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Slide 1 EE40 Fall 2009 Prof. Cheung EE40 Lec 20 MOS Circuits Reading: Chap. 12 of Hambley Supplement reading on MOS Circuits http://www.inst.eecs.berkeley.edu/~ee40/fa09/handouts/ EE40_MOS_Circuit.pdf

EE40 Lec 20 MOS Circuits

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EE40 Lec 20 MOS Circuits. Reading: Chap. 12 of Hambley Supplement reading on MOS Circuits http://www.inst.eecs.berkeley.edu/~ee40/fa09/handouts/EE40_MOS_Circuit.pdf. OUTLINE. Bias circuits Small-signal equivalent circuits Examples: Common source amplifier Source follower - PowerPoint PPT Presentation

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Page 1: EE40 Lec 20 MOS Circuits

Slide 1EE40 Fall 2009 Prof. Cheung

EE40 Lec 20

MOS Circuits

Reading: Chap. 12 of Hambley

Supplement reading on MOS Circuitshttp://www.inst.eecs.berkeley.edu/~ee40/fa09/handouts/EE40_MOS_Circuit.pdf

Page 2: EE40 Lec 20 MOS Circuits

Slide 2EE40 Fall 2009 Prof. Cheung

OUTLINE

–Bias circuits–Small-signal equivalent circuits –Examples:Common source amplifier Source follower Common gate amplifier

–Digital Gates–CMOS

Page 3: EE40 Lec 20 MOS Circuits

Slide 3EE40 Fall 2009 Prof. Cheung

Bias Circuits

• Use load line to find Quiescent operating point.• Remember no current flow through the gate.

VDD

RDR1

R2 RS

VDD

RD

VG+vin

Fixed-plus Self-Bias CKT

Page 4: EE40 Lec 20 MOS Circuits

Slide 4EE40 Fall 2009 Prof. Cheung

Steps for MOSFET Circuit Analysis

• 1) Look at DC case to find Q point– Use load line technique– All capacitors are open circuit, Inductors are

short circuit

– Determine Q-point, get gm and rd for small signal AC model

• 2) AC Small signal analysis– DC source is ac ground (because there is no

AC signal variation).– All capacitors are approximated as short

circuit (unless otherwise specified).

Page 5: EE40 Lec 20 MOS Circuits

Slide 5EE40 Fall 2009 Prof. Cheung

Example: Common Source Amplifier

VDD

RDR1

R2 RS C

C

RL

+

-voC

+

-vin

+

-v(t)

VG

Page 6: EE40 Lec 20 MOS Circuits

Slide 6EE40 Fall 2009 Prof. Cheung

Step 1: find Q point

VDD

RDR1

R2 RS C

C

RL

+

-voC

+

-vin

+

-v(t)

VGVDS

2

1 2

( )

G DD

GS G D S

DD D D S DS

RV V

R R

V V I R

V I R R V

Not connectedfor DC component

Not connectedfor DC component

Page 7: EE40 Lec 20 MOS Circuits

Slide 7EE40 Fall 2009 Prof. Cheung

Load line to determine Q Point by graphical method

From load lines, we get ID and hence gm and rd

S

GSGD R

VVI

Loadline to determine VGSQ

S

GSGD R

VVI

S0

DSDDD RR

VVI

S0

DSDDD RR

VVI

Loadline to determine VDSQVGSQ

Page 8: EE40 Lec 20 MOS Circuits

Slide 8EE40 Fall 2009 Prof. Cheung

Load line to determine Q Point by analytical method

From load lines, we get ID and hence gm and rd

2tGSQDQ

S

GSQGDQ

)VV(KI

R

VVI

Solve VGSQ assume saturation region first

IDQ is known, then solve VDSQ

DSQSDDQDD V)RR(IV

Check VDSQ value is consistent with saturation region ( i.e. VDS> VGSQ-Vt)

