Capitolo 12
12.1
( ) ( )( ) ( ) ( )( )
( ) ( )
( ) ( )
( ) HzfMATLABjA
Hzf
Hzfs
rads
ssAs
ssssFA
ssssA
Lv
L
LLLv
Lmidv
80.4 : | 302
50
79.4020223021
77.4230
2 | 30 |
3050 | 30= Si,
0,0 : Zeri| 2,-30- :Poli | 302
| 50 | 302
50
2222
2
2222
22
−=++
=
=−−+=
=≅=≅+
≈−
++==
++=
ωωωω
π
ππωω
12.2
( ) ( )( ) ( ) ( )( )
( ) ( )
( ) ( )
( ) z 100 :MATLAB | 6198.80
200
4.9902026198.8021
5.982619 | 619 |
619200
619= | esufficient è 1:5 divisione una Si, | 0 0, : Zeri| s
619,-80.8- :Poli
8.80619 | 200 |
8.80619200
000,10014002400
2222
2
2222
22
2
2
HjA
Hzf
Hzfs
rads
ssA
sradss
ssFAss
sssssA
v
L
LLv
Lmidv
++=
=−−+≅
=≅≅+
≈
−
++==
++=
++=
ωωωω
π
πω
12.3
( ) ( )( )( ) ( ) ( )
( )( )
( ) ( )
( )s
rad1.17 | z 72.2f :MATLAB | 2012
15150
54.115202201221
i.distanziat poco sono zeri gli e poli i o, | s
rad 15- 0, : Zeri| s
rad 20- 12,- :Poli
201215 | 150 |
201215150
LL2222
22
2222
==++
+=
=−−+≅
+++
=−=++
+−=
ωωω
ωωω
π
HjA
Hzf
N
sssssFA
sssssA
v
L
Lmidv
Si noti che ωL =12.1 rad/s non soddisfa l’ipotesi usata per ottenere l’Eq. (12.15), e la stima usando l’Eq. (12.15) è piuttosto scarsa.
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-1
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.4
( ) ( )( )( )
( )
( )
( )( ) ( )
zk 58.1 :MATLAB | 1010
103
58.11212101
101
21
59.1210 | 10 |
110
300 :Sì | s
,-1010- :Poli
110
110
1 | 300
110
110
300=1
101
10
1010103103103.33
109
H252242
11
1222
5
2
4
44
4
54
54
5454
5411
952
11
HfxjA
kHzf
kHzfs
radssArad
sssFA
ssssx
xsxsxsA
v
H
HHv
Hmid
v
=++
=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎠⎞
⎜⎝⎛
∞−⎟
⎠⎞
⎜⎝⎛
∞−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛≅
=≅≅+
≈
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
==
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
=++
=
−
−−
ωωω
π
πω
12.5
( )( )
( )
( )
( ) ( )( ) ( )
zM 59.1 :MATLAB | 1010
103102
59.11210312
101
101
21
59.1210 | 10 |
101
300 :Si
,103- : Zeri10- ,10- :Poli |
101
101
1031
| 300
101
101
1021
300=
101
10110
1031103
292272
2929
122
9
2
9
2
7
47
7
997
97
9
97
9
977
99
Hfxx
jA
MHzx
f
MHzfs
rads
sA
xss
xs
sFA
ssxs
ssxsx
sA
Hv
H
HHv
Hmid
v
=++
+=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎠⎞
⎜⎝⎛
∞−⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
=≅≅⎟⎠⎞
⎜⎝⎛ +
≈
∞⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
==
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
=
−
ωω
ωω
π
πω
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-2
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.6
( )( )( )
( )( )
( )
( ) ( )( ) ( )
kHzxx
xxjA
kHzxxx
f
radx
rad
xs
xs
xs
sFA
xs
xs
xs
xs
xsxx
xsxx
sA
v
H
Hmid
v
3.26 :MATLAB | 102103.1
105102
3.261210512
1021
105.11
21
bbero.interagire e iravvicinat molto sono zeri gli e poli i No, | ,s
105- :Zeri
s 2x10- ,1.5x10- :Poli |
1021
105.11
1051
| 3333
1021
105.11
1051
3333=
1021
103.11102105.1
1051105102
262252
2529
122
5
2
6
2
5
5
65
65
5
65
5
6565
559
++
+=
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎠⎞
⎜⎝⎛
∞−⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛≅
∞
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
==
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
=
−
ωω
ωω
π
12.7
( ) ( ) ( )( )
( ) ( )( ) ( ) ( )
( )
( ) ( ) ( ) ( )
HzHzf
HzHzf
xjA
sFsFssss
ssA
sssssxsA
H
L
v
HLv
v
133 :MATLAB | 14212122000
11000
121
380.0 :MATLAB | 356.002022121
2000100021106
No. | No. | , 0, 0, : Zeri| s
rad 2000- 1000,- 2,- 1,- Poli;
200
10001
5001
1 21
300
20001
10001
12120001000
106
2222
2222
22222222
28
2
28
=⎟⎠⎞
⎜⎝⎛
∞−⎟
⎠⎞
⎜⎝⎛
∞−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
=−−+=
++++=
∞∞
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎥⎦
⎤⎢⎣
⎡++
=
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +++
=
π
π
ωωωωωω
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-3
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.8
( ) ( )( ) ( )
( )( )( )( )
( ) ( )( )( )( ) ( ) ( )
( )( )
( ) ( ) ( ) ( ) ( )
Hz 0.61 :MATLAB | 5.121220012
3001
1001
1001
21
Hz 1.62 :MATLAB | 43.1021275321
300100753200110
alte. o basse frequenzeper sia dominante poloNessun
1067.6
3001
1001
2001
75311067.6
3001
1001
2001
7531
30010020010
122222
H
22222
2222222222
2222210
52
25
2
2
2
10
Hzf
Hzf
jA
sFsFxss
s
sssssxsA
ss
s
ssssssA
L
v
HLv
v
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎠⎞
⎜⎝⎛
∞−⎟
⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
=−−++=
+++++
++=
=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
⎥⎦
⎤⎢⎣
⎡+++
+=
⎟⎠⎞
⎜⎝⎛ +⎟
⎠⎞
⎜⎝⎛ +
⎟⎠⎞
⎜⎝⎛ +
++++
=
−
π
π
ωωωωωωωωω
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-4
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.9 Bassa frequenza:
1 MΩ2.43 M Ω
vi
2 k Ω 0.1 μF
43 k Ω
0.1 μF
10 μF13 k Ω
Rin
Rout
Centro banda:
vi
2 k Ω
1 MΩ
2.43 M Ω
43 k ΩRin
Rout
( ) ( )
( )( )
( )( ) ( )( )
( ) ( )( ) ( )( )
( ) ( ) ( ) ( )
( ) ( ) VVkmAVRRIVc
Hzf
Hzf
srad
kFsrad
kkFg
kF
srad
MkFsrad
kMF
MkmSMk
MA
kRrRRMR
mSVmA
VVIgA
RRRARRgRg
vvAb
DSSDDDD
L
L
Z
m
mid
DoDoutin
TNGS
Dmvt
inI
inmidoutmLm
g
dvt
2.165562.0 )(
58.769.727.4759.911.421 :ottiene si (12.15) Eq.l' Usando
59.72
:dominante è
69.71310
1 | 7.475.21310
111310
1
59.914310
1 | 11.4243.210
1
5.16143400.043.22
43.2
o.specificat ènon che momento dal 0= assumendo 43 | 43.2
400.012.022 | | )(
2222
33
555
3
7271
3
=+Ω=++=
=−++≅
=≅
=Ω
==ΩΩ
=
⎟⎟⎠
⎞⎜⎜⎝
⎛Ω
=
=Ω+Ω
==Ω+Ω
=
−=ΩΩ⎟⎠⎞
⎜⎝⎛
Ω+ΩΩ
−=
Ω=≅=Ω=
==−
=+
=−=−==
−−−
−−
π
πωω
ωω
ωω
λ
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-5
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.10
Bassa frequenza:
430 k Ω 560 k Ω
vi
5 k Ω 0.1 μF
220 k Ω43 k Ω
0.1 μF
10 μF13 k Ω
Rin
Rout
Centro banda:
220 k Ω43 k Ω
vi
5 k Ω
243 k Ω
Rin
Rout
( ) ( )
( )( )
( )( ) ( )( )
( ) ( )( ) ( )( )
( ) ( ) ( ) ( ) Hzf
srad
kFsrad
kkFg
kF
srad
kkFsrad
kkF
kkmSkk
kA
kRrRRkR
mSVmA
VVIgA
RRRARRgRg
vvA
L
Z
m
mid
DoDoutin
TNGS
Dmvt
inI
inmidoutmLm
g
dvt
5.1169.727.470.383.401 :(12.16) Eq.l' Usando
69.71310
1 | 7.475.21310
111310
1
0.382204310
1 | 3.40524310
1
1.1422043400.02435
243
o.specificat ènon che momento dal 0= assumendo 43 | 243
400.012.022 | |
2222
555
3
7271
3
=−++≅
=Ω
==ΩΩ
=
⎟⎟⎠
⎞⎜⎜⎝
⎛Ω
=
=Ω+Ω
==Ω+Ω
=
−=ΩΩ⎟⎠⎞
⎜⎝⎛
Ω+ΩΩ
−=
Ω=≅=Ω=
==−
=+
=−=−==
−−−
−−
ωω
ωω
λ
2π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-6
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.11
( ) ( )
( )
( ) ( ) ( )( )
( )( ) ( )( )
( ) ( ) ( ) ( )
( ) ( )
( )
( ) ( )( )
( )( ) ( )( )
( ) ( ) ( ) ( ) Hzf
srad
kkFsrad
kkF
srad
kFsrad
kkFFC
mSVmA
VVIgF
kkg
RC
sradf
Hzf
srad
MkFsrad
kMF
srad
kFsrad
kkFFCb
mSVmA
VVIgF
kkg
RC
sradf
L
Z
TNGS
Dm
mS
L
L
Z
TNGS
Dm
mS
L
1.503.5123180.383.4021 :ottiene si (12.15) Eq.l' Usando
0.382204310
1 | 3.40524310
1
3.51135.1
1 | 3185.213 5.1
1 | 5.1 Scegliendo
4.014.02 dove 52.1
5.2133141
11
314502 :dominante sia che Assumendo c
3.493.51231859.911.421 :ottiene si (13.16) Eq.l' Usando
59.914310
1 | 11.4243.210
1
3.51135.1
1 | 3185.213 5.1
1 | 5.1 Scelto
4.014.02 dove 52.1
5.2133141
11
314502 :dominante sia che Assumendo a
2222
7271
33
3
3
33
2222
7271
33
3
3
33
=−++≅
=Ω+Ω
==Ω+Ω
=
=Ω
==ΩΩ
==
==−
==ΩΩ
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
==≅
=−++≅
=Ω+Ω
==Ω+Ω
=
=Ω
==ΩΩ
==
==−
==ΩΩ
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
==≅
−−
−−
π
ωω
μω
μωμ
μω
πωω
π
ωω
μω
μωμ
μω
πωω
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-7
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.12
Bassa frequenza:
vi
100 Ω100 k Ω
4.3 k Ω
1.3 k Ω
4.7 μF 1 μFRI
RS
RD
R3
C1 C2
Rin Rout
( ) ( ) ( )( ) ( )
( ) ( )
( )( )
( ) ( )
Hzf
srad
kksrad
x
dBkkA
rkRrRRRRRg
RR
mSgRRgRR
RRgRR
RARR
RAc
RRCg
RRCss
sAsAb
mid
oDoDoutoutLm
Sin
moutminI
inLm
inI
invt
inI
inmid
D
mSI
midv
1242
:dominante è
59.91003.410
1 | 779173100107.4
1
3.221.131003.4005.0173100
173
) (assumendo 3.4 | | 1731
5 |
0= a zeri 2 | 1 | 1
1 |
1L1
6261
3
3
322
1
121
2
=≅
=Ω+Ω
==+
=
→+=ΩΩ⎟⎠⎞
⎜⎝⎛
Ω+ΩΩ
=
∞=Ω====Ω==
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=++
=
−−
πωω
ωω
ωωωωω
12.13
( ) ( ) ( )
( )
( ) ( )
( ) Hzfs
radkk
srad
xCb
Fx
Cg
RR
RRC
sradHz
mSin
inI
10402
:dominante è | 3.10752210
1
65401731001056.0
1 | F 56.0 Scegliendo
583.01731001028.6
1 | 17320013001
1 | 628010002 :dominante sia che Assumendo a
1L162
611
31
111L1
=≅=Ω+Ω
=
=+
==
=+
=Ω=ΩΩ==
+===≅
−
−
πω
ωω
ωμ
μ
ωπωωω
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-8
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.14
Bassa frequenza:
vi
200 Ω
4.3 k Ω
4.7 μF
2.2 k Ω
Rin
1 μF
51 k Ω
Rout
Centro banda:
vi
200 Ω
4300 Ω
Rin
51 k Ω
Rout
2.2 k Ω
( ) ( ) ( )( ) ( )
( ) ( ) ( )
( )( )
( ) ( )
( ) ( )
( )( )
( ) ( )
Hzf
srad
kksrad
kx
dBkkk
kA
kRrRRkRSAg
Hzf
srad
kksrad
x
dBkkA
kRrRRRRRg
RR
SmAgRRgRR
RRgRR
RARR
RAc
azeriRRC
gRRC
sssAsAb
mid
CoCoutinm
mid
CoCoutoutLm
Ein
moutminI
inLm
inI
invt
inI
inmid
C
mES
midv
6.122
:dominante è
37.151022010
1 | 1.7949.2200107.4
1
1.359.565102200004.049.2200
49.2
220 | 49.2 | 0004.01040 d
1512
:dominante è
8.18512.210
1 | 9469.24200107.4
1
4.1934.9512.204.09.24200
9.24
2.2 | | 9.241
04.0140 |
0= 2 | 1 | 1
1 |
1L1
6261
1L1
6261
3
3
322
1
121
2
=≅
=Ω+Ω
==Ω+
=
→+=ΩΩ⎟⎠⎞
⎜⎝⎛
Ω+ΩΩ
=
Ω===Ω===
=≅
=Ω+Ω
==+
=
→+=ΩΩ⎟⎠⎞
⎜⎝⎛
Ω+ΩΩ
=
Ω====Ω==
==⎟⎟⎠
⎞⎜⎜⎝
⎛+
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=++
=
−−
−−
πωω
ωω
μπ
ωω
ωω
ωωωωω
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-9
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.15
( ) ( )
( ) ( )
( )
( ) ( )
( )
( ) ( )
( ) ( )
( )
( ) ( )
Hzf
srad
kksrad
x
CFx
C
kkRSAgg
RRRR
C
sradHz
Hzfs
radkk
srad
xCb
Fx
C
RSmAgg
RRRR
C
sradHz
inmm
EininI
inmm
EininI
4932
:dominante è
8.18512.210
1 | 210024902001012.0
1
F 12.0 scelga Si | 118.024902001014.3
1
49.22500430 | 0004.01040 | 1 | 1
31405002 :dominante sia che Assumendo c
4722
:dominante è | 8.18512.210
1
29609.24200105.1
1 | F 5.1 Scegliendo
42.19.242001014.3
1
9.24254300 | 04.0140 | 1 | 1
31405002 :dominante sia che Assumendo a
1L1
6261
131
11
1L1
1L162
611
31
11
1L1
=≅
=Ω+Ω
==+
=
==+
=
Ω=ΩΩ====+
=
==≅
=≅=Ω+Ω
=
=+
==
=+
=
Ω=ΩΩ====+
=
==≅
−−
−
−
πωω
ωω
μμ
μω
πωωω
πωωω
ωμ
μ
ω
πωωω
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-10
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.16
( ) ( ) ( )
( )( )
( ) ( ) ( )( ) Hz
kxxkxf
kkkRRRkkrRRR
kkkRRRRkkkRRR
kmSkk
kRgRR
RA
kkkRRRkkkkrRRR
kmSg
mSmAIga
L
CSo
thES
IthinIS
LminI
inmid
CLin
m
oCm
1232
9.696675.381431011
15010101
0.131021
21
14310043 | 150101
3.14987151
9871300100 | 0.130.121 :SCTC
1941.3000.70.121
0.12
1.3010043 | 0.123.14300100
notonon V r | 3.1400.7100r | 00.7175.04040
766
332
211
321
Ao
=++
=⎥⎦
⎤⎢⎣
⎡Ω
+Ω
+Ω
≅
Ω=Ω+Ω=+=Ω=Ω+Ω
Ω=++
=
Ω=ΩΩΩ==Ω=Ω+Ω=+=
−=ΩΩ+Ω
Ω=
+=
Ω=ΩΩ==Ω=ΩΩΩ==
∞=Ω======
−−− ππ
β
β
π
π
π
(b) Si noti che il punto Q ipotizzato nella parte (a) non è completamente esatto.
