do an ocl don
TRANSCRIPT
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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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SVTH: M.TunV. Quyn T.Vin Trang- 2 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 4 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 5 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 6 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 7 -
+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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SVTH: M.TunV. Quyn T.Vin Trang- 8 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 9 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 10 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 11 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 12 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 13 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 14 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 15 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 16 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 17 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 18 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 19 -
(-)
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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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 20 -
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 21 -
+
-
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 22 -
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 23 -
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 .
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 24 -
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 (
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 25 -
+
-
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 26 -
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:
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 27 -
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.
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 28 -
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 30 -
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 31 -
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
)
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 32 -
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 33 -
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 34 -
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 35 -
= 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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 36 -
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
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n k thut mch
SVTH: M.TunV. Quyn T.Vin Trang- 37 -
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
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SVTH: M.TunV. Quyn T.Vin Trang- 38 -
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
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SVTH: M.TunV. Quyn T.Vin Trang- 39 -
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 :
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SVTH: M.TunV. Quyn T.Vin Trang- 40 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 41 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 43 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 44 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 45 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 46 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 47 -
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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SVTH: M.TunV. Quyn T.Vin Trang- 48 -
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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S nguyn l