to ra mch khuch i cng sut hng AB bng cch thm hai in

tr R1, R2 phn cc trc cho BJT, sao cho khi c tn hiu vo th BJT dn ngay.

Do , hiu sut ca khuch i cng sut hng AB l: A < AB < B.

T2

Uv

R1

UccR2

RL

T1

Hai mch thng dng sau ny l OTL v OCL.

3.3. Mch khuch i cng sut kiu OTL.


28/49

n k thut mch


RL

C2

ReRe2

Vcc

To

RcT1

T2

Re1

R1

R2

Uv

C1

R21k

C2

Ce

To

D2

D1

RL

Re

Vcc

RcT1

T2

R1

Uv

C1

3.3.1. Dng hai BJT mc theo

kiu y ko.

T0 : BJT khuch i o pha

T1 cng loi T2, mc theo kiu

y ko. ch tnh T0 c phn cc

sao cho:

IB1 = IB2 => IC1 = IC2

=>2CC

S

VU

ch ng:

bn k m(-) ca tn hiu vo T1 dn to ra dng IC1 np in cho t C2,i

t VCC -> T1 -> C2 -> RL -> mass.

bn k dng (+) ca tn hiu vo T2 dn to ra dng IC2 np in cho t

C2, i t C2(+) -> T2 -> mass -> RL -> C2(-).

Do tn hiu c trao y cho ti.

3.3.2. Dng hai BJT mc theo kiu b

ph.

T0 : BJT khuch i m.

D1: diode cng loi bn dn vi T1(npn)

D2: diode cng loi bn dn vi

T2(pnp).

T1,T2 :hai BJT mc theo kiu b ph.

ch tnh T0 c phn cc sao cho:

22 2121CC

SCCBBCC

P

UUIIII

UU

ch ng:

bn k m(-) ca tn hiu vo T1 dn to ra dng IC1 np in cho t C2,i

t VCC -> T1 -> C2 -> RL -> mass.

bn k dng (+) ca tn hiu vo T2 dn to ra dng IC2 np in cho t

C2,i t C2(+) -> T2 -> mass -> RL -> C2(-).

3.4. Mch khuch i cng sut kiu OCL.

IC2

IC1


29/49


30/49

n k thut mch


E

B

C

B2

E1

E2

Q2

Q1

Ta c:

IC = IC1= 1IB1 1IE2 1IC2 = 12IB2

Vy: Q c h s khuch i : =1.2

H s khuch i dng chung:

2

2

1

11

2

2

21

11

2

21

1..

fe

fe

E

B

fe

B

C

BB

BC

B

CC

B

C

feh

h

IIh

II

IIII

III

IIh

hfe = hfe1(1 + hfe2) + hfe2 = hfe1.hfe2

Tr khng vo: 212

2

212 1

1

feieie

fe

E

Bieieie hhh

h

I

Ihhh

5.2. S gi Darlington (Dng b).

H s khuch i dng:

2

1

1

1/

2

2

1/

2

2/

2

1/2/ .B

B

B

QBE

ie

B

QBE

B

QBE

B

QBEQBE

v

v

ieI

I

I

Uh

I

U

I

U

I

UU

I

Uh

2

1

E

B

B

C

feI

I

I

Ih

2

221

2

1111

B

Bfefe

B

Bfe

I

Ihh

I

Ih

hfe = (1+ hfe1) hfe2 hfe1.hfe2

Tr khng vo: 22

2/

ie

B

QBE

ieh

I

Vh

IC1

IB2

IC2=IB1

IE1

Q


31/49

n k thut mch


C1

R

ZL

Zb1b2

Q2

Q1

Q3

Q2

6. Hin tng mo xuyn tm (Crossover) v phng php khc phc.

6.1. Hin tng mo xuyn tm.

Khi tng khuch i lm vic ch B th im tnh Q nm gc to

O(0,0) trn c tuyn vo ca Transistor (1).Nhng ng vi Transistor (2) th c tuyn ny c vng cht cng ng vi khi

VBE IC = 0. Khi in p vo vt qu ngng V th mi xut hin

tn hiu u ra. Kt qu l tn hiu ra b mo vng tn hiu b. khc phc tnh

trng ny, ngi ta thng phn cc cho cc tip gip B-E di im tnh Q n gc

to . Lc d tng khuch ai cng sut lm vic ch AB.

