Lect. 9: Differential Amplifiers - Yonsei...
Transcript of Lect. 9: Differential Amplifiers - Yonsei...
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!