SPICE porta a: (144 μA, 3.67 V), Amid = 43.9 dB, fL = 91 Hz
c( ) VEQ = VCCR1
R1 + R2
=12 100kΩ100kΩ + 300kΩ
= 3V | REQ = R1 R2 =100kΩ 300kΩ = 75.0kΩ
IC = βF
VEQ −VBE
REQ + βF +1( )RE
=100 3− 0.775kΩ + 101( )15kΩ
=145μA
V = V − I R − I R =12 − 0.145mA( ) 43kΩ( )−101
CE CC C C E E 1000.145mA( ) 15kΩ( )= 3.57 V
Questi valori concordano con i risultati SPICE elencati sopra nella parte (b).
12.17 ( )
( )
( )
( ) ( )
( ) Hzfs
radx
C
Hzfs
rads
rads
radx
Cb
FR
C
srad
sradHz
L
S
LL
21802
133501.96225 | 133507.22103.3
1F 3.3 scelgo Hz, 2500 a superiore essere devenon se o
26502
163201.96225 | 225 | 1.96
163207.22107.2
1 | F 7.2 Scelgo
86.27.2215390
1=1
153901.9622515700 | 1570025002
dominante. sia che suppongo e 12.3.1, sezione della valorii Utilizzoa
L63
3
L12
633
333
213
3
=++
≅=Ω
=
=
=++
≅==
=Ω
==
=Ω
=
=+−=−−===
−
−
πω
μωπ
ωω
ωμ
μω
ωωωωπω
ω
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-11
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.18
(a) Centro banda: 21.5 k Ω
vi
1 k Ω
43 k Ω43 k Ω
100k Ω 300k Ω
75 k Ω
Rin
Bassa frequenza:
vi
1 k Ω
75 k Ω
5 μF
43 k Ω43 k Ω
1 μF
22 μF13 k Ω
( ) ( ) ( )
( )( )
( ) ( ) ( )( )
( ) ( ) ( ) VkmAVkmARIVRIc
Hzkxxkx
f
kkkRRRkkrRRR
kkkRRRRkkkRRR
kmSkk
kRgRR
RA
kkkRRRkkkkrRRR
kmSg
mSmAIgb
CCECEE
L
CSo
thES
IthinIS
LminI
inmid
CLin
m
oCm
0.1243164.079.213100101164.0V
0.502
6.112887.1486101
11581022
16.13105
121
864343 | 158101
2.15987131
9871300100 | 6.136.121 :SCTC
1315.2156.66.121
6.12
5.214343 | 6.122.15300100
notonon V r | 2.1556.6100r | 56.6164.04040
CC
666
332
211
321
Ao
=Ω++Ω⎟⎠⎞
⎜⎝⎛=++=
=++
=⎥⎦
⎤⎢⎣
⎡Ω
+Ω
+Ω
≅
Ω=Ω+Ω=+=Ω=Ω+Ω
Ω=++
=
Ω=ΩΩΩ==Ω=Ω+Ω=+=
−=ΩΩ+Ω
Ω=
+=
Ω=ΩΩ==Ω=ΩΩΩ==
∞=Ω======
−−− ππ
β
β
π
π
π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-12
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.19
SCTC : R1S = RI + RG =1kΩ +1MΩ =1.00 MΩ | ω1 =1
1.00MΩ 0.1μF( )=10.0 rad
s
R2S = RS1
gm
= 6.8kΩ1
1.5mS= 607Ω | ω2 =
1607Ω 10μF( )
=165 rads
R3S = RD + R3 = 22kΩ + 68kΩ = 90kΩ | ω3 =1
90kΩ 0.1μF( )=111rad
s
fL ≅10.0 +165+111( )
2π= 45.5 Hz
12.20
SCTC : R1S = RI + RG =1kΩ + 500kΩ = 501kΩ | ω1 =1
501kΩ 0.1μF( )= 20.0 rad
s
R2S = RS1
gm
=10kΩ1
0.75mS=1.18kΩ | ω2 =
11.18kΩ 10μF( )
= 84.8 rads
R3S = RD + R3 = 43kΩ +10kΩ = 53kΩ | ω3 =1
53kΩ 0.1μF( )=189 rad
s
fL ≅20.0 + 84.8 +189( )
2π= 46.8 Hz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-13
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.21
(a) Bassa frequenza
vi
2 k Ω
100 k Ω22 k Ω
12 k Ω
4.7 μF 1 μFRI
R4
R3 R7
C1
C2
892 k Ω 0.1 μF
C3
Centro banda:
vi
2 k Ω
12 k Ω 100 k Ω22 k Ω
Rin
( )
( )( ) ( )
( ) ( )
( ) ( ) ( ) Hzfs
radkkRRC
is
radkkxRRC
mSkk
kRgRR
RA
kkkRkkkg
RR
gmS
VmAg
L
ginI
LminI
inmid
Lm
Sin
mm
2.190.825.3821 | 0.82
221001011
!0 che dato importanon | 5.3853.32107.4
11
7.24dB 30.218k200.053.32
53.3
0.1810022 | 53.35121
5000=1 | 200.011.02
7733
3
261
1
=+≅=Ω+Ω
=+
=
===Ω+Ω
=+
=
=ΩΩ+Ω
Ω=
+=
Ω=ΩΩ=Ω=ΩΩ==
Ω==
−
−
πω
ωω
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-14
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.22
vi 75 k Ω
2 k Ω
13 k Ω 100 k Ω
4.7 μF 10 μF
Rin
vi75 k Ω
2 k Ω
11.5 k Ω
(a) bassa frequenza centro banda
b( ) Rin = R1 R2 rπ + βo +1( )RL | RL =13kΩ 100kΩ =11.5kΩ | rπ =100
40 0.25mA( )=10.0kΩ
Rin = R1 R2 rπ + βo +1( )RL[ ]=100kΩ 300kΩ 10.0kΩ + 101( )11.5kΩ[ ]= 70.5kΩ
Amid =Rin
RI + Rin
βo +1( )RL
Rin
= 0.972101 11.5kΩ( )
2 +10.0 +101 11.5( )[ ]kΩ= 0.963 | RB = R1 R2 = 75kΩ
R1S = RI + RB rπ + βo +1( )RL[ ]= 2kΩ + 75kΩ 10.0kΩ + 101( )11.5kΩ[ ]= 72.5kΩ
ω1 =1
72.5kΩ( )4.7x10−6 = 2.94 rads
R3S = R7 + RE
RB RI + rπ
βo +1( )=100kΩ +13kΩ
1.95kΩ +10.0kΩ101
=100kΩ
ω3 =1
10−5 105( )=1rad
s fL ≅
2.94 +1( )2π
= 0.627Hz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-15
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.23
(a) Bassa frequenza:
892 k Ωvi
2 k Ω
4.7 μF
100 k Ω
0.1 μF
12 k Ω
Centro banda:
892 k Ωvi
2 k Ω
10.7 k Ω
Rin
b( ) gm =2 0.1mA( )
0.75V= 0.267mS | Rin = R1 R2 = 892kΩ | RL =12kΩ 100kΩ =10.7kΩ
Amid =Rin
RI + Rin
gmRL
1+ gmRL
= 0.9980.267mS( )10.7kΩ( )
1+ 0.267mS( )10.7kΩ( )= +0.739 -2.62 dB( )
ω1 =1
C1 RI + Rin( )=
14.7x10−6 2kΩ + 892kΩ( )
= 0.238 rads
ω3 =1
C3 R7 + RS1
gm
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=1
10−7 100kΩ + 12kΩ1
0.267mS
⎛
⎝ ⎜
⎞
⎠ ⎟
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
= 97.2 rads
fL ≅1
2π0.238 + 97.2( )=15.5 Hz
c( ) VDD = VDS + IS RS = 8.8V + 0.1mA 12kΩ( )=12.0 V
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-16
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.24
( )
( )( ) ( )( )
( )
( ) C appendicedall' 15.0 152.05.2133140
1
50.21 | 400.012.022 |
11
dominante sarà | <<+
59.914310
1 | 11.4143.210
1
314050021 :richiede SCTC
3
3
3
3321
7271
3
1=
FFkk
C
kg
mSVmA
VVIg
gRC
srad
MkFsrad
kMF
srad
CR
mTNGS
Dm
mS
LL
i iisL
μμ
ω
ωωωωωω
ωω
πω
→=ΩΩ
=
Ω===−
=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
≅→
=Ω+Ω
==Ω+Ω
=
==≅
−−
∑
12.25
( )
( ) ( )( )
( )
( )
( ) ( )
( )( )
( )
( ). 100 che sicuri essereper F 0.68 usare voler Potremmo :Nota
C appendicedell' valoriai vicino 56.0 592.05.2430200628
1
5.2104011 |
11
dominante sarà | 37.151022010
1
62810021 :richiede SCTC
. 100 che sicuri essereper 2.8 scegliere Dobbiamo :Nota
C appendicedell' valoriai vicino 8.6 08.7253.4200628
1
25104011 |
11
dominante sarà | 8.18512.210
11
62810021 :richiede SCTC
1
5
1
1
1162
3
1=
1
3
1
1
11632
2
3
1=
Hzf
FFkk
C
kg
gRRC
srad
kkF
srad
CRb
HzfF
FFk
C
gg
RRC
srad
kkFRRC
srad
CR
L
m
mEI
L
i iisL
L
m
mEI
LLC
i iisL
≤
→=ΩΩ+Ω
=
Ω==
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
≅→=Ω+Ω
=
==≅
≤
→=ΩΩ+Ω
=
Ω==
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=
≅→<<=Ω+Ω
=+
=
==≅
−
−
−
−
∑
∑
μ
μμ
ω
ωωωω
πω
μ
μμ
ω
ωωωωω
πω
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-17
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.26
( ) ( )
( )
( ) ( ) ( )
( )( )
( ) C appendicedall' 68 5.631587.99
1
158101
2.15987131
| 7.996.117.14126
6.1186101
1 | 864343
7.146.13105
1 | 6.132.15751
1262021 :richiede SCTC
datanon V | 2.1556.6100 | 56.6164.04040
2
22
6333
611
3
1=
A
FFC
kkrRR
RRs
rad
kxkkkRRR
kxkkkkrRRR
srad
CR
rkmSg
rmSmAIg
o
IBES
CS
BIS
i iis
om
oCm
μμ
βω
ω
ω
π
β
π
π
π
→=≅
Ω=Ω+Ω
Ω=+
+==−−=
=Ω
=Ω=Ω+Ω=+=
=Ω
=Ω=ΩΩ+Ω=+=
==
∞=Ω======
−
−
∑
12.27
( )
( ). da fissato limite il sotto forzare possibile èNon
i.soddisfatt essere possononon progetto di obiettivi Gli | 28.60.821221011
12210022 caso, ogniIn
28.6121 :richiede SCTC
3
73
733
3
1=
Cfs
rads
radkx
kkkRRRs
radCR
L
S
i iis
>=Ω
=
Ω=Ω+Ω=+=
==
−
∑
ω
π
12.28
( )
( ) ( )
( )
( )[ ] C appendicel' outilizzand 15.0155.075.3121008.62
1
75.31.0275.01 |
1
1
dominante è | 238.08922107.4
11
892 | 8.621021 :richiede SCTC
3
73
3
3161
1
21
3
1=
Fs
radkkk
C
kmAV
g
gRRC
srad
kkxRRC
kRRRs
radCR
m
mS
L
LGI
Gi iis
L
μ
ωω
ωωωω
πω
→=ΩΩ+Ω
=
Ω==
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+
=≅
→>>=Ω+Ω
=+
=
Ω====≅
−
∑
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-18
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.29
( )
( )( )[ ] ( )[ ]
( )( )
( )
( ) C. appendicedell' valorii outilizzand 39.0 351.01005.281
100101
0.1095.1131001
5.2894.24.31 | 94.2107.45.72
15.725.111010.107521
0.1025.040
100 | 5.1110013
4.31521 :richiede SCTC
3
73
361
1
3
1=
FFk
C
kkkkkrRR
RRR
srad
srad
xk
kkkkkRrRRR
kmA
rkkkR
srad
CR
o
IBES
LoBIS
L
i iisL
μμ
β
ωω
β
πω
π
π
π
→=Ω
=
Ω=Ω+Ω
Ω+Ω=+
++=
=−==Ω
=
Ω=Ω+ΩΩ+Ω=+++=
Ω==Ω=ΩΩ=
==≅
−
∑
12.30
fT =1
2πgm
Cπ + Cμ
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ | Cπ =
gm
2πfT
− Cμ | gm = 40IC
IC fT Cπ Cμ 1/2πrxCμ
10 μA 50 MHz 0.773 pF 0.5 pF 1.59 GHz
100 μA 300 MHz 0.75 pF 1.37 pF 580 MHz
50 μA 1 GHz 2.93 pF 0.25 pF 3.19 GHz
10 mA 6.06 GHz 10 pF 0.500 pF 1.59 GHz
1 μA 3.18 MHz 1 pF 1 pF 795 MHz
1.18 mA 5 GHz 1 pF 0.5 pF 1.59 GHz
12.31
Cπ = gmτ F | Cπ =gm
ωT
− Cμ | VCB = 5 − 0.7 = 4.3V | Cμ =Cμo
1+VCB
φ jc
=2pF
1+4.3V0.9V
= 0.832pF
Cπ =40 2x10−3( )2π 5x108( )
− 0.832pF = 24.6 pF | τ F =Cπ
gm
=24.6x10−12
40 2x10−3( )= 0.308ns = 308 ps
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-19
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.32
fT =1
2πgm
CGS + CGD
⎛
⎝ ⎜
⎞
⎠ ⎟ | gm = 2KnID
ID fT CGS CGD
10 μA 15.9 MHz 1.5 pF 0.5 pF
250 μA 79.6 MHz 1.5 pF 0.5 pF
2.47 mA 250 MHz 1.5 pF 0.5 pF
12.33
a( ) fT =32
μn VGS −VTN( )L2 =
32
600 0.25V( )10−4( )2
cm2
V − s= 22.5 GHz
b( ) fT =32
μn VGS −VTN( )L2 =
32
250 0.25V( )10−4( )2
cm2
V − s= 9.38 GHz
c( ) NMOS: fT =32
μn VGS −VTN( )L2 =
32
600 0.25V( )10−5( )2
cm2
V − s= 2.25 THz
PMOS: fT =32
μn VGS −VTN( )L2 =
32
250 0.25V( )10−5( )2
cm2
