56
Phan 2 : MOT SO BAI TAP MAU
Chng 1: AP NG TAN SO THAP
1.1 a) S o mach va s o tn hieu be :
RERb CE
RL
i
>
iL
5uF
i4k
1k
100
iL
RLfe
E
h
CRb
i
>ic
.Re
=1K
5k .ib
ib
i(1+hfe)
1/
>
1k
sCc2
Re 5kRb
hie
1khfe
ri
ieu kien tnh :
==
=
V
BB
kkk
b
V
R
81,1110
10.20
10100//10
62
K
ie
e
fe
b
BBCQ hmA
Rh
R
VI 34,0
7,3
25.50)(7,3
7,0===
+
=
at
-3
5
1
110
10.51]
1//)[1(
+=+=ssCc
RhZ efe
-ssCc
RZ L
43
2
2
10.510
1+=+=
Ham truyen :
12
..
.ZhR
R
ZR
Rh
i
i
i
i
i
iA
ieb
b
c
efe
i
b
b
L
i
Li +++
===
Xay dng Ai bang cach khac :
iem zero : sradCcRe
/10001
;01
===
iem cc
srad
h
hRRCc
fe
iebe
a /1485
)]1
//([
1
1
=
+
+=
25][
1
2
=+
=Lc
bRRCc
o li tang gia
24. =++
==
feb
b
Lc
cfe
i
Lim
hR
R
RR
Rh
i
iA
)1485)(25(
)1000(.
21
500
+++
=ss
ssAi
63
0)( =++ BEEECQefe
b VVIRh
R
Suy ra
)(2,4
150
1
7,05mAI
KK
VV
CQ =+
=
nen : Kieh 3,02,4
2550 == Zin
=
=
20
1
fe
EQ
h
mAI
A i
24
16
0 1485103
25
suy ra :
1.7 T s o ta co :
ri
Rb
RE
Vcc
=
-5V
+
-
Vi
5V
Cc2
RL 10k1k
1k
Cc1
10
Ta co
KKKK
efeiebin RhhRZ 1)503,0//(1])1([' +=++=
nen :
FZ
CcZCcZrCc ininini
20010.5
111
)(
131
11
===+
=
1.9
R c
R bR E
=
=
20
1
fe
EQ
h
mAI
B B
iRL=
>iL
1 k
C c2
C e1 k
V cc
1 k
+V
1 ki
)1485)(25(
)1000(24
+++
=ss
ssAi
64
K
ieh 5,01
25.20 ==
>
1/
Lhfe/
Li
i
>ib
Vb
-
hie
Rbi sCehfeRe R
sCc1
Rchfeib
Ta co :
-e
fe
bibee
eC
h
RhRC
.75
1
)]//([
1
+=
-2
3
2
210.2
1
][
1
CcRRCc Lc=
+=
Xay ra iem cc kep tai srad /10= th srade /102 == nen
e tan so 3 db la 10 rad/s vi Ce=1333uF th neu khong ke en s gay meo cua Cc2 th sradeL /10== . Do o ta chon
sradL /12 =
65 hfe=100 hie = 1
K
12
..ZhR
R
ZR
Rh
i
i
i
i
i
iA
ieb
b
c
c
fe
i
b
b
L
i
Li +++
===
0)1)((
)1(.
1)(
..
2
2
+++
+
++=
sRChRR
CRR
sCcRR
sCcRhA
eeiebe
eeb
Lc
c
fei
]1)10(][)([
)1(..10
3
2
2
3
++++++
+=
sRCcRhRhsrRhC
ssRCRhCcA
Lbieefeebiee
eebfe
i
Thay so
)75,10)(1(
)75,0(67,6
+++
ss
ss
Cach khac
-iem zero sradCR ee
/75,010.1333.10
11,0
63===
-iem cc :- srad
h
RhRC
fe
bibee
/75,1010.1333.76,69
1
)]//([
161
+=
67,6. =++
=ieb
b
Lc
Lfe
imhR
R
RR
RhA
Vay : 1.11
)100093)(1(43
)10001333(20
+++
=ss
ssAi
)1)(7,10(
)75,0(.67,6
+++
=ss
ssAi
R1 R2
Ri
C
RC
ii 100k
10k
Vcc
1k
10k
10uF
66
ib
100K1K
i L1/
L'
L>hie nen co the bo qua ri o li vong T
''
0
'.
L
b
efe
L
b
efe
iL
L
v
i
i
iRh
v
iRh
v
vT
i
====
vi :
1)(
1
1
1
++=
++=
CshRRh
sC
sC
i
i
ieie
b
ieie
ie
ieL
hRRCshRCs
Cshr
hRsC
Rv
i
+++
++=
++=
2)(
1)(
)//(1
1'
nen
C iC
iL
Li
>i
b
R
hfei
b
R2R1
hierii
67
21011
10
]2)([
110
33
+
=+++
=shRshRCR
hTieie
fe
o li khong hoi tiep :
21011
)20(10
1//
1//
.3
0, +
+=
++
+===
=s
s
hsC
RR
sCRR
hi
i
i
i
i
iA
ie
fe
i
b
b
L
vi
Li
L
suy ra Tan so 3db )/(110 sradL = Gian o Bode
1-13
1,19
191
+=
sT
)110(11
)20(1000
1 ++
=
=s
s
T
AA iif
0
dbAif ,
24
39
20 L)(log scale
68
21 ,CcCc
Do Rg.>>ri nen : vg~vi
RgRs1
Rs2
Rd
RLri
L
VDD
VCc2
100k
100uF250
250100k
Cc1
+
-
Vi
5k
5k
S o tn hieu be :
ri
Rg 5K100k
250
250
D
gs
+
-
Vi1
5k
100k+
-
uv
rds
Rs1
RdRL
Rs2 100uF
Dung tng ng Thevenin nh tren ta co :
Lvg~vi
g~uvi
(u+1)
//Rl
V
Rd
Z
+
-
Vi1
5k
100k+
-
uv
rds
vi : sC
RRZ ss1
//21 +=+
25. ==+ mds gr
69
a) Xac nh ham truyen :
dsLd
Ld
i
Lv
rZRR
RR
v
vA
+++
==
)1(//
//
Do Rd
70
575,7)1(1
=+++
==
dsds
d
i
Lvm
RRr
R
v
vA
Do o ta cung co bieu thc giong tren:
