Do an KTMDT(Dung de Tham Khao Cach Lam)

7/27/2019 Do an KTMDT(Dung de Tham Khao Cach Lam)

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Mch Khuchi OTL Ng Von

5

PHNTNH TON THIT KMCH OTL NG

VO NCng sut: 60(W)Trkhng vo: Zi= 200 (k)Trkhng loa: PL= 8 ()in p vo: Vi= 0,5 (V)Mo phi tuyn: 0,30 %Bng thng: 30 Hz 15 kHz


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5

Tc dng cucc linh kin:Q1, Q3v Q2, Q4: cc cp BJT ghp Dalington khch i cng sutQ5: BJT khuch i thc.Q7: BJT khuch i u voQ

6, V

R2, D

4, D

5, D

6: to thnh ngun dng.

Q8, Q9: Cc BJT bo vqu ti, ngn mchR1, R2: in trn nh nhit v cn bng dng ra.R3, R4: in trrdng nhit.R15, R16, R17, R18,: in trphn cc cho Q8, Q9..D1, D2, D3, VR1: nh thin p cc BJT cng sut Q1, Q2lm vic chAB.R8: in trn nh nhit cho Q5.VR3, R9, C3: Thnh phn hi tip m mch n nh.R10, R11: Cu phn p cho Q5.R13, R14: Cu phn p cho Q7.

R12, CL: Mch lc ngun loi bcc thnh phn tn scao, chng hintng dao ng tkch trong mch.C1: Tlin lac ng voC2: Tlin lac ng raR20, C4: Thnh phn cn bng trkhng loa tn scao.


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5

1. Tnh ton tng ngun

1.1. Bin tn hiu ra loa:Tn hiu vo mch khuch i c dng: v= V.sint.

Nu hthng l tuyn tnh th tn hiu trn ti l:vL = VLP.sint + VLo

iL = ILP. sint + ILoTrong : VLP, ILPl bin in p v dng ra trn ti

VLo, ILol in p v dng DC trn ti.Do tng cng sut lm vic chAB nn dng tnh v in p tnh trn ti l

khng ng knn:v

L= V

LP. sint

iL = ILP. sintGi VL, ILl in p v dng hiu dng trn ti. Khi ta c:

VL=2

VLP ; IL=2

ILP

Cng sut trn ti l:

PL= IL2.R L= RL. 2

L

2L

R

V=

L

L2

R

V=

L

LP2

R2

V

VLP= 2. .L LP R

VLP= 8.60.2 = 31 (V)V ILP=

L

LP

R

V=

8

31= 3,87 (A).

1.2.in p ngun cung cp:Do Q1, Q2lm vic chAB nn chn hssdng ngun = 0,8.

trnh mo tn hiu ra chn Vcc2VLP.

2.VLP= .VCCVCC=

LPV2=

8,0

31.2= 77,5 (V)

Chn ngun cung cp 80 (V).

2. Tnh ton tng cng suttrnh mo xuyn tm, ng thi m bo hiu sut. Chn Q1, Q2lm vic

chAB. V mch lm vic chAB nn dng tnh collector nm trong khong2050mA. y, chn IEQ1= IEQ2 = 50 (mA).

Dng nh qua Q1, Q2l:IE1p= IE2p = ILp+ IEQ1= 3,87(A) + 0,05(mA) = 3,92(A).

2.1. Tnh ton R1, R2:

R1, R2c tc dng n nh nhit v cn bng dng cho Q1, Q2v tn hiu trnR1, R2cng l tn hiu qua loa:


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ie1= ie2= ILP.sint ( trong khong 0 )

ie

0 2 3 45 t6

Nu chn R1, R2ln th tn hao trn loa nhiu. Do phi chn sao cho tn hiura loa l ln nht.

trnh tn tht tn hiu ngi ta thng chn:

VR1P=20

1VLP= 31.

20

1= 1,55 (V)

Gi trin trR1, R2l:

R1= R2=P1E

P1R

I

V= )(40,0

92,3

55,1

Cng sut tiu tn trn R1, R2l:

PR1= PR2= R1.IL2

m IL2= =

=

0

2

0

222

42

2cos1

2

1sin

2

1 LPLPLP

Itd

tIttdI

( V dng qua R1chmt na chu k)

PR1= PR2= R1 )(49,14

)87,3(.40,0

4

22

WILP ==

Vy chn RR1, RR2l loi: R1= R2= 0,4()/3(W).