6.2. Cc phng php trnh mo Crossover.

Hnh sau minh ho bin php trnh mo Crossover bng cch phn cc trc

cho Q1 v Q2 lm vic ch AB.

Trong ZB1B2 c th l:

Diode v bin tr mc ni tip. Diode v bin tr mc song song. Mch Transistor.

6.2.1. Dng Diode vi bin tr mc ni

tip.

dng phn cc th din p trn diode hu nh khng i.Nu diode lm cng

vt liu vi cp BJT cng sut Q1, Q2 th khi dn V( ca diode xp x bng V ca

Transistor ).

VR dng chnh p phn cc cho ng yu cu.

(2

IB1

0

IB

VBEV

(1

)


32/49

n k thut mch


B2

Vr

B1

D2

D1

B1

D1

D2

B1

B2

VrR1

B1

B2

Vr

ZL

Zb1b2

Rc4

Rc3

C1

C2Q1

Q2

Q3

Phng php ny iu chnh im lm vic tnh Qnhng li b mt mt tn

hiu trn VR.

6.2.2. Dng diode vi bin tr mc song song.(Hnh b)

Khi tn hiu vo ln th dng s qua diode cn khi tn hiu b th dng

hu ht qua VR. Do , st p trn ZB1B2 hu nh khng i lm cho im lm vic

Q khng b x dch. Tuy nhin vi cch mc ny th s kh iu chnh VR hn.

6.2.3. Dng mch Transistor .(Hnh c)

Gi s dng vo Transistor nh. Ta c:

Khi lm vic VBE hu nh khng i, do VC1=VB1B2 khng i. Ngoi ra BJT

c nhim v gi cho im lm vic tnh Q khng b x dich khi nhit thay i.

7. Cc bin php nng cao h s khuch i.

7.1. Dng nguyn tc Bootstrap.

Mun tng h s khuch i th phi tng h s khuch i ca tng tng

khuch i nhng ch yu vn tng h s khuch i cho tng o pha (driver).

C hai bin php, mt l phi cung cp cho Q3 mt in p ln hn hoc phi

dng nguyn tc Bootstrap.Thng th ngi ta dng phng php th hai hoc dng

ngun dng.

Hinh a hinh b


33/49

n k thut mch


Zb

R1

Re

ZL

Bootstrap l bin php dng tng in tr ti cho tng o pha.V mt xoay

chiu RC4 mc song song vi ti ZL, con RC3 mc song song vi tr khang vao hie1 cu

Transistor Q1.

Ti ca tng o pha Q3 l:

Zt3= RC3//hie1+ (1+ hfe1)(ZL//RC4)Chn:

RC3 >> hie

RC4 >> ZL

=> Zt3 = (1+ hfe1)ZL

Trong thc t tng Zt3 ngi ta mc Darlington tng cui cng c hfe

ln v c RC3>> hie1 th ngi ta thay in tr RC3 bng in tr ng ca ngun

dng. Ni cch khc l ta thit k cho ti ca Q3 l ti tch cc do ngun dng to ra.

7.2. Dng ngun dng.

T s tng ng ta c h phng trnh:BieBE

dIhdV (1)

EEBBBEdIRdIRZdV )//( 1 (2)

CBEdIdIdI (3)

ieCEBfeChdVdIhdI (4)

Gii ra ta c ni tr ngun dng:

EieB

Efc

ieC

CE

iRhRZ

Rh

hdI

dVR

1//

.1

1

Ngun dng c ni tr Ri cng ln th cng n nh v cng gn vi ngun l

tng.Ta c:

nu Ri >>ZL th IL = I

Ngun dng c tr khng xoay chiu ln nn dng tng h s khuch i .V

cc tng khuch i mc ni tip nhau nn:

AV = AV1.AV2

IB IC

IL-IC ZL

Re

hie ZL

Re

hieZb//R1 hfeIB

ieh

1


34/49

n k thut mch


Trong : AV2 =Lxc

ie

feZ

h

h ,ZLxc : tr khng ti xoay chiu ca BJT.

ZLxc = Rc //ZinQ3, ZinQ3: tr khng vo tng khuch i cng sut.

ZinQ31 = hie1 + (1+ hfe13)RL

Do h s khuch i dng tng khuch i cng sut hfe13 rt ln nn ZinQ31rt ln, dn n ZLXC b nn AV2 khng ln -> AV khng ln -> RC khng th qu

ln.

khc phc thay RC bng ngun dng.