)7,55(
)40(575,7
++
=s
sAv
b)Neu ghep tu 2 au Rs=Rs1 + Rs2 th :
dsd
d
vrZR
RA
+++
=
')1(
vi Z'=(Rs1+Rs2)//1/sC
Thay so ta co :
46
)20(5,12
++
=s
sAv
->Bang thong giam va o li tang 1-12
R1
R2
R31K
RE
=10
V
E1k
1k
5k
5k
Vcc
V
Cc
+
-
Vi
ri 1K
1/
hfeRe
100k
hie
0.6k
Rb
3.5k
sCc
+
-
Vi1
ri
ieu kien tnh
KKKK
b
V
BB RV 5,35//51;51055
5=+==
+=
suy ra :
kieK
K
VV
CQ hmAo
I 6,015.4
2510015,4
1100
5,3
7,5===
+
=
71
Che o xoay chieu : do hfeRE = 100K>>hie->vE~VB
Ta co :
sCcrRhhR
RhhR
v
v
v
vA
iefeieb
efeieb
i
b
i
Ev 1
)//(
)//(
+++
++==
Thay so ta co ;
521.10.83,22
.10.62,175
5
+=
sCc
sCcAv
Tan so 3db :
CcCcL
83,22
521.10
10.83,22
521 5
5
==
Theo gia thuyet pipi 102 == LL f nen
)(26,710.83,22
521.10 5FCc
pi==
72
i
B=B1
b//rb'e b'e mv
c
iL
bi gC
RR
rbb'
bi b'e b'e M b'vm
LiL
LRg
V
CC1i r1R2
R41k
R31k
Vcc
+ Vbb
IiR
g
V
CC1i r1R2
R41k
R31k
Vcc
+ Vbb
Ii
Vi gia thiet Cb'c = 0 th CM = 0; rbb' = 0 : ngan mach B - B'.
S o ch con lai nh sau :
Theo gia thiet ta co :
ICQ = 2 mA suy ra :
hie = 2
25.feh = 12,5.hfe
ma : hie = rbb' + rb'e = rb'e (rbb' = 0) Do vay :
12,5.hfe = rb'e (1) gm = 5,12
1
'
=eb
fe
r
h= 0.08 mho
Tan so cao 3 dB :
fh = ebebMebeb CRCCR '''' 2
1
)(2
1
pipi=
+
vi : Rb'e = (ri // Rb + rbb') // rb'e = Rb // rb'e (rbb' = 0; ri = ) nen :
2-1
73
fh = ebebb CrR '' .//2
1
pi (2)
o li dong tan gia :
Aim = i
b
b
L
i
L
i
v
v
i
i
i '
'
.= = -gm.Rb // rb'e
Theo gia thiet : imA = 32 dB = 40
Suy ra :
Rb // rb'e = == 50008,0
40
m
im
g
A
vi :
Rb = 103 rb'e = =
50010
10.5003
3
103
T (2) ta co :
Cb'e = 500.10.800.2
1
//.2
1
3' pipi
=ebbh rRf
= 400 pF
T (1) suy ra :
hfe = 5,12
10
5,12
3
' =ebr
= 80
Vay : hfe = 80; rb'e = 1K; Cb'e = 400 pF 2-2 Cho s o mach nh sau :
Cac thong so : - r = 10
9 rad/s - hfe = 100 - Cb'c = 5pF - rbb' = 0 - IEQ = 10 mA i
CC
L
iLQ1
NPN
+V
i20uF
20uF
20uF
R
1k
1001k10k
1k10k
74
==
==
==
pFgC
mhoIg
KKKR
Tmeb
EQm
b
400/
4,040
9,010//1
'
i
Li
L
i
c1
b
b'e
b'e
ee
c2
mvb'e
+CM
c
B'
E
g
R
C
R
r
i C
C
C
RRr
a) Tnh o li tan gia Aim : cho ngan mach cac tu Cc1, Cc2, Ce va bo qua cac phan t Cb'e, Cb'c (cho h mach hai au cac phan t ay). Khi o :
Aim = i
eb
eb
L
i
L
i
v
v
i
i
i '
'
.=
vi : * Lc
c
m
eb
L
RR
Rg
v
i
+= .
'
* ebbi
bi
eb
i
eb
rRr
Rrr
i
v
'
'
'
//
//.
+=
Vay :
38)//(
).//(..
'
' =++
=ebbi
ebbi
Lc
c
mimrRr
rRr
RR
RgA
b) Tm tan so 3 dB fh : Ta co :
Cb'e = 40010
4,09==
T
mg
pF
75
Xet tan so cao ta se thay rang cac tu ghep ngoai Cc1, Cc2, Ce co tr khang rat be do rat ln ngan mach cac tu ghep ngoai. S o ch con :
bi b'eb'er
i//r
B' Li
cv
m+CM Lb'c
C
RR
gCRi
Va : CM = [1 + gm.(Rc//RL)]Cb'c CM = (1 + 0,4.500).5 pF CM = 1000 (pF) Tan so cao 3 dB :
h = )(
1
'' Mebeb CCR + (Rb'e = ri // Rb // rb'e = 196 )
h = 1210).1000400.(196
1+
= 3,64 (Mrad/s)
76
R = rbe + hfeRe, C = ''
1 em
eb
Rg
C
+
* o li tan gia : h mach Cb'c, C' :
Avm = ebiLefe
Lefe
rrRRh
RRh
')//(
)//(
++ = 0,9
* Tan so cao 3 dB : R' = rb'e + hfeRe' >> Re' : bo qua Re'. Do o tan so xay ra iem cc :
1 = )')((
1
)'('
1
'
'
'' CCRhrCCR cbefeebcb ++=
+
Thay so :
R' = hfe.CQI
25 + hfe.RE//RL = 10,5 K
C' = ''
'
1
/
1 em
Tm
em
eb
Rg
g
Rg
C
+=
+
= 2 pF
Suy ra : 1 = 12 Mrad/s : cho Zi e tnh tan so h ta xet hai trng hp sau :
= 12'' 10.40.500
1
.