2.2. Tnh chn Q1, Q

2:

Dng cung cp trung bnh cho Q1, Q2trong mt chu k

Itb=

LPLP

IttdI = sin2

1

0

Cng sut ngun cung cp: Pcc=Vcc.Icc= Vcc = Vcc.Itb= VccLPI

Cng sut tiu tn trn ti: PL= IL2RL= L

LP

LLP R

IR

I

2.)

2(

2

2 =

* Nu bqua cng sut tiu tn trn R1, R2th cng sut tiu tn trn tip xc Jcphn

cc ngc ca 02 BJT Q1, Q2l:


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5

2PttAC= PCC - PL = VCC .ICC- I2L RL= VCC L

LPLP RII

2

2

(2.1)

o hm hai vt(2.1) theo ILP ta c:

02

== LLPCC

LP

ttAC RIV

dI

dP

=>L

ccLP

R

VI

.

=

Thay vo (2.1) ta c cng sut tiu tn cc i trn mi BJT l:

)W(26,208..4

80

.4

1)

2

1(

2

12

2

2

2

2

2

2

2

max===

L

CC

L

CC

L

CC

ACtt R

V

R

V

R

VP

* Nu khng bqua cng sut tiu tn trn R1, R2th:

)W(49,1)87,3.(4,0.4

1

4

1)

2(. 221

21

2121 ====== LP

LPLRR IR

IRIRPP

Cng sut tiu tn trn 2 BJT Q1, Q2l:

21

2

1

4

1.2

222 LP

LPL

LPccRLccttAC IR

IR

IVPPPP ==

(2.2)

o hm 2 v(2.2) theo ILP:

0..2.4

1.2

2

22

1 == LPLPLcc

LP

ttAC

IRIRV

dI

Pd

=>

ccLPL

VIRR =+ max1)( => )( 1

max RR

VI

L

ccLP +

=

Thay vo (2.2) ta c:

Cng sut tiu tn cc i trn mi BJT l:

))(

.

2

1

)(.2

.

)((

2

12

1

2

21

2

1

2

2

1

2

2

max RR

VR

RR

VR

RR

VP

L

cc

L

ccL

L

ccttAC +

+

+

=

= )W(29,19)4,08(

80

4

1

)(4

12

2

12

2

=+

=+ RR

V

L

cc

Cng sut tiu tn trn mi BJT do dng tnh:

)W(2)(50.2

)(80.

2 ==== mAV

IV

IVP EQcc

EQCEQttDC

Vy cng sut tiu tn trn 1 BJT l:W)(29,21229,19 =+=+= ttDCttACtt PPP

Do Q1, Q2c chn sao cho thomn iu kin sau:

IC> IE1P= 3,92(A)VCEo > Vcc= 80(V)PC > , thng chn PttP C>2. = 2.21,29 = 42,58(W)ttP

Sau khi tra cu ta tm c lChn Q1, Q2l loi BJT 2SC5200, 2SA1943

BJT Pcmax (W) Ic (hfe) VCEo fgh T(0C) loi

2SC5200 150 15A 55160 230V 30MHz 150 SN2SA1943 150 15A 55160 230V 30MHz 150 SP

Bng thng tn hiu ti yu cu l: 30Hz15KHz nn tn slm vic ca BJTphi ln hn 16KHz.


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5

Di tn cho php ca 2SA1943 l:

)(5,187160

30KHZ

ff T ===

Nn BJT 2SA1943c BW = ( 0 187,5KHz) thomn yu cu bng thng(30Hz 15KHz).

2.3. Tnh chn in trrdng R3, R4:

* Chn hfe1= hfemin= 55

Dng )(89,0155

50

)1( min

11 mA

mA

h

II

fe

QEQB =+

=+

=

)(70)(07,0551

92,3

)1( min1

11 mAAh

II

fe

PEPB =+

=+

=

* Theo c tuyn vo ca BJT 2SC5200)(05,0)(501 AmAI QC == )(6,011 VV QEB =

)(92,31 AI PC = )(78,011 VV PEB = Ta c R3, R4l in trrdng nhit: Va n nh im lm vic tnh cho Q3,

Q4va lm tng tc chuyn mch cho Q1, Q2trong min tn sthp.i vi tn hiu 1 chiu: R3, R4cho i qua ddng, cn i vi tn hiu xoay

chiu th R3, R4cho i qua rt t khng btn hao tn hiu xoay chiu trn R3, R4.Do , chn R3, R4phi thomn cc iu kin sau:- Nhhn trkhng vo DC ca Q1, Q2rdng nhit, xin tch dkhi

cc transistor chuyn tdn sang tt.

