Chuong 1. mdt2
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Transcript of Chuong 1. mdt2
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7/25/2019 Chuong 1. mdt2
1/16
Chng 1:p ng tn sthp mch khuch i
Chng 1
P TUYN TN STHP CA MCH
KHUCH I GHP RC1.1. p ng tn sca mch khuch i Mi mch khuch i u c mt khong tn shot ng nht nh, gi
l bng thng (Band width) hot ng ca hthng.K hiu: BW = [fH fL] (Hz)
Mch khuch i c c trng bi hm truyn hskhuch i, cgi l Aihay Av. p tuyn bng thng ca mch khuch i
)(dBAx
f(Hz)
A
2
mA
fL fH
Midband
1.2. Phng php kho st p ng tn sca mch khuch i
Cc bc kho st:
i. Bc 1: Vmch tng ng vng tn shot ngii. Bc 2: Thit lp biu thc ca hm truyn hsK
iii. Bc 3: Vbiu Bode cho tn sv pha
V d: Cho mch in tng ng sau
0
Vo
I1ieC
R1
R2Rc
V1vi
+
-
Ta c
jwC
1R
jwC
1R
R
1Rc
v
i
i
Rci
Vi
VoA
2
2
1
i
e
e
e
v
+
+
===
1
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7/25/2019 Chuong 1. mdt2
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Chng 1:p ng tn sthp mch khuch i
)R//R(jwC1
)jwCR1(
RR
Rc
jwC
)RR(RR
)jwC
1R(
RcA21
2
2121
21
2
v ++
+=
++
+=
Vy)R//R(jwC1
)jwCR1(
RR
RcA
21
2
21
v ++
+=
t)R//R(C
1w,
CR
1w
21
2
2
1 ==
=>
)w
wj1(
)w
wj1(
RR
RcA
2
1
21
v
+
++=
Vy:2
2
2
1
21
v
)w
w(1
)w
w(1
RR
RcA
+
++= (1)
v )w
w(arctg)
w
w(arctg
21
= (2)
V u Bode cho tn stn hiu
Khai trin decibel ta c:
bi
( ) 22
2
121
vvlg20A + )
w
w(1lg20)
w
w(1)
RR
Rclg(20Alg20dB +++
== (dB)
Hay210v
AAA)dB(A ++=
Xp xgn ng
>>
=
=
+=
)ww(wwlg20
)Ww(dB3
)0w(dB0
)w
w(1lg20A
1
1
1
2
1
1
Biu Bode cho A1
2
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7/25/2019 Chuong 1. mdt2
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Chng 1:p ng tn sthp mch khuch i
Biu Bode cho cc A0, A1, A2
)(dBA1
w(rad/s)
Biu Bode tng ca Av
Biu Bode cho pha tn hiu
Ta c )w
w(arctg)
w
w(arctg =
21
t21 +=
Xp xgn ng
>
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7/25/2019 Chuong 1. mdt2
4/16
Chng 1:p ng tn sthp mch khuch i
Biu Bode cho gc pha tng
1.3. Kh BJT ghp RC
2 loi
ass cc E transistor (Emitter bypass capacitor). Kho
Lu
o st p ng tn sthp ca mch khuch i
Phng php kho st: n gin cho vic kho st ta tch raghp RC ring bit
Ghp tbypst p tuyn tn strn mch ny nhcc bc nu trn
Ghp tng vo v ra (Coupling capacitor)
:
Do c im chc nng ca mi loi tghp trong m n quyt nh
n cc gi trC coupling sao cho
V d: kho st p tuyn tn sthp c a mch khuch i sau
snh hng n hot ng ca mch khuch i. TEmitter quytnh tn sct di ca mch. Tcoupling chng vai tr l tlien
lc gia ng vo v ra.Khi thit kmch ta ch
)coupling(L)terBypassemit(Lff >>
45o
101w 1w 110w
10
2w2w 210w
w(Rad/s)
- 45o
- 90o
90o
101w 110w
102w
210ww(rad/s)
o
4
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Chng 1:p ng tn sthp mch khuch i
Vcc
0
Re
RcCc
R2 Ce
R1
Cb
ri RLi
1K
602k
2k
10k 100
Beta = 100hi e = 1K
a.
