Differential Amplifier Common & Differential Modes

ESE319 Introduction to Microelectronics

Differential Amplifier

Common & Differential Modes

●

Common & Differential Modes

●

BJT Differential Amplifier

●

Diff. Amp Voltage Gain and Input Impedance

●

Small Signal Analysis – Differential Mode

ESE319 Introduction to Microelectronics

Common and Differential Modes

Consider a circuit with the two-port configuration shown

Z

₁

Z

₃

Z

₂

I

₁

I

₂

V

₂

V

₁

Define (convention) the voltages:

V

_=V

_−V

V

_c

= “common mode voltage”

V

_d

= “differential mode voltage”

+

+

±

a

±

a

V

₌

V

2

1. Let V

₁

= V

₂

= V =>

V

_c

= V &

V

_d

= 0

2. Let V

₁

= -V

₂

= V =>

V

_c

= 0 &

V

_d

= 2V

ESE319 Introduction to Microelectronics

Replace V

₁

, V

₂

with CM and DM Sources

V

_c

₌

V

1

V

2

2

V

d

=V

1

−V

2

V

₁

_V

₂

₌

2V

_c

V

₁

_−V

₂

₌

V

_d

Adding: 2V

₁

=

2 V

_c

_

V

_d

V

₁

₌

V

_c

_

V

_d

_/2

Subtracting:

2V

₂

₌

2V

_c

₋

V

_d

V

₂

₌

V

_c

₋

V

_d

_/2

Z

Z

Z

I

I

V

V

/2

- V

/2

I

+ I

Simultaneous Equations:

+

+

+

V

V

±

a

±

a

Definition

±

.

ESE319 Introduction to Microelectronics

Common and Differential Mode Currents

Simultaneous Equations:

I

₁

_I

₂

₌

I

_c

I

₁

_−I

₂

₌

2 I

_d

Adding:

2 I

₁

₌

I

_c

_

2 I

_d

I

₁

₌

I

_c

_{/ 2}

_

I

_d

Subtracting:

2 I

₂

₌

I

_c

₋

2 I

_d

I

₂

₌

I

_c

_/2

₋

I

_d

Z

Z

Z

I

V

V

/2

- V

/2

I

= I

+ I

I

Define (convention):

I

_c

_=I

₁

_I

₂

I

_d

₌

I

1

−I

2

2

+

+

+

±

.

±

.

±

.

I

₃

= I

₁

+ I

₂

= I

_c

ESE319 Introduction to Microelectronics

CM-DM Circuit Description

Z

₁

Z

₃

Z

₂

I

_c

/2

V

_c

V

_d

/2

- V

_d

/2

I

_c

= I

₁

+ I

₂

I

_c

/2

I

_d

= I

₁

- I

₂

1. Voltage sources redefined as

common and differential mode

quantities.

2. Differential mode currents take

“blue”

path.

3. Common mode current takes

“red”

path.

OBSERVATIONS

i. No

DM I

flows through Z

.

ii. No

CM I

flows around Z

, Z

loop

+

+

+

±

.

±

.

±

.

I

I

I

₁

₌

I

_c

_{/ 2}

_

I

_d

I

₂

₌

I

_c

_/2

₋

I

_d

ESE319 Introduction to Microelectronics

Analyze Circuits by Superposition

Balanced Circuit Differential Mode (V

_C

= 0)

Z

₁

Z

₃

Z

₂

I

₁

=

I

_d

/2

V

_d

/2

- V

_d

/2

-I

₂

= I

₁

I

d

Z

_=Z

_=Z

V

_x

I

=0⇒

V

=0

I

₌

V

2

−

−V

2



Z

_Z

=

V

2 Z

+

+

±

.

±

.

I

I

=

V

2

−

V

Z



−V

2

−

V

Z

=

−2

V

Z

=0

balanced circuit =>

+

V

Z1

-+

- V

_Z2

Z

₌

V

I

=2 Z

NOTE: If , then =>

Z

=Z

 Z

I

≠ 0

V

≠ 0.

