bai giang chuong 2

8/6/2019 bai giang chuong 2

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CHNG 2

MCH KHUCH I CNG SUT M TN


2/38

1. MCH KCS M TN LP A (nhc li)

u im:

Mo phi tuyn t do chn c on c tuyn lmvic ca transistor.

Nhc im:

Cng sut tn hiu ra nh do mch ch lm vic vitn hiu nh.

Hiu sut b do phi phn cc DC trc chotransistor gy tiu tn DC khng mong mun.

Lp A: Transistor ch lm vic trong c chu k ca tnhiu ng vo phi phn cc DC cho transistor.


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2. MCH KCS M TN LP B

Mch KCS m tn lp B transistor ghp yko (push pull) dng bin p.

Mch KCS m tn lp AB transistor ghp bph: mch OTL, mch OCL.

Vn nng cng sut cho mch KCS m tn.

Lp B: Transistor ch lm vic trong 1 bn k catn hiu ng vo Vi tn hiu xoay chiu c 2 bnk ta phi dng 2 transistor.


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2.1. Mch KCS lp B transistor ghp y kodng bin p (tt)

dng ti khng mo

IP1 = IP2 = IP

Hot ng ca 2 transistor phii xng Q1Q2, cc bin p T1v T2 phi c im gia cun thcp v cun scp tng ng.(center-tapped transformer)

Dng trung bnh ca nguncung cp:

IP1

-IP2

IP1

iT2

iT1

Vi

iL

t

t

t

t

iS t

T/2 T 3T/2 2T

IP2

IP1 IP2

P

T

PSAV

IwtdtI

TI

2sin

22/

0


6/38


Cng sut trung bnh phn phi trn ti PL:

LPL

L

PLPLPLL RI

R

VIVP

22

2

1

2

1

2

1

Vi VPL, IPL l bin in p nh v bin dng nh ca ti:

P

S

PPL

P

P

SPL

IN

NI

VN

NV VP , IP l bin p v bin

dng khng mo ng ra cacc transistor.

LP

S

P

L

P

P

SPPL RI

N

N

R

V

N

NIVP

2

22

2

2

1

2

1

2

1

Suy ra:


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Cng sut trung bnh phn phi trn ti cc i:


LP

S

P

L

P

P

SPPL RI

N

N

R

V

N

NIVP

2

max

22

max

2

maxmaxmax2

1

2

1

2

1

T ng ti mt chiu DCLL,ta c bin p ng ra cami transistor cc i:

VPmax = VCC

Suy ra:L

CC

P

SL

R

V

N

NP

22

max2

1

Trng hp NP = NS L

CCL

R

VP

2

max2

1


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Cng sut trung bnh ca ngun cung cp PS:


PCCSAVCCS

IVIVP

2

Cng sut trung bnh cc i ca ngun cung cp PSmax:khi ti tiu th cng sut cc i PLmax.

maxmax

2max

PCCSS

IVPP

LP

Vi:L

CC

P

S

L

PL

P

SPL

P

SP

RV

NN

RV

NNI

NNI

2

maxmaxmax

Suy ra:

L

CC

P

SS

R

V

N

NP

22

max

2


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Cng sut tiu tn ca cc transistor:


LSndissipatiodtntiuttC PPPPP )()(Cng sut tiu tn ca 1 transistor: Pd/2

Cng sut tiu tn cc i ca cc transistor:

)(2

12 222

P

L

P

P

S

L

P

P

SCCLSC Vf

R

V

N

N

R

V

N

NVPPP

Ly o hm, kho st cc tr ta suy ra cng sut tiu tncc i ca cc transistor:

CCP VV

2 th

L

CC

P

SC

R

V

N

NP

22

2max

2


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CC

P

L

P

P

SCC

L

P

P

S

S

L

V

V

R

V

N

NV

R

V

N

N

P

P

42

2

1

2

22

Hiu sut cc i khi PLmax

v PSmax

VPmax

= VCC

%5.784max

maxmax

S

L

P

P

Hiu sut:


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u im: Do mi transistor lm vic 1 bn k tn hiu vo nnmch c th hot ng vi tn hiu c bin lncngsut ra trn ti ca mch ln.

