highered.mheducation.comhighered.mheducation.com/sites/dl/free/8838615551/1096611/...Capitolo 12...

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


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



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



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



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



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



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



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



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



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



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======

−−− ππ

β

β

π

π

π



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



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



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



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



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


. 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


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



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


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====≅

−

∑



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



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


32

250 0.25V( )10−4( )2

cm2

V − s= 9.38 GHz

c( ) NMOS: fT =32


32

600 0.25V( )10−5( )2

cm2

V − s= 2.25 THz

PMOS: fT =32


32

250 0.25V( )10−5( )2

cm2

V − s= 938 GHz

d( ) NMOS: fT =32


32

600 0.25V( )2.5x10−6( )2

cm2

V − s= 36.0 THz

PMOS: fT =32


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



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



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



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



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



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



( ) ( )( )

( ) ( )

( )( )

( )[ ] ( )[ ]

( )( ) ( )( )( ) ( ) ( )( ) 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



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



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



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======

−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



* .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



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



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



( ) ( ) ( ) ( )

( )[ ] ( )( )

( )( )

( )( )( ) ( )[ ] ( )( )

( )

( )( ) 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



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



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



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



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.



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



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



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



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



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



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



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



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



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



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


= 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



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


= 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



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.



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



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



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



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.



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:



( ) ( )( )( )( ) ( )

( )( )( )( )

( )

( ) ( )

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



( )

( )

( )

( )

( ). 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

=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+++++

+=

−



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



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.



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



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



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



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



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



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



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



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


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


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



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



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



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



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



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



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



( )( )( )

( ) ( )( )( )

( )( )( )

( )( )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



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



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



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



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



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



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