I

_c

=

I

₁

+ I

₂

= 0

I

_c

_=I

₁

_{ I}

₂

₌

I

d

2

−

I

_d

2

=0

ESE319 Introduction to Microelectronics

Analyze Circuits by Superposition

Balanced Circuit Common Mode (V

_d

= 0)

Z

₁

Z

₃

Z

₂

I

_c

/2

V

_c

I

_c

I

_c

/2

I

_d

=I

₁

– I

₂

= 0

Z

_=Z

_=Z

I

=

_Z

V

∣∣Z

Z

=

_Z

V

2

Z

+

_±

_.

balanced circuit =>

2I

_=I

_−I

₌

I

2

−

I

2

=0⇒ I

=0

I

₁

I

₂

Z

₌

V

I

=

Z

2

Z

ESE319 Introduction to Microelectronics

Quick Review

<=>

Definitions:

V

₌

V

2

V

=V

−V

I

_c

_=I

₁

_I

₂

I

_d

₌

I

1

−I

2

2

V

₁

₌

V

_c

_

V

_d

_/2

V

₂

₌

V

_c

₋

V

_d

_/2

I

₁

₌

I

_c

_{/ 2}

_

I

_d

I

₂

₌

I

_c

_/2

₋

I

_d

I

_I

I

I

All

V's

&

I's

have

differential

&

com-mon

mode

ESE319 Introduction to Microelectronics

Quick Review

DM

(

V

_c

= 0

)

CM

(

V

_d

= 0

)

-I

I

– I

=

I

I

=I

I

I

₌

I

−I

2

I

₌

I

_{/ 2}

_

I

I

₌

I

_/2

₋

I

Z

=

V

I

=Z

 Z

Z

=

V

I

=Z

∥Z

Z

V

= 0

V

= ?

I

=

I

/2

ESE319 Introduction to Microelectronics

Summary

In a

balanced

differential

circuit (Z

₁

= Z

₂

= Z):

1. Differential mode voltages result in differential mode currents.

2. Common mode voltages result in common mode currents.

The

differential mode input impedance

of a balanced circuit is:

Z

_{i n}

_−dm

_=Z

₁

_Z

₂

_{=2 Z}

The

common mode input impedance

of a balanced circuit is:

Z

_{i n}

_−cm

_=Z

₁

_∥Z

₂

_Z

₃

₌

Z

ESE319 Introduction to Microelectronics

BJT Differential Amplifier – Bias View

I

_{= I}

_/2

_{ I}

/2

I

I

I

₌

I

2

_{1}

R

_C1

_=R

_C2

_=R

_C

R

_B1

_=R

_B2

_=R

_B

Collector bias path

inherently

common mode

.

I

_=

I

2

≈

I

2

Choose

I

_cm

and R

_C

for approximate

½ V

_CC

drop across R

_C

.

balanced

circuit

I

Q1

_=Q2

I

I

I

₌

I

2

R

R

R

R

ESE319 Introduction to Microelectronics

Voltage Gain & Input Impedance

v

/2

+

-v

r

i

r

v

+

-r

v

i

r

+

-+

-

_v

r

=

v

i

A

=

v

v

r

=

v

i

A

₌

v

v

v

_v

Single-ended outputs: v

, v

Differential output: v

=

v

-

v

Single-ended outputs:

v

,

v

Differential output:

v

=

v

-

v

v

_v

v

/2

R

R

R

R

I =

I

I

R

R

R

R

A

=

v

v

A

=

v

v

(diff)

(s-e)

(diff)

(s-e)

i

_i

ESE319 Introduction to Microelectronics

BJT Differential Amplifier – Small-signal View

I

current source

output impedance

i

i

i

i

i

i

Z

Z

Z

v

/2

v

V

v

v

/2

_v

/2

NOTE:

1. r

for amplifier Q1 & Q2 is ignored.

2. v

/2 and

v

are ac small signals

Z

_=Z

_{=1r}

Z

_{=1r}

R

_R

R

R

βi

r

r

r

βi

V

-v

/2

r

_=r

_=r

R

_=R

_=R

R

_=R

_=R

ESE319 Introduction to Microelectronics

Superposition Small-signal Analysis

Differential Mode (v

_i-cm

= 0)

DM Path (ignoring both

R

):

.