V − s= 938 GHz
d( ) NMOS: fT =32
μn VGS −VTN( )L2 =
32
600 0.25V( )2.5x10−6( )2
cm2
V − s= 36.0 THz
PMOS: fT =32
μn VGS −VTN( )L2 =
32
250 0.25V( )2.5x10−6( )2
cm2
V − s=15.0 GHz
12.34
a( ) rπ =125 0.025V( )
1mA= 3.13kΩ | Rin = 7.5kΩ rx + rπ( )= 2.44kΩ | RL = 4.3kΩ 100kΩ = 4.12kΩ
gm = 40 10−3( )= 40mS | Amid = −Rin
RI + Rin
gmRL = −2.44kΩ
1kΩ + 2.44kΩ
⎛
⎝ ⎜
⎞
⎠ ⎟ 40mS( )4.12kΩ( )= −117
b( ) Rin = 7.5kΩ rπ = 2.21kΩ | Amid = −2.21kΩ
1kΩ + 2.21kΩ
⎛
⎝ ⎜
⎞
⎠ ⎟ 40mS( )4.12kΩ( )= −113
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-20
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.35
rπ =125 0.025V( )
1mA= 3.13kΩ | gm = 40 1mA( )= 40mS | RL = 3kΩ 47kΩ = 2.82kΩ
Rin = RB rx + rπ + βo +1( )RL[ ]=100kΩ 0.35kΩ + 3.13kΩ + 126( )2.82kΩ[ ]= 78.2kΩ
a( ) Amid = Amid =Rin
RI + Rin
βo +1( )RL
rx + rπ + βo +1( )RL
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=78.2kΩ
1kΩ + 78.2kΩ
126 2820( )350 + 3130 +126 2820( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
= 0.978
b( ) Rin = RB rπ + βo +1( )RL[ ]=100kΩ 3.13kΩ + 126( )2.82kΩ[ ]= 78.2kΩ
Amid =78.2kΩ
1kΩ + 78.2kΩ
126 2820( )350 + 3130 +126 2820( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
= 0.978
12.36
rπ =125 0.025V( )
0.1mA= 31.25kΩ | gm = 40 0.1mA( )= 4mS
Rin = RErx + rπ
βo +1= 43kΩ
200Ω + 31.25kΩ126
= 248Ω | RL = 22kΩ 75kΩ =17.0kΩ
a( ) Amid =Rin
RI + Rin
gmRL =248Ω
100Ω + 248Ω4mS( )17.0kΩ( )= 48.5
b( ) Rin = RErπ
βo +1= 43kΩ
31.25kΩ126
= 247Ω | Amid =247Ω
100Ω + 247Ω4mS( )17.0kΩ( )= 48.4
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-21
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.37
( )
( )
( ) ( )
( )
( )
( )( )
( ) ( )
( )errore 50% 34%, | 400 ,200
2200600
2800004600600
133600
80000s | 6001
600s
80000600s5.0400003000.5s
errore 11% | 1000 ,1006
270033006
1033433003300
9.903300
103s | 11003
3300s | 30000033003s
errore 14% | 300 ,502
2503502
150004350350
9.42350
15000s | 3501
350s
15000350s2300007002s
errore 2% | 5000 ,1002
490051002
105451005100
0.985100
105s | 51001
5100s | 5000005100s
2
21
22
52
5
212
2
21
22
52
5
212
−−→±−
=−±−
=
−=−≅−=−≅
++=++
−−→±−
=−±−
=
−=−≅−=−≅++
−−→±−
=−±−
=
−=−≅−=−≅
++=++
−−→±−
=−±−
=
−=−≅−=−≅++
s
ssd
xs
xsc
s
ssb
xs
xsa
12.38
( )( )( sin errore di 10% , sin errore di 11% | 10 ,100 ,1000
10001001010000001110001110s+s :epolinomial il ndofattorizza
01.91110001000000s | 100
1110111000s | 1110
11110s
10000001110001110s+s
31
23
321
23
−−−=+++=++
−=−≅−=−≅−=−≅
++
sssss
s
)
In MATLAB: roots([1 1110 111000 1000000])
12.39 ( )( )
( )( ) 5- 15,- 20,- 100,- :reali radici quattro trovanosi calcolo, di foglioun ndo Utilizza|
398000513900174690190287106300000398000256950582304757142
'1
2345'
23456
i
iii
sfsfss
ssssssfsssssssf
−=
+++++=
++++++=
+
Utilizzando MATLAB: roots([1 142 4757 58230 256950 398000 300000]) ans = -100, -20, -15, -5, -1+i, -1-i
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-22
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.40
Cμ
r π
g vm 1
C π1v 2
v
rx
+
-
+
-
RI
vi
1 k Ω 300 Ω
2500 Ω 4.3 k Ω 100 k Ω
RB
7.5 k Ω
RinRC R3
a( ) rπ =100 0.025( )
0.001= 2500Ω | Cμ = 0.75 pF | Cπ =
40 10−3( )2π 5x108( )
− 0.75pF = 12.0 pF
Rin = 7.5kΩ rx + rπ( )= 2.03kΩ | RL = 4.3kΩ 100kΩ = 4.12kΩ | gm = 40 10−3( )= 40mS
Amid = −Rin
RI + Rin
⎛
⎝ ⎜
⎞
⎠ ⎟
rπ
rx + rπ
⎛
⎝ ⎜
⎞
⎠ ⎟ gmRL = −
2.03kΩ1kΩ + 2.03kΩ
⎛
⎝ ⎜
⎞
⎠ ⎟
2500Ω300Ω + 2500Ω
⎛
⎝ ⎜
⎞
⎠ ⎟ 40mS( )4.12kΩ( )= −98.6
ωH =1
rπoCT
| rπo = rπ rx + RB RI( )[ ]= 2500 300 + 7500 1000( )[ ]= 803 Ω
CT =12.0 + 0.75 1+ 40 10−3( )4120( )+4120803
⎡
⎣ ⎢
⎤
⎦ ⎥ =140 pF | fH =
12π 803( )1.4x10−10( )
=1.42 MHz
b( ) GBW = 98.6 1.42MHz( )=140 MHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-23
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.41
(a)
5 k Ω650 Ω
15 k Ω2.15 k Ω
+12 V
R1
R2 RC
RE
( )( )
( )( )
( ) ( )VmAVkkmARIRIVV
mAkk
VRR
VVII
kkkRVkk
kVV
EECCCCCE
EFEQ
BEEQFBFC
EQEQ
71.2 , 31.3 :Q Punto | 71.265.010010115.231.312
31.365.010175.3
7.031001
75.3155 | 3155
512
=⎟⎠⎞
⎜⎝⎛ Ω+Ω−=−−=
=Ω+Ω
−=
++−
==
Ω=ΩΩ==Ω+Ω
Ω=
βββ
(b)
Cμ
rπ
g vm 1
C π1v 2
v
rx
+
-
+
-
RI
vi
1 k Ω 300 Ω
2.15 k Ω 100 k Ω
RB
3.75 k Ω
RinRC
R3
Nota: Come progettisti, siamo liberi di modificare il progetto dell’amplificatore, ma tipicamente non possiamo cambiare le caratteristiche delle resistenze di ingresso e di uscita.
rπ =100 0.025( )
3.31mA= 755Ω | Cμ = 0.75 pF | Cπ =
40 3.31x10−3( )2π 5x108( )
− 0.75 pF = 41.4 pF
Rin = 3.75kΩ rx + rπ( ) = 823Ω | RL = 2.15kΩ 100kΩ = 2.11kΩ
gm = 40 3.31x10−3( )=132mS
Amid = −Rin
RI + Rin
rπ
rx + rπ
⎛
⎝ ⎜
⎞
⎠ ⎟ gmRL = −
823Ω1000Ω + 823Ω
⎛ ⎝ ⎜
⎞ ⎠ ⎟
755Ω300Ω + 755Ω
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 132mS( ) 2.11kΩ( )= −90.0
ω H =1
rπoCT
| rπo = rπ rx + RB RI( )[ ]= 755Ω 300 + 3.75kΩ 1kΩ( )[ ]= 260Ω
CT = 41.4 + 0.75 1+ 132mS( ) 2.11kΩ( )+2.11kΩ
0.260kΩ⎡ ⎣ ⎢
⎤ ⎦ ⎥ = 312pF | f H =
12π 260Ω( ) 3.12x10−10 F( )
=1.96 MHz
c( ) GBW = 90.0 1.96MHz( )=176 MHz | GBW ≤1
2π1
r Cx μ
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ =
12π 300Ω( ) 0.75 pF( )
= 707 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-24
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.42
(a)
500 k Ω65 k Ω
1.5 M Ω215 k Ω
+12 V
R1
R2 RC
RE
(b)
Cμ
rπ
g vm 1
C π1v 2
vR3
rx
+
-
+
-
RI
vi
50 k Ω 300 Ω
215 k Ω 5 M Ω
RB
375 k Ω
RinRC
VEQ =12V 500kΩ500kΩ +1.5MΩ
= 3V | REQ = 500kΩ 1.5MΩ = 3.75kΩ
IC = βF IB = βF
VEQ −VBE
REQ + βF +1( )RE
=1003− 0.7( )V
375kΩ + 101( )65kΩ( )= 33.1 μA
VCE = VCC − IC RC − IE RE =12 − 33.1μA( ) 215kΩ +101100
65kΩ⎛
⎝ ⎜
⎞
⎠ ⎟ = 2.71 V | Q - Point : 33.1 μA, 2.71 V( )
Nota: Come progettisti, siamo liberi di modificare il progetto dell’amplificatore, ma tipicamente non possiamo cambiare le caratteristiche delle resistenze di ingresso e di uscita. In ogni caso le affermazioni del problema indicano modifiche in tutte le resistenze
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-25
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
( ) ( )( )
( ) ( )
( )( )
( )[ ] ( )[ ]
( )( ) ( )( )( ) ( ) ( )( ) MHz
pFCrGBWMHzkHzGBWc
kHzFxk
fpFkkkmSC
kkkkRRrrrCr
kmSk
kkk
kRgrr
rRR
RA
mSxgkMkRkrrkR
C
pFpFxxCpFCk
Ar
x
HT
IBxoTo
H
LmxinI
inmid
mLxin
70775.03002
1121 | 86.39.24155=
9.241028.20.282
1 | 22886.8
21121132.11 75.00
0.28503753005.75 | 1
0.15521132.15.75300
5.751.6350
1.63
32.1101.3340 | 2115215 | 1.63=375=
0 Fisso correnti. basseper non vale costante f a modello Il
possibile. ènon - 329.0=75.01052101.3340 | 75.0= | 5.75=
1.33025.0100
10
6T
8
6
=Ω
=⎟⎟⎠
⎞⎜⎜⎝
⎛≤=
=Ω
==⎥⎦⎤
⎢⎣⎡
ΩΩ
+Ω++=
Ω=ΩΩ+Ω=+==
−=Ω⎟⎠⎞
⎜⎝⎛
Ω+ΩΩ
⎟⎠⎞
⎜⎝⎛
Ω+ΩΩ
−=⎟⎟⎠
⎞⎜⎜⎝
⎛++
−=
==Ω=ΩΩ=Ω+Ω
=
−−=Ω=
−
−
−
ππ
π
ω
πμ
μ
πππ
π
π
π
π
πμπ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-26
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.43
Rin = R1 R2= 4.3MΩ 5.6MΩ = 2.43MΩ | RL = 43kΩ 470kΩ = 39.4kΩ
gm =2ID
VGS −VTN
=2 0.2mA( )
1= 0.400mS |
Amid = −Rin
RI + Rin
gmRL = −2.43MΩ
2kΩ + 2.43MΩ0.400mS 39.4kΩ( )= −15.7
fH =1
2πrπoCT
| rπo = R1 R2 RI = 2.00kΩ
CT = 2.5pF + 2.5 pF 1+ 0.400mS( )39.4kΩ( )+39.4kΩ
2kΩ
⎡
⎣ ⎢
⎤
⎦ ⎥ = 93.7 pF
fH = 12π 2kΩ( )93.7x10−12 F( )
= 849 kHz
12.44 *Problema 12.44 – Amplificatore a source comune VDD 7 0 DC 0 VS 1 0 AC 1 RS 1 2 2K C1 2 3 0.1UF R1 3 0 4.3MEG R2 3 7 5.6MEG RD 7 5 43K R4 4 0 13K C3 4 0 10UF C2 5 6 0.1UF R3 6 0 1MEG *Modello a piccolo segnale del FET GM 5 4 3 4 0.4MS CGS 3 4 2.5PF CGD 3 5 2.5PF * .AC DEC 20 1 10MEG .PRINT AC VM(6) .PROBE .END Risultati: Amid = -15.7, fL = 8.52 Hz, fH = 866 MHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-27
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.45
(a)
430 k Ω
R1
R2 RD
RS
560 k Ω10 k Ω
3.9 k Ω
VDD = 12 V
( )
( )
( )( )
( )( ) corretta. è attiva regione La - 78.29.31066312.k 10 a Riduco off!pinch in ènon re transistoIl
795.09.31366312
663 e 629.212
5.09.321.5
| 2
:attiva regionein ntofunzioname il suppone Si
243560430 | 21.5560430
43012
2
2
VkkARIRIVVR
VkkARIRIVV
AIVVVmAkV
RIVVVVKI
kkkRVkk
kVV
SSDDDDDS
D
SSDDDDDS
DGSGSGS
SDGSEQTNGSn
D
EQEQ
=Ω+Ω−=−−=Ω
=Ω+Ω−=−−=
==→−⎟⎠⎞
⎜⎝⎛Ω+=
+=−=
Ω=ΩΩ==Ω+Ω
Ω=
μ
μ
μ
(b)
CGD
g vm 1
CGS1
v 2v
R3+
-
+
-10 k Ω 100 k Ω
RI
vi
1 k Ω
RG
243 k Ω
Rin RD
Rin = R1 R2= 430kΩ 560kΩ = 243kΩ | RL =10kΩ 100kΩ = 9.09kΩ
gm =2ID
VGS −VTN
=2 0.663mA( )
1=1.33mS
Amid = −Rin
RI + Rin
gmRL = −243kΩ
1kΩ + 243kΩ1.33mS( ) 9.09kΩ( )= −12.0
f H =1
2πrπoCT
| rπo = R1 R2 RI = 243kΩ 1kΩ = 0.996kΩ
CT = 2.5pF + 2.5pF 1+ 1.33mS( ) 9.09kΩ( )+9.09kΩ0.996kΩ
⎡ ⎣ ⎢
⎤ ⎦ ⎥ = 58.1pF
f H =1
2π 0.996kΩ( ) 58.1x10−11F( )= 2.75 MHz
c( ) GBW =12.0 2.75 MHz( )= 33 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-28