Page 9: EE40 Lec 20 MOS Circuits

Slide 9EE40 Fall 2009 Prof. Cheung

Determination of gm and rd graphically

Example: Q point is known to be VGS=2.5V, VDS=6V

kror

SiemensV

mA

v

i

r

d

DS

D

d

20

1005.0)214(

)3.29.2(1 3

Page 10: EE40 Lec 20 MOS Circuits

Slide 10EE40 Fall 2009 Prof. Cheung

Determination of gm and rd by Analytical Models

DQDS

Dd

DQtGSGS

Dm

2tGSD

iv

ir/1

iK2)Vv(K2v

ig

)Vv(Ki

factorulationmodchannelL

W

2

KPK

In Saturation Region

In Triode Region

]v2)Vv(2[Kv

ir/1

Kv2v

ig

]vv)Vv(2[Ki

DSQtGSQDS

Dd

DSQGS

Dm

DS2

DStGSD

Page 11: EE40 Lec 20 MOS Circuits

Slide 11EE40 Fall 2009 Prof. Cheung

Small Signal Model

1 2

1 2

, 0

( )

g in s gs in

L Do m gs

L D

o L Dv m

in L D

inin

in

v v v v v

R Rv g v

R R

v R RA g

v R R

v R RR

i R R

For output impedance Rout:1. Turn off all independent

sources.2. Take away load impedance

RL

0, 0, 0in gs m gs

d Dout

d D

v v g v

r RR

r R

Inverting

Page 12: EE40 Lec 20 MOS Circuits

Slide 12EE40 Fall 2009 Prof. Cheung

Example: Source Follower

VDD

R1

R2 RS

C

RL

+

-vo

C

+

-vin

+

-v(t)

VG

Page 13: EE40 Lec 20 MOS Circuits

Slide 13EE40 Fall 2009 Prof. Cheung

Step 1: find Q point

2

1 2G DD

GS G D S

DD D S DS

RV V

R R

V V I R

V I R V

VDD

R1

R2 RS

C

RL

+

-vo

C

+

-vin

+

-v(t)

VG

Page 14: EE40 Lec 20 MOS Circuits

Slide 14EE40 Fall 2009 Prof. Cheung

Small Signal Model

1 1 1

1 2

1 2

1

(1 )

1

Ld S L

gs in o

o m gs L

in gs m L

o m Lv

in m L

inin

in

Rr R R

v v v

v g v R

v v g R

v g RA

v g R

v R RR

i R R

For output impedance Rout:1. Turn off all independent sources.2. Take away RL

3. Add Vx and find ix

1

1 1

, 0,

, ( )

1

x s g gs x

d s xs x m x x s m

d s s

outm d s

v v v v v

r R vR i g v v R g

r R R

Rg r R

Non-inverting, Voltage Gain 1Rin highCurrent gain can be high

Rout is small

Page 15: EE40 Lec 20 MOS Circuits

Slide 15EE40 Fall 2009 Prof. Cheung

Example: Common Gate Amplifier

VDD

RD

RS

C

RL

+

-vo

C+

-vin

+

-v(t)

VG

-VSS

Page 16: EE40 Lec 20 MOS Circuits

Slide 16EE40 Fall 2009 Prof. Cheung

Step 1: find Q point

VDD

RD

RS

C

RL

+

-vo

C+

-vin

+

-v(t)

VG

-VSS

0

( )GS D S SS

DD SS D D S DS

V I R V

V V I R R V

Page 17: EE40 Lec 20 MOS Circuits

Slide 17EE40 Fall 2009 Prof. Cheung

Load line

The only difference in all three circuits are the intercepts at the axes.Again from load lines, we get ID and hence gm and rd

Page 18: EE40 Lec 20 MOS Circuits

Slide 18EE40 Fall 2009 Prof. Cheung

Small Signal Model

1 1

1

1

( )