8. Khuch i vi sai.

Khi ng vo c tn hiu Vi do tnh phn p ca R-R nn s tc ng vo mi

Transistor lm cho Transistor Q1 dn v Transistor Q2 tt. ng ra in p trn cc C

ca Q1 gim i mt lng v in p trn cc C ca Q2 tng ln mt lng.

Ta c: V0 =V01 - V02 .

iu kin RB2 >>rBE.

li ca mi tng c xc nh nh sau:

)1(101

1)1(

011 . i

BEB

C

i

VKVrR

R

V

VK

)2(202

1)2(

022 . i

BEB

C

i

VKVrR

R

V

VK

Q2Q1

R8R

R

Rb1

Rb2RcRc

Rb2

Vcc

S tng vi sai yu cu cc Transistor Q1 v Q2 phi c cc tham s ging

nhau => K1 =K2.

V0 = V01 - V02= K1(Vi(1) - Vi(2))

V01V02

V0


35/49

n k thut mch


= K1.Vi

1

01

BBE

C

iRr

R

V

VKK

(*)

H thc (*) cho thy khi mch khuch i vi sai dng hai Transistor nhng

li in p bng mt tng khuch i thng thng ca mt Transistor.H s khuch i hiu:

Vi gi thit trng hp cn ly tn hiu trn mt u ra so vi t, ta c h s

khuch i i vi mt u ra:

22

121

ud

d

rd

d

r

udud

K

U

U

U

UKK

Vy h s khuch i i vi mt u ra bng mt na h s khuch i hiu

khi ly in p ra i xng.

H s nn tn hiu ng pha:

CM

ud

K

KCMRRG 1)( [dB]

vi KCM l h s khuch i tn hiu ng pha.

Trong ch khuch i tn hiu ng pha RE c tc dng hi tip m ln lm

gim h s khuch i tn hiu ng pha KCM. Do gim h s KCM ng thi

tng h s CMRR th ta phi chn RE ln. Tuy nhin , khng th chn RE qua ln sao

cho IE.RE nh hn (10-15)V m bo iu kin v cng sut

Tn hao trn hai in tr v iu kin v ngun cung cp. V vy, trong thc t

thay RE bng ngun dng c ni tr ln v h p trn n nh.

Hin tng tri:

Trong thc t mch khng hon ton i xng, do s thay i nhit gy

ra mt in p hiu (tri in p lch khng).

U0=UBE1 - UBE2

Ta c:

)1ln(

)1ln(

22

11

Enh

C

TEB

Ebh

C

TBE

I

IUU

I

IUU


36/49

n k thut mch


Mt s s c bn ca mch khuch i vi sai:

* * *

+V

+V+V

+V+V

9. Bin php nhm ci thin c tnh ca mch.

Khuych i vi sai Darlington Khuych i vi sai Darlington b

Khuych i vi sai c hitip m dng Khuych i vi sai

Khuych i vi sai Kaskode


37/49

n k thut mch


Trong tng cng sut s taie Emitter chung d xut hin nguy c qu ti nu

tr khng ti qu nh. khc phc hin tng c th mc thm mch hn dng

vo s nh sau:

Bnh thng Q5, Q6 ngt, VBE ca chng ph thuc vo p trn R1, R2

(R1 = R2). Khi dng in ra qu ln,VBE tng lm Q5,Q6 dn, lm VBE ca Q1,Q2

gim v dng ra gim.R1 c chn sao cho dng ra b hn ch trong trng hp n

vt qu tr s thng thng tnh ton(trongvng lm vic).

UrR61k

UvR51k

Q7

R41k

D2

D1

R31k

R2

R1

Q6Q2

Q5

Q1

+Vcc

-Vcc


38/49

n k thut mch


THIT K

CC THNG S YU CU THIT K MCH KHUYCH ICNG SUT OCL NG VON DNG BJT

Cng sut loa : 75W - 100Win tr loa : 8Tr khng vo : 100K - 120Kin p vo : 0,775VBng thng : 20Hz 15Khz mo : 2%

TC DNG CC LINH KINQ6, Q7, Q8, Q9 : L cc cp BJT ghp Dalington khuyt i cng sutQ2,VR1 ,D4, R5 : L cc phn t ca ngun dngQ4, Q5 : L cc BJT bo v qu ti v ngn mchR1, R2 : n nh nhit v cn bng dng raR3, R4 : in tr r dng nhitD1,D2,D3,VR2 : nh thin p cc BJT cng sut hot ng ch