- Ln hn trkhng vo AC ca Q1, Q2gim tn tht tn hiu. Ngha l:Zin ACQ1


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5

2.4. Tnh chn Q3, Q4:

Dng tnh qua R3: 81,2220

02,06,0

3

111

3=

+=

+=

R

VVI QRQEBQR (mA)

Dng cc i qua R3: 6,10220 56,178,03111

3 =+=+= R VVIPRQEB

PR (mA)

Dng tnh qua Q3:

7,389,081,2133

=+=+= QBQRQE III (mA)

Dng cc i qua Q3:

= 10,6+70 = 80,6 (mA)PBPRPE III 133 +=

Cng sut tiu tn ca Q3do tn hiu xoay chiu to ra:

3

max 2

2

4 tQCCttACR

VP = (2.3)

*Vi LfeinACQtQ RhZRR )1()//( 113 3 ++=

R3// ZinACQ1= )(6,2124220

24.220=

+

Nn )(6,4698)551(6,213

=++=tQR

Thay vo (2.3) ta c cng sut tiu tn ca Q3do tn hiu xoay chiu to ra:

)(345)(345,06,469.4

802

2

max mWWPttAC ===

Cng sut tiu tn DC ca Q3:

QECEttDC 33I.VP =

Vi )(38,39220.81,240.22 3333

VRIV

VV

V QRcc

QRcc

CE ====

Nn = 107,57 (mW)66,3.39,29=ttDCP

Do Ptt= 7,4907,145345max

===+ ttDCttAC PP (mW)

Chn Q3, Q4thoiu kin:IC> IE3P= 80,6 (mA)VCEo > Vcc= 80 (V)PC > , thng chn PttP C>2. = 2.490,7 = 981,4(mW) = 0,981(W)ttP

Sau khi tra cu ta chn c lChn Q3l, 2SC2344 Q4l 2SA1011

Loi BJT0CE

V CI PCmax (W) hfe fT( MHz) Tj(oC)

2SC2344 160V 1,5A 25 60200 100 150


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2SA 1011 160V 1,5A 25 60200 100 150

3. Tng li v mch phn cc ngun dng

3.1. Tnh chn cc diode D1, D

2, D

3v

1RV :

V cc BJT Q1, Q2, Q3, Q4lm vic chAB nn phi dng cc diode phn

cc cho cc BJT

Q1, Q2lm vic chdng tnh 50I QE1 = (mA) th in p VBEca cc BJT

thp chtnh l: 0,6 (V).

QBEQBEQRQRQEBQEBBB VVVVVVV 442221113343 +++++=

= 0,6 + 0,6 + 0,02 + 0,02 + 0,6 + 0,6

= 2,44 (V)t in p phn cc ny, ngi ta c thdng 4 diode D1, D2, D3, D4. Mt

khc, dng qua ngun dng Q6v cc diode t bnh hng bi dng base ca Q3,

Q4lc tn hiu ln. Nn chn PBCQ II 35 >>

m )(31,1)601(

6,80

1 min3

33 mAh

II

f

PEPB =+

=+

=

Q5lm vic n nh v t gy mo ta chn

= PBCQ II 35 )103( chn )(1,1331,1.10.10 35 mAII PBCQ ===

chn cc diode loi 1N914

Tc tuyn ca diode vi ID= 13,1(mA) VD= 0,72 (V)

thay i p phn cc cho cc BJT cng sut, ngi ta dng thay cho D1R

V 4

== DBBR V3VV 431 2,44 - 3.0,72 = 0,28 (V)

Vy )(211,13

28,0

5

11 ===

CQ

VRVR I

VR ,

Chn VR1 l bin tr100 (

) sau hiu chnh li cho thch hp3.2. Tnh chn Q5:

Q5c chn lm vic chA li cc BJT cng sut tng khuch i y

ko. Tng li Q5c ghp trc tip vi tng cng sut, dng tnh c cp bi ngun

dng Q6Do Q5 lm vic chA nn cng sut tiu tn c tnh ch tnh tc l

cng sut tiu tn mt chiu.