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.46
Cμ
rπ
g vm 1
C π1v 2
vR3
rx
+
-
+
-
RI
vi
1 k Ω 300 Ω
43 k Ω 43 k Ω
RB
75 k Ω
RCRin
( ) ( )
( )( )
( )( ) ( )
( )
( )( ) MHzFxk
f
pFkkkmSpFpF
rRRgCCC
kkRRRrrrCr
pFHzx
mSCgC
kmSkk
kRgRR
RA
kkkRRRkkkkrRRR
nVrkmSg
rmSmAIg
H
o
LLmT
IxoTo
H
T
m
LminI
inmid
CLin
Aom
oCm
10.11021.119.12
1
12119.1
5.215.2156.6175.034.11
19.1987300 2.15= | 1
34.1=75.01052
56.6
1315.2156.66.121
6.12
5.214343 | 6.122.15300100
notaon | 2.1556.6100 | 56.6164.04040
10
21
8
321
=Ω
≅
=⎥⎦⎤
⎢⎣⎡
ΩΩ
+Ω++=⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
Ω=+Ω+==
−=−=
−=ΩΩ+Ω
Ω−=
+−=
Ω=ΩΩ==Ω=ΩΩΩ==
∞=Ω======
−π
ω
πω
β
πμπ
πππ
μπ
π
π
12.47 *Problema 12.47 – amplificatore a emettitore comune VCC 7 0 DC 0 VS 1 0 AC 1 RS 1 2 1K C1 2 3 5UF R1 3 0 300K R2 3 7 100K RC 5 0 43K R4 7 4 13K C2 7 4 22UF C3 5 6 1UF R3 6 0 43K *Modello per piccolo segnale del BJT GM 5 4 8 4 6.56MS RX 3 8 0.3K RPI 8 4 15.24K CPI 8 4 1.34PF CU 8 5 0.75PF
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-29
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
* .AC DEC 100 1 10MEG .PRINT AC VM(6) .PROBE .END Risultati: Amid = -128, fL = 47 Hz, fH = 1.10 MHz
12.48 ( )
( ) ( )( ) ( )
( )( )
( )[ ] ( )[ ]( ) ( ) ( ) ( )
( )( )
( )ricondensato
trei tratiindipenden tensionidue solo sono ci e anello,un formano ricondensato treI
1
1
1
s=
0
12.96).-(12.88 eq. le vedanoSi a
2
1
2
2
1
c
CCCC
gCCCCC
gCgggCgCrRCCRgCCr
gCgggCgCggb
gggCgggCgCsCCCCC
sVsV
gCCsgsCsCgCCssI
LL
m
LL
oLLomLP
o
LLLmo
oLLomL
oLP
oLoLLomLLL
LLm
oS
+⎟⎟⎠
⎞⎜⎜⎝
⎛+
=++
++++≅
⎥⎦
⎤⎢⎣
⎡++++
=++++
≅
++++++++Δ
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡++−−
−++=⎥
⎦
⎤⎢⎣
⎡
μπ
μμπ
ππμπ
πμμππ
ππμπ
π
πππμπμμπ
μμ
μπμπ
ω
ω
12.49
CT = Cπ + Cμ 1+ gmRL( )= 20 pF +1pF 1+ 40 1mA( )1kΩ( )[ ]= 61 pF
fT =1
2πgm
Cπ + Cμ
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ =
12π
40 1mA( )20 pF +1pF
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
= 303 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-30
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.50
( ) ( ) ( ) ( ) ( )
( ) ( )( )
( ) ( )( ) ( ) ( ) ( ) Ω+=Ω+=
Ω+=
++
≅++
+=
++
=+
==
=+=+=+=+
=+
= −
kjxjZkjjZ
jjZ
ss
ss
ssA
sZY
Zb
FFACCAsC
sC
AsZAYa
inin
in
inin
inin
5.155.97102 | 6.2898.810
28.695.42000 :MATLAB oUtilizzand
101010
10101010
10101
101
1
10101101 | 1111
65
65
65
6
5
510
ππ
π
μ
12.51
a( ) Av s( )=
1RC
⎛
⎝ ⎜
⎞
⎠ ⎟
A s( )1+ A s( )
s +1
RC 1+ A s( )[ ] | A s( )= 10Ao
s +10 | Av s( )= 1
RC⎛
⎝ ⎜
⎞
⎠ ⎟
10Ao
s +10
1+ 10Ao
s +10
s +1
RC 1+10Ao
s +10⎛
⎝ ⎜
⎞
⎠ ⎟
Av s( )=1
RC⎛
⎝ ⎜
⎞
⎠ ⎟
10Ao
s2 + s 1+ Ao( )10 + s +10RC
=
10Ao
RC⎛
⎝ ⎜
⎞
⎠ ⎟
s2 + s 1RC
+10 1+ Ao( )⎡
⎣ ⎢
⎤
⎦ ⎥ +
10RC
Av s( )=
106
RC
⎛
⎝ ⎜
⎞
⎠ ⎟
s2 + s 1RC
+106⎡
⎣ ⎢
⎤
⎦ ⎥ +
10RC
≅
106
RC
⎛
⎝ ⎜
⎞
⎠ ⎟
s +106( ) s +1
105 RC
⎛
⎝ ⎜
⎞
⎠ ⎟
; ωL =1
105 RC
b( ) Av s( )=
107
RC
⎛
⎝ ⎜
⎞
⎠ ⎟
s2 + s 1RC
+107⎡
⎣ ⎢
⎤
⎦ ⎥ +
10RC
≅
106
RC
⎛
⎝ ⎜
⎞
⎠ ⎟
s +107( ) s +1
106 RC
⎛
⎝ ⎜
⎞
⎠ ⎟
; ωL =1
106 RC
c( ) Ao →∞lim Av s( )=
10Ao
RC⎛
⎝ ⎜
⎞
⎠ ⎟
10Aos=
1sRC
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-31
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.52
250 Ω
2500 Ω
CTrπ
rx
vi
( )
( ) ( )( )( ) ( )
( ) (
( ) ( )( )( )
)
( ) (
( ) ( )( )( )
)
( )
( ) ( Ω−Ω−=+
+=
−==
Ω−Ω−=+
+=
Ω−==
Ω−Ω−=+
+=
−==
=⎥⎦⎤
⎢⎣⎡ +++=Ω=ΩΩ=
1040836 :SPICE | 100075225002500250 :MATLAB zzando Utili
53.12127102
1 MHz, 1At
2262730 :SPICE | 247273025002500250 :MATLAB zzando Utili
1051.21271052
1 kHz, 50At
56.42750 :SPICE | 99.4275025002500250 :MATLAB zzando Utili
1025.1127102
1 kHz, 1At
1272272500250004.01115 | 2272502500
6
44
63
jjZ
ZZ
jpFj
Zc
jjZ
ZZ
xjpFxj
Zb
jjZ
ZZ
xjpFj
Za
pFCr
C
C
C
C
C
C
C
C
C
To
π
π
π
π
)
(d) *Problema 12.52 – Amplificatore a emettitore comune IS 0 1 AC 1 RX 1 2 0.25K RPI 2 0 2.5K CPI 2 0 15PF CU 2 3 1PF GM 3 0 2 0 40MS RL 3 0 2.5K .AC LIN 1 1KHZ 1KHZ *.AC LIN 1 50KHZ 50KHZ *.AC LIN 1 1MEG 1MEG .PRINT AC VR(1) VI(1) VM(1) VP(1) .END Si noti che l’approssimazione di CT non fornisce una buona stima di Zin alle alte frequenze (si noti la discrepanza a 1 MHz).
12.53 Amid = 39.2 dB, fL = 0 Hz, fH = 5.53 MHz
12.54
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-32
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
( ) ( ) ( ) ( )
( )[ ] ( )( )
( )( )
( )( )( ) ( )[ ] ( )( )
( )
( )( ) MHzFx
f
pFkkkmSpFpFC
kkkRRrrrb
MHzFx
f
pFkkkmSpFpF
rRRgCCC
CrfpFpF
HzxmSCgC
kkkRRRkkkRRrrr
nVrkmAI
VrmSmAIga
H
T
IBxo
H
o
LLmT
ToH
T
m
CLIBxo
AoC
ToCm
64.11041.16882
1
141688.012.412.40.40175.00.12
68815.703.13=
19.11040.19592
1
140959.012.412.40.40175.00.121
21 | 0.12=75.0
10520.40
12.41003.4 | 95915.75003.13=
notaon | 13.31
025.0125 | 0.4014040
10
10
8
3
=Ω
≅
=⎥⎦⎤
⎢⎣⎡
ΩΩ
+Ω++=
Ω=ΩΩ+Ω+=
=Ω
≅
=⎥⎦⎤
⎢⎣⎡
ΩΩ
+Ω++=⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
≅−=−=
Ω=ΩΩ==Ω=ΩΩ+Ω+=
∞=Ω======
−
−
π
π
ππω
β
ππ
πμπ
πμπ
ππ
π
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-33
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.55
Cμ
rπ
g vm 1
C π1v 2
vR3
rx
+
-
+
-
RI
vi
1 k Ω 400 Ω
43 k Ω 100 k Ω
RB
75 k Ω
RinRC
( ) ( ) ( ) ( )
( )[ ] ( )( )( )
( )( )
( )( )
( )( )( ) ( ) ( )( ) MHz
pFCrMHzMHzGBWb
MHzFxk
f
pFkkkmSpFpF
rRRgCCC
CrfpFpF
HzxmSCgC
kmSkk
kRgRR
RA
kkkRRRkkkkrrRRR
kkkkRRrrr
nVrkmAI
VrmSmAIga
x
H
o
LLmT
ToH
T
m
LminI
inmid
CLxin
IBxo
AoC
ToCm
53175.04002
1 21 | 12812.1114=
12.11009.131.12
1
10931.1
1.301.3000.4175.0523.01
21 | 523.0=75.0
105200.4
1141.3000.40.191
0.19
1.3010043 | 0.194.25300100
31.117540025=
notaon | 251.0
025.0100 | 00.41.04040
10
8
321
=Ω
==
=Ω
≅
=⎥⎦⎤
⎢⎣⎡
ΩΩ
+Ω++=⎟⎟⎠
⎞⎜⎜⎝
⎛+++=
≅−=−=
−=ΩΩ+Ω
Ω−=
+−=
Ω=ΩΩ==Ω=ΩΩΩ=+=
Ω=ΩΩ+Ω+=
∞=Ω======
−
ππ
π
ππω
β
μ
πμπ
πμπ
π
ππ
π
12.56
Rin = R1 R2= 4.3MΩ 5.6MΩ = 2.43MΩ | RL = 43kΩ 470kΩ = 39.4kΩ
gm =2ID
VGS −VTN
=2 0.2mA( )
1= 0.400mS |
Amid = −Rin
RI + Rin
gmRL = −2.43MΩ
2kΩ + 2.43MΩ0.400mS 39.4kΩ( )= −15.7
fH =1
2πrπoCT
| rπo = R1 R2 RI = 2.00kΩ
CT = 5pF + 2 pF 1+ 0.400mS( )39.7kΩ( )+39.7kΩ
2kΩ
⎡
⎣ ⎢
⎤
⎦ ⎥ = 78.5pF
fH = 12π 2kΩ( )78.5x10−12 F( )
=1.01 MHz | GBW = 15.7 1.01 MHz( )=15.9 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-34
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.57 Il Problema 12.57 dovrebbe riferirsi alla Fig. 12.34, o alla Fig. 1631.
( ) ( )( )
( ) ( )
( ) ( )
( )( ) MHzGBWMHzpFCr
f
pFCA
kRR
MHzMHzGBWA
RkRRS
R
pFpFpF
CCC
rgR
rRRgCCC
pFMHz
CCCr
f
ToH
Tmid
LC
mid
CCLL
T
omL
o
LLmT
TTTo
H
156= | 16.50.476562
12
1
0.47656813813064.015.09.19 | 2.30
1560250882813100
813100820 | 820 è 5% al valorepiù vicino Il
16059.31 = | 9.311560250882
858100
865100 | 858
6561064.
2.56
2.5615.0
9.195.4811 | 1
5.4856562
1 | 6562
12
1
=Ω
==
=⎥⎦⎤
⎢⎣⎡ +++=−=
Ω+Ω+ΩΩ
−=
Ω=ΩΩ=Ω=
=−=Ω+Ω+Ω
Ω−=
Ω=→Ω=Ω=⎟⎠⎞
⎜⎝⎛
Ω+
=
=−−
=−−
=⎟⎟⎠
⎞⎜⎜⎝
⎛+⎥
⎦
⎤⎢⎣
⎡+++=
=Ω
=Ω
==
ππ
πππ
π
μ
π
ππμπ
π
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-35
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.58
75 k Ω
43 k Ω
3 V
13 k Ω
+12 V
3 k Ω
43 k Ω 100 k Ω 100 Ω
30.1 k Ω
vi
75 k Ω
Rin
IC =100 3 − 0.775kΩ +101 13kΩ( )
= 0.166mA | VCE =12 − 43kΩIC −13kΩIE = 2.70 V
gm = 40 0.166mA( )= 6.64mS rπ 0 =100
6.64mS=15.1kΩ Cπ =
gm
ωT
− Cμ = 3.02 pF
Costanti di tempo di cortocircuito
R1S =100Ω + 75kΩ 300Ω +15.1kΩ +101 3kΩ( )[ ]= 60.8kΩ
R2S =10kΩ 3kΩ +15.1kΩ + 99.9Ω
101⎛
⎝ ⎜
⎞
⎠ ⎟ = 2.40kΩ
R3S = 43kΩ +100kΩ =143kΩ
fL ≈1
2π1
60.8kΩ( )1μF( )+
12.40kΩ( )2.2μF( )
+1
143kΩ( )0.1μF( )⎡
⎣ ⎦ ⎢ ⎢
⎤ ⎥ ⎥
= 43.9Hz
Costanti di tempo di circuito aperto
( )[ ]
( )( )( )( )
( )( )
( )( )
( ) ( )( )( )( ) ( ) MHzHzMHzGBW
kmSkmS
kkb 0.889.4327.954.9 54.9
364.611.3064.6
8.607.60A mid =−=−=
Ω+Ω
⎟⎠⎞
⎜⎝⎛
ΩΩ
−=
MHzpF
fpFC
kkmSkmSpF
kmSpFC
kkr
HTB
TB
27.90.443902
1 | 0.44
3901.30
364.611.3064.615.0
364.6102.3
390751003001.15 :12.2 tabelladella risultati i oUtilizzand 0
=Ω
==
⎥⎦
⎤⎢⎣
⎡ΩΩ
+Ω+Ω
++Ω+
=
Ω=Ω+Ω=
π
π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-36
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.59
( )[ ] ( )( )
( )( )( )
( )
( )( )
( )( )( )
( )
( )( )( )
( ) MHzGBWkk
kA
MHzpF
f
pFkkmS
kmSpFkmS
pFC
kRR
MHzGBWkkA
R
pFkRmSkmSpF
RmSpFC
MHzkkr
mid
H
TB
E
mid
E
EETB
227= | 6.3082.01011.153009.99
1.30100999.0
41.71.553902
1
1.55390
1.3082.064.611.3064.615.0
82.064.6102.3
12 e 820 sono 5% al più vicini valoriI
220= | 3.298621011.153009.99
1.30100999.0
862 :MATLAB oUtilizzand
4.54390
1.3064.61
1.3064.615.064.6102.3
4.543905.72
1C 390751003001.15
12.58 Prob. del e 12.2 tabelladella risultati i oUtilizzand
6
TB0
−=Ω+Ω+Ω+Ω
Ω−=
=Ω
=
=⎥⎦
⎤⎢⎣
⎡ΩΩ
+Ω+
Ω++
Ω+=
Ω=Ω=
−=Ω+Ω+Ω+Ω
Ω−=
Ω=
=⎥⎦
⎤⎢⎣
⎡ΩΩ
++
Ω++
+=
=Ω
=Ω=Ω+Ω=
π
ππ
Nota: Il punto Q cambierà leggermente e questo è stato trascurato.