1
=ebeb Cr
= 50 (Mrad/s)
a) Neu < : bo qua C'. Khi o :
Av = i
b
b
e
i
e
v
v
v
v
v
v '
'.=
Trong o :
* 1'
b
e
v
v
* 1/
1'
+=
ii
b
sv
v
vi i = '
'
1
icb RC ='iR ri + rbb'
77
h = 230 Mrad/s Aim 1
Do ta ang xet < ma > . Vay : co the lay xap x h = 1' 2-6 Cho s o mach nh sau :
- T = 109 rad/s
- Cb'c = 6p
78
- rbb' = 0 - IEQ = 1 mA - hfe = 20
Transistor Q2 c ghep dang B chung vi Q1 nham muc ch lam giam ien dung Miller CM . Tan so 3 dB ngan mach B chung f fT. Do o ta ch xet Q2 tan so thap va Q1 tan so cao nen Re1 b ngan mach.
S o thay the :
b ngan mach (noi mass). ay ta a bo qua mot tu Cbe mac song song Rc2 do qua nho. Trong o :
CM = (1 + gm.Rc1 // 1
'
+fe
eb
h
r).Cbc = 11,85 pF 12 pF
a) Tnh Aim : ngan mach cac tu ghep ngoai, h mach cac tu ghep trong ta se co s o sau e tnh Aim va ngan mach cac ien tr trong tr ien tr rbe. Ta co :
Aim = i
eb
eb
e
e
L
i
L
i
v
v
i
i
i
i
i1'1
1'1
..=
Aim = -hfb. )//.(.1
1
2
2ieb
ibc
cm
Lc
c hRhR
Rg
RR
R
++
Aim = ))((
)//(..
1
12
ibcLs
iebcmc
hRRR
hRRgR
++
Thay so :
Aim )2510)(1010(
)500//10.(10.04,0.10333
333
++
= -6,5.
Vay : Aim = -6,5 b) Tm tan so cao 3 dB h :
79
h = 12'' 10).1240.(33,333
1
))(//(
1+
=+ Mebebb CCrR
= 57,7 Mrad/s
Vay : h = 58 Mrad/s
* Ta thay tu CM co gia tr rat nho la do Rc1 // 1
'
+fe
eb
h
rma
1
'
+fe
eb
h
r= hib
co gia tr nho. Neu khong mac them Q2 vao th : CM = (1 + gm.Rc1 // RL).Cbc =120 pF Khi o :
h = 1210).12040(33,333
1+
= 18,75 Mrad/s : nho hn trng
hp tren rat nhieu. Do o ghep them Q2 lam tang bang thong cua mach. 2-9
i
iL
Z
Z
i
o
Q22N4223
100k
Vcc20V
20uF
1k4k
r6 20uF
+
-
v1
rds = 5K, Cgs = 6p, Cgd = 2p, gm = 0,003 mho
gmrds = =15 Gia s FET c phan cc che o tnh vi cac thong so nh tren. * Xet s o mach tan so thap :
a) Tr khang vao :
80
=+= igc
i ZRsC
Z1
' 1
'iZ khi 0
Neu mach hoat ong tan so > 2
1
cLCR= 50 Hz th
1//
//
++
=
ds
Ls
Ls
g
s
rRR
RR
v
v= 0,72
81
5,01
1
+=
++=
s
s
sCrR
R
v
v
cig
g
i
g
Vay :
Av = )5,0)(50(
72,0
1)(
..72,0.
1.
.
1
1
2
2
++=
+++ sssCrR
sCR
sCR
sCR
cig
cg
cL
cL
* Tan so gia : a) Tr khang vao : Zi = Rg =100K
b) Tr khang ra : Zo = Rs // 1+
dsr = 0,29K
c) o li ap : Avm =
1//
//
++
==
ds
Ls
Ls
g
L
i
L
rRR
RR
v
v
v
v= 0,72
Tan so thap 3 dB : L = 50 Hz (Tren thc te ham truyen Av co 3 iem cc song co 2 iem cc xap x bang nhau). * Tan so cao : ngan mach cac tu ghep ngoai :
1) Neu
82
Z = ri // 1
1
+=
sCr
r
C gdi
i
gd
Z1 = gsC
1
Suy ra :
+=
++=
sg
osgms
vZZ
Zv
ZZivvgv
1
1)]()([
T o suy ra :
Zo = 1
1
1 Zg
ZZ
i
v
mo
s
+
+=
hay :
Zo = )1)(/1(
)(1[.
1
gdimgs
igd
m CrjgCj
rCgsCs
g ++
++
Z0 = 28
811
)10.5(
)10.25,1(10.67,6
+
+
s
s
o li ap : a) < gm/Cgs : bo qua R. Khi o :
1210.95,3.1
1
]).1(
[1
1+
=
+++=
sRg
CCs
v
vA
m
gsgd
i
g
v
Tan so cao 3 dB : h = 2531.10
2 Mrad/s > gm/Cgs = 500 Mrad/s : loai b) >> gm/Cgs : bo qua C.
Av = 8
8
10.24,121
10.5
]).//(1[
//
+=
+=
gdii
i
i
g
CrRsr
rR
v
v
h = 12,24.108 rad/s = 1224 Mrad/s >> gm/Cgs
Gian o Bode :
83
1 = RC '.1
2 = RCC gd1.1
'1
+
Zi =
+'
1//
1
sCR
sCgd
2-10 S o mach : Cac thong so : + ri = 1K, Rg = 100K + Rs1 = 1,5K; Rs2 = 2,5K + RL = 1K + Cgd = 2p, Cgs = 6p + gm = 0,003 mho, Rds = 5K = 15 Ta xet cac khoang tan so sau :
a) Tan so thap : S o tng ng nh sau : Trong o :
+ Zi =
K
K
K
RR
R
ss
g
4
5,2.
16
151
100
]/[1
1 2 =
+
= 241K : rat ln so vi
ri.
