Applied Electronics II...Applied Electronics II Chapter 2: Di erential Ampli er School of Electrical and Computer Engineering Addis Ababa Institute of Technology Addis Ababa University

Applied Electronics II

Chapter 2: Differential Amplifier

School of Electrical and Computer EngineeringAddis Ababa Institute of Technology

Addis Ababa University

Daniel D./Abel G.

April 4, 2016

Chapter 2: Differential Amplifier (AAIT) Chapter Two April 4, 2016 1 / 29

Overview

1 Introduction

2 The MOS Differential PairOperation with a Common-Mode Input VoltageOperation with a Differential Input VoltageLarge-Signal Operation

3 Small-Signal Operation of the MOS Differential PairDifferential GainThe Differential Half-CircuitThe Differential Amplifier with Current-Source LoadsCascode Differential AmplifierCommon-Mode Gain and Common-Rejection ratio (CMRR)Differential versus Single-Ended OutputCurrent Source, Biasing Techniques

4 Exercise


Introduction

Introduction

The purpose of a differential amplifier is to amplify the differencebetween two signals.The differential-pair of differential-amplifier configuration is widelyused in IC circuit design.

One example is input stage of op-amp.Another example is emitter-coupled logic (ECL).

Technology was invented in 1940s for use in vacuum tubes the basicdifferential-amplifier configuration was later implemented withdiscrete bipolar transistors.

However, the configuration became most useful with invention ofmodern transistor / MOS technologies.

Vo

V1

V2

Differntial

Amplifier


The MOS Differential Pair

The MOS Differential Pair

Two matched transistors (Q1 and Q2) joined and biased by a constantcurrent source I. MOSFET’s should not enter triode region of operation.

Figure: The basic MOS differential-pair configuration.


The MOS Differential Pair Operation with a Common-Mode Input Voltage

Operation with a Common-Mode Input VoltageConsider case when two gate terminals are joined together.

Connected to a common-mode voltage (VCM).vG1 = vG2 = VCM

Q1 and Q2 are matched.Current I will divide equally between the two transistors.

ID1 = ID2 = I/2,VS = VCM − VGS

where VGS is the gate-to-source voltage.



Operation with a Common-Mode Input Voltage

Neglecting channel-length modulation, VGS and I/2 are related by

I

2=

1

2k

′n

W

L(VGS − Vt)

2

in terms of the overdrive voltage VOV ,

I

2=

1

2k

′n

W

LV 2OV or VOV =

√I

k ′nWL

The voltage at each drain will be

vD1 = vD2 = VDD −I

2RD

As long as Q1 and Q2 remain in the saturation region, the current I willdivide equally and the voltages at the drains will not change. Thus thedifferential pair does not respond to (i.e., it rejects) common-mode inputsignals.



Operation with a Common-Mode Input Voltage

An important specification of a differential amplifier is its inputcommon-mode range.This is the range of VCM over which thedifferential pair operates properly.The highest value of VCM is limited by the requirement that Q1 and Q2

remain in saturation, which means vDS ≥ VOV

max(VCM) = Vt + VDD −I

2RD

The lowest value of VCM is determined by the need to allow for a sufficientvoltage across the current source I for it to operate properly. If a voltageVCS is needed across the current source, then

min(VCM) = −VSS + VCS + Vt + VOV


The MOS Differential Pair Operation with a Differential Input Voltage

Operation with a Differential Input Voltage

If vid is applied to Q1 and Q2 is grounded, following conditions apply:vid = vGS1 − vGS2 > 0iD1 > iD2

if vid is positive, vGS1 will be greater than vGS2 and hence iD1 will begreater than iD2 and the difference output voltage (vD2 − vD1) will bepositive.




The differential pair responds to difference-mode or differential inputsignals by providing a corresponding differential output signal between thetwo drains.To find the vid that causes the entire bias current I to flow in one of thetwo transistors.

vGS1 reaches the value that corresponds to iD1 = I ,

vGS2 is reduced to a value equal to the threshold voltage Vt , atwhich point vS = −Vt .

The vGS1 can be found as

I =1

2

(k

′n

W

L

)(vGS1 − Vt)

2

vGS1 = Vt +√

2I/k ′n(W /L) = Vt +

√2VOV




The corresponding max(vid) is

max(vid)= vGS1 + vS

max(vid)= Vt +√

2VOV − Vt =√

2VOV

To steer the current completely to one side of the pair, a difference inputvoltage vid of at least

√2VOV (4VT for bipolar) is needed.