Hiu sut cao.

Nhc im: Mo xuyn tm do ngng dn ca transistor.

Bin p cng knh, t tin. tn hiu ng ra khng mo th cc bin p trongmch phi c cun scp (T2) v th cp (T1) i xng.

Mo tn hiu cun thc cp bin p khi tn hiu vo

cun scp ln do hin tng t tr.



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Bi tp p dng:

Mt mch khuch i cng sut m tn lp B transistorghp y ko dng bin p c dng collector nh v inp nh ng ra mi transistor l 4 (A) v 12 (V). Nguncung cp 24 (V), t s bin p Np:Ns = 1:1. Gi s b qua

cc tn hao dy qun ca cc bin p. Hy tm:a. Cng sut trung bnh phn phi trn ti.

b. Cng sut trung bnh c cung cp t ngun DC.

c. Cng sut tiu tn trn mi transistor.

d. Hiu sut ca mch trong trng hp ny.e. Gi s ti 8 (), tnh cng sut cc i phn phi

trn ti.



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Bi gii:Theo bi: VP = 12 (V), IP = 4 (A), Np/Ns = 1.

a. Cng sut trung bnh phn phi trn ti: PL= 0.5VPIP = 24 (W)

b. Cng sut trung bnh c cung cp t ngun DC:PS = VCCISAV = VCC(0.636IP) = 61.115 (W)

c. Cng sut tiu tn trn mi transistor:

Pd = PS PL = 37.115 (W) 1 transistor: Pd/2 = 18.5575 (W)

d. Hiu sut: = PL/PS = 39.27 %

e. Cng sut cc i phn phi trn ti nu RL = 8 ():

PLmax = 0.5(VCCVCC)/RL = 36 (W).

Bi tp v nh: 2.13, 2.14, 2.15, 2.16, 2.17, 2.18.



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Lp AB: Transistor ch lm vic trong 1 bn k catn hiu ng vo nhng trnh mo xuyn tm taphi phn cc trc cho mi transistor in p mi ni

VBE v in p mi ni VEB ln (0.7 V) khi ctn hiu xoay chiu ng vo th transistor s dn ngay.

Do hn ch ca mch KCS T dng bin p nn trnh cc hn ch th ta khng dng bin p trongcc mch KCS T na mch KCS T khngdng bin p ng ra dng OTL (OutputTransformerLess).

2.2. Mch KCS lp AB transistor ghp b ph


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2.2.1. Mch KCS m tn OTL

iL

iT2

iT1

Re2

D1

Q1

D2

M

Re1

RL

R1

Vi

R2

+Vcc

B1

Q2

Co

CiB2

Phn tch mch v nguyn lhot ng:

in trR1, R2, diode D1, D2: tophn cc trc cho transistor Q1,Q2. Cc dng khc:

VRQ3

Ry

B1

Rx

B2

D1VR

D2

B1

B2

D1

VR

B1

B2

VR

B1

B2


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2.2.1. Mch KCS m tn OTL (tt)

iL

iT2

iT1

Re2

D1

Q1

D2

M

Re1

RL

R1

Vi

R2

+Vcc

B1

Q2

Co

Ci

B2

Cc in trRe1, Re2: l cc in trn nh nhit cho Q1, Q2.

T in Co: cch ly DC ti vi ngra tng cng sut (M) v ng vai trngun cung cp cho Q2 hot ng bn k m ca Vi nn gi l t xutm.

Bn k dngca Vi:Q1 dn, Q2khng dn i

T1

, iT2

= 0.

Bn k m ca Vi:Q1 khng dn,Q2 dn iT2, iT1= 0.

Dng ti: iL = iT1- iT2Dng ngun: i

S= i

T1


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dng ti khng mo

IP1 = IP2 = IP

Hot ng ca 2 transistor

phi i xng Q1Q2 (chntheo cp), R1 = R2, Re1 = Re2, inth ti im gia VM = Vcc/2.

