EECE488: Analog CMOS Integrated Circuit Design Set 4 - Courses

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 Columbiashahriar@ece.ubc.ca

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