+ Z16
5
1
' Krdso =+=
= 0,32K
+ A +
=1
'
v 1
* Tan so cat thap :
1 = 361 10].1241[10.20
1
][
1
+=
+ iic rZC= 0,2 (rad/s)
84
2 = ]320//10.410[10.20
1
]//[
1336'
2 +=
+ osLc ZRRC= 39 (rad/s)
Suy ra : L 38 (rad/s) * Dan nap ngo ra :
Yo = 33' 10.32,0
1
10.4
111+=+
os ZR= 3,38(kmho)
* o li ap :
Av = i
g
g
s
s
L
i
L
v
v
v
v
v
v
v
v..=
Trong o :
501.10.20.10
.10.20.10
1.
..
1 63
632
2
+=
+=
+=
+=
s
s
s
s
sCR
CsR
sCR
R
v
v
sL
cL
cL
L
s
L
39
)50(714,0
1//
1//
'
2
2
++
=+
+
+
=s
s
ZsC
RR
sCRR
v
v
oc
Ls
cLs
g
s
2,0.10.20.10.2421
.10.20.10.241
1 123
123
1
+=
+=
++=
s
s
s
s
sCrZ
Z
v
v
cii
i
i
g
Suy ra :
Av = )2,0)(39(
714,0 2
++ sss
Cung co the tnh bang cach khac : - Da vao s o mach ta thay Av co iem zero kep tai = 0. - iem cc :
+ 1 = 1)(
1
cii CrZ += 0,2 (rad/s)
85
+2 = 6332
' 10.20).320//10.410(
1
]//[
1+
=+ cosL CZRR
=39 (rad/s)
+ Aim = g
L
i
g
g
L
i
L
v
v
v
v
v
v
v
v=
.
(do Zi >> ri) = '//
//
oLs
Ls
ZRR
RR
+= 0,714
Vay :
Av = Aim.)39)(2,0(
714,0
))(( 21
2
++=
++ sssss
b) Tan so cao : hay rut gon con : Trong o : bo qua Rg do Rg >> ri .
+ C= 670.10.31
10.6
1 3
12
+=
+ Rg
C
m
gs= 2p
+ R = rds//Rs//RL = 0,67K
* Neu
86
- o li :
Av = ]10.2).10//670(1[10
10//670
]).//(1[
//1233
3
+=
+ sCrRsr
rR
gdii
i
Av = )1(
4,0
h
s+
- Dan nap ngo vao :
Yo = ssCrRr
gd
isds
2
33310.2
10
1
10.4//10.5
11
//
1 ++=++
Yo = 0,00145 + 2.10-12s, iem zero : 725 Mrad/s
Gian o Bode : 2-11 Cac thong so mach : - T = 10
9 rad/s, hfe = 100, Cbc = 5p, Cbe = gm/T = 40ICQ/T - ICQ = 10 mA Cbe = 400 pF, R1 = 10K, R2 = 1K Rb 1K - rbb = 0, ri = 10K, RL = 1K
- Cc1 = Cc2 = Cc3 = 20 F; rbe = hie = 100.10
25= 0,25K
S o tng ng tong quat : Trong o : + Rb1 = Rb2 = 1K; rbe1 = rbe2 = 0,25K + Rc1 = Rc2 = Rc =1K at R1 = ri // Rb1 //rbe1 =10K // 1K // 0,25K = 0,2K R2 = Rc1 // Rb2 //rbe2 =1K // 1K // 0,25K = 0,17K R3 = Rc2 // RL = 1K // 1K = 0,5K S o thay the bi tu Miller : Trong o : CM = (1 + gm. Rc2 // RL).Cbc2 = (1 + gm.R3).Cbc2 = (1 + 0,4.500).5p = 1000p CM _ phan anh tr khang tang sau ve tang trc. + C2 = Cbe2 + CM = 1400p
87
+ C = R2gmCbc1 = 0,17.103.0,4.5p = 340 pF
+ R = 1'
2
cbmCg
C= 1400p/(0,4.5p) = 700 = 0,7K
* Tan so 3 dB tren : Ta co :
= 1
''
C
CC cbeb +
C1 = Cb'e + (1 + gmR2)Cb'c C1 = 400p + (1 + 0,4.170).5p = 745p
nen :
= p
pp
745
5100 + = 0,544
1 = 200.10.745
1112
11
=
CR = 6,71 (Mrad/s)
2 = 170.10.1400
1112
22
=
CR= 4,2 (Mrad/s)
nen :
h =
+
+++
++ 2
2
2
1
1
2
2
1
1
2
2
21 4)1(2)1(22
= 2544,0.2
2,4.71,6
+
++ )544,01(2
2,4
71,6
71,6
2,4
+
+
++ 2
2
544,0.4)544,01(22,4
71,6
71,6
2,4
= 47,61[-(3,13) + 18,1)13,3( 2 + ]
h = 2,95 (Mrad/s) * o li tan gia : tan so cao ta co :
88
Av =
.
111
1.
)(
).(
21
2
21
2
2122
ss
RR
RRRg
Lc
cm
+
++
+
Suy ra :
Aim = Lc
cmv
s RR
RRRgA
+=
2
2122 ..
lim
Thay so :
Aim = 33
32
1010
190.200.10.4,0
+ = 2720
* Neu bo qua anh hng cua tu Miller, dung cong thc ghep n tang (n = 2).
