Differential Amplifiers (Ch. 10)p

김 영 석김 영 석

충북대학교 전자정보대학

2012.9.1.9.

Email: kimys@cbu.ac.kr

전자정보대학 김영석 10-1

Contents10.1 General Considerations

10.2 Bipolar Differential Pair

10.3 MOS Differential Pair

10 4 C d Diff ti l A lifi10.4 Cascode Differential Amplifiers

10.5 Common-Mode Rejection

10.6 Differential Pair with Active Load1 .6 Differential Pair with Active Load

전자정보대학 김영석 10-2

10.1 General ConsiderationsHumming Noise in Audio Amplifier from 60Hz Power Line

CCCCout IRVV −=

10-3전자정보대학 김영석

10.1 General ConsiderationsHow to suppress the Hum ?

Vout should not to be referenced to GND

Since both node X and Y contain the ripple, their difference will be free of ripple

CCrCCout IRvVV −+= )(

will be free of ripple

But, wasting current in Q2

rinvX vvAv +=

) ( rinvYXout

rY

vNOvAvvVvv

=−==

10-4전자정보대학 김영석

10.1 General ConsiderationsTo utilize the wasting current in Q2

Apply the inputs differentially

• the outputs are 180° out of phase

h i V t h d diff ti ll• enhancing Vout when sensed differentially

But, Common-Mode (CM) problem: increase VCM => increase gm => increase gaing g

rinvX

AvvAv +=

invYXout

rinvY

vAvvVvvAv

2=−=+−=

10-5전자정보대학 김영석

10.1 General ConsiderationsTo have the differential-mode gain independent of input CM level

Add a tail current

10-6전자정보대학 김영석

10.2 Bipolar Differential PairCommon-Mode Response

21 = BEBE VV

CM gain

221

==

==

EE

EECC

IRVVV

III

oft independen2

∂

−==

CM

CCCYX

VV

RVVV

0=∂∂

=∴CM

outCM V

VA

CM i tCM input range

EECCCCM

IRVV −< Region)Activein(Q1

BECM

CCCCM

VVEEV

RVV

+>

< Region)Activein (Q2 1

10-7전자정보대학 김영석

Differential Pair Characteristics

(a)

EEC

III =0

1EEC II =2

(b)

CCY

EECCCX

C

VVIRVV

I

=−=

= 02

CCX

EECCCY

C

VVIRVV

I

=−=

= 01

(b)( ) (b)(b)

(a)(a)

10-8전자정보대학 김영석

Small-Signal Analysis

EEEE

mEEEE

C

VgIIII

VgIIII

Δ−=Δ−=

Δ+=Δ+=221

mCCY

mCCX

mC

VgRIRVVgRIRV

VgII

Δ+=Δ+=ΔΔ−=Δ−=Δ

Δ=Δ=222

in

mCYXout

mCCY

VVV

VgRVVVΔ=

Δ−=Δ−Δ=2

2

Cmin

outv Rg

VVA −==

Virtual Ground: 0=Δ PV

10-9전자정보대학 김영석

10.2.2 Large Signal Analysis

Tin VV 4max, ≈Δ

EECC III 21 =+

inin

SSTinSC

TinSC

VVI

IIVVIIVVII

21

21222

111

exp

, /exp/exp

−==

=

T

inin

TEE

C

VVV

VI

I21

1

exp1

exp

−+

=

T

inin

EEC

VVV

II21

2

exp1 −+

=

T

ininEEC

outout

VVVIR

VV

2tanh 21

21

−−

=−

10-10전자정보대학 김영석

TV2

Linear/Nonlinear Regions

The left column operates in linear region, whereas the right column operates in nonlinear regionright column operates in nonlinear region.

10-11전자정보대학 김영석

10.2.3 Small-Signal Model

10-12전자정보대학 김영석

Half Circuits

mCout

ggvgRvvgRv

=−=−= 111 π

inin

mmmCout

vvvvvv

ggvgRv

−+=+=

==

21

2211

21222

,,

ππ

π

Cminin

outout Rgvvvv

−=− 21

21

Since VP is grounded, we can treat the differential pair as two CE “half circuits”, with its gain equal to one half circuit’s single-ended gain.