p ng ca tBypassBqua nh hng ca cc tCoupling bng cch ni tt chng, xt mchtng ng tn hiu nhnhsau:
iibi
bi
i
i
Rc1k
Re60
hie
Rb1k
ri10k
RL100Ce
iibi b
i
i
ri10k
hie
Rb1k
Rc1k
Re60
CeR10
bi
ib
iL
iBi iRr )//(Rc1k
Re60
ri//Rb hie
RL100Ce
5
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Chng 1:p ng tn sthp mch khuch i
Rc
1k
RL
100Re60 Ce
iL
ie
iBi iRr )//(
)//( ieBi hRr +
Ta c,i
bii
bii
e
e
L
i
L
ii
)R//r(i
)R//r(i
i
i
i
i
iA ==
)R//r(
Cj
1
R
Cj
1R
)hr//R(
1
RR
Rbi
ee
e
e
ieiB
LC
C
+
+
+
+=
e
e
e
i
ei
e
e
LC
biC
Cj
R
Cj
RRR
Cj
1R
RR
)R//r(R
++
+
+
= , vi
=+
= 20)hr//R(
R ieiBi
[ ])R//R(jC1)RR(RCj1
RR
)R//r(R
eieei
ee
LC
biC
++
+
+=
Hay,[ ])R//R(jC1
RCj1
)RR)(RR(
)R//r(RA
eie
ee
eiLC
biC
i ++
++
=
2
1
j1
j1
)6020)(100K1(
)K1//K10(K1
+
+++
= , Vi
==
==
s/Rad23.133C)R//R(
1
s/Rad3.33CR
1
eie
2
ee
1
6
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7/25/2019 Chuong 1. mdt2
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Chng 1:p ng tn sthp mch khuch i
Vy,
2
1i
j1
j1
33.10A
+
+= , hoc
2
1i
f
fj1
f
fj1
33.10A
+
+= , vi
=
=
Hzf
Hzf
25.21
3.5
2
1
2
2
2
1
dBi1log201log2020A
++
++= (dB)
li trung tn (hskhuch i trung tn):c nh ngha lgi trca hskhuch i ti tn sct di ca mch, hay
.)f(AALiim
=
Biu Bode
20
)(dBAi
33.3
20dB/decade
133.23
Mid-bank gain
32dB
L
w(rad/s)
)(dBAi
w(Rad/s)
w1 10w1
20
20dB/decade
w2 10w2
A1
A2
A0
Theo tnh cht ca biu Bode tn sct di ca mch
, hays/Rad23.133L= Hz25.21f
L=
Biu pha
45o
101w 1w 110w
10
2w
2
w2
10ww(Rad/s)
90o
- 45o
- 90o
Biu pha tng hp
7
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7/25/2019 Chuong 1. mdt2
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Chng 1:p ng tn sthp mch khuch i
ng ca tcouplingb.pypass bng cch ni tt n, ta xt mch tngBqua nh hng ca tB
ng tn hiu nhnhhnh di.
Dng php bin i tng ng Thevenin ta c
(error: ngun p, thiu k hiu i0)
Ta c,i
ii
ii
0bbLLi
iiiiiA
0bbi i
ir
iriiii
==
i
ieb
b
iieB
B
C
LC
C r
h//RCj
1r
1
hR
R
Cj
1RR
R
+++
++
=
[ ] [ ])h//Rr(Cj1)RR(Cj1CjCj
hR
rRR
iebibCLC
bC
ieb
ibC
++++
+=
45o
10
1w 1w
110w
10
2w
2w 210w w(rad/s)
90
- 45o
- 90o
1
2
o
027
Rc1k
R1
Rb1k
ri10k
hie
Cb Ceib
bi
iL
ii
Cb Ceri
Rc1k
hie RL
100Rb
1k
10k
ib
bi
iL
iiir
8
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7/25/2019 Chuong 1. mdt2
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Chng 1:p ng tn sthp mch khuch i
n gin ta chcn xt cc tn sct di:
)RR(CCLC
1L +1
= , v)h//Rr(C
iebib
2L +1
=
Do mc ch thit k l cc tCoupling chng vai tr tlin lcgia ng vo v ra. N mch hot ng n nh, tc cc tn sct do
ca tan
tcoupling skhng nh hng n tn sct di ca mch (fLng vitBypass), th:
2L1L,
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7/25/2019 Chuong 1. mdt2
10/16
Chng 1:p ng tn sthp mch khuch i
0
1MEGVi RdRs
r i
RgCs
gsmvg
G S
rds
D iLi0
Mch tng ng thevenin
0
1MEGVi Cs
Rs Rd
r i
Rg
D iLi0G S rds
gsdsm vrg
+ _
Ta c, , v dtsggs
vvv =dsm
rg=
0
RdRLCs
Rs
S rds D iLi0
iv
+ _+_
Sv
Dng php bin i tng ng Thevenin cho on mch MO