Aim = (-gmRbe)n/2 = (-gmRbe)
2/2 vi : Rbe = Rc // Rb // rbe = 1K // 1K // 0,25K = 0,17K Suy ra :
Aim = (-0,4.170)2/2 = 2312 (
Lc
c
RR
R
+=
2
1)
Tan so 3 dB tren cua mot tang :
1 = 12'' 10.400.170
11
=ebeb CR
= 14,7 (Mrad/s)
Suy ra tan so 3 dB tren khi ghep 2 tang :
h = 0,64.1 = 9,4 (Mrad/s) (0,64 = 122/1 )
Ta thay ro rang hai ket qua qua chenh lech nhau. ieu nay ung bi le trong cong thc ghep n tang, ngi ta xem Cbc = 0 trong khi theo gia thiet Cbc = 5p. 2-12 S o mach : + ri = 1K, Rg = 1M + Rd = 10K, RL = 10K
89
+ Cgs = 6p, Cgd = 2p + gm = 0,03 mho, rds = 5K S o thay the : at : R1 = Rg // ri 1K R2 = Rd // rds // Rg Rd // rds = 3,33K R3 = Rd // rds // RL = 2,5K Tng t nh BJT : bien oi s o thanh : Trong o :
R = mgd gC
C 1.2
C = R2.gm.Cgd1 C2 = Cgs2 + CM = Cgs2 + (1 + gmR3)Cdg2 * o li ap : Ta tnh cac tham so : C1 = Cgs + (1 + gmR2)Cgd C1 = 6p + (1 + 0,03.3,33.10
3).2p = 206p C2 = 6p + (1 + 0,03.2,5.10
3).2p =158p nen :
= p
pp
C
CC gdgs
206
62
1
+=
+ = 0,039
1 = 31211 10.10.206
11
=RC
= 4,85 (Mrad/s)
2 = 31222 10.33,3.10.158
11
=RC
= 1,9 (Mrad/s)
Suy ra :
+
+++
++= 2
2
1
2
2
1
1
2
2
1
2
212 4)1(2)1(22
h
Ta tnh :
90
a = 22
21
039,0.2
9,1.85,4
2=
= 3029 (Mrad/s)2
b = )039,01(285,4
9,1
9,1
85,4)1(2
1
2
2
1 ++=++
= 4,87
c = =+ 22 4a 22 039,0.487,4 + = 4,870624
Suy ra :
2h = a(-b + c)
nen :
h = 000624,0.3029 = 1,375 (Mrad/s)
* o li tang gia :
tan so cao :
Ai =
21
2
21
212
111
1.
])//[(
)//.()(
ss
RrR
RRrRg
Ldsd
dsdm
+
++
+
nen :
Aim = Ldsd
dsdmv
RrR
RRrRgA
+=
)//(
).//.(lim 21
2
0
Aim = 33
3332
10.1010).5//10(
10.33,3.10.10).5//10.()03,0(
+
Aim = 748 Avm = Aim.i
L
r
R
91
2-13 Cac s o mach nh sau : + rbe = 1K + Cbe = 1000p + Cbc = 20p + gm = 0,05 mho
hfe = 50. a) S o tan so thap : Tan so 3 dB thap :
fL = )]///([2
1
febibee hRhRC +pi
fL = )]20020//(10[10.20.2
136 +pi
Suy ra : fL = 44 Hz * o li tan gia :
Aim = ieb
bfe
i
bfe
i
L
hR
Rh
i
ih
i
i
+=
=
Aim = -50.KK
K
110
10
+ = -45
* tan so cao ta co s o nh sau : Trong o :
CM = (1 + gmRc).Cbc CM = (1 + 0,05.10
3).10p = 500p Tan so 3 dB tren :
fh = )1000//1000.(10).1000500(2
1
)//)((2
112
''+
=+ pipi ebbebM rRCC
fh = 116 KHz Suy ra :
GBW = )4410.21,0(25)( 6 = Lhim ffA = 5,249 (MHz)
92
Vay : GBW = 5,249 MHz
b) S o tan so thap : Tan so 3 dB thap :
fL = )]//([2
1
1 efeiebic RhhRrC ++pi
Thay so :
fL = )]10.5010//(1010[10.20.2
133536 ++pi
= 0,23 (Hz) = fL
* o li tan gia :
Aim = efeiebi
bii
i
L
RhhRr
Rrr
i
i
++=
//
)//.(
Aim = KKKK
KKK
501100//1
)100//1.(1
++= 0,02
* tan so cao : bo qua Rb do Rb >> ri , ta co s o : Vi : R 'i = ri + rbb ri
R = rbe +hfe 'eR = 1K +50.1K = 51K
'eR = Re = 1K
C = 3''
10.1.05,01
1000
1 +=
+
p
Rg
C
em
eb = 20p
Ta xet hai kha nang sau : Neu < th Av 1 GBW = : loai.
Neu >> = 123'' 10.1000.10
1
.
1
=ebeb Cr
= 1 Mrad/s :bo
qua R. Khi o :
Av = i
b
v
v ' (do ve < 'bv )
93
Av = )'...(].').[(1
'.1
''2
'''
'
cbeicbiie
e
CCRRsCRCRRs
CRs
++++
+
Av = )10.8,22/1)(10.2,2/1(
10.5/1
10.2.10.5.1
10.2.177
7
1628
8
ss
s
ss
s
++
+=
++
+
h = 22 (Mrad/s) (h >> : thoa) fh = 3,5 (MHz) Vay : GBW = imLh Aff ).( 3,5 (MHz)
S o ghep C chung co bang thong ln hn s o ghep E chung nhng o li lai nho hn.
2-14 Cbc = 10p, gm = 0,05 mho Cbe = 1000p; rbe = 1K * Cach 1 : bo qua Cbc, coi tan so fL rat nho. Ap dung cong thc ghep 2 tang giong nhau, ta co :
Aim = (-gm.Rbe)(-gm.Lc
c
RR
R
+).rbe
vi : Rbe = ri // Rb // rbe = 1K // 10K // 1K = 0,5K = 2
1rbe
nen : Aim = (-0,05.0,5.10
3)2 = 625
Tan so 3dB cao cua mot tang :
f1 = 123'' 10.1000.10.5,0.2
1
2
1
=pipi ebeb CR
= 0,318 (MHz)
* Cach 2 : tnh toan chnh xac bang s o tng ng : a) Xet tan so thap :
94
vi Z = hie + hfe.
e
esC
R1
//
95
Chng 3 : KHUECH AI CONG SUAT AM TAN ____________
3-16
V IC = 0,2A nen ay la iem Q
bat ky. a) o doc cua ng tai AC :
50
1
20.)58,1(
11122
===LAC RNR
= 0,02 b) Phng trnh ng tai AC
trong he toa o tong quat :
iC ICQ = -ACR
1(vCE VCEQ)
+ Cho vCE = 0 maxCi = ICQ +
50
202,0 +=
AC
CEQ
R
V= 0,6A
+ Cho iC = 0 maxCE
v = VCEQ + ICQ.RAC = 20 + 0,2.50 = 30V
c) Gia tr nh cc ai cua ien ap collector khi khong b sai dang :
maxCmV = min[VCEQ, ICQ.RAC] = min[20 , 0,2.50] = min[20,10] = 10V
iC(A)
vCE(V) 0
Q 0,2
iCmax= 0,6
VCC= 20 vCEmax=30
DCLL( )
ACLL(-0,02)
20b
+CC
=
1,58:1
L
V 20V
R
R
96
d) PL = 50
10.2
1.2
1.2
1 2
2
22
==L
Cm
L
Lm
RN
V
R
V = 1W
e) PCC = ICQ.VCC = 0,2.20 = 4W
= 4
1=
CC
L
P
P= 0,25 = 25%
97
= 24.14,3
4.2 = 61,15W
c) 2PC = PCC PL = 61,15 24 = 37,15W
PC = 2
15,37 = 18,575W
d) = 15,61
24=
CC
L
P
P = 39,24%
3-20
ien ap collector nh trong moi transistor la cc ai : === CCCmp VVV maxmax 24V; maxmax Cmp II = = 8A
(v dong nh tang cung ty le ln gap 2 lan so vi 3-19)
a) 2
1.
