7. differential amplifiersaries.ucsd.edu/najmabadi/class/ece102/12-f/notes/ece102_f12 … ·...

7. Differential Amplifiers

ECE 102, Fall 2012, F. Najmabadi

Sedra & Smith Sec. 2.1.3 and Sec. 8 (MOS Portion)

(S&S 5th Ed: Sec. 2.1.3 and Sec. 7 MOS Portion & ignore frequency-response)

F. Najmabadi, ECE102, Fall 2012 (2/33)

Common-Mode and Differential-Mode Signals & Gain

Differential and Common-Mode Signals/Gain


Consider a linear circuit with TWO inputs

2211 vAvAvo ⋅+⋅=

By superposition:

Define: 12 vvvd −=

2

21 vvvc+

=

Difference (or differential) Mode

Common Mode

Substituting for and in the expression for vo: 2

1d

cvvv −=

2 2

dc

vvv +=

( ) dcd

cd

co vAAvAAvvAvvAv ⋅

−

+⋅+=

+⋅+

−⋅=

22212

2121

ddcco vAvAv ⋅+⋅=

2

2

2

1

dc

dc

vvv

vvv

+=

−=

Differential and common-mode signal/gain is an alternative way of finding the system response


2

2

1221

2112

AAAvvv

AAAvvv

dc

cd

−=

+=

+=−=

2211 vAvAvo ⋅+⋅= ddcco vAvAv ⋅+⋅=

Differential Gain: Ad Common Mode Gain: Ac Common Mode Rejection Ratio (CMRR)*: |Ad|/|Ac|

* CMRR is usually given in dB: CMRR(dB) = 20 log (|Ad|/|Ac|)

2

2

2

2

22

11

dcd

c

dcd

c

AAAvvv

AAAvvv

+=+=

−=−=

To find vo , we can calculate/measure either A1 A2 pair or Ac Ad pair


Difference Method (finding Ad and Ac ): 1. Set vc = 0 (or set v1 = − 0.5 vd & v2 = + 0.5 vd)

compute Ad from vo = Ad vd 2. Set vd = 0 (or set v1 = + vc & v2 = + vc )

compute Ac from vo = Ac vc 3. For any v1 and v2 : vo = Ad vd + Ac vc

Superposition (finding A1 and A2 ): 1. Set v2 = 0, compute A1 from

vo = A1 v1 2. Set v1 = 0, compute A2 from

vo = A2 v2

3. For any v1 and v2 : vo = A1 v1 + A2 v2

Both methods give the same answer for vo (or Av ).

The choice of the method is driven by application: o Easier solution o More relevant parameters

)(5.0 2112 vvvvvv cd +=−=

Caution


In Chapter 2.1.3, Sedra & Smith defines vd = v2 − v1 But in Chapter 8, Sedra & Smith uses vd = v1 − v2 While keeping vo = vo2 − vo1 as before (this is inconsistent)

21d

cvvv −=

22d

cvvv +=

21d

cvvv +=

22d

cvvv −=

Here we use vd = v2 − v1 and vo = vo2 − vo1 throughout

Therefore, Ad (lecture slides) = −Ad (Sedra & Smith) for difference Amplifiers.

Use Lecture Slides Notation!

21d

cvvv −=

22d

cvvv +=


Differential Amplifiers: Fundamental Properties

Differential Amplifier


o For now, we keep track of “two” output, vo1 and vo2 , because there are several ways to configure “one” output from this circuit.

Identical transistors. Circuit elements are symmetric about the mid-plane. Identical bias voltages at Q1 & Q2 gates (VG1 = VG2 ). Signal voltages & currents are different because v1 ≠ v2.

Load RD: resistor, current-mirror, active load, …

RSS: Bias resistor, current source (current-mirror)

Q1 & Q2 are in CS-like configuration (input at the gate, output at the drain) but with sources connected to each other.

Differential Amplifier – Bias


ID ID

ID ID

2ID

21

21

21

21

DSDSDS

DDD

OVOVOV

GSGSGS

VVVIII

VVVVVV

====

====

SSS

GGG

VVVVVV

====

21

21

and Since

ooo

mmm

rrrggg

====

21

21Also:

This is correct even if channel-length modulation is included because 2211 DSDDDSDD VRIVRI +=+

Differential Amplifier – Gain


Signal voltages & currents are different because v1 ≠ v2

We cannot use fundamental amplifier configuration for arbitrary values of v1 and v2.

We have to replace each NMOS with its small-signal model.

