eece488: analog cmos integrated circuit design set 4 differential
TRANSCRIPT
EECE488 Set 4 - Differential Amplifiers 1SM
EECE488: Analog CMOS Integrated Circuit Design
Set 4
Differential Amplifiers
Shahriar MirabbasiDepartment of Electrical and Computer Engineering
University of British [email protected]
EECE488 Set 4 - Differential Amplifiers 2SM
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
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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
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Higher Immunity to Noise Coupling
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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).
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Example
• Sketch the small-signal gain of a differential pair as a function of its input common-mode level.
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Analysis of Differential Amplifier
TH
oxn
DGSGSGSinin V
LWC
IVVVVV +=−=−µ
2,2121
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Analysis of Differential Amplifier
LWC
I
LWC
IVVoxn
D
oxn
Dinin
µµ
2121
22−=−
( ) )2(22121
221 DDDD
oxn
inin IIII
LWC
VV −+=−µ
( ) 212
21 221
DDSSininoxn IIIVVLWC −=−−µ
( ) ( ) 212
2124
212 4)(
41
DDininoxnSSSSininoxn IIVVLWCIIVV
LWC =−−+− µµ
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Analysis of Differential Amplifier
2212121 )(4)(
21
inin
oxn
SSininoxnDD VV
LWC
IVVLWCII −−−=−
µµ
Using:2
2122
212
2121 )()()(4 DDSSDDDDDD IIIIIIIII −−=−−+=
We have:
( ) ( )221
2421
221 )(
414 ininoxnSSSSininoxnDD VV
LWCIIVV
LWCII −−+−= µµ
and
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Analysis of Differential Amplifier
/4
/4
21
2
2
idoxn
SS
idoxn
SS
oxnmid
d
vLWC
I
vLWC
I
LWCG
vi
−
−==
µ
µµ
∂∂
vid vid
2421
id
oxn
SSidoxnd v
LWC
IvLWCi −=
µµ
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Analysis of Differential Amplifier
• For small vid:
• We have:
11 mmSSoxnid
dm ggI
LWC
viG ==== µ
∂∂
Dminin
outout
ininmDDDDoutout
RgVVVV
VVGRIIRVV
121
21
212121 )()(
=−−
−=−=−
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Differential Gain
Av(diff ) =Vout1 − Vout 2
Vin1 − Vin2
= gmRD
Dm
v RgEndedSingleA2
)( =−
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Differential Pair as a CS and CD-CG Amplifier
Av = − gmRD
1+ gmRS
= −gm
2RD , (S.E.)
= gmRD , (diff )
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Common-Mode Response
Av = Vout
Vin ,CM
= − RD / 21/(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)
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Common-Mode Response
VX − VY
Vin ,CM
= −gm1 − gm 2
(gm1 + gm 2 )RSS +1RD
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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 ≈ gm 1[(gm 3ro3ro1 ) || (gm 5ro5ro7 )]
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Gilbert Cell
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Gilbert Cell
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Gilbert Cell
VOUT = kVinVcont