Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Differential Amplifier

If vG1 = vG2, vo = vD1-vD2 = 0

If vG1 > vG2, vo < 0

If vG1 < vG2, vo > 0

Non-zero output only for input differenceDifferential amplifier

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Consider Common-Mode and Differential-Mode separately.

1 1, 2 2id id

G CM G CMv v

v V v V= + = −

1 21 2,

2G G

CM id G Gv v

V v v v+

= = −

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Common-Mode: vG1= vG2 = vCM

1 2 0 (No CM output)D Dυ υ− =

vCM, max?

DS GS tv v V≥ −

( ) ( )2DD D CM GS GS tIV R v v v V− − − ≥ −

,max 2CM t DD DIV V Rυ∴ = + −

2DD D t CMIV R V v− + ≥

vCM, min?

Input common-mode range

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Differential-mode small-signal analysis (assume vid < VCM)

No voltage changesince left and right areanti-symmetric

1 2 = 2 2id id

od o o m D m D m D idv v

v v v g R g R g R v= − = − − −

odd m D

id

vA g R

v= = −

( || )d m D oA g R r= −With , or

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Differential-mode small-signal half-circuit

1

1

( || )/ 2

2

( || )

om D o

id

odo

odm D o

id

vg R r

vv

v

vg R r

v

= −

=

∴ = −

Differential pair acts as CS amplifier for input difference!

Consider only half circuit

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Why a differential amplifier?

- Input stage for operational amplifier (Op-Amp)

- Not sensitive to noises that are commonto both input signals.

Common-Mode Rejection Ratio

CMRR = |Ad/Acm|

Ideally, infinite

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

What if the current source is NOT ideal?

Common-mode

open circuit(Left and Right aresymmetric)

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Common-Source with source resistanceCommon-mode half circuit

Differential output ( vo = vo1 – vo2), vo= 0

common-mode gain due to Rss for single-ended output

1 ?o

icm

vv

= 1 -o m gs Dv g v R=

(2 )gs icm m gs SSv v g v R= −

1 2icm

gsm SS

vv

g R=

+

1

1 2o m D

icm m SS

v g Rv g R

= −+ 2

D

SS

RR

−

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

What if the current source is NOT ideal? Differential-Mode

vid/2 - vid/2

With anti-symmetric input, the mid-point is still small-signal ground!

( || )d m D oA g R r= −

CMRR: Common-Mode Rejection Ratio(Single-ended output)

( || )CMRR=

2

d m D o

Dcm

ss

A g R rRAR

CMRR for differential output?

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

MOS differential pair is based on the symmetry

Anything that breaks the symmetry affects the circuit performance(CMRR, input offset)

Device mismatches cause non-ideal performance Eliminate RD

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

Active-loaded MOS differential pair(single-ended)

It can be shown with finite , 1

2due to non-ideal current source.

ss

ocm

icm m ss

Rv

Av g R

= −

vicm vicm

Common mode

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

2o m o

did

v g rA

v= =

( || )/ 2

om o o

id

vg r r

v= −

−

But with more detailed analysis,

What is wrong with the half-circuit analysis?

Differential mode small-signal analysis

Assume it is (almost) symmetric

4o m o

id

v g rv

∴ =

Source terminals are grounded

Half circuit analysis is possible

Electronic Circuits 2 (07/1) W.-Y. Choi

Lect. 11: Differential Amplifiers

2

12

12

m od

m o sscm

m ss

g rACMRR g r R

Ag R

∴ = =

The current mirror action is not considered!

Current mirror doubles the transconductance

Twice voltage gain!

12

od m o

id

vA g r

v∴ = =