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-37
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.60
( ) ( ) ( )
( )
( ) ( ) pFCrpFxmSCmSmAg
kmA
rVV
VIkIkVmAkk
Ia
xm
F
CCCEC
0.1 | 300 | 8.51=11022
4.66 | 4.66 66.140
51.1= 66.1025.0100 | corretto. è attiva regionein ntofunzioname Il 7.069.2
69.23.13.412 | 66.13.11015.7
7.03100
8 =Ω=−===
Ω=≥
=⎟⎟⎠
⎞⎜⎜⎝
⎛Ω−Ω−==⎥
⎦
⎤⎢⎣
⎡Ω+Ω
−=
μπ
π
π
α
7.5 k Ω
4.3 k Ω
3 V
1.3 k Ω
+12 V
200 Ω
4.3 k Ω 47 k Ω 250 Ω
3.94 k Ω
vi
7.5 k Ω
Rin
Rin = R1 R2 rx + rπ + βo +1( )RE1[ ]=10kΩ 30kΩ 0.350kΩ +1.51kΩ + 101( )200Ω[ ]= 5.60 kΩ
Rth = 7.5kΩ 250Ω = 242Ω | RL = 4.3kΩ 47kΩ = 3.94kΩ
Amid = −Rin
RI + Rin
βoRL
rx + rπ + βo +1( )RE1
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
= −5.60kΩ
350Ω + 5.60kΩ
100 3.94kΩ( )22.0kΩ
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
= −16.9
(b) Utilizzando le costanti di tempo di cortocircuito:
R1S = 250Ω + 7.5kΩ 350Ω +1.51kΩ +101 200Ω( )[ ]= 5.85kΩ
R2S = 4.3kΩ + 43kΩ = 47.3kΩ
R3S =1.1kΩ 200Ω +1.51kΩ + 350 + 242Ω
101⎛
⎝ ⎜
⎞
⎠ ⎟ =184Ω
fL ≅1
2π1
5.85kΩ( )5μF( )+
147.3kΩ( )1μF( )
+1
184Ω( )4.7μF( )⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=193Hz
(c) Utilizzando le costanti di tempo di circuito aperto:
( )( )( )( )
( )( )
( )( ) MHzfpFC
pFkmS
kmSpFmS
pFC
rRr
LTB
TB
xth
08.91 6.29
6.2959294.3
2004.66194.34.6611
2004.6618.51
592350242 : 12.2 tabelledella risultati i oUtilizzand 0
===
=⎥⎦
⎤⎢⎣
⎡ΩΩ
+Ω+
Ω++
Ω+=
Ω=Ω+Ω=+≅π
pF6.295922 Ωπ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-38
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.61
( )( )
( )( )( )
( )Ω=
=⎥⎦
⎤⎢⎣
⎡ΩΩ
++
Ω++
+=
=Ω
=Ω=Ω+Ω=+≅
305 :MATLAB oUtilizzand
4.2259294.3
4.66194.34.6611
4.6618.51
4.22125922
1 | 592350242
:12.60 Prob. del valorii e 12.2 tabellanella risultati i oUtilizzand
0
E
EETB
TBxth
R
pFkRmSkmSpF
RmSC
pFMHz
CrRrππ
( )( )( )
( )
( )( )( )[ ] ( )[ ]
( )( ) 8.11
1.3294.3100
08.625008.6
1
94.3473.4 | 2422505.7
08.630010151.1300.030101
9.116.225922
1
6.2259294.3
3004.66194.34.661 1
3004.6618.51
1 e 300 sono 5% al più vicini valoriI
1
121
6
−=⎥⎦⎤
⎢⎣⎡
ΩΩ
Ω+ΩΩ
−=⎥⎦
⎤⎢⎣
⎡++++
−=
Ω=ΩΩ=Ω=ΩΩ=
Ω=Ω+Ω+ΩΩΩ=+++=
=Ω
=
=⎥⎦
⎤⎢⎣
⎡ΩΩ
+Ω+
Ω++
Ω+=
Ω=Ω=
kk
kk
RrrR
RRRA
kkkRkR
kkkkkRrrRRR
MHzpF
f
pFkmS
kmSpFmS
pFC
kRR
Eox
Lo
inI
inmid
Lth
Eoxin
H
TB
E
ββ
β
π
π
π
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-39
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.62
R2SR1S
1 k Ω
10 k Ω
1 k Ω
1 k Ω
R1OR2O
1 k Ω
1 k Ω
1 k Ω
10 k Ω 10 k Ω1 k Ω
(a) SCTC:
R1S =10kΩ 1kΩ = 909Ω | R2S =1kΩ 1kΩ = 500Ω | ωL =1
909 10−6( )+
1500 10−5( )
=1300 rads
(b) OCTC:
R1O =10kΩ 2kΩ =1.67kΩ | R2O =1kΩ 11kΩ = 917Ω | ω H =1
1670 10−6( )+ 917 10−5( )= 92.3 rad
s (c) Ci sono due poli. La tecnica SCTC suppone che entrambi siano a bassa frequenza e si ottiene il polo più grande. La OCTC suppone che entrambi siano ad alta frequenza e si ottiene il polo più piccolo.
d( ) sC1 + G1 + G2( ) −G2
−G2 sC2 + G2 + G3( )⎡
⎣ ⎢
⎤
⎦ ⎥
V1
V2
⎡
⎣ ⎢
⎤
⎦ ⎥ = 0
Δ = s2C1C2 + s C2 G1 + G2( )+ C1 G2 + G3( )[ ]+ G1G2 + G2G3 + G1G3
Δ = s210−11 + s 1.30x10−8( )+1.20x10−6
Δ = s2 +1300s +1.20x105 → s = −1200,−100 rads
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-40
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.63
vi
100 Ω
1.3 k Ω 100 k Ω4.3 k Ω
Rin
RL = 4.3kΩ 100kΩ = 4.12kΩ | CGS = 3.0 pF | CGD = 0.6 pF
Rin = RS1
gm
=1.3kΩ1
5mS=173Ω
Amid =Rin
RI + Rin
gmRL =173Ω
100Ω +173Ω5ms( )4.12kΩ( )= +13.1
fH ≅1
2π1
CGD RL
⎛
⎝ ⎜
⎞
⎠ ⎟ =
12π
10.6 pF 4.12kΩ( )
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ = 64.4 MHz
12.64 *Problema 12.12 – Amplificatore a gate comune – modello ac per piccolo segnale VI 1 0 AC 1 RI 1 2 100 C1 2 3 4.7UF RS 3 0 1.3K RD 4 0 4.3K C2 4 5 1UF R3 5 0 100K G1 4 3 0 3 5mS .OP .AC DEC 100 0.01 100MEG .PRINT AC VM(5) VP(5) .END Risultati: Amid = +13.3, fL = 123 Hz, fH = 64.4 MHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-41
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.65
4.3 k Ω
2.2k Ω 51k Ω
200 Ω
vi2.11 k Ω
Rin
gm = 40 1 mA( )= 0.04S | rx = 300Ω | rπ =
100 0.025( )1 mA
= 2.50kΩ | Cμ = 0.6pF
Cπ =40 10−3( )
2π 5x108( )− 0.6 =12.1 pF | Rth = 4.3kΩ 200Ω =191Ω | RL = 2.2kΩ 51kΩ = 2.11kΩ
Rin = RErx + rπ( )βo +1
= 4.3kΩ0.3kΩ + 2.50kΩ( )
101= 27.6Ω
Amid =Rin
RI + Rin
βoRL
rx + rπ
⎛
⎝ ⎜
⎞
⎠ ⎟ =
27.6Ω200Ω + 27.6Ω
100 2.11kΩ( )2.80kΩ
= +9.14
ω H =1
191 12.1pF1+ 0.04 191( )
1+300191
⎛ ⎝ ⎜
⎞ ⎠ ⎟ + 0.6 pF 300Ω( )1+
0.04 2110( )1+ 0.04 191( )
⎡
⎣ ⎢
⎤
⎦ ⎥ + 0.6pF 2110Ω( )
f H =1
2π1
6.876x10−10 +1.938x10−9 +1.266x10−9
⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 40.9MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-42
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.66 Per prima cosa si stimano i parametri SPICE richiesti:
Cμ =CJC
1+VCB
PHIE⎛ ⎝ ⎜
⎞ ⎠ ⎟
ME | CJC = CJC = 0.6pF 1+2.80.75
⎛ ⎝ ⎜
⎞ ⎠ ⎟
0.333
≅1.01pF
τ F =Cπ
gm
=1
ωT
−Cμ
gm
=1
109π−
0.6pF40 1mA( )
= 303 ps
*Figura P12.14 – Amplificatore a base comune VCC 6 0 DC 5 VEE 7 0 DC -5 VI 1 0 AC 1 RI 1 2 200 C1 2 3 4.7UF RE 3 7 4.3K Q1 4 0 3 NBJT RC 4 6 2.2K C2 4 5 1UF R3 5 0 51K .MODEL NBJT NPN BF=100 RB=300 CJC=1.01PF TF=303PS .OP .AC DEC 50 1 50MEG .PRINT AC VM(5) .PROBE .END Risultati: Amid = 19.1 dB, fL = 149 Hz, fH = 43.8 MHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-43
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.67
4.3 k Ω
2.2k Ω 51k Ω
200 Ω
vi2.11 k Ω
Rin
IC = αF IE =100101
−0.7 − −10( )4300
⎡
⎣ ⎢
⎤
⎦ ⎥ = 2.14 mA | VCE =10 − 2.14mA( ) 2.2kΩ( )− −0.7( )= 5.99 V
gm = 40 2.14mA( )= 85.6mS | rx = 300Ω | rπ =100 0.025( )
2.14 mA=1.17kΩ | Cμ = 0.6pF
Cπ =85.6mS
2π 5x108( )− 0.6 = 26.7 pF | Rth = 4.3kΩ 200Ω =191Ω | RL = 2.2kΩ 51kΩ = 2.11kΩ
Rin = RErx + rπ( )βo +1
0.3kΩ +1.17kΩ( )101
= 4.3kΩ =14.5Ω
Amid = Rin
RI + Rin
βoRL
rx + rπ
⎛
⎝ ⎜
⎞
⎠ ⎟ =
14.5Ω200Ω +14.5Ω
100 2.11k( Ω)1.47kΩ
= +9.70
ω H =1
191 26.7 pF1+ 0.0856 191( )
1+300191
⎛ ⎝ ⎜
⎞ ⎠ ⎟ + 0.6 pF 300Ω( )1+
0.0856 2110( )1+ 0.0856 191( )
⎡
⎣ ⎢
⎤
⎦ ⎥ + 0.6pF 2110Ω( )
17.556x10−10 + 2.054x10−9 +1.266x10−9f H = 1
2π⎛ ⎝ ⎜
⎞ ⎠ ⎟ = 39.1 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-44
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.68
vi
2 k Ω
12 k Ω 100 k Ω22 k Ω
Rin
Rth =12kΩ 2kΩ =1.71kΩ | RL = 22kΩ 100kΩ =18.0kΩ | CGS = 3.0 pF | CGD = 0.6 pF
gm =2 0.1mA( )
1V1
gm
10.200mS
= 0.200mS | Rin =12kΩ =12kΩ = 3.53kΩ
Amid =Rin
RI + Rin
gmRL =3.53kΩ
2kΩ + 3.53kΩ0.200ms( )18.0kΩ( )= +2.30
fH =1
2π1
CGS
Gth + gm
+ CGD RL
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟
=1
2π1
3.0 pF0.5848 + 0.200( )mS
+ 0.6 pF 18.0kΩ( )
⎛
⎝
⎜ ⎜ ⎜ ⎜
⎞
⎠
⎟ ⎟ ⎟ ⎟
=10.9 MHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-45
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.69 ( )
( )
( )( )
( ) ( ) ok 37.9 | 254.0 | 73.467.22
2.012-30.7
| 8922.25.1
30.7182.25.1
5.1 :18 = per Q punto il trovoOra
2.011.022 | 67.21
67.3121.0 | 87.4122.25.1
5.112.21 problema sul basandomi K e V trovocosa primaPer
2
22
nTN
VVmAIVVVmSkV
RIVVkMMR
VVMM
MVVV
mSmAVV
IKVVVVV
VVkmAVVVVMM
MV
a
DSDGSGSGS
SDGSGGGG
GGDD
TNGS
DnTNTNGS
GSGSGGGG
===→−⎟⎠⎞
⎜⎝⎛Ω=
=−Ω=ΩΩ=
=Ω+Ω
Ω=
==−
==→=−
=→Ω=−=Ω+Ω
Ω=
vi
2 k Ω
12 k Ω 100 k Ω22 k Ω
Rin
( )( )
( )( )
( ) ( )
.f a cecontribuisnon quindi ,Cin corrente di segnale èc'non che noti Si
6.201121
12222100 | 48.4MOSFET. del f bassa della causa a
o trascuratessere puònon ingresso di polo del contributo il che noti Si
3.110.186.0
32.05848.00.3
1211
21
19.30.18320.048.22
48.2
48.2320.0
112112 | 320.067.226.4
254.026.0 | 0.3 | 0.1810022 | 71.1212
L2
3311
731
T
HzCRCR
f
kkkRRRkRRR
MHzkpF
mSpFRC
gGCf
kmskk
kRgRR
RA
kmS
kg
kRmSV
mAg
pFCpFCkkkRkkkR
SSL
outSinIS
LGDmth
GSH
LminI
inmid
minm
GDGSLth
=⎟⎟⎠
⎞⎜⎜⎝
⎛+=
Ω=Ω+Ω≅+=Ω=+=
=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
Ω++
=
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
++
=
+=ΩΩ+Ω
Ω=
+=
Ω=Ω=Ω==−
=
==Ω=ΩΩ=Ω=ΩΩ=
π
ππ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-46
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.70
gm = 40 0.25 mA( )=10.0mS | rx = 300Ω | rπ =100 0.025( )
0.25 mA= 10.0kΩ
Cμ = 0.6 pF | Cπ =0.01
2π 5x108( )− 0.6 = 2.58 pF | RB =100kΩ 300kΩ = 75.0kΩ
RL =13kΩ 100kΩ =11.5kΩ | Rth = 75kΩ 2kΩ =1.95kΩ
Rin = RB rx + rπ + βo +1( )RL[ ]= 75.0kΩ 300Ω +10.0kΩ + 101( )11.5kΩ[ ]= 70.5kΩ
Amid =Rin
RI + Rin
⎛
⎝ ⎜
⎞
⎠ ⎟
βo +1( )RL
rx + rπ + βo +1( )RL
= 0.972101 11.5kΩ( )
0.300 +10.0 +101 11.5( )[ ]kΩ= 0.964
fH ≅1
2π1
1950 + 300( ) 2.58 pF1+10mS 11.5kΩ( )
+ 0.6 pF⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
=1
2π1
2250( )0.622 pF( )=114 MHz
(b) Calcolando i parametri SPICE richiesti:
Cμ =CJC
1+VCB
PHIE⎛ ⎝ ⎜
⎞ ⎠ ⎟
ME | CJC = 0.6pF 1+11.80.75
⎛ ⎝ ⎜
⎞ ⎠ ⎟
0.333
≅1.54 pF
τ F =Cπ
gm
=1
ωT
−Cμ
gm
=1
109π−
0.6pF40 0.25mA( )
= 260 ps | TF = 260 ps
*Problema 12.70 – Amplificatore a collettore comune VCC 6 0 DC 15 VS 1 0 AC 1 RS 1 2 2K C1 2 3 4.7UF R1 3 0 100K R2 6 3 300K Q1 6 3 4 NBJT R4 4 0 13K C3 4 5 10UF R7 5 0 100K .MODEL NBJT NPN BF=100 TF=260PS CJC=1.54PF RB=300 .OP .AC DEC 100 0.1 200MEG .PRINT AC VM(5) VP(5) .END Risultati: Amid = 0.962, fL = 0.52 Hz, fH = 110 MHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-47
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.71
VBB = 9V 100kΩ100kΩ + 300kΩ
= 2.25V | RB =100kΩ 300kΩ = 75.0kΩ
IC =1002.25 − 0.7( )V
75.0kΩ +101 13kΩ( )= 0.251mA
gm = 40 0.251mA( )=10.0mS | rx = 300Ω | rπ =100 0.025( )0.251mA
= 9.96kΩ
Cμ = 0.6pF | Cπ =0.01
2π 5x108( )− 0.6 = 2.58 pF
RL =13kΩ 100kΩ =11.5kΩ | Rth = 75kΩ 2kΩ =1.95kΩ
Rin = RB rx + rπ + βo +1( )RL[ ]= 75.0kΩ 300Ω + 9.96kΩ + 101( )11.5kΩ[ ]= 70.5kΩ
Amid =Rin
RI + Rin
⎛
⎝ ⎜
⎞
⎠ ⎟
βo +1( )RL
rx + rπ + βo +1( )RL
= 0.972101 11.5kΩ( )
0.300 + 9.96 +101 11.5( )[ ]kΩ= 0.964
f H ≅1
2π1
1950 + 300( ) 2.58pF1+10mS 11.5kΩ( )
+ 0.6pF⎡
⎣ ⎢
⎤
⎦ ⎥
=1
2π1
2250( ) 0.622pF( )=114 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-48
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.72
( )
( )( )( )
( )
( )( )( )( ) ( )
( ) ( )( )
( )( ) ( )
Hzf
HzF
mSkk
fHzFk
f
MHzpF
kmSpFkk
f
dBkmS
kmSRg
RgRR
RA
kkkRkRRRmSVmAg
VmAmA
VVIKVVVVV
VVkmAVVVVMM
MV
mAIK
L
PP
H
Lm
Lm
inI
inmid
Linm
TNGS
DnTNTNGS
GSGSGGGG
Dn
5.15
5.151.0
267.01121002
1 379.07.48942
1
9.576.0
7.10267.01389222
1: 12.2 tabellaDalla
62.2- 739.0+=7.10267.01
7.10267.0998.01
7.1010012 | 892 | 267.075.01.02
356.075.01.022 | 10.275.0
85.2121.0 | 05.4102.25.1
5.1 1.0per richiesto di valoreil trovocosa primaPer
21
21
222
≅
=
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛Ω+Ω
==Ω
=
=
⎥⎦
⎤⎢⎣
⎡+
Ω+ΩΩ
=
Ω+Ω
=++
=
Ω=ΩΩ=Ω====
==−
==→=−
=→Ω=−=Ω+Ω
Ω=
=
μπμπ
π
S
Si noti che a bassa frequenza, RHP zero rende il calcolo di fH una stima molto bassa per il caso del FET. Si veda l’analisi nel Prob. 13.73 che mostra ωz = gm/CGS. *Problema 12.72 – Amplificatore a drain comune VDD 6 0 DC 10 VS 1 0 AC 1 RS 1 2 2K C1 2 3 4.7UF R1 3 0 1.5MEG R2 6 3 2.2MEG M1 6 3 4 4 NFET R4 4 0 12K C3 4 5 0.1UF R7 5 0 100K .MODEL NFET NMOS VTO=2.10 KP=0.356MA CGSO=30NF CGDO=6NF .OP .AC DEC 100 1 500MEG .PRINT AC VM(5) VP(5) .END Risultati: Amid = 0.740, fL = 15.5 Hz, fH = 195MHz – Si noti che ci sono dei picchi nella risposta.