Q5lm nhim vkhuch i in p tn hiu cho tng cng sut th philn.

5Q/LZ


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Trkhng ti ca Q5:

Lfefe1be3feieQ/L R)h1)(h1())Rr//(R)(h1(hZ 131335 ++++++=

vi 4,4057,3

25.60

3

33 3 ====

QE

Tbeie I

Vrh ()

5,2750

25.55.

1

111====

QE

Tfebeie I

Vhrh ()

= 405,4 + (1 + 60)(220//(27,5 + 0,4)) + (1+60)(1+55).85Q/L

Z

= 15,57 (K)

Do Q5c ti ln nn dri vo vng bo ho gy mo tn hiu nn phi c R8l

in trhi tip n nh im lm vic. R8l in trn nh nhit cho Q5, R8cng

ln th n nh nhit cng tt nhng tn hao cng sut DC ca n ln nn nh hng

n ngun cung cp.

Chn =8R

V 240

80

40

1

10

18 == RV (V)

(mA)1,1358 == CQR II

R8= )(1531,13

2

8

8 ==R

R

I

V(), chn R8= 150 () 2 (V)

8RV

* Cng sut tiu tn DC trn Q5:

PttDC= 55 CQCEQ I.V

8225 4//2 RQEBQEBQR

ccCEQ VVVV

VV =

= 40 - 0,02 - 0,6 - 0,6 - 2 = 36,78 (V)

PttDC= 36,78.13,1 = 481,81 (mW)

Chn Q5thoiu kin

==>>

=>

=>

W)(963,0)(63,96381,481.2.2

)(1,13

)(80

5

0

mWPPPP

mAII

VVV

ttDCttDC

CQC

CCCE

Sau khi tra cu ta chn c lChn Q5l 2SD401A

Loi BJT P(W) VCE(V) T(0C) fT(MHz) IC(A)

2SD401A 25 150 150 120 2 90400

Chn = 150

3.3. Tnh chn ngun dng Q6:


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* Chn , D2R

V 4, D5

Ngun dng c ni trln c tc dng n nh dng in cho Q5v tng ti cho

Q5. Mun ni trngun dng ln th chn Q6l BJT c ln v dng tnh

56 CQCQ II = = 13,1 (mA)- Chn 2 diode l loi 1N914

- Chn = 13,1 (mA) V654 CQDD

III == D= 0,72 (V)

62 Q/EBVR VV + = 2VD

62 /2 QEBDVR VVV = = 2.0,72 - 0,6 = 0,84 (V)

VR2 = )(64)(064,01,13

84,02 == mVR .

Chn VR2 l bin tr220 () ri hiu chnh li cho thch hp.- Do Q6lm vic chA nn cng sut tiu tn chyu l cng sut 1 chiu.

PttDC= VCE66. ICQ6

VCE6= VCC- VVR2- VBE/Q3- VBE/Q1- VR1-2

VCC

=2

VCC - VBE/Q3 - VBE/Q1- VVR2-VR1Q

= 40 - 0,6 - 0,6 - 0,84 - 0,02 = 37,9 (V)

PttDC= VCE6.ICQ6= 37,9. 13,1 = 497 (mW)

Chn Q6thoiu kin:

==>>

=>

=>

W)(994,0)(994497.2.2

)(1,13

)(80

5

0

mWPPPP

mAII

VVV

ttDCttDC

CQC

CCCE

Saukhi

tracu ta chn c l

Loi BJT P(W) VCE(V) IC(A) T(0C) fT(MHz)

2SB546 25 150 2 150 5.0 40200

Chn Q6l 2SB546

VR6= VCC - (VD4+VD5) = 80 - 2.0,72 = 78,56 (V)

IR6= 13,1(mA)


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R6= )(61,13

56,78= K ,

Chn R6= 5,7 (k).

4. Tnh tng vo

ICQ5= 13,1(mA) chn hfe5= 150

IBQ5= )(09,0150

1,135 mAICQ ==

Chn ICQ7>> IBQ5khng nh hng n VAv n nh im lm vic cho Q5

Chn ICQ7= 10.IBQ5= 10.0,09 = 10,9(mA)

4.1. Tnh R9, R10, R11:R9cng ln th tc dng hi tip m dng mt chiu cng ln, im lm vic

ca Q7cng n nh.

in p 1 chiu VR9c chn: VR9= (10

2

10

1 ) VA

Chn VR9=10

1VA=

2

V

10

1 CC = 4(V)

IR9= ICQ7R9= )(44,49,0

4= K ,

Chn R9= 4,4(k) VR94(V).