ABR6 ,R18, C6 : n nh nhit cho Q3R10, VR3 ,C2 : B lc thng caoR9, R10 : in tr phn cc cho Q1R8, R17 : Cu phn p cho Q3R13, R14 : Cu phn p cho Q5R15, R16 : in tr phn cc cho Q4C1, C2 : T lin lc ng vo v ng raC4 , C5 : T thot cao tng, chng dao ng k sinhR0, C1 : Mch lc zobel thnh phn cn bng tr khng

TNH TON

I. Tnh ton ngun:a. Bin tn hiu raKhi tn hiu vo c dng hnh sin: Vl=Vlp th tn hiu ra c dng:

Vl= Vlpsin + VECQI l = I lpSin + ICQ

ILhd =2

LI

; VLhd =2L

V

Cng sut trn loa:

2 2*80*8 35,7( )LP L L

U PR V Chn PL = 80 W


39/49

n k thut mch


Khi 36 4,5( )8

LPLP

L

UI A

R

Chn h s ngun l = 0,935,7

2 2* 80( )0,9

LPCC

UV V

Vy chn VCC

= 80(V)2 236

812 16

LPLP

L

UP

R (W)

4,52 2 80 229LP

CC CC

IP V

(W)

n% = 80 *100229

LP

CC

P

P % = 35,37(%)

II. Tnh ton tng cng sut :Tng cng sut c nhim v khuch i tn hiu a ra loa.

trnh mo phi tuyn ta chn cc BJT Q8, Q9 lm vic ch AB.

a. Chn R1 , R2:y l hai in tr c nhim v cn bng dng v n nh nhit choQ8, Q9. trnh tn hao cng sut trn R1, R2 th ta chn in p ri trn R1, R2nh hn nhiu so vi in p nh :

VR 1 = VR2 =36

1,8( )20 20LPV V

11 2

1,80,4( )

4,5R

LP

VR R

I

Vy chn R1 = R2 = 0,5( )2

1 2 1 1

4,5* * *0,5 0,7LPR R tb

IP P I R R

(w)

Chn 1 20,5( )

1(W)

R R

P

b. Tnh v chn Q8,Q9:Cng sut tiu tn trn 2 BJT Q8 v Q9 :

2PQ8 = Pcc PL 2 PR12 2

1*8 2

2 * * 22

2cc LP L LP LP

Q

V I R I R I P

M ILP =1

2

2 2*806,2( )

4 4*0,58

10

CC

L

VA

RR

82 158QP (W) 8 79QP (W)PQ8dc = VCEQ* ICQ = Vcc * ICQ = 80 * 50* 310 = 4(W)PQ8 = 4 + 79 = 83 (W)

Vy chn Q8, Q9: - P0 > 2PQ8 = 283(W) = 166(W)- VCE > 2VCC = 160(V)- Ic > ILP = 4,5(A)

Vy chn cp b ph l : 2SC5200 v 2SA1943 c cc thng s sau :


40/49

n k thut mch


BJT PC

(W) hfe

IC

(A)Tf (MHz) V CEQ (V)

2SA1943 150(W) 55 - 160 15(A) 30 230(V)

2SC5200 150(W) 55 - 160 15(A) 30 230(V)

c. Chn R3 ,R4 :Chn ICQ8 = 50 (mA) , chn hfe8 = 100

IBQ8 = 350*10

0,5100

mA

8acR < R3 , R4 < Rdc8

Rac8 = hie8 + R1* hfe8 =3 3

81 8 3

*26*10 100*26*10* 102( )

0,5*10fe

fe

CQ

hR h

I

Rdc8 = 8 1 8 38

0,7* 0,5*100 1450( )

0,5*10be

fe

bQ

VR h

I

102 < R3, R4 < 1450Vy chn R3, R4 = 390 ( )

d. Tnh v chn Q6, Q7VR3 = VbeQ8 +R1 * Icq8 = 0,70+ 0,5 * 50mA = 0,725(V)

IR3 = 33

0,7252( )

390R

VmA

R

IcQ6 = IbQ8 + IR3 = 0,5 + 2 = 2,5(mA)Xt AC:

