Differential Amplifiers
•Single Ended and Differential Operation
•Basic Differential Pair
•Common-Mode Response
•Differential Pair with MOS loads
Hassan Aboushady
University of Paris VI
References
• B. Razavi, “Design of Analog CMOS Integrated Circuits”,
McGraw-Hill, 2001.
H. Aboushady
University of Paris VI
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Differential Amplifiers
•Single Ended and Differential Operation
•Basic Differential Pair
•Common-Mode Response
•Differential Pair with MOS loads
Hassan Aboushady
University of Paris VI
Single Ended and Differential Operation
• Single Ended Signal:
- Measured with respect to a fixed potential, usually ground.
• Differential Signal:
- Measured between 2 nodes that have equal and opposite
excursions around a fixed potential.
- The center potential is called “Common Mode” (CM).
H. Aboushady
University of Paris VI
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Rejection of Common Mode Noise
• Single Ended Signal:
- Due to capacitive coupling,
transitions on the clock line
corrupt the signal on L1 .
• Differential Signal:
- If the clock line is placed
midway, the transitions disturb
the differential signals by equal
amounts, leaving the difference
intact.
H. Aboushady
University of Paris VI
Rejection of Power Supply Noise
• Maximum Output Swing:
Vout max = VDD − (VGS − VTH )
H. Aboushady
• Maximum Output Swing:
VX max − VY max = 2[VDD − (VGS − VTH )]
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Differential Pair
Differential circuit
sensitive
to the input CM level.
if
Differential pair
minimal
dependence on input CM level.
Vin1 ≠ Vin 2
⇒ I D1 ≠ I D 2
⇒ g m1 ≠ g m 2
I SS = I D1 + I D 2
if
Vin1 = Vin 2
I SS
2
I
Output CM = VDD − RD SS
2
⇒ I D1 = I D 2 =
H. Aboushady
University of Paris VI
Differential Pair: Qualitative Analysis
Vin1 << Vin 2
I D 2 = I SS
Vin1 = Vin 2
I
I D1 = I D 2 = SS
2
Vin1 >> Vin 2
I D1 = I SS
H. Aboushady
M1 OFF, M2 ON
Vout1 = VDD
Vout 2 = VDD − I SS RD
M1 ON, M2 ON
Vout1 = Vout 2 = VDD −
I SS
RD
2
M1 ON, M2 OFF
Vout1 = VDD − I SS RD
Vout 2 = VDD
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Differential Pair: Qualitative Analysis
• Maximum and minimum levels are well-defined and
independent of the input CM: VDD and VDD - RD ISS
• The small signal gain (the slope of Vout1-Vout2 vs Vin1-Vin2)
is maximum for Vin1=Vin2 (equilibrium).
H. Aboushady
University of Paris VI
Differential Pair: Common-Mode Behavior
To study Common-Mode
For proper operation:
• M3 in saturation
• M1 & M2 in saturation
H. Aboushady
Vin1 = Vin 2 = Vin ,CM
Vin ,CM ≥ VGS 1 + (VGS 3 − VTH 3 )
I
Vin ,CM ≤ VDD − RD SS + VTH
2
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Differential Pair: Output Voltage Swing
For M1 & M2 in saturation:
Vout − VP ≥ Vin ,CM − VP − VTH
Vout ≥ Vin ,CM − VTH
Output Voltage Swing:
VDD ≥ Vout ≥ Vin ,CM − VTH
To increase output swing, we choose a low
H. Aboushady
Vin ,CM
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Differential Pair: Quantitative Analysis
VP = Vin1 − VGS1 = Vin 2 − VGS 2
Vin1 − Vin 2 = VGS1 − VGS 2
1
Assuming M1 & M2 in saturation:
(VGS − VTH ) 2 =
VGS =
2I D
µ nCox
2I D
W
µ nCox
L
W
L
+ VTH
From eq. 1 & 2 :
Vin1 − Vin 2 =
2
Squaring the 2 sides, and since:
(Vin1 − Vin 2 ) 2 =
H. Aboushady
2 I D1
2I D 2
−
W
W
µ nCox
µ nCox
L
L
I SS = I D1 + I D 2
2
W
µ nCox
L
( I SS − 2 I D1 I D 2 )
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Differential Pair: Quantitative Analysis
Previous equation can be written:
µ nCox W
2
L
(Vin1 − Vin 2 ) 2 − I SS = −2 I D1 I D 2
Squaring the 2 sides, and since:
4 I D1 I D 2 = ( I D1 + I D 2 ) 2 − ( I D1 − I D 2 ) 2
2
= I SS
− ( I D1 − I D 2 ) 2
We arrive at:
2
1⎛
W
W⎞
( I D1 − I D 2 ) = − ⎜ µ nCox ⎟ (Vin1 − Vin 2 ) 4 + I SS µ nCox (Vin1 − Vin 2 ) 2
4⎝
L
L⎠
2
I D1 − I D 2 =
µ nCox W
2
L
4 I SS
− (Vin1 − Vin 2 ) 2
W
µ nCox
L
(Vin1 − Vin 2 )
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3
University of Paris VI
Differential Pair: Quantitative Analysis
Let ∆Vin = Vin1 − Vin 2
and ∆I D = I D1 − I D 2
Deriving eq. 3 with respect to
∆Vin
4 I SS
− 2∆Vin2
∂∆I D µ nCox W µ nCoxW / L
Gm =
=
∂∆Vin
2 L
4 I SS
− ∆Vin2
µ nCoxW / L
For
∆Vin = 0 , Gm = µ nCox
Since:
4
W
I SS
L
Vout1 − Vout 2 = VDD − I D1 RD1 − VDD − I D 2 RD 2
∆Vout = Gm ∆Vin RD
∆Vout = ∆I D RD
The small signal
differential voltage gain:
H. Aboushady
Av =
∆Vout
W
= Gm RD = µ n Cox I SS RD
∆Vin
L
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Drain Currents and Overall Transconductance
∆Vin1 is when I D1 = I SS
I D1 − I D 2 =
µ nCox W
2
L
∆Vin1 = VGS1 − VTH 1
(Vin1 − Vin 2 )
2 I SS
W
µ nCox
L
∆Vin1 =
4 I SS
− (Vin1 − Vin 2 ) 2
W
µ nCox
L
3
4 I SS
− 2∆Vin2
µ C W µ nCoxW / L
Gm = n ox
2 L
4 I SS
− ∆Vin2
µ nCoxW / L
H. Aboushady
4
University of Paris VI
∆ID vs ∆VD
Plot the input-output characteristics
of a differential pair as the device
width and the tail current vary:
∆Vin1 =
W
↑↑
L
2 I SS
W
µ nCox
L
I SS ↑↑
H. Aboushady
University of Paris VI
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Differential Pair: Small Signal Gain
Av =
∆Vout
W
= Gm RD = µ n Cox I SS RD
∆Vin
L
In equilibrium, we have
I D1 = I D 2 =
Av = g m RD
I SS
2
Where gm is the
transconductance of M1 & M2.
H. Aboushady
University of Paris VI
Calculating Small Signal Gain by Superposition
Set Vin2 =0
M1 forms a common source
stage with a degeneration
resistance
Av =
VX
g R
= − m1 D
Vin1
1+ g m1 RS
Neglecting channel length
modulation and body effect
RS = 1 / g m 2
VX
g m1 RD
=−
Vin1
1 + g m1 / g m 2
VX
RD
=−
1
1
Vin1
+
g m1 g m 2
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5
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Calculating Small Signal Gain by Superposition
Replacing M1 by its Thévenin equivalent:
VT = Vin1
RT = 1 / g m1
VY
RD
=
1
1
Vin1
+
g m1 g m 2
From eq. 5 & 6, we get:
(V X − VY ) due to V =
in1
H. Aboushady
Rin 2 = 1 / g m 2
6
− 2 RD
V
1
1 in1
+
g m1 g m 2
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Calculating Small Signal Gain by Superposition
(V X − VY ) due to V =
in1
Since:
g m1 = g m 2 = g m
− 2 RD
V
1
1 in1
+
g m1 g m 2
(V X − VY ) due to V = − g m RDVin1
in 1
Similarly we can say that:
(V X − VY ) due to V = g m RDVin 2
in 2
The small signal
differential voltage gain:
H. Aboushady
(V X − VY ) total
Vin1 − Vin 2
= − g m RD
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The concept of Half Circuit
If a fully symmetric differential pair senses differential
inputs then the concept of half circuit can be applied.
