5. differential amplifier

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)

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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).

17. EECE488 Set 4 - Differential Amplifiers 17
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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

22. EECE488 Set 4 - Differential Amplifiers 22
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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

23. EECE488 Set 4 - Differential Amplifiers 23
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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 )

26. EECE488 Set 4 - Differential Amplifiers 26
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Common-Mode Response
Av =
Vout
Vin,CM
= −
RD /2
1/(2gm) + RSS

27. EECE488 Set 4 - Differential Amplifiers 27
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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