VR11= VBE/Q5+ VR8= 0,6 + 2 = 2,6 (V)

IR110,9(mA)

R11=9,0

6,2

11

11 =R

R

I

V= 2,88(k)

Chn R11= 4,4 (k) VR114(V).

VR10 =2

VCC - VR9- VCE/Q7- VR11

* Q7khuch i khng bmo v bin in p ra ln chn Q7hot ng

chA, im tnh nm gia ng ti ng:

VCE/Q7= 204

80

4 ==CC

V(V)

)(1242044010 VVR ==


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5

)(4,139,0

1210 == KR

Chn R10= 14,4 (k) VR1013(V) VCE/Q7= 19(V).

4.2. Tnh Q7:Q7lm vic chA nn cng sut tiu tn l cng sut tiu tn mt chiu.Nn ta c:

)(1,179,0.19. 77 mWIVP CQCEQttDC ===

* Chn Q7 thoiu kin:

CEV >2

VCC = 40 (V)

IC> ICQ7= 0,9 (mA)P > PttDC , thng chn P > 2. PttDC= 2.117,1 = 34,2 (mW)Sau khi tra cu ta chn c l

Chn Q7l 2SA1015Loi BJT P(mW) fT(MHz) VCE(V) IC (mA) T(

oC) 2SA1015 400 80 50 150 125 240

4.3. Tnh R12, R13, R14:)(3644097/ VVVV RAQE ===

)(3,357,0367/7/14 VVVVV EBQEQBR ====

)(9,07 mAICQ =

)(10.75,3240

9,09,0 3

77 mAh

Ife

BQ===

* Chn IR14=10.IBQ7 = 10.3,75.10-3 (mA) = 375.10-3 (mA)

R14= 0375,03,35

14

14 =R

RI

V = 941 (K).

Chn R14= 900 (k).

Ta c

Zin=

1413

1413.

RR

RR

+= 200 (K)

)(1,257700

180000180000200900 131313 === KRRR

Chn R13= 280 (k).

* VR12= Vcc- VR13- VR14= 80 - (280 + 560).375.10-3

= 35,75 (V)

R12=0375,0

75,35

1312

12 == RRR

II

V = 953 (K )

5. Tnh hskhuch i in p, trkhng vo


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5

5.1. Hskhuch i in p ca Q5:

B5

rbe5

R8

rco5 rco6 ZinB3B4 rco5 rco6 ZinB3B4rbe5*

r*be 5 = rbe 5 + (1 + hfe 5 ).R8= rbe 5 + (1 + 150). 153

Vi: rbe 5 = hfe 5 )(290

1,13

25150

5

=CQ

T

I

V

rco 5 = rce 5

++

85

85.1Rr

Rh

be

fe . Chn rce 5 = 104250C.

rco 5 = 104 )(53010.80,52)

153290

153.1501( 4 =

++ K

rco 6 = rce 6

+++

266

26

//2

.1

Rbed

Rfe

VrRr

Vh. Chn rce 6 = 10

4250C.

Vi rbe 6 =

6

6.

CQ

T

I

V = 150. 286

1,13

25 () ; )9,1( ==

D

Td

I

Vr

rco 6 = 104

++

+64286)(7,5//8,3

64.1501

K

104

++

+642868,3

64.1501 = 281,3 (k)

* ZinB3B4= ZL/Q 5 =30,87(k) Zr/Q5= ZinB3B4 // rco5// rco630,87 (k).

Vy hskhuch i in p ca Q5

Av5= hfe 5 . 5,19015,2487,30

.1505

55 ==

be

r

r

QZ

5.2. Hskhuch i in p ca Q7:


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5

rbe 7 = hfe 7 )(94,69,025

.2507

== kI

V

CQ

T

Zin= R13// R14= 5,213900280

900.280=

+(k)

* Khi cha c hi tip: ZinQ7= (R13// R14) // rbe 7V (R13// R14) >> rbe 7 ZinQ7rbe 7 = 6,94 (k)ZVQ5= r be

5

+ (1 + hfe5

).R8= 290+(156).15324,15 (k)(R11// ZvQ 5 ) + R10= (4,4 // 24,15) + 14,4 18,12 (k)V rce 7 7* Hskhuch i ca Q

