20-4-2012

Vin=2V, Vdd=5V, VTHn=1V, VTHp=-1V, b1= 400mA/V2, b2=200mA/V2, bF3=50 , Vg=0,6V, VT=25mV, RC=100 W, RB=10 KW

Vdd

Vin

Vout

M1

M2

Q3

Vx Vy

Rc

Rb

2 211

2 222

2 1

: 1, 2

4002 1 200

2 2

2005 2 1 400

2 2

200

10

4

0.6

0.6 10000 200 0.6 2 2.6

D in THn

D dd A THp

B D D

C B

out DD C C

y

x y B B

Ipotesi M M saturi

I V V A

I V V V A

I I I A

I I mA

V V R I V

V V

V V R I

ipotesi di saturazione verificat

bm

bm

m

b

m

e

20-4-2012 Vdd

Vin

Vout

M1

M2

Q3

Vx Vy

Rc

Rb

gm2 (-vin)

gm1 vin

vin

gm3 vbe

vout

r

Rb

Rc

p

gm2 vingm3 vy

voutvin

Rb

r

Rc

gm1 vin

vx vy

1 2

3

1 2 3

( )

( )

m m in y

out m C y

outm m m C

in

r g g v v

v g R v

vr g g g R

v

p

p

gm1=400(2-1)=400 mS, gm2=200(5-2-1)=400 mS, gm3= IC3/ VT= 400mS, rp3= b3/ gm3=125W, A V= - 3.2 V

20-4-2012

R1=4 kW, R2=2 kW, R3=2 kW, Vin=0.5 V

1 2 3

2 3 2 3

1 1 2 3

2 20.25

( ) 4 (2 2)

out x

in out out

out

in

V V

V V V V V V

R R R

R RV R R

V R R R R

IR1= 125 mA, IR2= -62,5mA, IR3=-62,5mA, V out= V x= -0.125 V

20-6-2012

Vin1=4V, Vin2=4V, Vdd=5V, VTHn=1V, bA1= bB1= 20mA/V2, bC1= 40mA/V2, bA2= bB2= 40mA/V2, bC2= 80mA/V2, R=20 kW.

Vout

Vdd

RMA1

Vx1

Vin1 Vin2

Vx2

MB1 MC1 MB2MC2

MA2

2 21 11 1 1 1 1

1 1 1

11

2 211 1 1

2 211 1

22

2 2

1

:

2 2

22

202 1 10

2 2

402 1 20

2 2

...

22

20

4

A BDA in x THn DB x THn

in x THn x THn

inx

BDA DB x THn

CDC x THn

inx

DA DB

DC

Ipotesi MOSFET saturi

I V V V I V V

V V V V V

VV V

I I V V A

I V V A

VV V

I I A

I

b b

b mm

b mm

m

2 1

0

60

10

5 20 *60 3,8

R DC DC

C B

out DD R

A

I I I A

I I mA

V V RI K V

ipotesi di saturazione verificate

m

m

b

m

RgmA1(vin1-vx1)

vx1

vout

gmB1gmC1 vx1

gmA2(vin2-vx2)

vx2

gmB2gmC2 vx2

11 1 1

1

11 1

1 1

22 2

2 2

1 1 2 2

1 21 1 2

1 1 2 2

1 1 1

1 1 1

( )

( )

20 *40 0,82

mAx in x

mB

mAx in

mA mB

mAx in

mA mB

out mC x mC x

mA mAout mC in in

mA mB mA mB

out mC mA mC

in mA mB

gv v v

g

gv v

g g

gv v

g g

v R g v g v

g gv R g v v

g g g g

v g g gR R k

v g gm

gmA1= gmB1=20(2-1)=20 mS, gmC1=40(2-1)=40 mS, gmA2= gmB2=40(2-1)=40 mS, gmC2=80(2-1)=80 mS, goA1= goB1= goC1= goA2= goB2= goC2=0

