Chapter 5

Differential amplifier

By -ISMET-

Edited by Nazirah Mohamat Kasim & Shahilah Nordin

• An arrangement of transistors which allows the difference between two signals source to be amplified.

• The output is proportional to the difference between these two inputs.

• The best direct coupled stages available to the IC designer.

• Differential amplifiers are used in low and high frequency amplifiers, analog modulators and digital logic states.

• The basic differential amplifier is shown in Figure 5.1:-

Figure 5.1

For the dc analysis of differential amplifier circuit, all inputs are set to zero as shown in Figure 5.2.

Figure 5.2

From Figure 5.2,

CT

CCC

T

EET

BEBE

B

TC

R 2

I-V Vvoltage, Collector

R

0.7V-VI current, Emitter

0.7V0.7V0V-V Vtherefore,

0 Vand2

II

Three possible input signal combinations for differential amplifier:

Single-Ended Mode

Differential mode

common mode

•Single ended mode – an active signal is applied to only one input while the other is grounded.

•Differential mode – two opposite polarity active signals are applied to the amplifier.

•Common mode

-Two signals of the same amplitude, frequency and phase are applied to the differential amplifier.

- The output of the amplifier is ideally zero when measured the difference between the output terminals.

Definition

Single ended output

The input is in differential mode and the output is measured between the

differences of two output terminals.

The input is applied and the output is measured at

one of the output terminal.

Double

ended

output

The input signal is applied to one input with the other input is connected to the ground, as shown in Figure 5.3:

Figure 5.3

The ac equivalent circuit of Figure 5.3 is shown in Figure 5.4.

Figure 5.4

Vi1

Vo 1 Vo 2

RT

RC IbIb RC

-

+r?

r?

Assume:

CQ

T

C

i

O

Co

b1i

b1i

21

bb2b1

IV

2

R

V

V Av

gain, Voltage

Therefore,

IbR- V

be willcollector the at output The

r2I V

0r2I- V

obtain weKVL using and largevery is RT Assume

βββ and

III

CQ

T

I

Vr,

The mode of this operation is assumed by letting a single sine wave signal is connected between the two inputs. Refer Figure 5.5 (Double ended differential amplifier circuit) Figure 5.5

The ac equivalent circuit of Figure 5.5 is shown in Figure 5.6.

Figure 5.6

Vi1 Vi2

Vo 1 Vo 2

RT

RC IbIb RC

+

--

+r?

r?

Figure 5.6 shows an ac connection of a differential amplifier. We separate the input signals as Vi1 and Vi2, with separate outputs resulting as VO1 and VO2 with the transistors replaced with its ac equivalent circuit.

CQ

T

C

d

O

Cbo

bd

d2i1i

bb2i1i

T

IV

2

R

V

V Av

gain, voltage alDifferenti

Therefore,

RI- V

terminal, collector the at output The

r2I VTherefore,

VV- V

rIrIV- V

circuit open becomes it and largevery is R Assume

operation, ended single the for asSimilarly

Refer Figure 5.6,

2rZ

Therefore,

II and

r2I V

I

V)Z impedance Input

i(diff)

bi

bd

i

di(diff

The same input signal is applied to the two input terminals of differential amplifier with the same magnitude and phase.An ac connection showing common input to both transistors is shown in Figure 5.7

Ideal differential amplifier-The output voltage Vo is expected to be zero.-because the difference between the two outputs at each collector is opposite to each other and they are cancels out each other.

Figure 5.7

RC

Q2Q1

RC

RT

VCMV01

-VEE

+VCC

However…practically there is

an output at the collectors but the value is small.

we treat the amplifier as symmetry, that is RT is made to be parallel and be replaced with 2RT as shown in Figure 5.8

To analyze,

RC

Q1

2RT

V01

-VEE

+VCC

VCM-

+

Figure 5.8

To determine the common mode gain, the ac equivalent circuit of figure 5.8 is drawn as given in Figure 5.9

T

C

T

C

i

OCM

CM

Co

TCQ

Tb

TEbiCM

2R

R

2R

R

V

V A

amplifier, aldifferenti

the of Again, mode common the Therefore,

IbR- Vand

2RI

VI

2RIrIV V

5.9, Figure of circuit the From

CQ

T

IV

Figure 5.9

RC2RT

r bβIVCM Vo

The ratio of the magnitude of its differential gain, Ad to the magnitude of its common mode gain, ACM.