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-49
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.73
( )
( )( )( )
( ) ( )
( )( )
( )( )( )( ) ( )
( ) ( )( )
MHzpF
kmSpFkk
f
dBkmS
kmSRg
RgRR
RA
kkkRkRRRmSV
mAg
VVmAIVVVV
mAkV
RIVVkMMRVVMM
MV
VVVmAmA
VVIKVVVVV
VVkmAVVVVMM
MV
VRIVVKVV
H
Lm
Lm
inI
inmid
Linm
DSDGSGSGS
SDGSGGGGGG
DD
TNGS
DnTNTNGS
GSGSGGGG
SDDSDDnTNDD
4.756.0
7.10519.01389222
1: 12.2 tabellaDalla
46.1- 846.0+=7.10519.01
7.10519.0998.01
7.1010012 | 892 | 519.010.256.3
379.02
ok 5.15 | 379.0 | 56.310.22356.012-11.8
| 8922.25.1 | 11.8202.25.1
5.1. 20 = con Q punto nuovo il trovoOra
356.075.01.022 | 10.275.0
85.2121.0 | 05.4102.25.1
5.1 10 :12.22 Prob. dal e , trovocosa primaPer
21
22
222
=
⎥⎦
⎤⎢⎣
⎡+
Ω+ΩΩ
=
Ω+Ω
=++
=
Ω=ΩΩ=Ω===−
=
===→−⎟⎠⎞
⎜⎝⎛Ω=
=−Ω=ΩΩ==Ω+Ω
Ω=
==−
==→=−
=→Ω=−=Ω+Ω
Ω=
=+=
π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-50
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.74
v x
g m vC μ C π
+
-
v
RL
ix
r π
i1
( ) ( )( )
( ) ( )
( )
( ) ( )
( )
( )
( ) ( )
( ) ( )
. generatore del resistenza dalla e ingresso di capacità dalla adeterminat è
1 e 1
11
1 ,per Quindi
11 a 1111
1 1
1
11
1
11
11
11
11
1
|
H
1
11
11
11
11
xthin
LoinLm
in
LoLmT
TmL
mLLm
L
o
Lm
L
LoLm
Lo
L
LoLo
LoLLoLLx
Lmxxx
rRC
RrRRg
CCC
RrRgCsY
gRC
mgRCRg
RC
per
RgRCs
RrRgsC
RrRrCs
RrRrrsC
ZY
rsCRrRrsC
rsCRrRRrsC
IVZ
RgsC
IgIgsC
IVIVsCI
+
++=+
+=
+++
+≅<<
>⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+<<→<<
+
>>+
+
+++
+≅
+++
+++
++==
++++
=+
+++==
⎟⎟⎠
⎞⎜⎜⎝
⎛+
+++
=+=
ω
β
βωω
ωωω
ββ
β
ββ
ββ
ππ
μ
π
π
ππ
π
π
π
π
π
ππ
ππ
ππ
ππ
πππ
ππ
πππ
ππππμ
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-51
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.75
v xgm vCGD
RL
CGS+
-
v
v s
ix
( ) ( ) ( )
( )( )
( )
GS
mz
Lm
GSGDINT
TGS
m
m
GS
Lm
LGS
Lm
LGS
Lm
LGSLm
GSGD
x
x
xLGSLm
LmGSS
LGSLm
xGSxGDx
LGSLm
xLmGSxSxGSxGDx
Cg
RgCCC
Cg
gC
RgRC
RgRC
RgRCsRg
CCsVI
VRsCRg
RgsCVNoteRsCRg
VsCVsCI
RsCRgVVRVgVsCVVVVsCVsCI
−=+
+≈
>=≈+
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+++
+=
+++
=++
+=
++=++=−+=
ωωω
ω
a in zero lo noti Si | 1
:<< Supponendo
& 1
|
11
11
1 : |
1
1 | |
SV
12.76
Av = −141 43dB( ) | fH = 6x106 Hz | fT ≥ 2 141( )6x106( )=1.69 GHz
GBW ≤1
rxCμ
| rxCμ ≤1
2π 1.69x109 Hz( )= 94.2 ps
12.77
Av =100 40dB( ) | fH = 40x106 Hz | fT ≥ 2 100( )4x107( )= 8.00 GHz
GBW ≤1
rxCμ
| rxCμ ≤1
2π 8x109 Hz( )=19.9 ps
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-52
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.78
( )
( ) ( ) ( )
( ) ( ) ( ) 0.5pF,760 177 0.75pF, :, erealistichpiù àpossibilit Altre
.0con pF 0.884 eccedere puònon ideale toreun transisPer | 1001.11 884.0
1084.8108.12
1101001 | 1=1
11=
1
1
1
1 | suppone Si
10000.1100 | 00.1
100100 |
3
86
5
ΩΩ
=+
=
==++++
++⎟⎟⎠
⎞⎜⎜⎝
⎛++
≅
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+++
=
≅=>>
Ω=≅==Ω
==
−
−
xu
xx
xH
LLmx
LLmx
x
LLmx
x
LLmx
H
xxox
m
omLmmid
rC
rCrx
pFC
xx
CCrCRRgCr
CRRgCrrRRgCr
rRRgCCr
rrrrrr
kmSg
rmSk
gRgA
μμ
μμμμ
μμμμπ
πππ
π
πω
ω
β
12.79
( )
( )
.1420 che dato attraverso fornire possiamonon che noti Si
672202
1642
| 12.2100164.0120
164=103100120
110252
1001201 | 1
| 1
22
12
6
DDLDLD
n
mDL
m
m
mI
mmidL
Im
LmLm
inI
inmid
GDLH
VVRIRI
mA
VmS
mSKgIkR
mSgFx
g
x
gR
gAR
RgRgRg
RRRA
CR
>>=
=⎟⎠⎞
⎜⎝⎛
==Ω=⎟⎠⎞
⎜⎝⎛ +=
→
⎟⎟⎠
⎞⎜⎜⎝
⎛Ω+
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+=⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
+≅
+=≅
−
π
ω
Una possibilità:
vi
RI
RS RL
C1 C2
+V DD
L
Questo non è davvero un progetto realistico. La corrente e la potenza sono troppo alte. Dobbiamo trovare un FET con un elevato Kn.
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-53
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.80
( ) ( )( )
( ) ( )
( )
0.28 | 87.101.
33.9
. 5.4=05.015.0= .015.0= Scelgo 0.01S. di maggiore essere deve g
. 78125.02025.025.0
vogliamo valido),sia quadratico modello il (affinchè inversione forteer
2005.02
| 93301.033.9
5117
10100102521
100 |
100517100
110252
1001512100
1
1
1
2
m
2
222
1266
=Ω=−
=
=≥→≥−
===Ω≤→−=
⎥⎦
⎤⎢⎣
⎡−=+
⎥⎦
⎤⎢⎣
⎡⎥⎦⎤
⎢⎣⎡ ++
=
⎥⎦
⎤⎢⎣
⎡⎥⎦⎤
⎢⎣⎡ +++
=
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡+++
=
−
Lmm
L
Dm
DTNGS
mm
n
mDL
Lm
LLm
LLm
LLm
th
LLmGDGSth
H
Rgkg
R
mAISg
AIVVV
P
ggKgIR
Rg
xRRg
RRgpFpFx
RRgpFpFRRRgCCR
μ
ππ
ω
12.81
fH ≤
12πRLCμ
=1
2π 12kΩ 47kΩ( )2 pF( )→ fH ≤ 8.33 MHz
12.82
rπ1
gm1 v1
cu1
cπ1
rx1
RL
+
-
v1
rx2 + r π2Cπ1 & Cμ1
Q2
RL
1gm1
ib
ibQ1
Q2
Cπ2 & C μ2
Utilizzo l’approccio della costante di tempo a circuito aperto:
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-54
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
( ) ( )( )( )( ) ( )
( )( )( )( )
( )
( ) ( )
Ω=⎟⎠⎞
⎜⎝⎛ Ω
+Ω⎟⎠⎞
⎜⎝⎛ +
=−==−=
Ω=Ω==
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++++⎟⎟
⎠
⎞⎜⎜⎝
⎛+++⎟
⎠⎞
⎜⎝⎛ +
=
⎟⎟⎠
⎞⎜⎜⎝
⎛++=+=
⎟⎠⎞
⎜⎝⎛ +=⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
+≅=
++
+≅
++⎥
⎦
⎤⎢⎣
⎡+++
++−≅
<<
⎟⎠⎞
⎜⎝⎛ ++===≅=≅
<<⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+
++≅
−−⎟⎟⎠
⎞⎜⎜⎝
⎛+
−⎥⎦
⎤⎢⎣
⎡+++
++−≅
−−
45110
50.230050.2=10
=
7.205.01061040 | 62.15.0
1061040
0.25250=
2= | 50.2
1025.0100
21
11101
1110
12
12
+
:uscitaall' indietro tornache /2i=i per che eccetto C di calcolo ilper usato stesso lo è circuito l :
10 1
C di calcolo ilper usato stesso lo è circuito l :11
10101
11
:con zionesovrapposi la uso e i Suddivido :11101 | 10 | 100 | 10
. supponendo 1
11 :
2222
8
3
28
4
1
22
22222
1211
211
2222222
xb
T2
222
1222
T2
211
2221
11
12221
221
22x1
21121
121
22221
211
22221
22111
kkrrrR
pFpFx
CpFpFx
C
kmAVrRUsek
mAVr
RRRgCCR
rRRgrCrrC
RRRgRRRgRRR
IR
rrrg
rrR
IR
rrivR
ggi
rrrri
ggi
rrrrrriv
rrR
RgrRivR
rrgrr
rrRgrr
rrRiR
RiRrrrrr
rrririvRa
xO
oL
O
LLmO
x
LLmx
xH
O
LLmO
LLmOOO
O
xm
xO
O
x
x
xO
x
oxx
m
x
xo
xoxxx
xO
LmxLx
xO
omo
xLmo
oxLxO
LxLx
o
xo
xoxxxxxO
πππ
ππ
π
πμππμ
ππ
ππππμ
μ
ππππ
π
ππ
ππππ
π
πππ
π
ππ
μππ
ππ
πππ
πμ
πππ
πμ
ππ
ω
βββ
ββ
ββ
ββ
β
f H =1
2π
1.62pF 3kΩ + 2.5kΩ11
⎛ ⎝ ⎜
⎞ ⎠ ⎟ + 0.5pF 300Ω( )1+ 40mS 25kΩ( )+
25kΩ300Ω
⎡ ⎣ ⎢
⎤ ⎦ ⎥
+451Ω 20.7pF + 0.5pF 1+ 40mS 25kΩ( )+25kΩ902Ω
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎡
⎣ ⎢ ⎤
⎦ ⎥
⎧
⎨ ⎪ ⎪
⎩ ⎪ ⎪
⎫
⎬ ⎪ ⎪
⎭ ⎪ ⎪
−1
= 393 kHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-55
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
( )
( )
( )
( )
( ). 741 leoperazionanell'
E-C/C-C cascata della usodell' ragione una è migliorata passante banda La dMiller. di zionemoltiplica alla soggetto ènon C perchè Darlington ioneconfiguraz alla
rispetto migliore edecisament passande banda una offre E-C/C-C cascata La
640
45125254015.07.20451
3005.011
5.2362.1
21
111
101
.Cper usicitaall' ritornanon /2i=i la e, solo vede :icambiament importanti dueper che eccetto stesso lo tepraticamen è circuito Il
1
1
2222211
211
2xb
11
μ
πμππμ
ππ
μ
μ
π
ω
c
kHzkkmSpFpF
pFkkpFf
RRRgCCRrCrrC
rCb
H
O
LLmOx
xH
x
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛
ΩΩ
+Ω++Ω+
Ω+⎟⎠⎞
⎜⎝⎛ Ω+Ω
=
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+++++
+=
−
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-56
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.83
rπ
g mvbe
Cu
Cπ
RC
+
-
vbevcm ro
2REECEE
2
Si suppone che RL rappresenti un carico di modo differenziale tra i collettori. La degradazione del CMRR inizia con lo zero nella funzione di trasferimento del guadagno di modo comune.
( )
( ) ( ) ( )
[ ]
collettore al base la unisce C quando dB 6- raggiunge CMRR Il
annullato. e 2
5.0 e
2
1
è aledifferenzi modo di risposta nella dominante polo il nulla, r base di resistenza unaPer 2
11 tivamenteapprossima è comune modo di guadagno il dc,In
.1 è dominante polo Il . di prossimitàin è altopiù polo Il
222=
da ideterminat sono dominanti terminiI
2
211
2
12
11
2
2
:dominante radice della zionefattorizza la oUtilizzand222
tivamenteapprossima è numeratore il dominanti, terminii Mantenendo
22
PcmPdmPdm
x
PcmT
2
2
μ
μ
μ
μππμπμ
μπ
μπμ
π
ππμπμ
μμ
ππππ
ωωω
β
ωω
β
μβ
ω
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
=
+⎥⎦
⎤⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛ +=+⎟
⎠⎞
⎜⎝⎛ +++⎟
⎠⎞
⎜⎝⎛ +Δ
⎟⎠⎞
⎜⎝⎛ −
⎟⎟⎠
⎞⎜⎜⎝
⎛−
≅+⎟
⎠⎞
⎜⎝⎛ −
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
+⎟⎠⎞
⎜⎝⎛ −
−≅
−+⎥⎦
⎤⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛ −+⎟
⎠⎞
⎜⎝⎛ +≅
++++−=−
−⎥⎦
⎤⎢⎣
⎡++++⎟
⎠⎞
⎜⎝⎛ +=++
LCmdm
LC
EEooCcm
C
CmEE
CmCEE
mEE
EE
EEoo
f
EEEEooom
EE
EEmo
Z
EEmoom
EEEE
cmCoeomcmm
coeEE
omEE
cmm
RRgACRR
RrRA
CR
GsCgCCsGgGCCsgsCCCCs
CC
RrCCCRrgCgCC
Gggg
GggggCgCCsCCCsN
vGgsCvggvgsC
vgvGgggCCsvggsC
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-57
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
gm = 40IC = 4x10−3 ro =50 +10.1
10−4 = 611kΩ
Acm = RC1
βoro
−1
2REE
⎛
⎝ ⎜
⎞
⎠ ⎟ = 6kΩ 1
100 611kΩ( )−
120 MΩ
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
= −2.02x10−4 (-73.9dB)
Adm = −0.5 4x10−3( )6kΩ100kΩ
2
⎛
⎝ ⎜
⎞
⎠ ⎟ = −10.7 CMRRdB = 94.5 dB
fZ ≅1
2π
1βoro
− 12REE
⎛
⎝ ⎜
⎞
⎠ ⎟
Cμ −CEE
2⎛
⎝ ⎜
⎞
⎠ ⎟
= 12π
-3.36x10-8
-0.2 pF
⎛
⎝ ⎜
⎞
⎠ ⎟ = 26.8 kHz
ωPdm =1
RLCμ
= 88.4 MHz ωPdm =1
RCRL
2
⎛
⎝ ⎜
⎞
⎠ ⎟ Cμ
= 99.0 MHz
Si veda il grafico nel Prob. 12.84.