Dng trung bnh ca ngun

cung cp:

P

T

PSAV

IwtdtI

TI

2/

0

sin1


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Cng sut trung bnh phn phi trn ti cc i khi: VP = Vpmax

Xt ng ti DCLL:

QAB 0

VCC/2

2max

CCP

VV

L

CC

Le

LL

R

V

RR

RP

22

max8

1

Nu chn Re


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PCCSAVCCS

IVIVP

Cng sut trung bnh cc i ca ngun cung cp PSmax

:

khi ti tiu th cng sut cc i PLmax.

maxmax

max

PCCSS

IVPP

LP

)(2maxmax

Le

CC

Le

PPRR

V

RR

VI

)(2

2

maxmax

Le

CCSS

RR

VPP

LP


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Cng sut tiu tn ca mt transistor:

LSndissipatiodtntiuttC PPPPP )()(

CC

P

Le

L

eL

PCC

L

P

Le

L

S

L

V

V

RR

R

RR

VV

R

V

RR

R

P

P

21

2

1 22

Hiu sut cc i khi Plmax v PsmaxVPmax = VCC/2

B

Le

L

S

LAB

RR

R

P

P

4max

maxmax

Hiu sut:


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u im:

Mch khng dng bin p nn khc phc cc nhcim ca mch KCS T dng bin p.

Tn hiu ra khng mo xuyn tm.

Hiu sut cao.

Nhc im:

Mch dng t xut m nn lm suy hao tn hiu.

Do suy hao ca t khng ng u theo tn s nn dngt xut m s hn ch nhng tn hiu c tn s thp mch hn ch tn hiu siu trm. Tn s ct thp ca mch:

oLe

CCRR

f)(2

1


23/38


Bi tp p dng:Cho mch KCS T dng OTL c ngun cung cp 20 (V), cc in trR1 = R2 =10 (K), Re1 = Re2 = 1(), RL = 8 (). Diode dng loi Si. T Co = 500 (uF). Gis mch c thit k i xng. Hy tm:

a. Cc dng in qua cc in trR1, R2.b. Cc in th ti cc nt B1, B2.

c. Nu bin p nh ng ra ca mi transistor l 8 (V), tnh cng sut trungbnh phn phi trn ti trong trng hp ny.

d. Cng sut trung bnh ca ngun cung cp cu c.

e. Hiu sut cu c.

f. Cng sut trung bnh cc i phn phi trn ti.

g. Nu Vi c gi tr 5 (Vrms), tnh cng sut trung bnh phn phi trn ti.

h. Tn s ct thp ca mch.


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Bi gii:

a. V DC, cc transistor c phn cc in p B-E ln nn ta cth xem IB1, IB2


25/38


Bi gii: (tt)

d. Cng sut trung bnh ca ngun cung cp:

)(66.5)(

WRR

VV

IVIVP

Le

PCC

PCCSAVCCS

e. Hiu sut: %84.55S

L

P

P

f. Cng sut trung bnh trn ti cc i VPmax = VCC/2 = 10 (V)

)(94.4

8

1 22

max W

R

V

RR

RP

L

CC

Le

LL

g. Cng sut trung bnh trn ti nu Vi = 5 (Vrms) VL= 5 (Vrms)

h. Tn s ct thp: )(37.35)(2

1Hz

CRRf

oLe

C

Bi tp v nh: 2.22, 2.23


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2.2.2. KCS m tn OCL (tt)

IP1

-IP2

IP1

iT2

iT1

Vi

iL

t

t

t

t

iS t

T/2 T 3T/2 2T

IP2

IP1 IP2

dng ti khng mo

IP1 = IP2 = IP

Hot ng ca 2 transistor

phi i xng Q1Q2 (chntheo cp),R1 = R2, Re1 = Re2, VCC = VEE(ngun i i xng) in th tiim gia VM = 0.