The MOS Differential Pair Large-Signal Operation

Large-Signal Operation

Objective is to derive expressions for drain current iD1 and iD2 in terms ofdifferential signal vid = vG1 − vG2.Assumption taken

Differential pair is perfectly matched

Channel-length Modulation is Neglected (λ = 0)

The circuit maintains Q1 and Q2 in the saturation region of operationat all times.

Load Independence

Step 1 Expression drain currents for Q1 and Q2.

iD1 =1

2k

′n

W

L(vGS1 − Vt)

2 and iD2 =1

2k

′n

W

L(vGS2 − Vt)

2

Step 2 Take the square roots of both sides of both√iD1 =

√1

2k ′n

W

L(vGS1 − Vt) and

√iD2 =

√1

2k ′n

W

L(vGS2 − Vt)




Step 3 (vGS1 − vGS2 = vG1 − vG2 = vid) Subtract and perform appropriatesubstitution . √

iD1 −√iD2 =

√1

2k ′n

W

Lvid

Step 4 Squaring both sides and substituting for iD1 + iD2 = I

2√iD1iD2 = I − 1

2k

′n

W

Lv2id

Step 5 Replacing iD2 = I − iD1 , squaring both sides and solving the

quadratic and substituting VOV =√I/(k ′nWL

)iD1 =

I

2+

(I

VOV

)(vid2

)√1−

(vid/2

VOV

)2

iD2 =I

2−(

I

VOV

)(vid2

)√1−

(vid/2

VOV

)2




The Transfer characteristics are nonlinear due to the term involving v2id

Figure: Normalized plots of the currents in a MOSFET differential pair.

Since Linear amplification is desirable vid will be as small as possible. Fora given value of VOV , the only option is to keep vid/2 much smaller thanVOV .



Large-Signal OperationThe approximation is

iD1 uI

2+

(I

VOV

)(vid2

)and iD2 u

I

2−(

I

VOV

)(vid2

)

Figure: The linear range of operation of the MOS differential pair can beextended by operating the transistor at a higher value of VOV .


Small-Signal Operation of the MOS Differential Pair

Small-Signal Operation of the MOS Differential Pair

Figure: Small-signal analysis of MOS deferential amplifier.Chapter 2: Differential Amplifier (AAIT) Chapter Two April 4, 2016 15 / 29

Small-Signal Operation of the MOS Differential Pair Differential Gain

Differential GainFrom Figure (a) vG1 = VCM + 1

2vid and vG2 = VCM − 12vid causes a

virtual signal ground to appear on the common-source (common-emitter)connection

where VCM denotes a common-mode dc voltage

where vid denotes a differential input applied complementarily (or balanced)

Also note that each of Q1 and Q2 is biased at a dc current of I/2 and isoperating at an overdrive voltage VOV .Assuming vid/2� VOV , the drain current will be

id1 =

(I

VOV

)(vid2

)and id2 = −

(I

VOV

)(vid2

)The transconductance of MOSFET is

gm =2IDVOV

=2(I/2)

VOV=

I

VOV

Combining the equations

id1 = gm(vid

2

)and id2 = −gm

(vid2

)Chapter 2: Differential Amplifier (AAIT) Chapter Two April 4, 2016 16 / 29

Small-Signal Operation of the MOS Differential Pair Differential Gain

Differential Gain

The output can be taken between the drain and the ground,refereed assingle-ended outputs vo1 and vo2.

vo1 = −id1 × RD = −gm(vid

2

)RD and vo2 = −id2 × RD = gm

(vid2

)RD

The output can e taken between the two drain terminals, refereed asdifferential output vod

vod = vo2 − vo1 = gmvidRD

The differential gain

Av =vodvid

= gmRD

When the output resistance of the MOSFET is taken into account

Av =vodvid

= gm[RD ‖ ro ]


Small-Signal Operation of the MOS Differential Pair The Differential Half-Circuit

The Differential Half-Circuit

The performance can be determined by considering only half the circuitsince the circuit is symmetrical and balanced. It is easier for analysis.

Q1 is biased at I/2 and isoperating at VOV .

This circuit may be used todetermine the differentialvoltage gain of the differentialamplifier

Av = gm[RD ‖ ro ]

Figure: Half-circuit of the differential amplifier.