2
1maxmaxmax
== CmCmL IVP .24.8 = 96W
b) PCC = ITB.VCC = CCC
VI
.2
max
pi =
14,3
8.2.24 = 122,3W
c) = 3,122
96=
CC
L
P
P = 78,5%
e PC max ta co :
ICm = pi
28.
2max pi
=CmI = 5,1A
2PC = PCC - CmCmCmCCL VIIVP .2
1.
2 2max
=pi
= pi
2.24.5,1 -
2
1.5,1.24
77,96 61,2 = 16,76W PC = 8,38W
98
3-21 He so ghep bien ap :
N = 40
160=
s
p
N
N= 4
ILm = 8
40.22=
L
L
R
P= 3,16A VLm = LLRP2
VLm = 8.40.2 = 25,3V VCm = VLm.N = 101,2V
Vay minCC
V = 101,2V
3-23 Goi sai dang khi cha co hoi tiep la D = 10%. Goi sai dang khi co hoi tiep la Df = 1%.
Ta co :
Df = A
D
+1 = 10% 1 + A = 10 A = 9 = T
A = 02,0
9=
T = 450
3-24
Avf = v
v
A
A
+1 = 20 1 + Av =
20
80=
vf
v
A
A= 4
Mat khac :
Df = + vA
D
1 D = Df(1 + Av) = 0,1.4 = 0,4 = 40%
3-26
99
a) LL
CCLLL R
R
VRIV ..
maxmax== = 20V
V mach khuech ai mac collector chung nen gia tr nh ngo vao ln nhat la : Vin = VL = 20V (Av 1)
b) 12
20.2
1.2
1 22
max
max==
L
Lm
LR
VP = 16,67W
c) Vdt = 2VCC = 2.20 = 40V
3-28
L
c1in
+
c2 L
E1
E2
B
B
1
2
30V
R680
R
680
+
C
500uF
R
8
R
0.5
R
0.5
+
Cv
T1
T2
a) Ipa = 680.2
4,130
2
2 =
R
VV DCC = 21mA
-
L
CC
CC
=20V
=-20V
c1 Y h
+
C vin
T2
V
+V
R 12
T1
100
b) 1B
V = VCC Ipa.R = 30 21.10-3.680 = 15,7V
2B
V = 1B
V 2VD = 15,7 1,4 = 14,3V
c) PL = 8
10.2
1.2
1 22
=L
p
R
V= 6,25W
PCC = ITB.VCC = 69,26
300
)85,0(
30.10
)(
.. =
+=
+=
pipipi LE
CCp
CCCm
RR
VVV
I
= 11,24W
= 24,11
25,6=
CC
L
P
P = 55,6%
d) fL = ==+ 7,266
10
10.5.5,8.28,6
1
)(2
1 4
4CEL CRRpi
37,5 Hz
3-29
a) RL + RE = =cLCfpi2
1
471
10
15.10.5.28,6
1 4
4=
= 21,23
'LR = 21,23 0,5 = 20,73
b) +
=+
=)73,205,0(2
30
)(2 '1
max
LE
CCLm
RR
VI 0,7A (= 0,7065A)
73,20.)7,0.(2
1.)(
2
1 2'2maxmax
== LLL RIP = 5,174W
PCC = ITB.VCC = =2
.2
max CCL VI
pi30.
14,3
7065,0= 6,75W
= 75,6
174,5max =CC
L
P
P = 76,65% ( =
EL
L
RR
R
+'
'
.4
pi = 76,65%)
Bai mau :
101
Cho mach ien nh hnh ve : Q3, Q4 co hfe = 20; rbe = 10; Q1, Q2 co hfe = 50; rbe = 100 a) Tm
maxmaxmax,, CCCL PPP
tren moi BJT, bo qua tieu tan cua thanh phan phan cc. b) Hay ve o th bieu dien PCC, PC, PL, theo ICm cua Q3, Q4. Bai giai : V Q1 va Q2 giong nhau, Q3 va Q4 giong nhau nen ta
ch can tnh cho Q1 va Q3. a) e PL max ta co :
41
15maxmax3max4 +
=+
===LE
CCLmCmCm
RR
VIII = 3A
20
3
3
3
3==
fe
cm
bmh
II = 0,15A
max 3,4 max
2 21 1. .3 .42 2
L Lm LP I R= = = 18W
3maxC
P tai ICm = pimax
2 CI= 1,91A
2 4.)91,1.(2
115.91,1.
2 2max
=pi
CP = 18,256 7,296 = 10,96W
maxC
P = 5,5W
333,4
22 . .
Cm
CC TB CC CC
IP I V V
pi
= = = 28,66W
c1
Yh
+
B
B
1
2
in
c2
VCC=
CCV- =
LR
4
-15V
R7
1
R6
1
Q4
Q3
v
C
C
R2
15V
R3
R1
R5
100
R4
100
Q1
Q2
102
28,66
P
Pcc
PL
PC
1,91 3
Icm 5,33 5,5
18
0
= max 3,4
3,4
18
28,66
L
CC
P
P= = 62,8%
6' .. 3334 RIrIV CmebbmR += = 0,15.10 + 3,1 = 4,5V
100
5,4
4
4
4==
R
VI
R
R = 0,045A = 45 mA
341 bmRCm
III += = 45 +150 =195 mA 0,2A
50
10.195 3
1
1
1
==fe
cm
bmh
II = 3,9 mA
+=+= 23]20.110//[100]//[ 36341 feieAC hRhRR
WRIP ACcmL 46,023.)2,0.(2
1..
