3.single stage amp i

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EE6326 Analog IC Design – Fall 2007

Topic 3. Single-stage Amplifiers

- CS

J. Liu

Associate Professor

UT Dallas



Single Stage Amplifier

Common source (inverting) amplifiers

Source follower (common drain)

Common gate stage

Cascode stage*



Characterization of Single Stage Amplifiers

Large signal transfer function

Large signal swing limitations Small signalGain

Input resistanceOutput resistanceFrequency response

OthersNoisePower dissipation



Common Source Amplifiers

AKA

Inverters

Inverting Amplifiers

ConfigurationsResistive load

Diode connected load

Current source loadTriode load

Source degeneration



Inverting Amplifier with Resistive Load

G

D

S

VDD

R D

VinVout

Vin iDiD

Voltage across R D

Vout



Inverting Amplifier with Resistive Load- Large Signal Transfer Function

G

D

S

VDD

R D

VinVout

iD

Vin

VoutVDD

VT

If Vin <VT, M1 is in cutoff regioniD = 0, Vout=VDD

If Vin >VT, M1 is in either triode or

saturation region

M1




G

D

S

VDD

R D

VinVout

iD

VTM1

Criteria for Saturation of M1:VDS1>VGS1-VT

Vout>Vin- VT, area above line

Vout= Vin- VT

Vin

VoutVDD

VDD

Sat.

Triode

Cutoff




G

D

S

VDD

R D

Vin

Vout

iD

M1

In saturation region:

DT oxn

T oxn D

T GS oxn D

D D

RV Vin L

W

C VDDVout

V Vin L

W C i

V V L

W C i

Vout RiVDD

2

'

2

'

2

'

)(2

1

)(2

1

)(2

1

−−=

−=

−=

+=

µ

µ

µ




G

D

S

VDD

R D

Vin

Vout

iD

M1

T

DT oxn

V VinVout

RV Vin L

W

C VDDVout

−=

−−=

11

)1(2

1

1

2

'µ

Vin

VoutVDD

VDD

Sat.

Triode

Cutoff

Vin1

Vout1

When Vin >Vin1, M1 is in triode region




G

D

S

VDD

R D

VinVout

iD

M1

When Vin >Vin1, M1 is in triode region

DT oxn

T oxn D

DS DS T GS oxn D

D D

RVout Vout V Vin L

W C Vout VDD

Vout Vout V Vin L

W

C i

V V V V L

W C i

Vout RiVDD

]2

1)[(

]2

1

)[(

]2

1)[(

2

2

2

−−+=

−−=

−−=

+=

µ

µ

µ




G

D

S

VDD

R D

VinVout

iD

VTM1

Vin

VoutVDD

VDD

Sat.

Triode

Cutoff

Vout= Vin- VT

Input swing range:VT<Vin<Vin1 for saturation

Output swing range:

Vout1<Vout<VDD for sat.