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-58
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.84
Le tre curve dall’alto al basso sono: CMRR, -20log(Acm), 20log(Adm/2). Alle alte frequenze (vicino a 100 MHz), il polo di Adm annulla il polo di Acm.
12.85
ω H =gm1
CGS1 + CGS 2 + CGD2 1+ gm1ro2 + gm2ro2( )
gm1 = gm 2 = 2 25x10−6( ) 51
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 10−4( )=158 μS | ro2 ≅
50V0.1mA
= 500kΩ
CGS1 = 3pF | CGS 2 = 3pF | CGD1 = 0.5pF | CGD 2 = 0.5pF
f H =1
2π158μS
3pF + 3pF + 0.5pF 1+ 2 0.158mS( )500kΩ[ ]= 294 kHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-59
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.86
ω H =gm1
CGS1 + CGS 2 + CGD2 1+ gm1ro2 + gm2ro2( ) | ID 2 = 5ID1 =1.00mA | ro2 =
50V1mA
= 50kΩ
gm1 = 2 25x10−6( ) 51
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 2x10−4( )= 224μS | gm 2 = 2 25x10−6( ) 25
1⎛ ⎝ ⎜
⎞ ⎠ ⎟ 1x10−3( )=1.12mS
CGS & CGD ∝W : CGS1 = 3pF | CGS 2 =15pF | CGD1 =1pF | CGD2 = 5 pF
f H =1
2π0.224mS
3pF +15pF + 5pF 1+ 1.12mS + 0.224mS( )50kΩ[ ]= 99.3 kHz
12.87 La risposta più probabile è
ω H =gm1
Cπ1 + Cπ 2 + Cμ 2 1+ gm1 + gm 2( )ro2[ ] | IC 2 ≅10IC1 =1.00mA | ro2 =
60V1.00mA
= 60kΩ
Cπ1 =40 10−4( )
1.2x109π− 0.5pF = 0.561pF | Cπ 2 =
40 10−3( )1.2x109π
− 0.5pF =10.1pF
f H =1
2π40 10−4( )
0.561pF +10.1pF + 0.5pF 1+ 40 1.1mA( )60kΩ[ ]= 478 kHz
In ogni caso, C� dovrebbe essere approssimativamente proporzionale all’area di emettitore:
Cμ2 =10Cμ1 = 5.00 pF | Cπ 2 =40 10−3( )
1.2x109π− 5.00pF = 5.10pF
f H =1
2π40 10−4( )
0.561pF + 5.10pF + 5.00pF 1+ 40 1.1mA( )60kΩ[ ]= 48.2 kHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-60
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.88 Con l’aggiunta di rx, dobbiamo rivedere le costanti di tempo di circuito aperto.
g m1 v1
+
v1
-
r x1
r π1cπ1
cμ1
r o1
g m2 v2
cμ2
cπ2
ro2
+
v2
-
rπ2
r x2
Si suppone: rx << ro
( )
( )[ ]
( )2221
21
1
22221
2111
11
11
112211
11
111
11
2211
111
111
1
21
122222
111 :dominante sarà termineultimoL'
111
| =on 111 :
11
| 11 :
11con comune emettitore a stadio uno di parti sono &
omm
xmH
omm
xmx
xm
xH
xoxo
mo
xo
mxm
xo
xox
mxo
m
xm
mxo
rgCg
rg
rgCCg
rgrCrg
rC
rRrrrRc
Rg
rRC
grgrR
rrgg
gggRC
grg
grrrCC
++
≅
⎭⎬⎫
⎩⎨⎧
+++
+++
=
≅+++
=
≅+
≅
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+++
++=
+≅⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
−
−
μ
μπμπ
μπμμ
π
π
πππ
ππμπ
ω
ω
β
La risposta più probabile è
IC 2 ≅ 4IC1 =1.00mA | ro2 =50V
1.00mA= 50kΩ | gm 2 = 40 0.001( )= 40mS
Cπ1 =40 2.5x10−4( )
109π− 0.3pF = 2.88pF | Cπ 2 =
40 10−3( )109π
− 0.3pF = 9.73pF
f H ≅1
2π1
1+ 0.01S 175Ω( )0.01S
0.3pF 1+ 40mS 50kΩ( )[ ]= 964 kHz
In ogni caso, C� dovrebbe essere approssimativamente proporzionale all’area di emettitore:
Cμ2 = 4Cμ1 =1.2pF | f H ≅1
2π1
1+ 0.01S 175Ω( )0.01S
1.2pF 1+ 40mS 50kΩ( )[ ]= 241 kHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-61
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.89
ω H =gm1
Cπ1 + Cπ 2 + Cμ2 1+ gm1 + gm 2( )ro2[ ] | IC 2 ≅ IC1 =100μA
ro2 =60V
100μA= 600kΩ | Cπ 2 = Cπ1 =
40 10−4( )108π
− 2 pF =10.7pF
f H =1
2π40 10−4( )
10.7pF +10.7pF + 2pF 1+ 2 40( ) 0.100mA( )600kΩ( )⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
= 66.2 kHz
12.90 L’analisi utilizza l’approccio della costante di tempo in circuito aperto.
M 2
IO
IREF
C3
C 2
C1
M1
M3
C1:
g m1 vx
+
vx
-
r o
ix
+
-
v1
g m v 1
g m
1
C1 = CGS1 + CGS 2 | C2 = CGD2 + CGS 3 | C1 = CGD3
R1O : vx = ix + gmv1( ) 1gm
| v1 = −μ f vx − vx | R1O =vx
ix
=1
gm μ f + 2( )
C2:
gm1 vx
vx
r o
ix
+
- g m v x
gm
1
i
i
C3:
gm1 vx
v x
r o
ix
+-
g m v 1
gm
1
i
i
r ovg
+
-
v 1
vs
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-62
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
R2O : vx = ix − i( )ro − gmvx − ix( ) 1gm
| i = gmvx − ix
2vx = ix ro +1
gm
⎛
⎝ ⎜
⎞
⎠ ⎟ − ro gmvx − ix( ) | R2O =
vx
ix
=2ro +
1gm
μ f + 2≅
2gm
R3O : vx = ix − gmv1( )ro +ix
gm
− ix + i( )ro | i = ix | v1 = −2ixro −ix
gm
R3O =vx
ix
= 2μ f ro + 4ro +1
gm
≅ 2 μ f + 2( )ro ≅ 2μ f ro
ω H ≅1
2CGS
gmμ f
+ 2 CGS + CGD
gm
+ 2μ f roCGD
≅1
2μ f roCGD
=1
2gmro2CGD
f H ≅1
2π1
2 2 2.5x10−4( )2.5x10−4( ) 502.5x10−4
⎛ ⎝ ⎜
⎞ ⎠ ⎟
2
10−12( )= 5.63 kHz
Nota: R3O trascura ogni resistenza collegata. Se esiste un carico, essenzialmente tutta la ix passerà attraverso il carico RL, e la risposta di frequenza sarà migliorata in maniera significativa. Per questo caso, R3O ≈ RL + ro ≈ ro.
12.91
a( ) rπ =100 0.025( )
15x10−6 =167kΩ | Cμ = 0.5 pF | Cπ =40 15x10−6( )2π 75x106( )
− 0.5pF = 0.773 pF
rπo = rπ = 499 Ω | gm = 40 15x10−6( )= 0.6mS | ω H =1
rπoCT
rx =167kΩ 500Ω |
CT = 0.773+ 0.5 1+ 0.6mS 430kΩ( )+430kΩ499Ω
⎡ ⎣ ⎢
⎤ ⎦ ⎥ = 561pF | f H =
12π 499( ) 5.61x10−10( )
= 568 kHz
b( ) rπ =100 0.025( )
5x10−5 = 50.0kΩ | Cμ = 0.5 pF | Cπ =40 5x10−5( )2π 75x106( )
− 0.5pF = 3.74 pF
rπo = rπ = 495 Ω | gm = 40 5x10−5( )= 2.00mS | ω H =1
rπoCT
rx = 50kΩ 500Ω |
CT = 3.74 + 0.5 1+ 2.0mS 140kΩ( )+140kΩ495Ω
⎡ ⎣ ⎢
⎤ ⎦ ⎥ = 285pF | f H =
12π 495( ) 2.85x10−10( )
=1.13 MHz
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-63
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.92
a( ) Cμ =1 pF | Cπ =40 125x10−6( )2π 100x106( )
−1pF = 6.96 pF | rx = 500Ω | gm = 40 125x10−6( )= 5.00mS
CT = 6.96 +1.0 2 +5.00mS 62kΩ( )
2+
62kΩ0.500kΩ
⎡
⎣ ⎢
⎤
⎦ ⎥ = 288pF | f H =
12π 500( ) 2.88x10−10( )
=1.11 MHz
b( ) Cπ =40 1x10−3( )
2π 100x106( )−1pF = 62.7 pF | rx = 500Ω | gm = 40 1x10−3( )= 40.0mS
CT = 62.7 +1.0 2 +40.0mS 7.5kΩ( )
2+
7.5kΩ0.500kΩ
⎡
⎣ ⎢
⎤
⎦ ⎥ = 230pF | f H =
12π 500( ) 2.30x10−10( )
=1.39 MHz
12.93
a( ) Cμ =1 pF | Cπ =40 100x10−6( )2π 100x106( )
−1pF = 5.37 pF | rx = 500Ω | gm = 40 100x10−6( )= 4.00mS
rπ =100 0.25( )
10−4 = 25kΩ | rπo = 500Ω 25kΩ = 490Ω
f H =1
2π 490Ω( ) 5.37 + 2( )pF + 500 + 75kΩ( )1pF[ ]= 2.01 MHz
b( ) Cμ =1 pF | Cπ =40 1x10−3( )
2π 100x106( )−1pF = 62.7 pF | rx = 500Ω | gm = 40 1x10−4( )= 40.0mS
rπ =100 0.25( )
10−3 = 2.5kΩ | rπo = 500Ω 2.5kΩ = 417Ω
f H =1
2π 417Ω( ) 62.7 + 2( )pF + 500 + 7.5kΩ( )1pF[ ]= 4.55 MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-64
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.94
( )[ ]( ) ( )
( )
( )( )( )
( ) ( )[ ]
( ) ( )
( )[ ]
( ) ( )
( )
( ) ( )
( ) Hzsxsxsx
f
sxpFkpFkmS
k
kkkRpFpFpFCIgC
sxkkkmSpFpFCR
kkkkk
f
HzFFk
f
kkkRkkkR
kkkkkf
MMA
AkmSA
kkkkRA
krkRkRA
H
thCT
m
To
H
L
Sth
L
v
vtvt
Lvt
EBmid
k 5761052.11054.11007.12
1
1052.1125071.3101217.04.1591
25071.3
71.32.547.49.25 | 1011512C
1054.1610.007.307.38.671139 610
07.325065.1815009.257.42.54R
12.10.5 sezione alla riferito ,Per
533223001
10.151234037231999
21
30081
5.071.365.1250 | 71.32.547.49.25
0.1525065.1815009.252.547.4R 12.10.5 sezione alla riferito ,Per
dB 59.5or 943972.020578.401.11
0.972=21781+500
21781= | 20526.38.62
26.325065.1815009.257.4 | cambianon
5002
1= ,65.1= ,25.9k=28.51= :10.8 sezione alla riferito ,Per
877
8
33
722
L2
63
5S
32
21
333
=++
=
=Ω+Ω+Ω+
Ω+Ω
Ω=ΩΩΩ==−=→∝=+
=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛
ΩΩ
+Ω++Ω=
Ω=ΩΩ+ΩΩΩΩ=
=⎟⎟⎠
⎞⎜⎜⎝
⎛Ω
+Ω
++++=
Ω=Ω+Ω
Ω+=Ω=ΩΩΩ=
Ω=ΩΩ+ΩΩ+ΩΩ=
+=−−Ω
Ω=
−=Ω−=
Ω=ΩΩ+ΩΩΩ==
Ω=Ω
ΩΩΩ
−−−
−
−
π
ω
μμπ
πμπ
π
π
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-65
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.95
( )( )
( )( )( )
( )
( )
( )
( ) ( )
( )
( )
( ) ( )
( ) ( )
( ) kHzsxsxsx
f
sxpFkpFkmS
k
kkkkR
sxkkkmSpFpFCr
pFpFpFCIgCC
kkkRkkR
kRrkR
sxSpFpFkCRkR
f
Hzf
kkkRkkkR
kRRrRRR
kkRkkkkR
f
dorM
MA
nAkkmSAkkR
mSAkkRkkr
kRRkRkRA
H
th
To
CT
m
inL
Soth
TthL
H
L
Sth
inBoC
SS
thS
L
v
vtvtI
vtI
CEmid
5261031.81095.11092.92
1
1031.8125008.250232.06.791
25008.2
08.28.5127.4
22.54
1095.1544.002.202.21361179544
791402
02.28.1925.227.48.51
5962505442
39.2 | 54422.12598
1092.9990039939901.0115 9.9 | 399
239.2598
12.10.5 sezione alla riferito ,Per
8331584.6139876103199921
28881
00.108.23.3250 | 08.21.2735.28.51
3.1622
| 4.11151
1195552750
5522.1262022.17 | 64.1
239.2
22.172.12620
12.10.5 sezione alla riferito ,Per
B 4.53 46795.012791.301.11
cambiaon | 1278.1925.28.62 | 25.28.5135.2
91.320.157810 | 5781139620 | 40.12
39.2=
35.2= ,750= ,11= ,39= :10.8 sezione alla riferito ,Per
978
9
3
722
2
22
422
811
63
3322
54
3
322
112
22121
=++
=
=Ω+Ω++
Ω+Ω
Ω=ΩΩΩ
=
=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛
ΩΩ
+Ω++=
=−=→∝=+
Ω=ΩΩ=Ω
Ω=
Ω=Ω+ΩΩ
==Ω=Ω
Ω=
=⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛
ΩΩ
+Ω++Ω=Ω=Ω
Ω=
=+++++=
Ω=Ω+Ω
Ω+=Ω=ΩΩΩ=
Ω=+⎟⎟⎠
⎞⎜⎜⎝
⎛=Ω=
+=
Ω=ΩΩΩ
=Ω=⎟⎟⎠
⎞⎜⎜⎝
⎛ ΩΩ+ΩΩ=
+=−−Ω
Ω=
−=ΩΩ−=Ω=ΩΩ=
−=ΩΩ−=Ω=ΩΩΩ=Ω=Ω
ΩΩΩΩ
−−−
−
−
−
π
ω
π
π
πμπ
π
π
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-66
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.96
fo =1
2π LCGD
=1
2π 10−5 5x10−12( )= 22.5 MHz
gm =2ID
VGS −VTN
=0.02
2= 0.01S ro =
10.0167
+10
0.01= 6.99kΩ
Av = −gm ro RL( )= −0.01S 6.99kΩ 10kΩ( )= −41.1
BW =1
2πRPCGD
=1
2π 4.11kΩ( ) 5pF( )= 7.75 MHz Q = 22.5
7.75= 2.90
12.97
a( ) fo =1
2π C + Cμ( )L→ C =
12πfo( )2 L
− Cμ =1
2π 10.7x106 Hz( )[ ]210−5 H
− 2pF = 20.1pF
b( ) ro =75V +10V
10mA= 8.50kΩ | BW =
12π 8.5kΩ( ) 22.1pF( )
= 847kHz | Q = 10.70.847
=12.6
c( ) Q =100 | BW =fo
Q=107kHz | ro =
1ωo C + Cμ( )
=1
2π 107kHz( ) 22.1pF( )= 67.3kΩ
n2 =67.3kΩ8.50kΩ
= 7.918 | n = 2.81
d( ) Cμ' =
Cμ
n2 =2pF7.918
= 0.253pF | C = 22.1− 0.253 = 21.9pF
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-67
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.98
VC 6 μH
220 pF
CD
CD =20pF
1+VC
0.9
(a) CD =20pF
1+0
0.9
= 20pF | C =20 220( )20 +220
pF =18.3pF
fo =1
2π LC=
12π 6μH( ) 18.3pF( )
=15.2MHz
(b) CD =20pF
1+100.9
= 5.75 pF | C =5.75 220( )5.75 +220
pF = 5.60 pF
fo =1
2π LC=
12π 6μH( ) 5.60 pF( )
= 27.5MHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-68
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.99(a)
C1 C2
4
1
r π
1 pF
5 pF
RL
gm v
+
-
v
20 pF
CEQ
REQ
20 pF
8.24 pF
5 μΗn 2 REQ
3.33 k Ω
CEQ
n 2
CEQ = Cπ + Cμ 1+ gmRL[ ]= 5pF +1pF 1+ 40 1mA( ) 5kΩ( )[ ]= 206 pF
CP = 20pF +CEQ
n2 = 20pF +206pF
52 = 28.2pF | fo =1
2π 5μH( ) 28.2pF( )=13.4 MHz
REQ = rπRL
1+ gmRL( ) ωRLCμ( )2 = 2.5kΩ5000
1+ 200( ) 2π 13.4MHz( ) 5kΩ( ) 1pF( )[ ]2 =2.5kΩ 140Ω =133Ω
RP = n2REQ = 25 133Ω( )= 3.33kΩ | BW =1
2π 3.33kΩ( ) 28.2pF( )=1.70 MHz | Q = 13.4
1.70= 7.88
Si noti il grande errore che viene introdotto dall’uso della sola rπ come resistenza di ingresso.