Differential Amplifier – Gain


0)()(

0)(

0)(

323113233

32322

31311

=−−−−−

+−

+

=−+−

+

=−+−

+

vvgvvgr

vvr

vvRv

vvgr

vvRv

vvgr

vvRv

mmo

o

o

o

SS

mo

o

D

o

mo

o

D

o

Node Voltage Method:

Node vo1:

Node vo2:

Node v3:

Above three equations should be solved to find vo1 , vo2 and v3 (lengthy calculations)

322

311

vvvvvv

gs

gs

−=

−=

Because the circuit is symmetric, differential/common-mode method is the preferred method to solve this circuit (and we can use fundamental configuration formulas).

Differential Amplifier – Common Mode (1)


Because of summery of the circuit and input signals*:

Common Mode: Set vd = 0 (or set v1 = + vc and v2 = + vc )

dddoo iiivv === 2121 and

We can solve for vo1 by node voltage method but there is a simpler and more elegant way.

id

id id 2id

* If you do not see this, set v1 = v2 = vc in node equations of the previous slide, subtract the first two equations to get vo1 = vo2 . Ohm’s law on RD then gives id1 = id2 = id



Because of the symmetry, the common-mode circuit breaks into two identical “half-circuits”.

id

id id 2id

id

0

id

* 23 SSd Riv =

* Vss is grounded for signal



oDSSm

Dm

c

o

c

o

rRRgRg

vv

vv

/2121

++−==

CS Amplifiers with Rs

0

The common-mode circuit breaks into two identical half-circuits.

Differential Amplifier – Differential Mode (1)


32

31

5.0

5.0

vvvvvv

dgs

dgs

−+=

−−=

0)5.0()5.0(

0)5.0(

0)5.0(

3313233

3322

3311

=−+−−−−−

+−

+

=−++−

+

=−−+−

+

vvgvvgr

vvr

vvRv

vvgr

vvRv

vvgr

vvRv

dmdmo

o

o

o

SS

dmo

o

D

o

dmo

o

D

o

Node Voltage Method:

Node vo1:

Node vo2:

Node v3:

Differential Mode: Set vc = 0 (or set v1 = − vd /2 and v2 = + vd /2 )

022)(11321 =

+−+

+ vg

rvv

rR mo

oooD

( ) 02211321 =

−+++− vg

rRvv

r moSS

ooo

Node vo1 + Node vo2 :

Node v3: 0 0

3

2121

=−=⇒=+

vvvvv oooo

Only possible solution:

Differential Amplifier – Differential Mode (2)


Because of the symmetry, the differential-mode circuit also breaks into two identical half-circuits.

21213 and 0 ddoo iivvv −=⇒−==

v3 = 0

id id

v3 = 0

CS Amplifier

)||(5.0

, )||(5.0

21Dom

d

oDom

d

o Rrgv

vRrgv

v−=

+−=

−

id id

0

id id

Concept of “Half Circuit”


Common Mode Differential Mode

For a symmetric circuit, differential- and common-mode analysis can be performed using “half-circuits.”

Common-Mode “Half Circuit”


id id

id id

0

Common Mode Half-circuit 1. Currents about symmetry line are equal. 2. Voltages about the symmetry line are equal (e.g., vo1 = vo2) 3. No current crosses the symmetry line.

21 oo vv =

Common Mode circuit

21 ss vv =

Differential-Mode “Half Circuit”


Differential Mode circuit

Differential Mode Half-circuit 1. Currents about the symmetry line are equal in value and opposite in sign. 2. Voltages about the symmetry line are equal in value and opposite in sign. 3. Voltage at the summery line is zero

21 oo vv −=

021 == ss vv

id id

id id

Constructing “Half Circuits”


Step 1: Divide ALL elements that cross the symmetry line (e.g., RL) and/or are located on the symmetry line (current source) such that we have a symmetric circuit (only wires should cross the symmetry line, nothing should be located on the symmetry line!)

Constructing “Half Circuit”– Common Mode


Step 2: Common Mode Half-circuit 1. Currents about symmetry line are equal (e.g., id1 = id2). 2. Voltages about the symmetry line are equal (e.g., vo1 = vo2). 3. No current crosses the symmetry line.

coco vv ,2,1 =

0

Constructing “Half Circuit”– Differential Mode


Step 3: Differential Mode Half-Circuit 1. Currents about symmetry line are equal but opposite sign (e.g., id1 = − id2) 2. Voltages about the symmetry line are equal but opposite sign (e.g., vo1 = − vo2) 3. Voltage on the symmetry line is zero.

dodo vv ,2,1 −=

“Half-Circuit” works only if the circuit is symmetric!


Half circuits for common-mode and differential mode are different.

Bias circuit is similar to Half circuit for common mode.

Not all difference amplifiers are symmetric. Look at the load carefully!

We can still use half circuit concept if the deviation from prefect symmetry is small (i.e., if one transistor has RD and the other RD + ∆RD with ∆RD << RD). o However, we need to solve BOTH half-circuits (see slide 30)


Why are Differential Amplifiers popular?