Small-Signal Operation of the MOS Differential Pair The Differential Amplifier with Current-Source Loads

The Differential Amplifier with Current-Source LoadsTo obtain higher gain, the passive resistances (RD) can be replaced with currentsources. The current sources are realized with PMOS and biased to conduct I/2.

Av = gm1[ro1 ‖ ro3]


Small-Signal Operation of the MOS Differential Pair Cascode Differential Amplifier

Cascode Differential Amplifier

Gain can be increased via cascode configuration.

The differential Gain

Av = gm1[Ron ‖ Rop]

where

Ron = [gm3ro3]ro1

Rop = [gm5ro5]ro7


Small-Signal Operation of the MOS Differential Pair Common-Mode Gain and Common-Rejection ratio (CMRR)

Common-Mode Gain and CMRR




In practice there is no ideal current source or symmetrical matching.Non-ideal current source: Assuming the current source have a finiteoutput resistance RSS , and a small common-mode signal vicm is add onVCM . In the ideal case the drain voltage will not change or thecommon-mode gain is zero.since RSS is very large we can assume Q1 and Q2 are operate at a bias current of

I/2.

vicm =i

gm+ 2iRSS and i =

vicm1/gm + 2RSS

The drain voltage

vo1 = vo2 = −RD i = − RD

1/gm + 2RSSvicm

since 2RSS � 1/gmvo1

vicm=

vo2

vicmu − RD

2RSS




vo1 and vo2 are corrupted by vicm ,still common-mode signal is rejected

vod = vo2 − vo1 = 0

Effect of RD Mismatch:Assume Q1 load is RD and Q2 load is(RD + ∆RD). The drain voltage

vo1 u − RD

2RSSvicm and vo2 u −RD + ∆RD

2RSSvicm

Thus

vod = vo2 − vo1 = −∆RD

2RSSvicm

The common-mode gain

Acm =vodvicm

= −∆RD

2RSS= −

(RD

2RSS

)(∆RD

RD

)∴ Mismatch in the drain resistances causes the differential amplifier tohave a finite common-mode gain.




Common-mode rejection ratio (CMRR)

CMMR =|Ad ||Acm|

CMMR for drain resistance mismatch of ∆RD

CMMR =2gmRSS

∆RD/RD

For high CMMR ↑ RSS

Effect of gm Mismatch: Assume gm1 = gm + 12 ∆gm , gm2 = gm − 1

2 ∆gmfrom the figure on the next slide

i1

(1

gm1

)= i2

(1

gm2

)and i1 + i2 = i1

(1 +

gm2

gm1

)







vicm = i1/gm1 + (i1 + i2)RSS = i1/gm1 + i1

(1 +

gm2

gm1

)RSS

rearranging to expressi1 and i2 in terms of vicm

i1 =gm1vicm

1 + (gm1 + gm2)RSSand i2 =

gm2vicm1 + (gm1 + gm2)RSS

The differential output voltage vod = vo2 − vo1 = −i2RD + i1RD

vod =(gm1 − gm2)RD

1 + (gm1 + gm2)RSSvicm =

∆gmRD

1 + 2gmRSSvicm

The common-mode gain

Av =∆gmRD

1 + 2gmRSSu(

RD

2RSS

)(∆gmgm

)The corresponding CMMR

CMMR = (2gmRSS)/(∆gmgm

)


Small-Signal Operation of the MOS Differential Pair Differential versus Single-Ended Output

Differential versus Single-Ended Output

Differential Output:

It decreases the common-mode gain and increases the common-moderejection ratio (CMRR) dramatically

It increases the differential gain by a factor of 2 (6 dB) because theoutput is the difference between two voltages of equal magnitude andopposite sign.

Single-Ended Output:

Needed to connect it to an off-chip load.

Advantage of Differential Amplifier:

The differential transmission of the signal on the chip also minimizesits susceptibility to corruption with noise and interference.

Enables us to bias the amplifier and to couple amplifier stagestogether without the need for bypass and coupling capacitors.


Small-Signal Operation of the MOS Differential Pair Current Source, Biasing Techniques

Current Source, Biasing Techniques

The current source is implemented using a current mirror. Q3 and Q4 isthe current mirror implementation.


Exercise

Exercise

The following questions in the text book are exercises to be done for thetutorial session.

8.1

8.6

8.17

8.21

8.25

Reading AssignmentBJT Differential Amplifier


Documents

Applied Electronics II...Applied Electronics II Chapter 2: Di erential Ampli er School of Electrical and Computer Engineering Addis Ababa Institute of Technology Addis Ababa University