1

LL D

gs in

o m gs L

ov m L

in

gsin m gs

s

inin

in m s

RR R

v v

v g v R

vA g R

v

vi g v

R

vR

i g R

For output impedance Rout:1. Turn off all independent sources.2. Take away RL

3. Add Vx and find ix

, 1 0

s

s

xx m gs

D

gs m gs m gs

out D

RRR

R R

vi g v

R

v g v R but g R v

R R

Non-inverting

Page 19: EE40 Lec 20 MOS Circuits

Slide 19EE40 Fall 2009 Prof. Cheung

Logic Gates : Pull-Up and Pull-Down

PMOS or Resistor

NMOS or Resistor

Page 20: EE40 Lec 20 MOS Circuits

Slide 20EE40 Fall 2009 Prof. Cheung

Inverter = NOT Gate

VoutVin

Vin

Vout

VV/2

Ideal Transfer Characteristics

Page 21: EE40 Lec 20 MOS Circuits

Slide 21EE40 Fall 2009 Prof. Cheung

VDD/RD

VDD

NMOS Inverter: Resistor Pull-Up

vDS

iD

0

vOUT

vIN0

VDD

RD

+

vDS = vOUT

iD

+

vIN

VDD

RD

+

vDS = vOUT

iD

+

vIN

Circuit: Voltage-Transfer Characteristic

VDD

VT

A F0 11 0

AF

increasingvGS = vIN > VT

vGS = vin VT

vIN = VDD

VDD

Page 22: EE40 Lec 20 MOS Circuits

Slide 22EE40 Fall 2009 Prof. Cheung

NMOS NAND Gate

• Output is low only if both inputs are highVDD

RD

A

B

F

A B F0 0 10 1 11 0 11 1 0

Truth Table

Page 23: EE40 Lec 20 MOS Circuits

Slide 23EE40 Fall 2009 Prof. Cheung

NMOS NOR Gate

• Output is low if either input is highVDD

RD

A B

F

A B F0 0 10 1 01 0 01 1 0

Truth Table

Page 24: EE40 Lec 20 MOS Circuits

Slide 24EE40 Fall 2009 Prof. Cheung

Disadvantages of NMOS Logic Gates

• Large values of RD are required in order to

– achieve a low value of VLOW– keep power consumption low

Large resistors are needed, but these take up a lot of space.

Page 25: EE40 Lec 20 MOS Circuits

Slide 25EE40 Fall 2009 Prof. Cheung

CMOS Inverter: Intuitive Perspective

VDD

Rn

VIN = VDD

CIRCUIT SWITCH MODELS

VDD

Rp

VIN = 0 V

VOUT VOUT

VOL = 0 V VOH = VDD

Low static power consumption, sinceone MOSFET is always off in steady state

VDD

VIN VOUT

S

D

G

G S

D

Page 26: EE40 Lec 20 MOS Circuits

Slide 26EE40 Fall 2009 Prof. Cheung

The CMOS Inverter: Current Flow

VIN

VOUT

VDD

VDD00

N: offP: lin

N: linP: off

N: linP: sat

N: satP: lin

N: satP: sat

A B D E

C

ii

I

S

D

G

GS

D

VDD

VOUTVIN

Page 27: EE40 Lec 20 MOS Circuits

Slide 27EE40 Fall 2009 Prof. Cheung

Power Dissipation: Direct-Path Current

VDD-VT

VT

time

vIN:

i:

Ipeak

VDD

0

0

i

S

D

G

GS

D

VDD

vOUTvIN

peakDDscdp IVtE Energy consumed per switching period:

tsc

Page 28: EE40 Lec 20 MOS Circuits

Slide 28EE40 Fall 2009 Prof. Cheung

CMOS NAND Gate

A B F0 0 10 1 11 0 11 1 0

A

F

B

A B

VDD

Notice that the pull-up network is related to the pull-down network by DeMorgan’s Theorem!

NMOS, Pull-down PMOS, Pull-up

Page 29: EE40 Lec 20 MOS Circuits

Slide 29EE40 Fall 2009 Prof. Cheung

CMOS NOR Gate

A

F

B

A

B

VDD A B F0 0 10 1 01 0 01 1 0

Notice that the pull-up network is related to the pull-down network by DeMorgan’s Theorem!

NMOS, Pull-down PMOS, Pull-up

Page 30: EE40 Lec 20 MOS Circuits

Slide 30EE40 Fall 2009 Prof. Cheung

Multiple Input NOR Gate

Page 31: EE40 Lec 20 MOS Circuits

Slide 31EE40 Fall 2009 Prof. Cheung

Features of CMOS Digital Circuits

• The output is always connected to VDD or GND in steady state Full logic swing; large noise margins Logic levels are not dependent upon the relative sizes of the

devices (“ratioless”)

• There is no direct path between VDD and GND in steady state no static power dissipation