sjtu zhou lingling1 chapter 5 differential and multistage amplifier

SJTU Zhou Lingling 1

Chapter 5

Differential and Multistage Amplifier


Outline

• Introduction• The CMOS Differential Pair• Small-Signal Operation of the MOS Differential

Pair• The BJT Differential Pair• The differential Amplifier with Active Load• Frequency Response of the Differential amplifier• Multistage Amplifiers


Introduction

• Two reasons of the differential amplifier suited for IC fabrication: IC fabrication is capable of providing matched devices.Utilizing more components than single-ended

amplifier:Differential circuits are much less sensitive to noise and

interference.Differential configuration enable us to bias the amplifier and

to couple amplifier stages without the need for bypass and coupling capacitors.


The MOS Differential Pair

• Basic structure of differential pair.

• Characteristics


The MOS Differential Pair


Operation with a Common –Mode Input Voltage


Operation with a Common –Mode Input Voltage

• Symmetry circuit.

• Common-mode voltage.

• Current I divides equally between two transistors.

• The difference between two drains is zero.

• The differential pair rejects the common-mode input signals.


Operation with a Differential Input Voltage

The MOS differential pair with a differential input signal vid applied.

With vid positive: vGS1 vGS2, iD1 iD2, and vD1 vD2; thus (vD2 vD1) will be positive.

With vid negative: vGS1 vGS2, iD1 iD2, and vD1 vD2; thus (vD2 vD1) will be negative.


Operation with a Differential Input Voltage

• Differential input voltage.• Response to the differential input signal.• The current I can be steered from one transistor to

the other by varying the differential input voltage in the range:

• When differential input voltage is very small, the differential output voltage is proportional to it, and the gain is high.

OVidOV VvV 22


Large-Signal Operation

Transfer characteristic curves

Normalized plots of the currents in a MOSFET differential pair.

Note that VOV is the overdrive voltage at which Q1 and Q2 operate when conducting drain currents equal to I/2.



• Nonlinear curves.• Maximum value of input differential voltage.

• When vid = 0, two drain currents are equal to I/2.

• Linear segment.• Linearity can be increased by increasing overdrive

voltage(see next slide).• Price paid is a reduction in gain(current I is kept

constant).



The linear range of operation of the MOS differential pair can be extended by operating the transistor at a higher value of VOV.


Small-Signal Operation of MOS Differential Pair

• Linear amplifier

• Differential gain

• Common-mode gain

• Common-mode rejection ratio(CMRR)

• Mismatch on CMRR


Differential Gain

a common-mode voltage applied to set the dc bias voltage at the gates.

vid applied in a complementary (or balanced) manner.


Differential Gain

Signal voltage at the joint source connection must be zero.


Differential Gain

An alternative way of looking at the small-signal operation of the circuit.


Differential Gain

• Differential gain Output taken single-ended

Output taken differentially

Advantages of output signal taken differentially• Reject common-mode signal• Increase in gain by a factor of 2(6dB)

Dmid

od Rg

v

vA 2

111

Dmid

od Rg

v

vA 2

122

Dmid

od Rg

v

vA


Differential Gain

MOS differential amplifier with ro and RSS taken into account.


Differential Gain

Equivalent circuit for determining the differential gain.

Each of the two halves of the differential amplifier circuit is a common-source amplifier, known as its differential “half-circuit.”


Differential Gain

• Differential gain Output taken single-ended


)//(211

1 oDmid

od rRg

v

vA )//(2

122 oDm

id

od rRg

v

vA

)//( oDmid

od rRg

v

vA


Common-Mode Gain

The MOS differential amplifier with a common-mode input signal vicm.


Common-Mode Gain

Equivalent circuit for determining the common-mode gain (with ro ignored).

Each half of the circuit is known as the “common-mode half-circuit.”


Common-Mode Gain

• Common-mode gain Output taken single-ended


ss

D

ssm

D

icm

o

icm

ocmcm R

R

Rg

R

v

v

v

vAA

221

2121

012

icm

oocm v

vvA


Common-Mode Rejection Ratio

• Common-mode rejection ratio(CMRR) Output taken single-ended


This is true only when the circuit is perfectly matched.

CMRR

ssmcm

d

cm

d RgA

A

A

ACMRR

2

2

1

1


Mismatch on CMRR

• Effect of RD mismatch on CMRR

• Effect of gm mismatch on CMRR

icmss

Dooo v

R

Rvvv

212

)()2(D

Dssm R

RRgCMRR

)()2(m

mssm g

gRgCMRR


Mismatch on CMRR

Determine the common-mode gain resulting from a mismatch in the gm values of Q1 and Q2.