Dng trung bnh ca nguncung cp:

P

T

PSAV

IwtdtI

TI

2sin

22/

0


28/38



LPL

L

PLPLPLL RI

R

VIVP

22

2

1

2

1

2

1

Vi VPL, IPL l bin in p nh v bin dngnh ca ti. Gi s VP , IP l bin p v bin dngkhng mo ng ra ca cc transistor. Ta xt Q1 dn:

IP = IPL

Q1

VpL

Vp

+

-

Re

RL

+

-

Le

PPPL

P

Le

LPL

RR

VII

VRR

RV

L

P

Le

LL

R

V

RR

RP

22

2

1


29/38


Cng sut trung bnh phn phi trn ti cc i khi: VP = Vpmax

Xt ng ti DCLL:

QAB 0

VCC

L

CC

Le

LL

R

V

RR

RP2

2

max2

1

Nu chn Re


30/38



PCCSAVCCS

IVIVP

2

Cng sut trung bnh cc i ca ngun cung cp PSmax

:

khi ti tiu th cng sut cc i PLmax.

maxmax

2max

PCCSS

IVPP

LP

Le

CC

Le

PPRR

V

RR

VI

maxmax

)(

2 2

maxmax

Le

CCSS

RR

VPP

LP


31/38


Cng sut tiu tn ca mt transistor:

LSndissipatiodtntiuttC PPPPP )()(

CC

P

Le

L

eL

PCC

L

P

Le

L

S

L

V

V

RR

R

RR

VV

RV

RRR

P

P

42

21

22

Hiu sut cc i khi PLmax v PSmaxVPmax = VCC

B

Le

L

S

LAB

RR

R

P

P

4max

maxmax

Hiu sut:


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Mt s bi tpyucu

Bi 1: Hy chn transistor cng sut, in tr nnh nhit, ngun cung cp cho mch OTL mtknh c cng sut 16 (W), ti loa 8 (), hiu sutthit k 0.6.

Bi 2: Hy chn transistor cng sut, in tr nnh nhit, ngun cung cp cho mch OCL mt

knh c cng sut 20 (W), ti loa 8 (), hiu sutthit k 0.6.

Bi 3: Cc bi tp 2.19, 2.20, 2.21.


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2.3. Vn nng cng sut chomch KCST


LPLL

PL

PLPLLRI

R

VIVP

22

2

1

2

1

2

1

Nng bin dng qua ti.

Nng bin p trn ti mch ghp transistordng cu BTL (Bridge (Balanced) Transistor Line-out).

Nng bin dng v bin p trn ti.


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2.3. Vn nng cng sut chomch KCST (tt)

Nng bin dng qua ti:

Ghp transistor dng Darlington.

Ghp song song transistor: thng s dng trn thc t.

Q2

R1

R4

D3

R3

D2

Q4

+Vcc

RL

Q3

Ci

D1

Q1

Vi

-Vcc

R2

Q2R1

R4

R8

D3

R3

R9

D2

Q4

+Vcc

RL

Q3

Ci

R6

R10

D1

Q1

Vi

R5

R7

-Vcc

R2


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Nng bin in p trn ti BTL:

R1

R4

D3

R4

R1

R3R3

Q1

R2

D3

D2

+Vcc

RL

Q3

-Vi

Ci

D1

Q1

Q3

D1

Vi

D2

Ci

-Vcc

R2

Phi dng mch khuch m o.


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Q2R1

Q4

R4

R8

R6D3

R4

R1

R3R3

R9

Q1

R9

R2

D3

D2

R8

Q4

+Vcc

RL

R7

Q3

R10 -Vi

Ci

R6

R10

D1

Q1

Q3

D1

Vi

D2

R5

Ci

R7

-Vcc

R2

Q2

R5


Nng bin in p v bin dng ti:

Phi dng mch khuch m o.


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Mt sbi tp yu cu:

Bi 1: L lun vic tng cng sut trn ti cho cc dng mchnng cng sut bng phng php nng dng, nng p, nngdng v nng p.

Bi 2: Tm hiu nguyn l mch iu chnh m sc, mchequalizer.

Bi 3: Tm hiu s khi v mch nguyn l ca 1 ampli honchnh.

Q & A

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