88

8

4,545( )

100cac

bac

fe

II mA

h

VacR3 = hie8 * Ibac8 +R1 * Icac8 =8

8 1 88

*26 * * *febac cac

CQ

h mVI R I

I

=3

33

100*26*10*45*10 0,5*4,5 5( )

50*10V

33

3

513( )

390acR

acR

VI mA

R

8max 3 13 45 58( )cQ acR bcRaI I I A

6 3QR R // 8 8 1 8* *ie fe L feh h R R h R3//8 *26 100*0, 5 8*10050

feh mV

mA

= 272( )

Cng sut tiu tn trn 6Q 2 2

6max 26

802,4(

* 10*272cc

cQ

Q

VP

P W)


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Chn 6 7,Q Q max 6max

ax 6

2 160( )

2 160( )

58(

2 4,8(W)

CE CC

CB CC

C CQ

m CQ

V V V

V V V

I I mA

P P

Vy BJT 6 7,Q Q l : 2SC3244 v 2SA1011 c cc thng s sau :BJT P

C(W) h

feI

C(A)

Tf (MHz) V

CEQ(V)

2SC324425 60 - 200 1,5 100 180

2SA101125 60 - 200 1,5 120 160

III. Tnh phn li: gim mo phi tuyn ta phi phn cc tnh mc ngng cho cc

BJT cng sut bng cch dung D1, D2, D3, RV2.

Chn 6 100feh

66

6

2.50,0025( )

100cQ

bQ

fe

I mAI mA

h

6, 7 6 32*( ) 2*(0,7 0,725) 2,85( )b b beQ RV V V V

a. Tnh 0 1,R C

Ta c ( )L L

R j L R // 0

0

0

1

1( )

1

L

L

L

j L R Rj C

R Rj C

j L R R j C

2 20 0

1 1

LL L L L L

R LR j L R R R R j L R j LR

j C C

2

1 0, 8( )L

L

L

LR

C R R

j LR j L

T 1C ngt mch tng s cao:

ax 15mf kHz do 1 0cX R

1 3ax 1

1 18 1( )2 * 2 *15*10 *8m

C Ff C

Vy chn 01

8( )

1( )

R

C F

b. Chn 1 2 3, ,D D D

Ta c 6, 7 2,85( )b bV V


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Vy chn Q 3 l BJT 2SC2383 gm cc thng s sau

BJTP

C(W) h

fe I

C(A)

Tf (MHz) V CEQ (V)

2SC2383 0,9 60 - 320 1 100 160

d. Tnh 7 8 17, ,R R R

1 40( )2CEQVcc

V V

Chn 8 310*R bQI I

1 6 17 8 3

611* 0,7( )

100EbQ cQ R R bQmA

I I I I R mA

Vy chn 7 4,7( )R k Chn 6 18 3( )R RV V V

6 18 38

1

3 0,71( )

0,7

R R beQ

CQ

V V VR k

I mA

Vy chn 8R = 1(k )

17 1 3 6 18 17 33, 3( )R CC ceQ beQ R R RV V V V V V V V

1717 3

33,347( )

0,7 0,7*10R

VR K

mA

Vy chn 17 47( )R k Mt khc

, 35,7 32,62 2 *0,775

1 32,6

Lhp LP

ht

V Vk

Vi Vi

K

3 73

3 7 3

4,732,6 1 32,6 148( )

32,6 1R

R

R R

V RV

V R V

Vy chn 3RV =500( )e. Tnh v chn 18R

6 18 3( )R RV V V

6 183

3 30,5( )

6CQR R k

I mA

Mt khc phi hp tr khng: 3 1VQ RQZ Z 3 17VQZ R // 18R 3 6 17( 1)beR R R // 18R

17 83 6

17 8

*26* 100* 2( )

6QR RmA

RmA R R

M 6 18 180,5( ) 500 10 490( )R R k R

Vy chn 186

680( )

10( )

R

R


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Tnh chn Q1:1 1 1 1* * 80*0,7 50( W)Q ceQ cQ CC cQP V I V I m

Chn Q10 12 0,122(W)

160( )

0,7( )

Q

ce

C

P P

V V

I mA

Vy chn Q1 l BJT 2SA1013 gm cc thng s sau :BJT

PC

(W) hfe

IC

(A)T

f (MHz) VCEQ

(V)

