Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: 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 difference Differential amplifier

Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: 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 (15/1) W.-Y. Choi

Lect. 9: Differential Amplifiers

Common-Mode: vG1= vG2 = vCM

1 2 0 (No CM output)D D

vCM, max?

DS GS THv v V

( ) ( )2DD D CM GS GS THIV R v v v V

,max 2CM TH DD DIV V R

2DD D TH CMIV R V v

vCM, min?

Input common-mode range

Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: Differential Amplifiers

Differential-mode small-signal analysis

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 (15/1) W.-Y. Choi

Lect. 9: Differential Amplifiers

Differential-mode small-signal half-circuit

1 ( || )/ 2

om D o

id

vg R r

v

Differential pair acts as CS amplifier for input difference!

Consider only half circuit

1 2od

ov

v

( || )odm D o

id

vg R r

v

Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: Differential Amplifiers

Why a differential amplifier?

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

Common-Mode Rejection Ratio

CMRR = |Ad/Acm|

Ideally, infinite

Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: Differential Amplifiers

What if the current source is NOT ideal?

Common-mode

open circuit(Left and Right aresymmetric)

Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: 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 (15/1) W.-Y. Choi

Lect. 9: 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

Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: 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 (15/1) W.-Y. Choi

Lect. 9: Differential Amplifiers

Active-loaded MOS differential pairfor single-ended output

vicm vicm

Common mode

Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: Differential Amplifiers

Differential mode small-signal analysis

Applying half-circuit analysis,

1 2

out

in in

vv v

( )2

mN ON OPg r r

But the given circuit is asymmetric

Half circuit analysis is not possible

With complicated analysis (Razavi 10.6.2)

1 2

1[1 ( )]

2 2

OP mP OPout mP

mN ONin in ON OP

r g rv gg r

v v r r

Small-signal circuit

( )mN ON OPg r r

Factor of 2 difference!

Electronic Circuits 2 (15/1) W.-Y. Choi

Lect. 9: Differential Amplifiers

The current mirror action is not considered with half-circuit analysis

Current mirror doubles the transconductance

Twice voltage gain!