A

A(dB) CMRR

dB, in given often is CMRR the of value The

A

ACMRR

CM

d

CM

d

10log20

Example 1.The circuit given in Figure 5.10 has the following parameters:

hfe1 = hfe2 = 120, VT = 26mV, VBE1 = VBE2 = 0.7V

Calculate:a) ICQ1, ICQ2 and VCE1

b) Differential gain, Ad

c) Common mode gain, ACM

d) CMRR in dBe) Differential input impedance, Zi(diff)

f) Output impedance, Zo

Figure 5.10

SolutionDC analysis:

Since VB = 0, VE = -0.7V

Using KVL around loop A: VE – IERE + 10V = 0

Then 1.66mA5.6k

(-0.7)10

R

V10I

T

EE

Solution a)

b) From the ac equivalent circuit of differential mode:

7.46V

0.7)(9k)(0.83m)(3.10

0V-RI-VV

0.83mA2

II ,I

EC1C1CCCE1

ECQ2CQ1

62.262(31.32)

3.9k AThen

31.32Ω0.83mA

26mV

I

V but ;

IV

2

R

V

V A

d

C

T

C

T

C

d

Od

Solution c) From the ac equivalent circuit of common mode:

d)

e)

f)

45dB0.35

62.2620log

A

A20logCMRR(dB) 10

CM

d10

7.5kΩ32)2(120)(31.I

V2Z

C

Ti(diff)

3.9kΩRZ CO

0.352(5.6k)31.32

3.9k

2RIV

R

V

V A

TC

T

C

i

OCM

Good differential amplifier should have high CMRR

High CMRR means the differential amplifier circuit has the ability to reject common mode signal (noise).

Ideally, ACM = 0 not amplify the noise signal.

CM

d

A

ACMRR

In practical way there is still small signal at the output of common mode signal.

to we have

To reduce common mode gain, we can increase Popular method to increase is by using :Constant Current Source

Practical current source is a current supply with a resistance.

CMA TR

TR

CMRR, CMA

TC

T

CCM

2RIV

RA

An ideal current source, R=∞

Whereas practical current source resistance, R is very large.

An ideal current source provides a constant current

regardless of the load connected to it.

Practical Current Source

RTIT Figure 5.11

Constant Current Source Constant Current Source circuit can be built using:circuit can be built using:

FET devices

BJT devices

Combination of FET & BJT devices

Three popular Constant-Current Three popular Constant-Current Sources for differential amplifierSources for differential amplifier

Bipolar Transistor Constant Current Source

Transistor / Zener Constant Current Source

Current Mirror circuit.

BJT Constant Current SourceBJT Constant Current Source Figure 5.12a shows that an npn transistor together with

resistors operate as a constant current source.

RER2R1

-VEE

ITC3

B3

E3

RT

Figure 5.12a

BJT Constant Current SourceBJT Constant Current Source Figure 5.12b and Figure 5.12c shows simplified circuit from Figure

5.12a

By using KVL at loop1 in Figure 5.12c

EE21

1TH V

RR

RV

RER2R1

VEE

ITC3

B3

E3

VEERT

Q3

RE

VEE

ITC3

B3

E3

RTH

VTH

RT

LOOP1

21TH R//RR

Figure 5.12b Figure 5.12c

ETH

BE3EETH

TH

R1

RV-VV

V-

3E

E3E3BETHB

I

0RIVRI 3E3CT III

ETH

BE3EETH

R1

RV-VV

TI

BJT Constant Current SourceBJT Constant Current SourceFigure 5.12d and Figure 5.12e shows ac equivalent circuit for circuit

in Figure 5.12a

where

1rR1r 3ceb3ceTR

bR

e

e

R

R

Ee RR

rrR//RR 'bb21b

Figure 5.12d Figure 5.12c

rbb'

birce3

rR1 R2

RE

b3 c3

e3

RT

ib

ie

rbb'

bi

r

R1//R2 RE

b3 c3e3

RT

ib ie

rce3

Example 2If R1 = 4.7kΩ, R2: 4.7kΩ, RE = 2.2kΩ and VEE = 20V in figure 5.12:

Calculate current, I.Answer: 4.23mA

RER2R1

-VEE

IT

Figure 5.12

Zener Constant Current SourceZener Constant Current Source•From BJT constant current source, R2 is replaced with zener diode as in Figure 5.13.