2SA1013 0,9 60 - 200 1 50 160

f. Tnh v chn R9, R10, R11, R12:

Chn 11

0,710* 10* 0,07( )

100cQ

PCQq

I mAI mA

10 12 7 1 80 3 0,7 76,3( )R R CC R beQV V V V V V

10 12

10 121

76,3

1,09( )0,07R R

PCQ

V V

R R MI

11 9 10 122* ( ) 160 76,3 83,7( )R R CC R RV V V V V V

11 9

83,71,19( )

0,07R R M

mA

Chn R11 = R12 9 10 1,19 1,09 0,1( )R R M

9 100,1R R Tr khng vo ton mch:

9 10 10 10

9 10 10

* (0,1 )100( )

0,1 2*in

R R R RZ k

R R R

210 10 10

2 910 10

10

10

0,1 100(0,1 2* )

100 10 0

161803,4( )

61803, 4( )

R R R

R R

R

R

Chn R10 = 220(K )M 9 10 90,1 0,1( ) 220( ) 261( )R R R M k k

11 1,19( ) 261( ) 929( )R M K k

Vy chn10

11 12

9

220( )1( )

330( )

R kR R M

R k

IV. Tnh ton v ch t C2, C3, C4, C5.

Ta c 2min

10 100,8( )

2 * * 2 *10*100inC F

f Z

Chn C2 = 1 F


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a. Tnh v chn C3:Phi chn C3 sao cho h s hi tip khng ph thuc vo in p

ri trn t m ch ph thuc vo R7 v VR3 do ta chn:3 3

3 3min 3 min 3

1 10 10

10 2 * * 10 2 * * 2 *20*500R R

C

R

V VX C

f C f V

3 159( )C F Vy chn C3 = 470( F )M C4, C5, C6 v R11, R12 to thnh mch lc thng thpChn tng s ct 15( )Cf hz

411

1 10.006( )

2 * * 2 *25*1C

C Ff R M

T C4 ngn dng 1 chiu thot xoay chiu nn:

4 99 4

1 1CX R

R C

49

1 10,019( )

2 * * 2 *25*330C

C Ff R

Chn C4 = C5 = 1 ( )F

6 18CX R ng vi min 20( )f hz

18 186 6

min 18 min 18

10 10

20 10 2 * * * *CR R

X Rf R f R

6 6

10 10117 234

2 *20*680 *20*680C F C F

Vy chn C6 = 220( F )V. Tnh h s khuch i ton mch:

a. H s khuch i in p t tng li n tng ra:S tng ng :

3TQ iZ R // 6, 8Q QR nt6 8(1 * )L fe feR h h

8, 6 3Q Q ieR h R // 8acR

M 626 26

* 100* 5200( ) 5,2( )0,5ie e

mV mV h k

I mA

6, 8

320*1025200 5( )

320 102Q Q

R k

Ni tr ngun dng :

18

2 1 6 5

1 *1

2fe R

ve R ie D

h VR

h V h Z ssR

2

264( )

6T

D

CQ

V mVR

I mA

2 242

1 1 26 2640( ) * 100* 433( )

10 6 6ie feve

k h h k h


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26 264( )

6D D

mVZ

I

100*50040 1 390( )

2*4*100500 5, 2

2*4 100

iR

3TQ iZ R // 6, 8 8 6(1 * )Q Q L fe feR R h h

390 // 5 8(1 100*100) 70( )k

b. H s khuch i t Q3 n loa :3 3

33 3 6

* 26100* 433( )

* 6fe TQ

VQ

ie fe

h ZA

h h R

3

100*705000

433 100*10VQk

A

c. H s khuch i t Q1 n tng li Q3Khi khng c hi tip th xem cc E1 xem nh ni t, nn cc

E1 th R12 ni // VR3.. Nn h s khuch i c tnh theo s tng :17 18 3

0 1 11 1 3 17

/ /*

( / / )VQ

Q

beQ R

R R RV

R V R

Vi 3 3 3 6 3 3 63

26(1 ) * (1 )

VQ beQ

CQ

mAR R R R

I

26100* (1 100)*10 10,5( )

6k

1 1

26* 4( )