Common-mode half circuit is not available due to mismatch in circuit.

The nominal value gm.


Mismatch on CMRR

• Effect of gm mismatch on CMRR

2

1

2

1

m

m

d

d

g

g

i

i

icms vv

ssm

icmmdd Rg

vgii

221

)()2(m

mssm g

gRgCMRR


The BJT Differential Pair

• Basic operation• Large-signal operation• Small-signal operation

Differential gainCommon-mode gainCommon-mode rejection ration


The BJT Differential Pair

The basic BJT differential-pair configuration.


Basic Operation

The differential pair with a common-mode input signal vCM.

Two transistors are matched.

Current source with infinite output resistance.

Current I divide equally between two transistors.

The difference in voltage between the two collector is zero.

The differential pair rejects the common-mode input signal as long as two transistors remain in active region.


Basic Operation

The differential pair with a “large” differential input signal.

Q1 is on and Q2 is off.

Current I entirely flows in Q1.


Basic Operation

The differential pair with a large differential input signal of polarity opposite to that in (b).

Q2 is on and Q1 is off.

Current I entirely flows in Q2.


Basic Operation

The differential pair with a small differential input signal vi.

Small signal operation or linear amplifier.

Assuming the bias current source I to be ideal and thus I remains constant with the change in vCM.

Increment in Q1 and decrement in Q2.



• Nonlinear curves.

• Linear segments.

• Maximum value of input differential voltages

• Enlarge the linear segment by including equal resistance Re

in series with the emitters.

Tid Vv 21

mvVv Tid 1004



The transfer characteristics of the BJT differential pair (a) can be linearized by including resistances in the emitters.


Small Signal Operation

The currents and voltages in the differential amplifier when a small differential input signal vid is applied.



A simple technique for determining the signal currents in a differential amplifier excited by a differential voltage signal vid; dc quantities are not shown.



A differential amplifier with emitter resistances.

Only signal quantities are shown (in color).


Input Differential Resistance

• Input differential resistance is finite.

The resistance seen between the two bases is equal to the total resistance in the emitter circuit multiplied by (1+β).

• Input differential resistance of differential pair with emitter resistors.

rri

vR e

b

idid 22)1(

)22)1( eeb

idid Rr

i

vR ＋（


Differential Voltage Gain

• Differential voltage gain Output voltage taken single-ended

Output voltage taken differentially

Cmid

od Rg

v

vA 2

111 Cm

id

od Rg

v

vA 2

122

Cmid

ood Rg

v

vvA

12


Differential Voltage Gain

• Differential voltage gain of the differential pair with resistances in the emitter leads Output voltage taken single-ended


The voltage gain is equal to the ratio of the total resistance in the collector circuit to the total resistance in the emitter circuit.

ee

C

id

od Rr

R

v

vA

2

111

ee

C

id

ood Rr

R

v

vvA

12

ee

C

id

od Rr

R

v

vA

2

122


Differential Half-Circuit Analysis

Differential input signals.

Single voltage at joint emitters is zero.

The circuit is symmetric.

Equivalent common-emitter amplifiers in (b).



This equivalence applies only for differential input signals.

Either of the two common-emitter amplifiers can be used to find the differential gain, differential input resistance, frequency response, and so on, of the differential amplifier.

Half circuit is biased at I/2.

The voltage gain(with the output taken differentially) is equal to the voltage of half circuit.



The differential amplifier fed in a single-ended fashion.

Signal voltage at the emitter is not zero.

Almost identical to the symmetric one.


Common-Mode Gain

The differential amplifier fed by a common-mode voltage signal vicm.


Common-Mode Gain

Equivalent “half-circuits” for common-mode calculations.


Common-Mode Gain

• Common-mode voltage gainOutput voltage taken single-ended


EE

C

icm

ocm R

R

v

vA

21

1

012

id

oocm v

vvA

EE

C

icm

ocm R

R

v

vA

22

2


Common-Mode Rejection Ratio

• Common-mode rejection ratioOutput voltage taken single-ended


This is true only when the circuit is symmetric.• Mismatch on CMRR

EEmRgCMRR

CMRR

C

C

EE

Ccm R

R

R

RA

2


Input Common-Mode Resistance

Definition of the input common-mode resistance Ricm.

The equivalent common-mode half-circuit.