>> 18,12 (k) ZrQ = 18,12 (k)

7khi cha c hi tip:

AV 7 = hfe 7 . 65394,612,18

.2507

7 =inQ

rQ

Z

Z

* Hskhuch i in p ca Q1, Q2, Q3, Q4Do hai cp Q1, Q2v Q3, Q4mc theo kiu C - C

AVQ 31 42* Hskhuch i vng hton mch= 1; AVQ = 1

Av= AV 7 . AV 5 . AV 1 3 42* Hskhuch i khi c hi tip

AV = 653.190,5 = 12396.

Avht=93

3

RV

V

R

R

+

* Hskhuch i ca mch

Av= 8,435,0.2

31

2.1 ====

+in

LP

in

L

vhtv

V

V

V

V

V

AA

A

Avht= 0,022 99

022,0.022,0022,033

3

3 RVVRV

V RRR

R +==+

rbe7

ZV/Q5

R14

B7

R Rr13 10ce7

R11

R9VR3


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5

VR3 = 99 ()chn VR3* Trkhng vo ca mch:

l bin tr470 ( ),Ri hiu chnh li cho thch hp.Khi cha c hi tip, trkhng vo ca mch chnh l trkhng vo ca Q7:

Zv= R13// R14// rbe 7 v R13// R14>> rbe 7 Zvrbe 7Khi c hi tip, trkhng vo tng (1 + K.Kht) ln

Zv= rbe 7 (1 + K.Kht) vi Kht= 022,044009999

93

3 =+

=+RV

V

R

R

Zv= 6,94 (1 + 2736,71) = 18,99(M) >> rbe 7

6. Mch bo vqu ti6.1. Trng hp qu ti: Mch qu ti khi Vin> 500(mV)

VLP> 31 (V)* Trng hp qu ti ln nht khi Q1, Q2xp xdn bo ho

VA= VLP=2

80

2 =CC

V= 40 (V)

Dng nh qua ti: ILP=

8

40=

L

LP

R

V= 5 (A)

Cng sut loa:

PL= L2

L RI = RL2

2LPI =RL.

8.2

40

22

22

2

2

==L

LP

L

LP

R

V

R

V= 100 (W)

Cng sut ngun cung cp:

PCC= VCC .Itb = 60.

5.80

'2=LP

I= 127,3 (W).

Cng sut tiu tn trn in trR1, R2

PR1= PR2= R2.8

'2LPI =0,48

52=1,25 (W)

* Do R1, R2l 0,4 /3W R1, R2khng bnh thng.* Cng sut tiu tn trn BJT Q1,Q2:

2PC= PCC - PR1- PR2 - PL= 127,3-2.1,25-100 = 24,8 (W)

PC=2

8,24=12,4(W) < 150(W)

Q1,Q2hot ng bnh thng.6.2. Trng hp ngn mch ti:

Khi ngn mch ti: R1, R2l ti ca mch.Trng hp nng nht l khi my ang lm vic bnh thng th ngn mch ti,

p xoay chiu cc i ln lt t ln R1, R2. Dng qua R1, R2 l:


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IR1= IR2=4,0.2

31

2 1=

R

VLP = 38,75 (A)

* Cng sut tiu tn trn R1, R2PR1= PR2= I

2R1.R1= (38,75)

2.0,4 = 600,62 (W)

* Cng sut do ngun cung cp:PCC= VCC.Itb= VCC.

LPI =

4,0.

31.80.

1 =

R

VV LPCC 1975 (W)

* Cng sut trn BJT Q1

PC=2

2.62,6001975

221 =

RRCC PPP =386,88(W) > 150(W)

Cc BJT Q1v Q2sbnh thng.PR1= PR2= 600,62 (W) R1, R2cng bnh thng.6.3. Tnh mch bo vQ8, Q9:

Bnh thng, mch bo vQ8, Q9 ngt khi mch khuch i cng sut lm

vic, khng nh hng n hot ng ca mch.Khi ngn mch, dng qua R1, R2tng lm Q8, Q9dn, dng ICQ8, ICQ9tng

IB3, IB4gim,

Dng nh qua Q1, Q2l 3,92(A)chn dng mch bo vhot ng: I'E1P= 3,92 + 10%.3,92 I'E1P= 4,31 (A)