v s
2
20 pF C A v1
SC A
n 2 REQ
5 μH v s
2 20 pF
C EQ
n 2
v1 = j2π 13.4MHz( ) 20pF( ) 12
⎛ ⎝ ⎜
⎞ ⎠ ⎟ 3.33kΩ( )vi = j2.80vs
vo = −gmRL( ) j2.80vs
5
= −40 10−3( )5kΩ( )j0.560vi | Av =112∠ − 90o
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-69
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
RLgm v v s
2.805
j
+
-
v
12.99(b)
C EQ
10 pF10 pF
133 Ω 206 pFg m v
RL
+
-
vv s
5 μH
REQ
CT = 5pF +1pF 1+ 40 1mA( ) 5kΩ( )[ ]= 206 pF | CP = 20pF + 206pF = 226pF
fo =1
2π 5μH( ) 226pF( )= 4.74MHz
REQ = rπRL
1+ gmRL( ) ωRLCμ( )2 = 2.5kΩ5000
1+ 200( ) 2π 4.74MHz( ) 5kΩ( ) 1pF( )[ ]2
REQ = 2.5kΩ 1.12kΩ =774Ω | BW =1
2π 774Ω( ) 226pF( )= 910 kHz | Q = 4.74
0.910= 5.21
vo = j2π 4.74MHz( ) 10pF( ) 774Ω( ) −gmRL( )vi
vo = j2π 4.74MHz( ) 10pF( ) 774Ω( ) −0.04mS 5kΩ( )[ ]vi | Av = 46.1∠ − 90o
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-70
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.100 (a) Riferendosi alle eq. (12.180-12.182) in Jaeger and Blalock,
fo =1
2π L CGD + C( )=
12π 10μH( ) 25pF( )
=10.1 MHz
ro ≅1
λID
=1
0.02 /V( ) 20mA( )= 2500Ω gm ≅ 2KnID = 2 0.005( )(0.02 =14.1 mS
BW = 12π ro( ) CGD + C( )
= 12π 2.5kΩ( ) 25pF( )
= 2.55 MHz
Q =10.12.55
= 3.96 | Amid = −gmro = −μ f = −35.4
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-71
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.100(b)
gmVS
ro
LC1
C2+
-
VOVS
V1
CGD
CGS
( ) ( )( ) ( )
( )
( )
( )
( )
( )
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
=⎟⎟⎠
⎞⎜⎜⎝
⎛−=−
<<⎟⎟⎠
⎞⎜⎜⎝
⎛+−≅>=
⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛+−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−
+−
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−−=
−
−−=
+=
⎥⎥⎦
⎤
⎢⎢⎣
⎡++−
+=Δ
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ++++=Δ
+=
Δ−
==⎥⎦
⎤⎢⎣
⎡⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
+−
−+++=⎥
⎦
⎤⎢⎣
⎡ −
2
22
22
12
1
121
2
22
2
2
2
22
2
2
2
22
2
2
2
22
1
21
22
221
1
1
1111==
:13.193)-(13.192 Eq. alle iRiferendos
per 1 So, . Ma
1111
111
1 | 1
1
Lascio
| 10
CC
CrQ
CC
CrLCCrLCC
gCg
QBW
CC
jACg
jC
Cj
CCCjA
gCj
LC
rg
LCCg
Cg
CCj
Cg
jA
LCCCC
LCCjg
Cgj
LCCCC
CCj
LCCCC
LCsCg
CgssCCs
CCC
gCjCjVVjA
VV
sLsCsC
sCgCCCsVgsC
EQEQoo
EQEQo
oEQoEQo
o
EQ
oo
TEQ
fovTGD
m
oEQf
o
EQ
EQ
fov
m
GDo
o
om
EQo
o
EQ
oo
EQ
GDo
EQ
mo
ov
EQ
EQo
EQ
o
EQ
o
EQ
EQEQ
EQ
EQ
EQ
o
EQ
oEQ
GDEQ
mGD
s
oV
o
oGDsmGD
ω
ωωω
ωωμωωω
ωωμ
ωωμ
ω
ω
ωωω
ωωω
ωω
ωωω
ωωω
2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-72
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
( )( )( )
( ) ( )( )( )
( )( )( )
( )( )o
GD
m
EQfmid
o
oEQ
EQ
MHzpFC
g
pFpF
CC
A
kHzMHzBW
pFkMHzkmA
r
MHzpFH
fpFpFCC
CC
ωππ
μ
π
μπ
4505
02.0005.0221
21 :zioneapprossimal' oVerificand
1.7540451250002.0005.021
6664.16
9.10=
4.16404514550.29.102=Q | 50.2
2002.01
9.102.21102
1 | 2.214045
4045
2
2
2
>>=⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
−=⎟⎟⎠
⎞⎜⎜⎝
⎛+−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−=
=
=⎟⎠⎞
⎜⎝⎛ +ΩΩ=≅
===+
=+
12.101 ( )
( )( ) ( )
( )( )
( )( )( ) 7.7050501250002.0005.021 | 9.15
635.01.10=
635
505015050.22
1 :13.113 prob. del risultati i oUtilizzand
50.22002.01 | 1.10
251021
45= | 50= | 2555
2
1
1221
21
−=⎟⎟⎠
⎞⎜⎜⎝
⎛+−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+−==
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+Ω
=
Ω====
=++
+=
pFpF
CCAQ
kHz
pFpFpFk
BW
kmA
rMHzpFH
f
pFCpFCpFCpFCCpFCC
fmid
oo
EQ
μ
π
μπ
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-73
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.102 *Problema 12.102(a) - Fig. P12.100(a) VS 1 0 AC 1 CGD 1 2 5PF GM 2 0 1 0 14.1MS RO 2 0 2.5K C1 2 0 20PF L1 2 0 10UH .AC LIN 400 8MEG 12MEG .PRINT AC VM(2) VP(2) .PROBE V(2) .END Risultati: Amid = 35.3, fo = 10.1 MHz, BW = 2.50 MHz *Problema 12.102(b) - Fig. P12.100(b) VS 1 0 AC 1 CGD 1 2 5PF GM 2 0 1 0 14.1MS RO 2 0 2.5K C1 2 0 40PF C2 2 3 40PF L1 3 0 10UH .AC LIN 400 8MEG 12MEG .PRINT AC VM(2) VP(2) VM(3) VP(3) .PROBE V(2) V(3) .END Risultati: Amid = 75.1, fo = 10.1 MHz, BW = 670 kHz *Problema 12.102(c) - Problema 12.101 VS 1 0 AC 1 CGD 1 2 5PF GM 2 0 1 0 14.1MS RO 2 0 2.5K C1 2 0 45PF C2 2 3 50PF L1 3 0 10UH .AC LIN 400 8MEG 12MEG .PRINT AC VM(2) VP(2) VM(3) VP(3) .PROBE V(2) V(3) .END Risultati: Amid = 70.7, fo = 10.1 MHz, BW = 640 kHz
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-74
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.103
( ) ( )
( )[ ] ( )
( )( )[ ]
( )( )( )
( )( ) ( ) ( )( )( )[ ]
( )( )
( )( )( )
( )( )( )( )( ) 4
3
P21
12
1
221111
1
21
22
31
1221
22
12
2
111111
1041.41000.103.582002.62
| 2401.25 02.6 | ingresso.all' k 58.3 e uscitaall'
100R che dato zioneapprossimaun' è questa che noti Si | 1.2512
0.39703.582
12
1= | 3.58140100
510002.6211100100
1
1 | 02.670102
12
1
65570 Richiede
70302020 | 30115200.10005.001.02 | corretta. è esaturazion di regione la13
301.0005.024 | 00.510
201.0
11
xkmSkpFMHzA
RgRvCvkHzMHzQ
kkHzBWBW
kHzpFkCR
BWkkkR
pFMHzk
CRRgRRR
gRMHz
pFHLCfb
pFpFpFCCCC
pFpFpFpFCCCCpFpFpFCmSggVVVV
VVmAII
ggCCRgCCCa
mid
DmPio
PPP
GDLLm
LGP
mL
Po
PGD
EQPEQ
mmSGDS
GSDD
m
mGDGSLmGDGSEQ
=Ω−Ω=
−===Ω
Ω==−≅
=Ω
=Ω=ΩΩ=
+Ω=
+=
====
=−=→=+
=++=++==++=
===→>+==
−=+−==−−==
⎟⎟⎠
⎞⎜⎜⎝
⎛++=++≅
π
ω
ππ
πω
μππ
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-75
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.104 *Problema 12.103 – Amplificatore cascode accordato VDD 5 0 DC 12 VS 1 0 AC 1 C3 1 2 20PF L1 2 0 10UH C1 2 0 20PF RG 2 0 100K M1 3 2 0 0 NFET1 CGS1 2 0 20PF CGD1 2 3 5PF M2 4 0 3 3 NFET2 CGS2 3 0 20PF CGD2 4 3 5PF L2 4 5 10UH C2 4 5 65PF RD 4 5 100K .MODEL NFET1 NMOS VTO=-1 KP=10M .MODEL NFET2 NMOS VTO=-4 KP=10M .OP .AC LIN 200 5.5MEG 6.5MEG .PRINT AC VM(2) VP(2) VM(3) VP(3) VM(4) VP(4) .PROBE .END Risultati: Amid = 279, fo = 6.10 MHz, Q = 24
Si noti che l’anello dei condensatori intorno a M1 scombina i risultati manuali basati sull’approssimazione di CEQ. L’approssimazione di CEQ può non essere abbastanza accurata da ottenere un’accordatura precisa. Si disegnano i grafici di V(2) e V(3) per vedere il problema. Il problema viene evidenziato anche dall’elevato errore nel guadagno di centro banda.
12.105
( ) ( ) ( )
( )( ) ( )( )
( )
40152
08.6= | 08.62
02.1 nuova La
15227.2302.602.0+
20.39
2+02.0+
2
7.233.67102
1 | 0.39=703.582
1
100 | 3.6253.67
3.6702.1
7002.102.12
12
1 :12.103 Prob. Dal
1
21
1
521
2222
221
2
1
22
==+
≅
=+=≅
=Ω
=Ω
=
Ω==−=−=
===
⎟⎟⎠
⎞⎜⎜⎝
⎛==
kHzMHzQMHzfff
kHzkHzMHzkHzBWfBWBW
kHzpF
BWkHzpFk
BW
kRpFpFpFCCC
pFpFC
LLC
LfC
ooo
o
pGDP
P
P
oP
ππ
ππ
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-76
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
12.106 *Problema 12.105 – Amplificatore cascode VDD 5 0 DC 12 VS 1 0 AC 1 C3 1 2 20PF L1 2 0 10UH C1 2 0 20PF RG 2 0 100K M1 3 2 0 0 NFET1 CGS1 2 0 20PF CGD1 2 3 5PF M2 4 0 3 3 NFET2 CGS2 3 0 20PF CGD2 4 3 5PF L2 4 5 10UH C2 4 5 62.3PF RD 4 5 100K .MODEL NFET1 NMOS VTO=-1 KP=10M .MODEL NFET2 NMOS VTO=-4 KP=10M .OP .AC LIN 200 5.5MEG 6.5MEG .PRINT AC VM(2) VP(2) VM(3) VP(3) VM(4) VP(4) .PROBE .END Risultati: Amid = 512, fo = 6.19 MHz, BW = 0.19 MHz, Q = 33
12.107
rπ
R L
g m v
+
-v
Cπ
Cμ
Yin
RL
gm v
+
-v
Cμ
Y1
Yin = gπ + sCπ + Y1 | sCμ − gm( )V = sCμ + GL( )Vo | Vo =sCμ − gm( )sCμ + GL( )
V
I = sCμ V −Vo( )= sCμgm + GL
sCμ + GL( )V | Y1 =
IV
= sCμgm + GL
sCμ + GL( )= sCμ
1+ gmRL
sCμRL +1( )Y1 jω( )= jωCμ
1+ gmRL
jωCμRL +1( )= jωCμ 1+ gmRL( )
1− jωCμRL
ωCμRL( )2+1
| For ωCμRL( )2<<1,
Y1 jω( )≅ jωCμ 1+ gmRL( )+1+ gmRL( )
RL
ωCμRL( )2
03/09/2007 ©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-77
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education
Dai risultati si vede che le capacità di ingresso sono rappresentate correttamente dalla capacità di ingresso totale di Miller, ma la resistenza di ingresso non è correttamente rappresentata dalla sola rπ:
Cin = Cπ + Cμ 1+ gmRL( ) | Rin = rπRL
1+ gmRL( ) ωCμRL( )2
( ) ( )[ ]
( )( ) ( )[ ] ( )( )( )[ ]
( )( ) infinito. da lontani molto sono valorii Entrambi | 2951081052
1 che inoltre noti Si
! 49710210521051
101
1081051261 )(
6
262
Ω==
Ω=ΩΩ+
Ω=
+=
=Ω++=++=
pFxX
kpFxkmSk
RCRgRR
pFkmSpFpFRgCCCb
inC
LGDLm
Lin
LmGDGSin
π
πω
Sebbene l’approssimazione di CT dia un’eccellente stima del polo dominante dell’amplificatore a emettitore comune, non fa un buon lavoro nel rappresentare l’ingresso alle alte frequenze. È necessario migliorare la stima per via dei problemi che seguiranno.
rπ
R L
g m v
+
-v
Cπ
Cμ
Yin
RL
gm v
+
-v
Cμ
Y1
Yin = gπ + sCπ + Y1 | sCμ − gm( )V = sCμ + GL( )Vo | Vo =sCμ − gm( )sCμ + GL( )
V
I = sCμ V −Vo( )= sCμgm + GL
sCμ + GL( )V | Y1 =
IV
= sCμgm + GL
sCμ + GL( )= sCμ
1+ gmRL
sCμRL +1( )Y1 jω( )= jωCμ
1+ gmRL
jωCμRL +1( )= jωCμ 1+ gmRL( )
1− jωCμRL
ωCμRL( )2+1
| For ωCμRL( )2<<1,
Y1 jω( )≅ jωCμ 1+ gmRL( )+1+ gmRL( )
RL
ωCμRL( )2
Dai risultati si vede che le capacità di ingresso sono rappresentate correttamente dalla capacità di ingresso totale di Miller, ma la resistenza di ingresso non è correttamente rappresentata dalla sola rπ:
Cin = Cπ + Cμ 1+ gmRL( ) | Rin = rπRL
1+ gmRL( ) ωCμRL( )2
©Richard C. Jaeger and Travis N. Blalock - 3/10/07 12-78
Jaeger, Blalock, Microelettronica, 4e - ©2017 McGraw-Hill Education