2
1 21
2
1max 1==
WVI
P cccm
CC 91,115.14,3
2,0.22 11max ===
pi
%;2491,1
46,01max == Pcmax1 tai mA
II cmcm 2,124
2 max11 ==
pi
WP
P
C
c
5,0
23.)10.2,124(2
115.10.2,124.
22
1max
233
max1
=
=
pi
b) Neu Vi = 10V hay tnh cac bc nh tren . Gia thiet R1 , R2 , R3 rat ln nen bo qua:
103
at R=R4hfe1//[rb'e3 + R6hfe1 +R6hfe1hfe3] =0,5//[0,5K +1k] = 225
' ' .hfe1
hfe3
hfe3R1hfe14K
R6hfe11KR4hfe1500
rbe3
500
rbe1
100
+
-
Vi
925,04225,1,0
4
311
31
'
=++
=++
==KKOKhhRRr
hhR
V
VA
K
fefeLeb
fefeL
i
Lv
AR
VII
VVAV
L
LcmLm
iVL
31,24
25,9
)(25,910.925,0.
3 ====
===
Ah
II
fe
cm
bm 1155,020
31,233 ===
WR
VRIP
L
Lm
LLmL 7,104
)25,9(
2
1
2
1.
2
1 2223max ===
WVI
VccIPcc cccm
TB 2215.14,3
31,2.2.2. 333 ===
pi
AIIP
Pcmcm
CC
L 47,12
%;6,4822
7,1033
'
3
=====pi
103
fo=30Mhz rb
,e=1
K rbb'=0 hfe=100 fT=500Mhz Cb'c=2p
Teb
fe
eb r
hC
1
'
' =
Cb'e=31,8pF
4_3 Cho mach khuech ai cong hng nh sau :
R2
R1
100
L
->oo
->oo
->ooi
Vcc
i L Ce
Cc1
RFC Cc2
R
1kRe1k
10k
S o thay the Rb=1
K//10K~1K,qm=hfe/rb'e= 1,0 mho
1K1K 1k
'B
b'e'b L
L
m 'bg v
ii L Rb r
C R
V
Ta co R=Rb//rb'e=0,5K Cb' = Cb'e + CM = 31,8p + (1+gmRL)Cb'c = 31,8p + (1+0,1.103)2p = 234pF Cong hng tai fo = 30 Mhz th
)(302
1
'
MhzLC
fb
o ==pi
suy ra : HfC
Lob
pipi
12,0)10.30(4.10.234
1
)2(
1262122
'
===
Bang thong :
)(36,110.234.500.2
1
2
112
'
MhzRC
BWb
===pipi
o li dong :
104
)(1
. '
'
oo
i
m
i
b
b
L
i
Li
jQ
Rg
i
v
v
i
i
iA
+===
Vi : 2210.234.10.5.10.3.28,6 1227' == boi RCQ
nen : Aim = -gmR = -0,1.500 = -50
)10.3
10.3(221
150
7
7
+
=
j
Ai
4_4 Thiet ke mach cong hng n co : Aim = 10db = 3,16 ri = 1
K fo = 40 Mhz RL = 1
K BW = 1Mhz Vcc = 10 V Cuon day co Q = 50; S o can thiet ke co dang nh sau :
iC
iL
1k 1ke'b L
L
m 'bg v
i irLi r
Rp C R
V
ay ta chon transistor co : gm=0,01 mho ; rbb'=0 ; hfe = 100
Cb'c = 10p, Cb'e=1000p 2
2
'10
10
==m
fe
ebg
hr
Ta co :
6102
1==
RCBW
pi
vi R = ri //Rp//rb'e C = Cb'e + (1+gmRL)Cb'c + C' : C' la tu ghep ngoai
105
va c
o
c
o
ebi Q
C
Q
C
rRprR
+=++=++= 3
'
10.4,12500
1
1000
11111
vi c
cr
LQ
= : he so ton hao cua cuon day
Do o :
)50
)'1110(10.4,1.(
)'1110(2
110 36
CpF
CpFBW o
++
+==
pi
hay
)50
)'1110(10.4,1[
10.2
1)'1110( 3
6
CpFCCpF o
++==+
pi
hay
6
3
6 10.2
10.4,1)
10.502
21(
pipipi
= of
C
suy ra :
111010.1,1
50
401
10.22,0 99
==
=
C pF
Ta nhan thay : C = 1110+C' = 1110pF -> transistor co cac thong so khong thoa. Chon lai transistor co Cb'c = 2p rb'e=2500 Cb'e = 32p hfe = 100
gm = 0,01 fT = 500Mhz Khi o : Cb'e + ( 1 + gmRL )Cb'c = 54 pF suy ra C' = 1100 -54 = 1046 pF = 0,001uF
S o nh sau : HC
Lo
0144,012
==
106
i i'
L
>i LL
V
RLi r
C
R = ri//Rp//rb'e Rp = Qc. Lo. = 1,81K
Aim = -gmR = -0,01.512 = -5,12 > 3,16 : cha toi u 4_5 :
RL
i i=1K
n:1 m:1
1kL2L1i r
Cac ieu kien : Aim cc ai tai fo = 30Mhz e n gian hoa ta gia s n=m S o can thiet ke :
RL
i i=1K
n:1 m:1
2N4223
Rg1k
L2L1i r
Cac thong so cua FET : rds = 5
K , Cgs = 6p , Cgd = 2p gm = 0,003 mho Phan khang tr khang t ai : R'L = n
2RL S o tng ng :
2N4223 FET
107
2i
a g vm gi 'L
a Cgd(1+gm(rds//R ))2 L'
>iL
RrdsCgsL1ri
at : Ci = a
2[Cgs + Cgd(1 + gm(rds//R'L))]
Gia s rds
108
Cho hfe = 50, hie = 1K , rbb' = 0
