This document discusses differential amplifiers, which measure the difference between two input signals and offer advantages like noise immunity. It describes the basic differential pair circuit and how loading it with resistors can improve linearity and differential gain. The document also covers analyzing differential amplifiers, including their differential and common-mode gains, as well as more advanced topics like using MOS loads and the Gilbert cell configuration.

1. EECE488 Set 4 - Differential Amplifiers 1
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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

2. EECE488 Set 4 - Differential Amplifiers 2
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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, 1903-
1942) 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

3. EECE488 Set 4 - Differential Amplifiers 3
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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).

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Board Notes (Differential Amplifiers)

5. EECE488 Set 4 - Differential Amplifiers 5
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Why Differential?
• Better immunity to environmental noise
• Improved linearity
• Higher signal swing compared to single-ended

6. EECE488 Set 4 - Differential Amplifiers 6
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Higher Immunity to Noise Coupling

7. EECE488 Set 4 - Differential Amplifiers 7
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Supply Noise Reduction

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Board Notes (Improved Linearity)

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Basic Differential Pair

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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
• Can we think of a solution?

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Differential Pair

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Common-Mode Response

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Common-Mode Input versus Output Swing

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Board Notes (“Half-Circuit” Concept)

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Board Notes (“Half-Circuit” Concept)

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Example
• Using the half-circuit concept, calculate the small-signal
differential gain of the following circuit (for two cases of λ=0 and
λ≠0).

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Example
• Using the half-circuit concept, calculate the small-signal
differential gain of the following circuit (for two cases of λ=0 and
λ≠0).

18. EECE488 Set 4 - Differential Amplifiers 18
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Example
• Sketch the small-signal gain of a differential pair as a function of
its input common-mode level.

19. EECE488 Set 4 - Differential Amplifiers 19
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Analysis of Differential Amplifier
TH
ox
n
D
GS
GS
GS
in
in V
L
W
C
I
V
V
V
V
V +
=
−
=
−
µ
2
,
2
1
2
1

20. EECE488 Set 4 - Differential Amplifiers 20
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Analysis of Differential Amplifier
L
W
C
I
L
W
C
I
V
V
ox
n
D
ox
n
D
in
in
µ
µ
2
1
2
1
2
2
−
=
−
( ) )
2
(
2
2
1
2
1
2
2
1 D
D
D
D
ox
n
in
in I
I
I
I
L
W
C
V
V −
+
=
−
µ
( ) 2
1
2
2
1 2
2
1
D
D
SS
in
in
ox
n I
I
I
V
V
L
W
C −
=
−
−
µ
( ) ( ) 2
1
2
2
1
2
4
2
1
2
4
)
(
4
1
D
D
in
in
ox
n
SS
SS
in
in
ox
n I
I
V
V
L
W
C
I
I
V
V
L
W
C =
−
−
+
− µ
µ

21. EECE488 Set 4 - Differential Amplifiers 21
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Analysis of Differential Amplifier
2
2
1
2
1
2
1 )
(
4
)
(
2
1
in
in
ox
n
SS
in
in
ox
n
D
D V
V
L
W
C
I
V
V
L
W
C
I
I −
−
−
=
−
µ
µ
Using:
2
2
1
2
2
2
1
2
2
1
2
1 )
(
)
(
)
(
4 D
D
SS
D
D
D
D
D
D I
I
I
I
I
I
I
I
I −
−
=
−
−
+
=
We have:
( ) ( )2
2
1
2
4
2
1
2
2
1 )
(
4
1
4 in
in
ox
n
SS
SS
in
in
ox
n
D
D V
V
L
W
C
I
I
V
V
L
W
C
I
I −
−
+
−
= µ
µ
and

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Analysis of Differential Amplifier
/
4
/
4
2
1
2
2
id
ox
n
SS
id
ox
n
SS
ox
n
m
id
d
v
L
W
C
I
v
L
W
C
I
L
W
C
G
v
i
−
−
=
=
µ
µ
µ
∂
∂
vid vid
2
4
2
1
id
ox
n
SS
id
ox
n
d v
L
W
C
I
v
L
W
C
i −
=
µ
µ

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Analysis of Differential Amplifier
• For small vid:
• We have:
1
1 m
m
SS
ox
n
id
d
m g
g
I
L
W
C
v
i
G =
=
=
= µ
∂
∂
D
m
in
in
out
out
in
in
m
D
D
D
D
out
out
R
g
V
V
V
V
V
V
G
R
I
I
R
V
V
1
2
1
2
1
2
1
2
1
2
1 )
(
)
(
=
−
−
−
=
−
=
−

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Differential Gain
Av(diff ) =
Vout1 − Vout 2
Vin1 − Vin2
= gmRD
D
m
v R
g
Ended
Single
A
2
)
( =
−

25. EECE488 Set 4 - Differential Amplifiers 25
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Differential Pair as a CS and CD-CG Amplifier
Av = −
gmRD
1+ gmRS
= −
gm
2
RD , (S.E.)
= gmRD , (diff )

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Common-Mode Response
Av =
Vout
Vin,CM
= −
RD /2
1/(2gm) + RSS

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Board Notes

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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)?

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Board Notes

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Common-Mode Response
∆VX
∆Vin,CM
= −
gm
1+ 2gmRSS
RD
∆VY
∆Vin,CM
= −
gm
1+ 2gmRSS
(RD + ∆RD)

31. EECE488 Set 4 - Differential Amplifiers 31
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Common-Mode Response
VX − VY
Vin,CM
= −
gm1 − gm2
(gm1 + gm2 )RSS +1
RD

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Common-Mode Response

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Differential Pair with MOS Loads
Av = −gmN (gmP
−1
|| roN || roP )
≈ −
gmN
gmP
Av = −gmN (roN || roP )

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MOS Loads

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MOS Loads
Av ≈ gm1[(gm3ro3ro1 ) || (gm5ro5ro7 )]

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Gilbert Cell

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Gilbert Cell

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Gilbert Cell
VOUT = kVinVcont