0
Cs
Rs
RL
Rd
Srds iv
+ _+_Sv
Di0 iLM
10
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7/25/2019 Chuong 1. mdt2
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Chng 1:p ng tn sthp mch khuch i
0
Rd
RLsR)1( +
)1( +sC
iv+ _ Di0 iLrdsM
Vy,i
i
i
0
0
LL
i
L
vv
v
v
i
i
iR
v
vA
==
( )
+++
+=
)]Cj
1//(R[1R//Rr
1
RR
RR
S
SLddsLd
dL
SS
SdLds
dL
SS
S
dLds
dL
CRj1
R
)1(
R//Rr
1)R//R(
)1(
CRj1
R)1(R//Rr
1)R//R(
++
+++
=
++
++=
t)1(
R//RrR dLds
i ++=
Ta c,
SSiSi
SS
dL
SS
S
i
dLvRCRRjR
CRj1)R//R(
)1(
CRj1
RR
1)R//R(
)1(A
+++
+=
+++
=
S
Si
iS
SS
Si
dL
SiSSi
SS
dL
CRR
RRj1
CRj1
RR
R//R
)1(CRRjRR
CRj1)R//R(
)1(
++
+++
=+++
+=
Hay,SSi
SS
Si
dLvC)R//R(j1
CRj1
RR
R//R
)1(A ++
++=
Vit gn li ta c:
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Chng 1:p ng tn sthp mch khuch i
( )
=
=
=
SSi
2
SS
1
dLdsmvm
C)R//R(1
CR
1
R//R//rgA
2
1
vmv s
sAA
+
+= , trong
b.p ng ca tghp cc mng
Mch tng ng tn hiu nhnhhnh v
L
d +
RV-
i
-
Vi
g
r
Lgs V
Vds
r
ggs R
m
+ Cd
R
Hm truyn ca mch:
i
gs
gs
L
i
L
vv
v
v
v
v
vA ==
ig
g
d
Ldds
d
L
Lm
vrR
R
sC
1R//R//r
sC
1R
RgA
+
+
+=
V rt ln do gR 1rR
R
ig
g +
L
vmvs
sAA+
= . Trong :
( )
( )
+
=
=
ddsLd
L
dLdsmvm
R//rRC
1
R//R//rgA
c.
p ng ca tghp cc cng
Mch tng ng tn hiu nhnhhnh v
12
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Chng 1:p ng tn sthp mch khuch i
L
i
d
+
R
V
Cd
- -
Vi
g
rL
gs
VV
ds
r
ggs R
m
+
R
Hm truyn ca mch:
( )
dig
g
Lddsmv
i
gs
gs
L
i
L
v
sC
1
rR
RR//R//rgA
v
v
v
v
v
vA
++
=
==
L
vmvs
sAA+
= . Trong :
( )
( )
+=
=
gid
L
dLdsmvm
RrC
1
R//R//rgA
Gi tr thng rt ln nngR L rt nhv vy Cd nh hng rt t n mo tn s
thp, mo chnh hng do Cs gy ra.
13
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Chng 1:p ng tn sthp mch khuch i
Bi tp chng 1
1.1Cho mch nhhnh:
I iRe
Rb RL
100
Ce
5uF4K 1K
IL
Cho bit .50h;K1h feie ==
a. Vmch tng tn hiu nhtn sthp
b. Tm hm truyn )s(i)s(i
Ai
Li =
c. Vbiu Bode cho bin v pha
1.2Vbiu Bode cho bin v pha ca hm truyn:
( )( )( )( )( )( )
++++++
=2000s12s2s
400s300s10s10A 4
1.3Cho mch nhhnh:
VCC
I i
I L
10uF4K
10K 10K1K
Cho bit .100h;K1h feie ==
a. Tm hm truyn)s(i
)s(iA
i
L
i=
b. Vbiu Bode cho bin v pha
1.4 Cho mch nhhnh:
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Chng 1:p ng tn sthp mch khuch i
1K
+
Vi
-
Vcc=10V
Ve
1K
Cc
1K
5K
5K
Bit .100hfe=
a. Tnh ton phn cc cho mchCE,CQ
VI
b. Vmch tng tn hiu nhtn sthp
c. Tm hm truyn)s(i
)s(iA
i
L
i=
d. Xc nh Cc tn sct thp 3dB l 5Hz.
1.5Cho mch nhhnh:
R1
10K Ce
Cb
RL
47K
180
Vcc (12V)
iri
0
Rc
Re 1K
Cc
100uF
1K
Bit gm= 5.10-3
, Cgs= 20pF, Cgd= 0.5pF, rds=17Ka. Vmch tng ng tn hiu nhtn sthp ca mch in trn
b. Tnhi
L
ii
iA =
c. Vbiu Bode v pha cho p ng min tn sthp ca mch
15
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16/16
Chng 1:p ng tn sthp mch khuch i
1.6Cho mch nhhnh:
-
Vdd
Q2FET N
5K
Vi100K 250
5K
250100K
LV
+
100uF
Bit -13
mds10.5g;K5r ==
a. Tmi
L
vV
VA =
b. Tmi
L
vV
VA = nu tBypass cc ngun ni song song chai in tr250
16