• A differential change in the inputs Vin1 and Vin2 is
absorbed by V1 and V2 leaving VP constant
H. Aboushady
University of Paris VI
Application of The Half Circuit Concept
Since VP experiences no change, node P can be considered
“ac ground” and the circuit can be decomposed into two
separate halves
Two common source amplifiers:
VX
= − g m RD
Vin1
VY
= − g m RD
Vin 2
VX − VY
= − g m RD
Vin1 − Vin 2
H. Aboushady
University of Paris VI
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The Half Circuit Concept : Example
Taking into account the output resistance
(channel length modulation)
Two common source amplifiers:
VX
= − g m (RD // rO1 )
Vin1
H. Aboushady
VY
= − g m (RD // rO 2 )
Vin 2
VX − VY
= − g m (RD // rO )
Vin1 − Vin 2
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Arbitrary Inputs to a Differential Pair
Conversion of arbitrary inputs to differential and common-mode
components:
H. Aboushady
Common
Mode
Differential
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Arbitrary Inputs to a Differential Pair: Example
Calculate VX and VY if Vin1 =Vin2 and λ=0
For differential mode operation:
V X = − g m (RD // rO1 )
(Vin1 − Vin 2 )
2
(
Vin 2 − Vin1 )
VY = − g m (RD // rO 2 )
2
VX − VY = − g m (RD // rO ) (Vin1 − Vin 2 )
For common mode operation:
I D1 = I D 2 = I SS / 2
Assuming fully symmetric circuit
and Ideal Current Source:
•ID1 and ID2 independent of VCM ,in
•VX and VY independent of VCM ,in
The Differential pair circuit:
Amplifies Vin1 − Vin 2 , Eliminates the effect of
H. Aboushady
VCM ,in
University of Paris VI
Common Mode Response: Non-Ideal Current Source
Assuming fully symmetric circuit
with finite output impedance
current source, RSS :
Equivalent circuit:
• Degenerated Common Source
Av ,CM =
Vout
RD / 2
=−
Vin ,CM
1 /( 2 g m ) + RSS
gm the transconductance of 1 transistor.
H. Aboushady
University of Paris VI
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Common Mode Response: RD Mismatch Effect
Assuming M1 & M2 identical:
∆V X = −∆Vin ,CM
RD
1 /( 2 g m ) + RSS
∆VY = − ∆Vin ,CM
RD + ∆RD
1 /( 2 g m ) + RSS
Common mode to differential conversion
Effect of CM noise in the
presence of resistor noise
H. Aboushady
University of Paris VI
Common Mode Response: M1-M2 Mismatch Effect
g m1 ≠ g m 2
I D1 = g m1 (Vin ,CM − VP )
I D 2 = g m 2 (Vin ,CM − VP )
VP = (g m 2 + g m 2 )(Vin ,CM − VP ) RSS
VP =
(g m 2 + g m 2 )RSS V
(g m 2 + g m 2 )RSS + 1 in,CM
V X = − g m1 (Vin ,CM − VP )RD
VX =
− g m1
RV
(g m1 + g m 2 )RSS + 1 D in,CM
VY = − g m 2 (Vin ,CM − VP )RD
VY =
− gm2
R V
(g m1 + g m 2 )RSS + 1 D in,CM
V X − VY =
(g m1 − g m 2 )RDVin,CM
(g m1 + g m 2 )RSS + 1
CM to DM Conversion Gain:
H. Aboushady
ACM − DM =
V X − VY
∆g m RD
=
(g University
)RParis
Vin ,CM
1
m1 + g m 2 of
SS +VI
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Common Mode Rejection Ratio
CMRR =
ADM
ADM : Differential Mode Gain
ACM − DM
ACM − DM : Common-Mode to
H. Aboushady
Differential Mode Gain
University of Paris VI
Cascode Differential Pair
Current Source Load:
Av = g mN (rON // rOP )
Low gain 10 to 20.
To increase the gain:
Cascode Differential Pair
Av = g m1 ( g m3 rO 3rO1 // g m5 rO 5 rO 7 )
H. Aboushady
University of Paris VI
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