EECE488: Analog CMOS Integrated Circuit Design
Set 4
Differential Amplifiers
Shahriar Mirabbasi
Department of Electrical and Computer Engineering
University of British Columbia
shahriar@ece.ubc.ca
SM
EECE488 Set 4 - Differential Amplifiers
1
Overview
•
The “differential amplifier” is one of the most important circuit
inventions.
•
Their invention dates back to vacuum tube era (1930s).
•
Alan Dower Blumlein (a British Electronics Engineer, 19031942) is regarded as the inventor of the vacuum-tube version of
differential pair.
•
Differential operation offers many useful properties and is widely
used in analog and mixed-signal integrated circuits
SM
EECE488 Set 4 - Differential Amplifiers
2
Single-ended and Differential Signals
• A “single-ended” signal is a signal that is measured with
respect to a fixed potential (typically ground).
• “Differential signal” is generally referred to a signal that
is measured as a difference between two nodes that have
equal but opposite-phase signal excursions around a
fixed potential (the fixed potential is called common-mode
(CM) level).
SM
EECE488 Set 4 - Differential Amplifiers
3
Board Notes (Differential Amplifiers)
SM
EECE488 Set 4 - Differential Amplifiers
4
Why Differential?
• Better immunity to environmental noise
• Improved linearity
• Higher signal swing compared to single-ended
SM
EECE488 Set 4 - Differential Amplifiers
5
Higher Immunity to Noise Coupling
SM
EECE488 Set 4 - Differential Amplifiers
6
Supply Noise Reduction
SM
EECE488 Set 4 - Differential Amplifiers
7
Board Notes (Improved Linearity)
SM
EECE488 Set 4 - Differential Amplifiers
8
Basic Differential Pair
SM
EECE488 Set 4 - Differential Amplifiers
9
Basic Differential Pair
•
Problem: Sensitive to input common-mode (CM) level
– Bias current of the transistors M1 and M2 changes as the
input CM level changes
– gm of the devices as well as output CM level change
•
SM
Can we think of a solution?
EECE488 Set 4 - Differential Amplifiers
10
Differential Pair
SM
EECE488 Set 4 - Differential Amplifiers
11
Common-Mode Response
SM
EECE488 Set 4 - Differential Amplifiers
12
Common-Mode Input versus Output Swing
SM
EECE488 Set 4 - Differential Amplifiers
13
Board Notes (“Half-Circuit” Concept)
SM
EECE488 Set 4 - Differential Amplifiers
14
Board Notes (“Half-Circuit” Concept)
SM
EECE488 Set 4 - Differential Amplifiers
15
Example
•
SM
Using the half-circuit concept, calculate the small-signal
differential gain of the following circuit (for two cases of λ=0 and
λ≠0).
EECE488 Set 4 - Differential Amplifiers
16
Example
•
SM
Using the half-circuit concept, calculate the small-signal
differential gain of the following circuit (for two cases of λ=0 and
λ≠0).
EECE488 Set 4 - Differential Amplifiers
17
Example
•
SM
Sketch the small-signal gain of a differential pair as a function of
its input common-mode level.