10-13전자정보대학 김영석

Example: Differential Gain

outout rgvv−=

− 21Om

inin

rgvv

=− 21

10-14전자정보대학 김영석

Half Circuit Example I

0=XV

( )1311 |||| RrrgA OOmv −=

10-15전자정보대학 김영석

Half Circuit Example II

( )1311 |||| RrrgA OOmv −=

10-16전자정보대학 김영석

Half Circuit Example III

Cv

RA 1−=

mE g

R 1+

10-17전자정보대학 김영석

Half Circuit Example IV

CRA −=

m

Ev

gRA

12+

mg

10-18전자정보대학 김영석

10.3 MOS Differential Pair’s Common-Mode Response

ICM gain = 0

The equilibrium overdrive voltage:

In ut CM Range

2SS

DDDYXIRVVV −==

( )

LWC

IVVVoxn

SSequilTHGS

μ=−=Δ

Input CM Range

VVV R i )S ii(M1

Loxnμ

THCMSS

DDD

THGD

VVIRV

VVV

−>−

−>

2

Region)Saturationin (M1

SSGSCM

THSS

DDDCM

VVVAnd

VIRVV

+>

+−<∴

,2SSGSCM,

10-19전자정보대학 김영석

Differential Response

(b)(a)

( )(a)

(b)(b)

(a) (b)

CH 10 Differential Amplifiers

10-20전자정보대학 김영석

Small-Signal Response

VΔ 0

Dmv

P

RgAV

−==Δ 0

Dmv

Similar to its bipolar counterpart, the MOS differential pair exhibits the same virtual ground node and small signal gain.

10-21전자정보대학 김영석

10.3.2 MOS Differential Pair’s Large-Signal Response

W1

THGSoxnD

THGSoxnD

VVL

WCI

VVL

WCI

−=

−=

222

211

)(21

)(21

μ

μ

( ) ( )241 SS VVIVVWCII

SSDD

THGSoxnD

IIIL

=+ 21

22 2

( ) ( )2211214

21

2 inin

oxn

SSinoxnDD VV

LWC

IVVLWCII

in−−−=−

μμ

( )LWC

IVVVVVoxn

DequilTHGSinin /

2 22max21 μ

=−=Δ=−

10-22전자정보대학 김영석

Contrast Between MOS and Bipolar Differential Pairs

In a MOS differential pair, there exists a finite differential input voltage to completely switch the current from one transistor to the other, whereas, in a bipolar pair that voltage is infinite.

And,

Tinin

inin

VVVBJT

VVVMOS

4:

2:

max21

max21

≈−

Δ=−

MOS Bipolar

10-23전자정보대학 김영석

The effects of Doubling the Tail Current

( )I

VVVV

D

equilTHGSinin

2

2max21

=

−=−

LWCoxn /

μ=

ISS is doubledSS

ΔVin,max increases by

ΔVout,max increases by

22

10-24전자정보대학 김영석

The effects of Doubling W/L

( )I

VVVV

D

equilTHGSinin

2

2max21

=

−=−

LWCoxn /

μ=

W/L is doubled

Vin max decreases by 2Vin,max decreases by

Vout,max same

10-25전자정보대학 김영석

Small-Signal Analysis of MOS Differential Pair

( ) ( )241 SS VVIVVWCII ( ) ( )

( ) ( ) ( )

21121

41

2 2 inin

oxn

SSinoxnDD

WIW

VV

LWC

VVL

CIIin

−−−=−μ

μ

( ) ( ) ( )2121214

21 ininmininSSoxn

oxn

SSininoxn VVgVVI

LWC

LWC

IVVLWC −=−=−≈ μ

μμ

When the input differential signal is small compared to 4ISS/μnCox(W/L), the output differential current is linearly proportional to it and small-signal model can be appliedproportional to it, and small signal model can be applied.

10-26전자정보대학 김영석

Virtual Ground and Half Circuit

PV =Δ 0

Cmv RgA −=

Applying the same analysis as the bipolar case, we will arrive at the same conclusion that node P will not move for small input signals and the concept of half circuit can

dbe used to calculate the gain.