They are much less sensitive to noise (CMRR >>1).

Biasing: Relatively easy direct coupling of stages: o Biasing resistor (RSS) does not affect the differential gain

(and does not need a by-pass capacitor). o No need for precise biasing of the gate in ICs o DC amplifiers (no coupling/bypass capacitors).

…

Why is a large CMRR useful?


A major goal in circuit design is to minimize the noise level (or improve signal-to-noise ratio). Noise comes from many sources (thermal, EM, …)

However, if the signal is applied between two inputs and we use a difference amplifier with a large CMRR, the signal is amplified a lot more than the noise which improves the signal to noise ratio.*

A regular amplifier “amplifies” both signal and noise.

noised

sigdccddo vCMRR

AvAvAvAv ⋅+⋅=⋅+⋅=

1 noisesig vvv +=

1 noisesigo vAvAvAv ⋅+⋅=⋅=

noisecsigd

noisesignoisesig

vvvvvvvvvvvv

==−=

++=+−=

&

5.0 & 5.0

12

21

* Assuming that noise levels are similar to both inputs.

Comparing a differential amplifier two identical CS amplifiers (perfectly matched)


Differential Amplifier Two CS Amplifiers

Comparison of a differential amplifier with two identical CS amplifiers – Differential Mode


)||(/ )||(

)5.0( )||( )5.0( )||(

,1,2

,2

,1

Domdodd

dDomdodood

dDomdo

dDomdo

RrgvvAvRrgvvv

vRrgvvRrgv

−==

−=−=

+−=

−−=

vo1,d , vo2,d , vod, and differential gain, Ad, are identical.

Half-Circuits

Identical

Differential amplifier Two CS amplifiers

Comparison of a differential amplifier with two identical CS amplifiers – Common Mode


Half-Circuits

NOT Identical

vo1,c & vo2,c are different! But voc = 0 and CMMR = ∞.

0/ 0

/21

,1,2

,2,1

==

=−=++

−==

cocc

cocooc

coDSSm

Dmcoco

vvAvvv

vrRRg

Rgvv

0/ 0

)||(

,1,2

,2,1

==

=−=

−==

cocc

cocooc

cDomcoco

vvAvvv

vRrgvv


Comparison of a differential amplifier with two identical CS amplifiers – Summary


∞=

==−==

CMRR

0 , )||( c

occDom

d

odd v

vARrgvvA

For perfectly matched circuits, there is no difference between a differential amplifier and two identical CS amplifiers. o But one can never make perfectly matched circuits!

∞=

==−==

CMRR

0 , )||( c

occDom

d

odd v

vARrgvvA

Differential Amplifier Two CS Amplifiers

Consider a “slight” mis-match in the load resistors


We will ignore ro in the this analysis (to make equations simpler)

“Slightly” mis-matched loads – Differential Mode


)5.0(/ )5.0(

)5.0( )( )5.0( )(

,1,2

,2

,1

DDmdodd

dDDmdodood

dDDmdo

dDmdo

RRgvvAvRRgvvv

vRRgvvRgv

∆+−==

∆+−=−=

+∆+−=

−−=

vo1, vo2, vod, and differential gain, Ad, are identical.

Half-Circuits

Identical


“Slightly” mis-matched loads – Common Mode


Half-Circuits

NOT Identical


vo1 and vo2 are different. In addition, voc ≠ 0 and CMMR ≠ ∞.

SSm

Dm

c

occ

cSSm

Dmcocooc

cSSm

DDmcoc

SSm

Dmco

RgRg

vvA

vRg

Rgvvv

vRg

RRgvvRg

Rgv

21

21

21

)( , 21

,1,2

,2,1

+∆

−==

+∆

−=−=

+∆+

−=+

−=

Dmc

occ

cDmcocooc

cDDmco

cDmco

RgvvA

vRgvvvvRRgv

vRgv

∆+==

∆+=−=

∆+−=

−=

)(

,1,2

,2

,1

A differential amplifier increases CMRR substantially for a slight mis-match (∆RD ≠0)


Dmc RgA ∆+=

)5.0( DDmd RRgA ∆+−=

DD RR /1CMRR

∆≈

Two CS Amplifiers

)5.0( DDmd RRgA ∆+−=

SSm

Dmc Rg

RgA21+

∆−=

DD

SSm

RRRg

/21CMRR

∆+

≈

Differential Amplifier

Differential amplifier reduces Ac and increases CMRR substantially (by a factor of: 1 + 2 gmRSS). The common-mode half-circuits for a differential amplifier are

CS amplifiers with RS (thus common mode gain is much smaller than two CS amplifiers).

We should use a large RSS in a differential amplifier!

* Exercise: Compare a differential amplifier and two CS amplifiers with a mis-match in gm

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