Cb'e = 10p , Cb'c = 1p RL= 100 , C" = 10p Mach cong hng tai fo = 10Mhz va BW = 1Mhz
2
310.5
10
50 ===ie
fe
mh
hg
)(63. MhzBWAGBW vm ==
4_7
100
n
L
L
L
"
n2 n1
nLV
R+
-
Vi
9K
C
Giai Mach dung cuon cam kep at a=n1/n2 S o thay the
9K
i'
B'
(n/n2)2C
)C"
+
- L_a_Vbe'
2a Cb m bg v
Vo1
Vo
00R2
R1
+V
R
R1
hay : ref1
2ref
1
2O V
)1(R
R]1
1
1[V
R
RV
+
=
+=
133
Gia thiet : RA = RB
va QC
C
R
R
2
1
2
1 ==
ay la OP-AMP khuech ai khong ao :
211R
R1k
A
B =+=+= (1)
7.17
V1
A
B
+VO/K Vo
C1
R1
+
-
Vi
R2
R
R
)1sCR(2
VV)2(
1sCR
V2V1sCR
1kV 22
o
1
22
11
22
O +=+=
+=
Tong dong tai nut 1:
0
sC
1
VV
R
Vk
V
R
VV
1
1O
2
1O
1
1i =
+
+
(3)
)4(0)1sCR(2
sCV
)1sCR(R2
V)1sCR(
R2
VsCV
R2
V
R
V
2211
22
2
O22
1
O1O
2
o
1
i
=+
++++
)5(02
sCVR
2
sRRCCV
R2
VR
2
sCVR
2
V
2
sCRVsRCV
R2
RVV
1O1
2
2121O
2
O1
2O1O22o
11O
2
1oi
=
++
)6(0sRRCCVQV
sCRVVsCRVsRCVQVV2
2
2121OO
21OO22O11Ooi
=
++
)7(1]CRCRRC[ssRRCC
2
V
VA
212211
2
2121i
O
v ++++==
134
a_Tm HoLp , o , Q
khi C1 = C2 =C; R1 = R2 = R3 = R4 = R AoL = -> v
+ = v-=0 -> V3 = 0 (1); Zi =
id = 0 ta co :
)8(RCQ
1
RCQ
1
QRRQCC
1
RRCC
122
2
O2
2
2
2
222122121
2
O ====
T (7) [ ]Q
sCRCRRCs
o=++ 212211
)9(1
22
11
21
21
11
22112122
2121
RC
RC
RC
CR
CR
CRRCCRCR
RRCCQ
+
=+
=
Thay vao v du 7 8 ta co : R1 = R2 = R; C1 = C2 = C ; fo = 1 khz
Vao 9 ta co oo fRC
Q pi 21
;1111
1===
+=
Chon C = 10nF kCf
Ro
1610.10.28,6
1
2
183===
pi
7_22
C1=C C2=C
=R2 =R =KR4
=R
V2 V3V1Vo
R3
KRR4R
+
-
Vi
R
22
4
3
4
32 kVVkR
V
R
V
kR
VV
R
VVO
oO ==
=
(2)
sC
V
R
VV
R
VV
R
Vv Oi
1
1
3
1
2
21
1
1 +
+
=
Do R nh nhau nen :
Vi = 3V1 -V2 -VO + sCRV1 = V1(3+sRC) + VO/k - VO (3)
135
Mat khac : 4
32
2
2
2
21
1 R
VV
sC
V
R
VV +=
Do V3 = 0; R2 = R4 = R C1 = C2 = C nen : V1 - V2 = sRCV2 + V2 -> V1 = V2 (2+sRC) = -(Vo/k)(2+sRC) (4)
Thay 4 vao 3 k
kVsRCsRC
k
VV o
Oi
+++=
1)3)(2(
222236)1()(
CRssRCsRCk
k
V
VsH
i
O
+++++
==
55
51
1
55)5( 222222
++
++
+=
+++
=
k
CRssRC
k
k
k
CRssRCk
k
RC
k
Q
sCRs
kk
kH O
o
OLP
5;
5
5;5
+==
++=
5
5
5
1
5
51
5
)5( +=
+
+=
+=
k
k
k
RC
kQ
O
136
C1 = C2 =C; HoLP > 1 V+ = V- = V2 (1)
AB
O
R
V
R
VV =
B
A
V1
V2Vo
R1
R
RR2
C2C1
+
-
Vi
22 )1()11
( kVR
RVV
RRV
R
V
A
BO
ABB
O =+=+= (2)
)1(1 2221
2
2
2
21 sCRVsCRVR
V
sC
VV+==
)1
1()1(
22
221
sCRk
V
sCR
CsRVV O +=
+= (3)
CsRVCsRVVsCRV
sC
VV
R
VV
sC
VVoi
Oi
1211121
1
11 )21(11
++=
+
=
CsRk
VsCR
sCR
sCR
k
VVCsRV oooi 11
2
21 )21(
)1(+
++=
21
22
12
2
21
2
)2()1()(
RRCsRRsCk
kCRRs
V
VsH
i
o
+++== (4)
Qk
RRC
RCR
CRR ool
1
1
)2()5(
11 12
2
2212
21
2 =
+==
7_20
137
kR
V
sRC
sCV O=+
1
1
)2(1
1
)1(1
+=
+=
sCRk
V
ksCR
sRCVV
o
O
)6()2(
1
12 RRC
kQ
o +
=
Thay 5 vao 6 ta c R2 can chng minh
7_23 . Cho C1 = C2 = C ; R1 = R2 =R
0=== + VVAOL (1); kR
V
sCR
V O=+
1
1
Vo
V+
V-V1
KR2=R=R==C
=C
+
-
Vi
R1C2
R2
C1
KR
SC
VV
SCR
V
R
VV oi11
111 ++
=
; oi SRCVSRCVSRC
SRCVVV +
+= 1
111
SRC
VRCSSCRVVCRSSCRVSRCVSRCVVV OOi +
+++=
1
222
1
222
1111
OOO V
kRCS
CRSkSRCSCRV
SCR
SCR
k
V
SRC
CRSSCR 222222 )1(11.
1
1 +++=
+
+
++=
222)1(1)(
CRSkSCR
kSCR
V
VsH
i
O
+++==
fo = 1khz ; Q = 10 ; k=1
138
k = 1 :
Qj
Q
j
CRSSRC
SRCsH
oO
o
+
=++
=2222
121
)(
QRC
RCCR OOO
1
;22
22
2 ===
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