10-27전자정보대학 김영석

MOS Differential Pair Half Circuit Example I

⎞⎛

≠

1

0λ

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 13

31 ||||1

OOm

mv rrg

gA

10-28전자정보대학 김영석

MOS Differential Pair Half Circuit Example II

0g

=λ

3

1

m

mv g

gA −=

10-29전자정보대학 김영석

MOS Differential Pair Half Circuit Example III

DDRA 20=λ

mSS

DDv gR

A12+

−=

10-30전자정보대학 김영석

10.4 Bipolar Cascode Differential Pair

( )[ ])1(|| 333131 OmOOmv rgrrrgA ++−= π

10-31전자정보대학 김영석

Bipolar Telescopic Cascode

( )[ ] [ ])||(|||| 575531331 ππ rrrgrrrggA OOmOOmmv −≈

10-32전자정보대학 김영석

Example: Bipolar Telescopic Parasitic Resistance

RrrgrR |||| 1 ⎟⎞

⎜⎛≈

[ ] opOOmmv

OmOop

RrrrggA

rrgrR

||)||(2

||||

31331

5755

π

π

−=

⎟⎠

⎜⎝

≈

10-33전자정보대학 김영석

MOS Cascode Differential Pair

1331 OmOmv rgrgA −≈

10-34전자정보대학 김영석

MOS Telescopic Cascode

( )[ ])(|| 7551331 OOmOOmmv rrgrrggA −≈ ( )[ ])(|| 7551331 OOmOOmmv ggg

10-35전자정보대학 김영석

Example: MOS Telescopic Parasitic Resistance

)||(

)]||(1[||

1331

155715

OmOopmv

OmOOop

rgrRgA

RrgrRrR

−≈

++=

10-36전자정보대학 김영석

10.5 CM RejectionEffect of Finite Tail Impedance

Δ

mEE

C

CMin

CMout

gRR

VV

2/12/

,

,

+−=

ΔΔ

If the tail current source is not ideal, then when a input CM voltage is applied, the currents in Q1 and Q2 and g pp 1 2 hence output CM voltage will change.

10-37전자정보대학 김영석

Input CM Noise with Ideal Tail Current

10-38전자정보대학 김영석

Input CM Noise with Non-ideal Tail Current

10-39전자정보대학 김영석

CM to DM Conversion, ACM-DM

Dout

RgR

VV

2/1 +Δ

=ΔΔ

EEmCM RgV 2/1 +Δ

If finite tail impedance and asymmetry are both present, then the differential output signal will contain a portion of input common mode signalinput common-mode signal.

10-40전자정보대학 김영석

Example: ACM-DM

A DMCM =−

[ ])||)(1(2131333 πrRrgr

R

OmO

C

+++

Δ

[ ])||)(( 313331

πgg OmO

m

10-41전자정보대학 김영석

CMRR

DMCM

DM

AACMRR =

DMCMA −

CMRR defines the ratio of wanted amplified differential input signal to unwanted converted input common-mode noise that appears at the output. pp p

10-42전자정보대학 김영석

10.6 Diff Pair with Active LoadDifferential to Single-ended Conversion

Many circuits require a differential to single-ended i h th b t l i t dconversion, however, the above topology is not very good.

10-43전자정보대학 김영석

Supply Noise Corruption

The most critical drawback of this topology is supply noise corruption, since no common-mode cancellation

h i i t Al l h lf f th i lmechanism exists. Also, we lose half of the signal.

10-44전자정보대학 김영석

Better Alternative

This circuit topology performs differential to single-ended conversion with no loss of gain.

10-45전자정보대학 김영석

Active LoadWith current mirror used as the load, the signal current produced by the Q1 can be replicated onto Q4.

This type of load is different from the conventional “static load” and is known as an “active load”.and is known as an active load .

10-46전자정보대학 김영석

Differential Pair with Active Load

Th i t diff ti l i d th t dThe input differential pair decreases the current drawn from RL by ΔI and the active load pushes an extra ΔI into RL by current mirror action; these effects enhance each other.

10-47전자정보대학 김영석

Active Load vs. Static Load

The load on the left responds to the input signal and p p g enhances the single-ended output, whereas the load on the right does not.

10-48전자정보대학 김영석

MOS Differential Pair with Active Load

Similar to its bipolar counterpart, MOS differential pair can also use active load to enhance its single-ended output.

10-49전자정보대학 김영석

Asymmetric Differential Pair

Because of the vastly different resistance magnitude atBecause of the vastly different resistance magnitude at the drains of M1 and M2, the voltage swings at these two nodes are different and therefore node P cannot be viewed as a virtual ground.

10-50전자정보대학 김영석

Thevenin Equivalent of the Input Pair

ininoNmNThev vvrgv )( 21 −−=

oNThev

ininoNmNThev

rRg

2)( 21

=

10-51전자정보대학 김영석

Simplified Differential Pair with Active Load

)||( OPONmNout rrgv

=21 inin vv −

10-52전자정보대학 김영석