₌

r

i

R

_=R

_=R

, R

_=R

_=R

, r

_=r

_=r

,

_

_=

_=

Balanced circuit =>

+

-

v

i

i

i

i

v

=2 r

i

=21r

i

Z

=1r

Z

_=Z

_{=1 r}

v

2

=

i

r



i

r

−

v

2

r

_{i n}

_−dm

_{=21r}

_e

ib1i

v

/2

_v

/2

i

i

R

r

βi

βi

R

R

R

r

r

⇒i

=i

=i

,i

=i

=i

, i

=i

=i

Z

_{=2 Z}

Recall:

i

i

i

i

i

i

₌

i

_=i

_=i

i

₌

i

_=i

_=i

i

₌

i

_=−i

_=−i

i

=

i

=−i

=−i

i

₌

i

_=i

_=i

i

₌

i

_=−i

_=−i

v

ESE319 Introduction to Microelectronics

Superposition Small-signal Analysis

Differential Mode (v

_i-cm

= 0) cont.

+

-

v

i

i

i

i

i

i

Balanced circuit

DM Differential voltage gain:

A

₌

v

v

=

R

_

1r

≈

R

r

v

_=−R

_−

i

_{−−R}

_

i

_

v

₌

v

₋

v

i

=

v

2

1

1r

A

=−

A

=

v

v

=

 R

2

1r

≈

R

2 r

v

=

2 R



1r

v

2

DM Single-ended voltage gains:

i

i

₌

i

_=i

_=i

i

₌

i

_=i

_=i

−

i

₌₋

i

_=i

_=i

v

/2

R

r

R

R

β

i



β

i

r

r

_R

v

v

i

i

i

i

i

−

i

₌₋

i

_=i

_=i

i

₌

i

_=i

_=i

−

i

₌₋

i

_=i

_=i

i

_=

i

v

/2

v

ESE319 Introduction to Microelectronics

Superposition Small-signal Analysis

Common Mode (v

_i-dm

= 0)

CM path: both r

are in parallel

v

=

2i

1



r

2

r



i

_r

v

2

r

≈

v

_r

v

+

-i

i

i

/2

i

/2

i

i

_/2

₌

i

_=i

_=i

v

_=[r

_∥r

_r

_]

i

Z

_=Z

_{=1r}

Z

_{=1r}

Balanced circuit

r

=

v

2i

=1

r

2

r

≈1r

i

=

i

=i

=i

R

β

i

β

i

r

r

r

R

R

_R

i

₌

2i

_{=1}

2i

Recall:

i

₌

i

_=i

_=i

i

_/2

₌

i

_=i

_=i

i

=

i

=i

=i

i

₌

i

_=i

_=i

v

Z

=Z∥ZZ

=

Z

2

Z

i

i

i

i

ESE319 Introduction to Microelectronics

Superposition Small-signal Analysis

Common Mode (v

_i-dm

= 0)

v

+

-i

i

i

i

i

i

/2

_i

/2

i

v

=−R



i

−−R



i

=0

i

₌

v

2

_{1}



r

2

r



≈

v

2

_{1r}

CM Differential voltage gain:

v

=

v

−

v

Balanced circuit

A

=

v

_v

=0

CM Single-ended voltage gains:

A

=

A

=

v

v

=−

R



2

1r

≈−

R

2 r

v

=

v

=−R



i

=

−R



2

1r

v

r

R

_R

R

r

r

R

β

i

β

i

i

_/2

₌

i

_=i

_=i

_i

_/2

₌

_i

_=i

_=i

i

₌

i

_=i

_=i

i

=

i

=i

=i

i

=

i

=i

=i

i

=

i

=i

=i

i

_=

i

v

i

_=

i

i

ESE319 Introduction to Microelectronics

Summary

Differential mode:

r

_{i n}

_−dm

_{=21r}

_e

A

_v

_−dm

₌

v

o

−dm

v

_i

_−dm

≈

R

_C

r

_e

Differential DM Voltage gain:

Common mode:

r

_{i n}

_−cm

_{=1}



r

e

2

r

o



≈1 r

o

Differential CM Voltage gain:

A

_v

_−cm

₌

v

o

−cm

v

_i

_−cm

=0

Single-Ended DM Voltage gain:

A

_v

_−dm1

₌₋

A

_v

_−dm2

₌

v

c1

−dm

v

_i

_−dm

≈−

R

_C

2 r

_e

A

v

−cm1

=

A

v

−cm2

=

v

_c1

_−cm

v

_i

_−cm

≈−

R

_C

2 r

_o