Vin1

Vout1



Inverting Amplifier with Resistive Load- Small Signal Characteristics

G

D

S

VDD

R D

VinVout

iD

M1

gmvgs gmbs

v bs

gds

G

S S

D

+

-

vgsR D

DC voltage supplies = gnd

DC current supplies = openvout

vin

D D

ds D

inmout

ds Dinmout

ds Dbsmbs gsmout

R g g g

v g

v

g Rv g v

g Rv g v g v

/1,

)/1||(

)/1||)((

=+

−

=

−=

+−=

v bs=0 vgs=vin



Inverting Amplifier with Resistive Load- Gain

G

D

S

VDD

R D

Vin

Vout

iD

M1

gmvgs gmbsv bs

gds

G

S S

D

+

-

vgsR D

vout

vin

ds D

m

in

out

ds D

inm

out

g g

g

v

v

g g

v g v

+

−=

+

−=

Low frequency gain



Inverting Amplifier with Resistive Load- Input Resistance

G

D

S

VDD

R D

Vin

Vout

iD

M1

gmvgs gmbsv bs

gds

G

S S

D

+

-

vgsR D

vout

vin

∞=inr Input resistance

v bs=0 vgs=vin



Inverting Amplifier with Resistive Load- Output Resistance

G

D

S

VDD

R D

Vin

Vout

iD

M1

gmvgs gmbsv bs

gds

G

S S

D

+

-

vgsR D vtest

vin=0itest

vgs=0, vbs=0

D D

ds D

out

out

ds D

test

test out

R g

g g r

g

g R

i

vr

/1

1

)/1(||

=

+==

==

Output resistance



Inverting Amplifier with Resistive Load- Small Signal Characteristics Summary

G

D

S

VDD

R D

Vin

Vout

iD

M1out mds Dm

in

out

ds Dout

in

r g g R g v

v g Rr

r

−=−=

=

∞=

)/1||(

)/1(||



Inverting Amplifier with Diode-connectedLoad

G1

D1

S1

VDD

Vin

VoutiD

M1

S2

D2

M2

G2

When the gate and the drain of a

MOSFET (N or P) are connectedtogether (diode-connected), it is always

in saturation region.

Saturation criteria:

vDS

>vGS

-VTWhen vD=vG, the above

criteria is always satisfied.



Inverting Amplifier with Diode-connectedLoad

Vin

Vin iD

vSG2

vout=VDD-vSG2

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

2

22 |)|(2

1T SGox p D V v

L

W C i −= µ

For M2,



Inverting Amplifier with Diode-connectedLoad - Large Signal Transfer Function

Vin

VoutVDD

VT

If Vin <VT, M1 is in cutoff regioniD = 0, M2 is in cutoff region

Vout=VDD-|VTP2|

If Vin >VT, M1 is on, iD is not zero,M2 turns on in sat.

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

VDD-|VTP2|M2 off

M1 of f

M2 satM1 sat

M1 triode

vin



l f h d d




G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

21

2

22

22

2

222

2

'1

2

11'1

|)|(2

1

|)|(2

1

)(

2

1

)(21

D D

T ox p D

SDSG

T SGox p D

T oxn D

T GS oxn D

ii

V Vout VDD L

W C i

Vout VDDvv

V v L

W C i

V Vin

L

W C i

V v LW C i

=

−−=

−==

−=

−=

−=

µ

µ

µ

µ

When M1 and M2 are in Sat.

l f h D d d




G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

T

D D

T ox p

T oxn

V vinvout

ii

V vout VDD LW C

V vin L

W C

−=

=

−−=

−

11

|)|1(21

)1(2

1

21

22

2

'

µ

µ

To find out the boundary between

the sat. and triode regions of M1,

Solve for vin1 and vout1.

I i A lifi i h Di d d




G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

2

2

2

|)|(2

1

]

2

1)[(

T ox p

T oxn D

V vout VDD L

W C

vout vout V vin

L

W C i

−−=

−−=

µ

µ

When vin>vin1, M1 is in triode region

And M2 is still in sat. region.

I i A lifi i h Di d d




Vin

VoutVDD

VTG1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2VDD-|VTP2|

M2 off

M1 of f

M2 satM1 sat

M1 triode



Plot transfer functions

Label X- and Y-axis

Label the operational regions of each device

Label the important transition points betweendifferent operational regions on both X- and Y-axis