V'R1= R1.I'E1P= 0,4.4,31 = 1,72 (V)* Chn I

CQ8= I

CQ9= 1mA

VCE8= VBE3+ VBE1+ VR1= 0,6 +0,6 +1.72 = 2,92 (V)PttDC= VCE8.IC8= 2,92.1 = 2,92 (mW)

* Chn Q8, Q9tho:

)(84,592,2.2.2

)(1

)(84,5)(92,2.2.2

8

8

mWPPPP

mAII

VVVV

ttDCttDC

CQC

CECE

==>>

=>

==>

Sau khi tra cu ta chn c lChn Q8:2SC458,Q9:2SA1029

Tn Loi BJT P(mW) fT(MHz) T(oC) VCE(V) IC(mA)

Q8 2SC458 200 230 150 30 100 100500Q9 2SA1029 200 230 150 30 100 100500

IBQ8=min

8

CQI =100

1= 0,01 (mA).

* Chn IR15>> IBQ8IR15= 15.IBQ8= 15.0,01= 0,15 (mA)* chlm vic bnh thng VLP= 31(V); VR1P= 1,55(V)* Q8, Q9ngt mch c ttChn VBEQ8= VBEQ9= 0,4 (V)

R15+ R17= VR1P/IR15= 1,55/0,15 = 10,3 (k).R17= VBEQ8/IR15= 0,4/0,15 = 2,67 (k), chn R17= 2,7 (k).


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R15= 10,3 - 2,7 = 7,6 (k)Chn R15= 7,6 (k ).

Chn R15= R16= 4,3 (k); R17= R18= 2,7 (k)

Q8, Q9dn bo ho khi VBE8= VBE9= 0,7 (V)VR1P=

67,2

)6,767,2.(7,0)(

17

17158 +=+

R

RRVBE = 2,69 (V)

IR1= )(72,64,0

69,2

1

1 AR

V PR ==

Khi dng tng ln 6,72 (A) th mch bo vlm vic.

7. Tnh cc t- TC1l tlin lc vi ng vo. tn hiu khng bgili trn t

Chn Xc1 = )(67,105,213.201.201 == KZin

Do tn sm thanh m n cho qua l 30Hz 15kHz nn tn sct ca lc phi nh

hn 30Hz.

Chn tn sct ca lc l 30Hz

)(5,0)(10.49,010.67,10.30.2

1

2

1)(10.67,10

2

1 63

11

3

11 FFfX

Cf

XCC

C

======

Chn C1= 0,5 F.- TC2v R12to thnh mch lc ngun, khghp k sinh gia tng ra, tng li

v tng vo n, n nh chlm vic ca mch, chng dao ng tkch.

Chn Xc2 = )(3,95)(953.10

1

10

112 == KKR .

Xc2= )(056,010.3,95.30.2

1

2

132

2

FCfC

==

Chn C2= 0,1 (F).

-TC3ct thnh phn AC cho cu hi tip VR3, R9. Chn C3 sao cho tshitip chphthuc vo VR3, R9v st p AC trn C3nhhn nhiu so vi VR3.

* Chn XC3= )(9,999.10

1

10

13 ==RV

* XC3= )(536)(10.35,59,9.30.2

1

2

1

2

1 4

33

3

mFFfX

Cf CC

====

* Chn C3= 510(F).

- TCLl tlin lc. tn hiu khng bgili trn CL tn sthp


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Chn XCL= )(28.4

1

4

1==

LR

XCL= )(10.65,22.30.2

1

.2

1

.2

1 3 FXf

CCf CL

LL

===

* Chn CL=2600(F).

8. Mch cn bng trkhng loaLoa c cu to l mt cun dy ng mnh nn trkhng loa l ZL=RL+jL

Trkhng loa phthuc tn s. tn scao, trkhng loa ln nn dpht sinh

dao ng. khc phc, ta mc thm mch Zobel gm R20v C4song song vi loa.

Thnh phn tn hiu c tn scao sthot qua tC4xung mass.

tn scao, XLtng nhng XC4gim nn RLkhng i.

ZL= (R20+ )Cj

14

// (RL+jL)

=

LjCj

1RR

C

L

Cj

RR.LjRR

LjRCj

1R

)LjR)(Cj

1R(

4

L20

44

L20L20

L

4

20

4

20 L

+++

+

++=

++

+

+

+

ZL khng phthuc vo tn sZL= RL.