0,7beQR k

0 1

100 47 1 / /10, 5974 100*(500/ /4,7)QV

Vy h s khuch i ton mch khi cha hi tip :3

0 1 3 1* 97*5000 485*10Q VQ VQA A A

VI. Tnh mch bo v :a. Kim tra kh nng chi ng ca tng cng sut.

Khi cha qu ti tc l mch lm vic ch thng Vin


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2 2max max 1* max

8 1 2

2* * * 2*2 2

2CC LP L LP LP

Q CC L R

V I P I R I P P P P

2 2

8

8

2 *80 *10 80*10 2*0,5*102 99, 2(W)

2 0,5* 2

50(W)

Q

Q

P

P

Vy khi mch dn bo ha th tn hiu ra loa s b mo dngnhng cc thng s dng in, cng sut tiu tn u tha mn cc iu kin cho ca Q8 , Q9 v R1, R2. Mch vn hot ng bnh thng.

Khi ngt mch ti (RL = 0) th lc ny R1, R2 cng chnh lti ca mch.

1max 2max

1

35,750, 5( )

* 2 0,5* 2LP

R R

VI I A

R

Cng sut tiu tn trn R1, R2 :2 2 3

1 2 1 1max*( ) 0,5*(50,5) 1275*10 (W)R R RP P R I b. Tnh mch bo v .

Tng khuch i cng sut Q89, Q67 b kha v c bo vDng nh qua Q8, Q9 l :

8 9 5( )LP E p E pI I I A

1 2 1* 5*0,5 2,5( )R R LPV V I R A Khi mch hot ng bnh thng Q4, Q9 ca mch bo v

khng hot ng. Lc dng nh qua Q1, Q9 ,R!, R2 l 5(A)Khi dng ti tng them 10% thi dng IbQ6 th cng tng thm

khong 10%6

6 6

6

0,510%* 0,1* 0.1* 0,05( )

100cQ

bQ bQ

II I mA

V

Mch bo v ta chn dng qua Q4, Q5 bng dng ti ln 6QIV

4 5 0,05( )cQ cQI I mA

Ta c : 4 6 8 1 0,7 0,7 2,5 3,9( )ceQ bQ beQ RV V V V V

Cng sut tiu tn trn Q4 :4 * 0,05 *3,9 0,195( W)TTQ C CE P I V mA m

Vy chn Q4, Q50

4

4

0, 39( W)

2* 0,1( )

7,8( )

TT

C CQ

ce ceQ

P P m

I I mA

V V V

Vy ch cp BJT Q4, Q5 l 2SC828 v 2SA564 :

BJTP C(W) h fe I C(A) Tf (MHz) V CEQ (V)

2SA564 0,8 60 - 120 0,2 302SC828 0,5 60 - 320 1 UNI 60


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c. Tnh v chn R13, R14, R15, R16 .1 13 15 8 1* * 5*0,5 2,5( )R R eQV V R I R V

4 34

4

0,050,5*10 ( )

100cQ

bQ

Q

I mAI mA

Chn dng phn p qua R14, R15 gp 30 ln IbQ63

6 30*0,5*10 0,015( )bQI mA

13 15

13 15 34 6

2,5167( )

0,015*10R R

Q Q

VR R k

I (1)

Mt khc : 14 413 15

RbQ beQ

VV V

R R

Chn VbeQ4 = 0.7(V)415

13 15 1

0,70, 28( )

2,5beQ

R

VR

R R V

(2)

T (1), (2)

13 15 13 15

1513 13 15

13 15

167( )167( )

0, 28( ) 0, 28( ) 167( )

R R KR R K

RR R R k

R R

13 15

13 15

1, 28 0, 28 167

167( )

R R

R R K

13

15

120( )

47( )

R k

R k

Vy chn 13 1415 16

120( )

47( )

R R k

R R k

VII. Tnh v kim tra mo phi tuyn

0BE BE BEMU U U sin t

Vi 8 0 1, 2 0,7 0,5( )BEM BEPQ BEU U U V

V thnh phn hi bc hai ln hn cc thnh phn hi bccao khc nn ta c th tnh gn ng :

0,5*100% 480%

4*26mV

Khi c hi tip t u vo th mo phi tuyn s gim i g =1+A0Aht ln :

Ta c : AV0 = 485*10 3 3

3 7

0,10,213

0,1 0, 7

Rht

R

VA

V R

3 301 * 1 482*10 *0,213 103306*10V htg A A

Mo phi tuyn khi c hi tip l :,

3

480%0,005%

103306*10g

, = 0,005 % < 2% . Tha mn yu cu bi.


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