Vout

Vdd

RMA1

Vx1

Vin1 Vin2

Vx2

MB1 MC1 MB2MC2

MA2

20-6-2012

RC=10 k W, RB=400 k W, RE=2 k W, RC=10 k W, VCC=5 V, Vin=1 V, VBE=0,6 V, b=100

-+

Vout

RE

IR

Vin

RB

VCC

RC

VBVC

VE

0

0,6 1 0,6 0,41

400 400

100

(1 ) 101

5 10 *100 4

2 *101 0,202

0,6 1 0,6 0,41

400 400

E

in BE E inB

B B

C B

R E B

C CC C C

E out E E

out E E

in BE E inB

B B

C

V V

V V V VI A

R R K K

I I A

I I I A

V V R I K V

V V R I

V R I K V

V V V VI A

R R K K

I

m

b m

b m

m

m

m

b

100

(1 ) 101

5 10 *100 4

0,6(1 ) (1 )

(1 )

B

R E B

C CC C C

inout E E E B E

B

Ev

B

I A

I I I A

V V R I K V

VV R I R I R

R

RA

R

m

b m

m

b b

b

12-9-2012

Vin=3V, Vdd=5V, Vcc=5V, VT=25mV, b2= 50 , R B=100 kW, R C=2 kW, VTHn=1V, b1=200mA/V2, R=20 kW

VVV 6,02 g

AKR

VVI

B

inB m24

100

6,0322

mAAII BC 2,124*50222 mb

VIRVV CCcc 6,22,1*253

AVVI thnD mb

25616,21002

22

31

1

VRIVV Dddout 12,012,55256,0*2051

R

VVVVVVI outddoutoutthnD

2

2

311 b VVout 89,0 AID m5,2051

M1 sat, Q2 normale diretta

Non in saturazione ma triodo, riprendo il circuito nellipotesi di triodo

V2

RB

VCC

RC

V3

Vout

R

Vdd

Q2M1

Vin

SVg outm mb 17889,0*200)(11

SVVVg outthno mb 142)89,06,1(*200)( 311

mSm

m

V

Ig

T

cm 48

25

2,12 W k

mgr

m

042,148

50

2

22

bp

Rgm2 v2

v2 vout

vin

gm1 v3

RB v3

RCrp go1

in

B

vrR

rv

2

22

p

p

in

B

mBmB vrR

rgRvgRv

2

22223

p

p

in

B

mBmomoout vrR

rgRgrRvgrRv

2

2211311 )||()||(

p

p

918,0042,1100

042,1482178,0)

1142,0*20

20()

1(

2

221

1

kk

kmkm

k

rR

rgRg

Rg

RA

B

mCm

o

v

p

p

12-9-2012 R1= R3= R5=10 k W, R2= R4= R6=20 k W, Vin1=1 V, Vin2=1 V

R4

R3-+

Vout

R6 R5

Vin1

Vin2

R2

R1

-+ VA

VB

VC

VVVR

RRV ininA 33 22

1

21

CBA VVV

3

1

4 R

VV

R

VV CinCout

1

3

4

3

431

3

4 )( inCCinCout VR

R

R

RRVVV

R

RVV

1

3

4

1

21

3

432 ininout V

R

R

R

RR

R

RRVV

VVVVVV ininininout 42610

20

10

2010

10

20101212

14-11-2012

Vin=3V, Vdd=5V, VTHn=1V, VTHp=-1V, I0 =100mA, b1=b2=b3=200mA/V2, b4=400mA/V2.

VxVout

VinM1

M4

M2

I0

M3

VDD

M1 M2 M3 M4 sat

2 21

1 100 5 3 1 1002

D DD in thpI V V V Ab

m

2 3 200D DI I Am

2

2

22,414Dx thn

IV V V

b 4

4

23Dout DD thp

IV V V V

b

1 1( ) 200m DD in thpg V V V Sb m 2 3 2( ) 282,8m m x thng g V V Sb m

4 4( ) 400m DD out thpg V V V Sb m 1 2 3 40o o o og g g g

gm2

gm4

gm1 (-vin)

vx

gm3 vx

vout

1 3

2 4

m mout in

m m

g gv v

g g

0,5vA

VxVout

VinM1

M4

M2

I0

M3

VDD

14-11-2012 R1= R2= R3=10 k W, R4= R5= R6 k W, VA=1 V, VB=2 V

-+

VzVAR1

R2

-+

VoutR3

R4

-+

R5

VB

Vx

Vy

Vk

2 1

2 2

1 1

4 4 2 2

3 3 1 1

1 3

1 2 4 6

y B A B

y B A

out z B A A B

V V V V

R R

R RV V V V

R R

R R R RV V V V V V V

R R R R