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 )]
University of Paris VI
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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 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
University of Paris VI
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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
University of Paris VI
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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
University of Paris VI
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 ) University of Paris VI
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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 )
H. Aboushady
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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 University of Paris VI
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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
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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 H. Aboushady
5 University of Paris VI
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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
University of Paris VI
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 University of Paris VI
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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
University of Paris VI
Arbitrary Inputs to a Differential Pair Conversion of arbitrary inputs to differential and common-mode components:
H. Aboushady
Common Mode
Differential University of Paris VI
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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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