EECE488 Set 4 - Differential Amplifiers
18
Analysis of Differential Amplifier
Vin1 − Vin 2 = VGS 1 − VGS 2 ,
SM
VGS =
EECE488 Set 4 - Differential Amplifiers
2I D
W
µ nCox
L
+ VTH
19
Analysis of Differential Amplifier
Vin1 − Vin 2 =
(Vin1 − Vin 2 )
2
=
2 I D1
−
W
µ nCox
L
2
W
µ nCox
L
2I D2
W
µ nCox
L
( I D1 + I D 2 − 2 I D1 I D 2 )
1
W
2
µ nCox (Vin1 − Vin 2 ) − I SS = −2 I D1 I D 2
2
L
W 2
1
W
4
2
(Vin1 − Vin 2 )2 = 4 I D1I D 2
( µ nCox ) (Vin1 − Vin 2 ) + I SS − I SS µ nCox
L
4
L
SM
EECE488 Set 4 - Differential Amplifiers
20
Analysis of Differential Amplifier
Using:
2
4 I D1 I D 2 = ( I D1 + I D 2 ) 2 − ( I D1 − I D 2 ) 2 = I SS
− ( I D1 − I D 2 ) 2
and
4 I D1 I D 2
1
W 2
W
4
2
(Vin1 − Vin 2 )2
= ( µ nCox ) (Vin1 − Vin 2 ) + I SS − I SS µ nCox
4
L
L
We have:
I D1 − I D 2
SM
4 I SS
1
W
= µ nCox (Vin1 − Vin 2 )
− (Vin1 − Vin 2 ) 2
W
2
L
µ nCox
L
EECE488 Set 4 - Differential Amplifiers
21
Analysis of Differential Amplifier
id =
W
1
µ nCox vid
L
2
∂id
1
W
= Gm = µ nCox
2
∂vid
L
4 I SS
2
− vid
W
µ nCox
L
4 I SS
− vid2
µ nCoxW / L
4 I SS
− vid2
µ nCoxW / L
vid
SM
EECE488 Set 4 - Differential Amplifiers
vid
22
Analysis of Differential Amplifier
• For small vid:
∂id
W
Gm =
= µ nCox
I SS = g m1 = g m1
∂vid
L
• We have:
Vout1 − Vout 2 = RD ( I D1 − I D 2 ) = RD Gm (Vin1 − Vin 2 )
Vout1 − Vout 2
= g m1 RD
Vin1 − Vin 2
SM
EECE488 Set 4 - Differential Amplifiers
23
Differential Gain
Vout1 − Vout 2
Av (diff ) =
= gm RD
Vin1 − Vin2
gm
Av ( Single − Ended ) =
RD
2
SM
EECE488 Set 4 - Differential Amplifiers
24
Differential Pair as a CS and CD-CG Amplifier
gm RD
Av = −
1+ gm RS
gm
= − RD , (S.E.)
2
= gm RD , (diff )
SM
EECE488 Set 4 - Differential Amplifiers
25
Common-Mode Response
Vout
Av =
Vin,CM
RD / 2
=−
1/(2gm ) + RSS
SM
EECE488 Set 4 - Differential Amplifiers
26
Board Notes
SM
EECE488 Set 4 - Differential Amplifiers
27
Example
•
In the following circuit assume that RSS=500Ω and W/L=25/0.5,
µnCox=50µA/V2, VTH =0.6, λ=γ=0 and VDD=3V.
a) What is the required input CM for which RSS sustains 0.5V?
b) Calculate RD for a differential gain of 5V/V.
c) What happens at the output if the input CM level is 50mV
higher than the value calculated in part (a)?
SM
EECE488 Set 4 - Differential Amplifiers
28
Board Notes
SM
EECE488 Set 4 - Differential Amplifiers
29
Common-Mode Response
gm
∆VX
=−
RD
∆Vin,CM
1+ 2gm RSS
gm
∆VY
=−
(RD + ∆RD )
∆Vin,CM
1+ 2gm RSS
SM
EECE488 Set 4 - Differential Amplifiers
30
Common-Mode Response
VX − VY
gm1 − gm 2
=−
RD
Vin,CM
(gm1 + gm 2 )RSS + 1
SM
EECE488 Set 4 - Differential Amplifiers
31
Common-Mode Response
SM
EECE488 Set 4 - Differential Amplifiers
32
Differential Pair with MOS Loads
−1
Av = −gmN (gmP
|| roN || roP )
gmN
≈−
gmP
SM
Av = −gmN (roN || roP )
EECE488 Set 4 - Differential Amplifiers
33
MOS Loads
SM
EECE488 Set 4 - Differential Amplifiers
34
MOS Loads
Av ≈ gm 1[(gm 3 ro3 ro1 ) || (gm 5ro5 ro7 )]
SM
EECE488 Set 4 - Differential Amplifiers
35
Gilbert Cell
SM
EECE488 Set 4 - Differential Amplifiers
36
Gilbert Cell
SM
EECE488 Set 4 - Differential Amplifiers
37
Gilbert Cell
VOUT = kVin Vcont
SM
EECE488 Set 4 - Differential Amplifiers
38