Input Common-Mode Resistance

• Input common-mode resistance

• Input common-mode resistance is very large.

)//2)(1(2 oEEicm rRR

)2

//)(1( oEEicm

rRR


Example


Example (cont’d)

Evaluate the following:• The input differential resistance.• The overall differential voltage gain(neglect the

effect of ro).

• The worst-case common-mode gain if the two collector resistance are accurate within ±1%.

• The CMRR, in dB.• The input common-mode resistance(suppose the

Early voltage is 100V).


The Differential Amplifier with Active Load

• Replace resistance RD with a constant current source results in a much high voltage gain as well as saving in chip area.

• Convert the output from differential to single-ended.


Differential-to-Single-Ended Conversion

A simple but inefficient approach for differential to single-ended conversion.


The Active-Loaded MOS Differential Pair

The active-loaded MOS differential pair.



The circuit at equilibrium assuming perfect matching.



The circuit with a differential input signal applied, neglecting the ro of all transistors.


Differential Gain of the Active-Loaded MOS Pair

• The output resistance ro plays a significant role in the operation of active-loaded amplifier.

• Asymmetric circuit.

• Half-circuit is not available.

• The gain will be determined as GmRo


Short-Circuit Transconductance

Determining the short-circuit transconductance Gm = io/vid


Short-Circuit Transconductance

mm gG


Output Resistance

Circuit for determining Ro. The circled numbers indicate the order of the analysis steps.


Output Resistance

Circuit for determining Ro.

The circled numbers indicate the order of the analysis steps.

42 // ooo rrR


Differential Gain

• The differential gain is determined as GmRo

• When

）42 //( oomomd rrgRGA

ooo rrr 42

20

21 A

rgA omd


Common-Mode Gain and CMRR

Analysis of the active-loaded MOS differential amplifier to determine its common-mode gain.

Power supplies eliminated.

Rss is the output resistance of the current source.



Asymmetric circuit.

Each of the two transistors as a CS configuration with a large source degeneration resistance 2Rss.

Common-mode gain:

CMRR

ssmm

o

sscm Rgrg

r

RA

3033

4

2

1

12

1

))(( ssmom RgrgCMRR


The Bipolar Differential Pair with Active Load

Active-loaded bipolar differential pair.


Determine the Transconductance

mm gG


Determine the output Resistance

42 // ooo rrR


Differential Gain

• The differential gain is determined as GmRo

• When

• Input differential resistance

）42 //( oomomd rrgRGA

20

21 A

rgA omd

ooo rrr 42

rRid 2



Common-mode gain:

CMRR

EE

o

EE

ocm R

r

R

rA

3

4

3

4 2

2

))(//(4

3402

o

EEom r

RrrgCMRR


Frequency Response of the Resistively Loaded MOS Amplifier

A resistively loaded MOS differential pair with the transistor supplying the bias current explicitly shown.

It is assumed that the total impedance between node S and ground, ZSS, consists of a resistance RSS in parallel with a capacitance CSS.



(b) Differential half-circuit. (c) Common-mode half-circuit.



common-mode gain



Differential Gain



CMRR with frequency.


Multistage Amplifier

• A four-stage bipolar op amplifier

• A two-stage CMOS op amplifier



• The first stage(input stage) is differential-in, differential-out and consists of Q1 and Q2.

• The second stage is differential-in, single-ended-out amplifier which consists of Q3 and Q4.

• The third stage is CE amplifier which consists of pnp transistor Q7 to shifting the dc level.

• The last stage is the emitter follower. • Biasing stage.



Equivalent circuit for calculating the gain of the input stage of the example.

Input differential resistance

Gain of first stage

krRid 2.202 1

4.22)//(

21

212011

ee

i

id rr

RRR

v

vA



Equivalent circuit for calculating the gain of the second stage of the example.

Gain of second stage

2.59//

54

33

01

022

ee

i

rr

RR

v

vA



Equivalent circuit for calculating the gain of the third stage of the example.

Gain of third stage

42.6//

47

54

02

033

Rr

RR

v

vA

e

i



Equivalent circuit for calculating the gain of the output stage of the example.

Gain of output stage

Output resistance

998.068

6

03

04

Rr

R

v

vA

e

152)1(// 586 RrRR eo


Two-Stage CMOS Op-Amp Configuration

sjtu zhou lingling1 chapter 5 differential and multistage amplifier

Documents

mos differential pair

differential circuits

differential configuration

differential output

differential input signal

bjt differential pair

mosfet differential

sjtu zhou lingling4