R20.RL+jL.R20 + LLLLL RLjcj

RRRR

C

L

cj

R..

4

220

44

+++=+

2

L

4

2

L

4 R

LCR

C

L==

jL. R20=jL.RL R20= RL = 8

V L ca loa thng rt nh, c0,1 H.

C4= ).(10.56,164

10.1,0 96

F

= Chn C4 = 1,3 (nF).

tn scao, tngn mch, cng sut trn R20ln nhng khng cn cng sut

chu ng ca R20ln v nu c th cng l nhng xung hp, bn nh.

Do tn scao, tngn mch, nn ngi ta thng chn R20ln hn RLmt t

trkhng ti khng i, chn R20= 8,4 ().

9. Kim tra mo phi tuyn Trong mch cc BJT lm vic chA, chc Q1, Q2lm vic chAB

nn mo phi tuyn trong mch chyu do Q1, Q2quyt nh.

Khi tn hiu vo hnh sin v Vin= 500(mV). Lc ny p t ln tip gip BE ca Q1:

vBE1(t) = VBE1Q+ VBE1m.Sim t


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Vi VBE1Q= 0,4 (V)

VBE1m= VBE1P- VBE1Q= 0,8 - 0,4 = 0,4 (V)

Dng ICca Q1, Q2: IC= ICo. TBE

V

v

e

ICOl dng rKhai trin biu thc ICdi dng chui Taylor:

.....!3!2

132

+++=xx

ex

....).2

sinsin1(.

2

22111 +++=

T

mBE

T

mBE

T

QBECoC V

tV

V

tV

V

VII

Do mo phi tuyn chyu l do cc thnh phn hi bc hai gy ra. Loi bcc

hi bc cao v thay sim2t =2

2cos1 t

)4

2cos.

4

sin1(. 2

21

2

2111

T

mBE

T

mBE

T

mEB

T

QBECoC V

tV

V

V

V

tV

V

VII

++=

Theo nh ngha mo phi tuyn

m

m

iim

I

I

1

1

2== trong : I1m: thnh phn dng cbn

Iim: l cc thnh phn hi* Mo phi tuyn chyu do thnh phn hi bc hai gy ra.

T

mBE

mBE

T

T

mBEm

mV

V

VVVVII .4.41

1

1 2

2

12/ ===

* Khi cha c hi tip:

410.25.4

4,0

4 3====

T

Bm

V

V

* Khi c hi tip:

KRg L ).1(

/

+=

:

Trong : K su hi tip.

28083,44

124396).1( ==+= VV

htV AAAAK

T

QE

QE

Tfe

fe

be

fe

V

I

I

Vh

h

r

hg 1

11

1

1

1 ===

00075,0

2808).8.21(

42

25

50=

+=== g < 0,30%

thomn yu cu thit k.


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10. Tnh ton btn nhit cho cc BJT cng sutKhi chuyn thnh cng c ch, mt phn cng sut slm nng cc BJT cng

sut. Nu nhit tng ln qu nhit cho php th cc BJT dbhng.Gisnhit mi trng xung quanh (bnh thng loa) l 500C

Nhit ton phn: K=1

max

Qtt

mtj

PTT

=

W)(2550150

0

CCo

= 4oC/W.

PttQ1= 21,29(W) Ly P= 25 (W)Nhit trK khi c cnh tn nhit

K = Kcm+ Kvc+ KtvKcm: nhit trtcnh n mi trngKvc: nhit trtvn cnhKtv: nhit trttip gip n v.

wcKtv

01150

150==

Chn ming m bng mica dy 0,4 mm.c kvc=2cw

Ptt= wcKK

P

TTK

KKK

TTvctv

tt

mttgcm

cmvctv

mttg 011225

50150=

=

=

++

Chn cnh tn nhit c hnh vung c din tch cnh tn nhit nhsau

S= 210001

1000cm=

Ta nhn thy din tch bkng knh v vy ta phi dng cnh tn nhit gm nhiu cnh

c xc nh nhsau.

Ttb=Ptt.Kcm=25.1=25c

Da vo c tuyn Kcm v L ta xc nh nhsau

mmLK WC

cm 13010

==

Scnh n n= 81,1425.15,01304,020

130=++

Vy ta chn phin tn nhit c

n =15 cnh


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L =130mm

Do an KTMDT(Dung de Tham Khao Cach Lam)

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