•Using KVL at loop 1:

E

BE3ZE3

EE3BE3Z

EE3BE3Z

R

V-V I

RIVV

0RIVV-

RER1

-VEE

IT

+

-VZ

Q3

C3

B3

E3VBE3

+

_IE3

loop1

E3C3T III

Figure 5.13a

E

BE3ZT R

V-V I

Figure 5.13b and Figure 5.13c shows ac equivalent circuit for circuit in Figure 5.13a

where

11 33 cebce rRrTR

bR

e

e

R

R

Ee RR

rrR 'bbb

Figure 5.13b Figure 5.13c

Zener Constant Current SourceZener Constant Current Source

rbb'

birce

rR1

RE

b3 c3

e3

RT

ib

ie

rbb'

bi

r

Re

b3 c3e3

ib ie

RT

rce3

Example 3If R1 = 3.3kΩ, RE: 2.2kΩ, and VEE = 15V in figure 5.13:

Calculate current, I.Answer: 2.64mA

RER1

-VEE

IT

+

-VZ

Figure 5.13

Current Mirror CircuitCurrent Mirror CircuitUsed primarily in IC.

Circuit in Figure 5.14 is an example of constant current source circuit that can be used for differential amplifier.

XCT

C

CX

BBCX

E4E3E

B4B3B

C4C3C

III Therefore

I

I I

so,

I But 2III

III

III

III

circuit, mirror current For

XXTC

C

C

II2

II

2

I2

I

RX

Q4 Q3

VBE

+VCC

+

-

IT=IC

IX

RT

IC 2IB

IB IB IEIE

Figure 5.14a

Figure 5.14b shows ac equivalent circuit for circuit in Figure 5.14a

3cerTR

Figure 5.13b

Current Mirror CircuitCurrent Mirror Circuit

rbb'3

3birce3

4r

RX

b3 c3

e3

RT

ib3rbb'4

4r

Ib4

4birce3

b4c4

e3

Example 4Given Rx = 1.5kΩ, VCC = 18V in figure 5.14, calculate the mirrored current, I in the circuit.Answer: 11.53mA

Using KVL around the loop:18 – IXRX – VBE = 0

IX = 11.53mA

Solution

R X

Q 1 Q 2

V BE

+18V

+

-

I

IX

Figure 5.14

Example 5For the differential amplifier of Figure 5.15, determine the following:

a) ICQ, VCEQ

b) ACM

c) Ad

d) CMRR in dBe) Differential input impedance, Zi(diff)

f) Output impedance, Zo

Given:rce = 1/1μS

β1 = β2 = β3 = 150

VT = 26mV

Q 2Q 1

V i1

V 0 V i2

+15V

4.7kΩ 4.7kΩ

Q 3

-15V

1.2kΩ

1.5kΩ

Figure 5.15

Solutiona) DC analysis:

14.1ΩI

V and

7.05V

V-RI- V

V- V V

1.84mA2

II

3.67mAR

VVI

CQ

T

ECCQC

ECCEQ

E3CQ

E

BEZE3

Solutionb)

6-CM

T

T

T

C

TC

T

C

i

OCM

1029.14- ATherefore

MΩ 80.635 R Then

R Where

2R

R-

2RIV

R-

V

V A

13cer

Solutionc)

d)

e)

f)

-166.67

IV

2

R

V

V A

C

T

C

d

Od

135.147dBA

A20logCMRR(dB)

CM

d10

4.23kΩ2rZ i(diff)

4.7kΩRZ CO

TUTORIAL 5

Question 1Figure 5.16 shows an emitter coupled pair differential amplifier with Vi1 and Vi2 as the input and VO as the output. Q1 and Q2 have the following parameters:

β1 = β2 = 150, VT = 26mV, VBE1 = VBE2 = 0.7V

Calculate:a) ICQ1, ICQ2

b) Differential gain, Ad

c) Common mode gain, ACM

d) Differential input impedance, Zi(diff)

e) Output impedance, Zo

Figure 5.16

Question 2Figure 5.17 is a differential amplifier with β1=β2= β3=100, VBE=0.7V, VT=26mV and 1/ro =40μS. Both diodes are made of silicon. Determine:

a) ICQ1, ICQ2

b) Ad, ACM and CMRR in dB

c) Zi(diff) and ZoQ 2Q 1

V i1

V 0 V i2

+15V

15kΩ 15kΩ

Q 3

-15V

470kΩ

2kΩ

Figure 5.17

Q 2Q 1

V i1

V 0 V i2

+12V

1.5k Ω 1.5k Ω

Q 3

-12V

1.5k Ω

Q 4

Question 3Figure 5.18 shows an emitter coupled pair differential amplifier with Vi1 and Vi2 as the input and VO as the output. Determine:

a) The quiescent point of the differential amplifier for transistor Q1

b) CMRR in dBc) Output resistance (Ro)

d) Differential input resistance, Ri(diff)

Figure 5.18

The End See you in the next lesson!Don’t forget to do revision.

-ISMET- edited by Nazirah Mohamat Kasim & Shahilah Nordin