Write the equation for each portion of the curve

S ll Si l E i l t f Di d C t d



Small Signal Equivalent of a Diode ConnectedMOSFET

GD

S

gm gds

v1

v1

VDD

gm is approximately 100 times go (gds) gm

gmvgs gds

v1

vgs

I ti A lifi ith PMOS Di d



Inverting Amplifier with PMOS Diode-Connected Load – Gain

2

1

2

1

2

1

221

1

1

1

m

m

in

out

in

m

m

m

inmout

dsmds

gsmout

g g

vv

v g

g

g v g v

g g g v g v

−≈

−=−≈

++

−=

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

vin

gm1vgs1 gmbs1v bs1

gds1

G

S S

D

+

-

vgs1 gm2+gds2

v bs1=v bs2=0v

gs1

=vinvgs2=vout

I ti A lifi ith PMOS Di d



Inverting Amplifier with PMOS Diode-Connected Load – Rin, Rout, and Gain

out m

dsmds

m

in

out

mdsmds

out

in

r g

g g g

g

v

v

g g g g r

r

1

221

1

2221

11

−=

++

−=

≈++

=

∞=

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

vin

gm1vgs1 gmbs1v bs1

gds1

G

S S

D

+

-

vgsgm2+gds2

S ll Si l E i l t f Di d C t d



Small Signal Equivalent of a Diode ConnectedMOSFET with Bulk Effect

G D

gm go

gmbs

S

gmVgs

Vgs=0-vx Vbs=0-vx

I ti A lifi ith NMOS Di d



Inverting Amplifier with NMOS Diode-Connected Load – Gain

η

η

+

−≈

+

−=

+

−≈

+++

−=

11

1

11

1

2

1

2

1

22

1

2221

1

m

m

in

out

in

m

m

mbsm

inmout

dsmbsmds

gsmout

g g

vv

v g

g

g g v g v

g g g g v g v

gm1vgs1 gmbs1v bs1

gds1

G

S S

D

+

-

vgsgm2+gmbs2+gds2

In tin Amplifi ith NMOS Di d



Inverting Amplifier with NMOS Diode-Connected Load – Rin, Rout, and Gain

out m

dsmbsmds

m

in

out

mdsmbsmds

out

in

r g

g g g g

g

v

v

g g g g g r

r

1

2221

1

22221 1

111

−=

+++

−=

+≈

+++=

∞≈

η

gm1vgs1 gmbs1v bs1

gds1

G

SS

D

+

-

vgsgm2+gmbs2+gds2

Inverting Amplifier with Current Source



Inverting Amplifier with Current SourceLoad - Large Signal Transfer Function

Vin

Vout

VDD

VT

M2 triode

M

1 of f

M2 sat

M1 triode

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

vin

VB

M2 Sat.: VDD-Vout >VDD-VB-|VTP2|

Vout < VB+|VTP2|

Assume M2 is on, VDD-VB>|VTP2|

VB+|VTP2|

Vout>Vin-VTN1

M1 sat




Inverting Amplifier with Current SourceLoad

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

vin

VB

Consider the small signal characteristics of

M2:

gm2vgs2

gds2

G2=0

S2S2

D2

+

-

Vgs=0

Equivalent to gds2




Inverting Amplifier with Current SourceLoad - Gain

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

vin

VB

gm1vgs1 gmbs1v bs1

gds1

G

S S

D

+

-

vgsgds2

21

1

21

1

1

dsds

m

in

out

dsds

inmout

g g

g

v

v g g

v g v

+

−≈

+

−=




Inverting Amplifier with Current SourceLoad – Rin, Rout, and Gain

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

vin

VB

gm1vgs1 gmbs1v bs1

gds1

G

S S

D

+

-

vgsgds2

out m

dsds

m

in

out

dsds

out

in

r g

g g

g

v

v

g g r

r

1

21

1

21

1

−=

+

−=

+=

∞≈




Inverting Amplifier with Current SourceLoad vs. Triode Load

G1

D1

S1

VDD

VoutiD

M1

S2

D2

M2

G2

vin

VB

If M2 operates in the saturation region,

M2 is a current source.

Calculate gds2 in saturation region.

If VB is very low, (Large VSG2), M2

is mostly likely work in the triode region.

Calculate gds2 in triode region.



HW #3

Due on 9/11

More exercise if desired (in Razavi book)

3.1, 3.2, 3.3, 3.4

3.15 (a) (c) (d)

3.16 (a) (b) (c)

3.21 (a) (b) (c)

3.single stage amp i

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