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BR Wiley/Razavi/Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 466 (1)

10

Differential Ampliﬁers

The elegant concept of “differential” signals and ampliﬁers was invented in the 1940s and ﬁrst

utilized in vacuum-tube circuits. Since then, differential circuits have found increasingly wider

usage in microelectronics and serve as a robust, high-performance design paradigm in many of

today’s systems. This chapter describes bipolar and MOS differential ampliﬁers and formulates

their large-signal and small-signal properties. The concepts are outlined below.

General

Considerations

Differential Signals

Differential Pair

Bipolar

Differential pair

Qualitative Analysis

Large−Signal Analysis

Small−Signal Analysis

Differential pair

Qualitative Analysis

Large−Signal Analysis

Small−Signal Analysis

MOS

Other Concepts

Cascode Pair

Common−Mode Rejection

Pair with Active Load

10.1 General Considerations

10.1.1 Initial Thoughts

In order to understand the need for differential circuits, let us ﬁrst consider an example.

Example 10.1

Having learned the design of rectiﬁers and basic ampliﬁer stages, an electrical engineering

student constructs the circuit shown in Fig. 10.1(a) to amplify the signal produced by a

microphone. Unfortunately, upon applying the result to a speaker, the student observes that the

ampliﬁer output contains a strong “humming” noise, i.e., a steady low-frequency component.

Explain what happens.

Solution

Recall from Chapter 3 that the current drawn from the rectiﬁed output creates a ripple waveform

at twice the ac line frequency (50 or 60 Hz) [Fig. 10.1(b)]. Examining the output of the common-

emitter stage, we can identify two components: (1) the ampliﬁed version of the microphone

signal and (2) the ripple waveform present on

V

CC

. For the latter, we can write

V

out

=

V

CC

,

R

C

I

C

;

(10.1)

noting that

V

out

simply “tracks”

V

CC

and hence contains the ripple in its entirety. The “hum”

originates from the ripple. Figure 10.1(c) depicts the overall output in the presence of both the

signal and the ripple. Illustrated in Fig. 10.1(d), this phenomenon is summarized as the “supply

466

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BR Wiley/Razavi/Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 467 (1)

Sec. 10.1 General Considerations 467

110 V

60 Hz

R

out

V

V

CC

C

Q

1

C

1

To Bias

t

V

CC

t

out

V

(c)

(a)

(b)

V

CC

Ripple

Signal

(d)

Voice Signal

Figure 10.1 (a) CE stage powered by a rectiﬁer, (b) ripple on supply voltage, (c) effect at output, (d) ripple

and signal paths to output.

noise goes to the output with a gain of unity.” (A MOS implementation would suffer from the

same problem.)

Exercise

What is the hum frequency for a full-wave rectiﬁer or a half-wave rectiﬁer?

How should we suppress the hum in the above example? We can increase

C

1

, thus lowering

the ripple amplitude, but the required capacitor value may become prohibitively large if many

circuits draw current from the rectiﬁer. Alternatively, we can modify the ampliﬁer topology such

that the output is insensitive to

V

CC

. How is that possible? Equation (10.1) implies that a change

in

V

CC

directly appears in

V

out

, fundamentally because both

V

out

and

V

CC

are measured with

respect to ground and differ by

R

C

I

C

. But what if

V

out

is not “referenced” to ground?! More

speciﬁcally, what if

V

out

is measured with respect to another point that itself experiences the

supply ripple to the same extent? It is thus possible to eliminate the ripple from the “net” output.

While rather abstract, the above conjecture can be readily implemented. Figure 10.2(a) illus-

trates the core concept. The CE stage is duplicated on the right, and the output is now measured

between nodes

X

and

Y

rather than from

X

to ground. What happens if

V

CC

contains ripple?

Both

V

X

and

V

Y

rise and fall by the same amount and hence the difference between

V

X

and

V

Y

remains free from the ripple.

In fact, denoting the ripple by

v

r

, we express the small-signal voltages at these nodes as

v

X

=

A

v

v

in

+

v

r

(10.2)

v

Y

=

v

r

:

(10.3)

That is,

v

X

,

v

Y

=

A

v

v

in

:

(10.4)

Note that

Q

2

carries no signal, simply serving as a constant current source.

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468 Chap. 10 Differential Ampliﬁers

out

V

V

CC

Q

1

To Bias To Bias

Q

2

XY

R

R

C2

C1

Ripple

V

CC

Q

1

Q

2

R

R

C2

C1

Ripple

A

1

(a) (b)

in

v

XY

Figure 10.2 Use of two CE stages to remove effect of ripple.

The above development serves as the foundation for differential ampliﬁers: the symmetric CE

stages provide two output nodes whose voltage difference remains free from the supply ripple.

10.1.2 Differential Signals

Let us return to the circuit of Fig. 10.2(a) and recall that the duplicate stage consisting of

Q

2

and

R

C

2

remains “idle,” thereby “wasting” current. We may therefore wonder if this stage can

provide signal ampliﬁcation in addition to establishing a reference point for

V

out

. In our ﬁrst

attempt, we directly apply the input signal to the base of

Q

2

[Fig. 10.3(a)]. Unfortunately, the

signal components at

X

and

Y

are in phase, canceling each other as they appear in

v

X

,

v

Y

:

v

X

=

A

v

v

in

+

v

r

(10.5)

v

Y

=

A

v

v

in

+

v

r

(10.6)

)

v

X

,

v

Y

=0

:

(10.7)

V

CC

Q

1

To Bias To Bias

Q

2

XY

R

R

C2

C1

in

V

V

r

V

CC

Q

1

Q

2

XY

R

R

C2

C1

in

v

+

in

v

−

(c)

(a) (b)

Figure 10.3 (a) Application of one input signal to two CE stages, (b) use of differential input signals, (c)

generation of differential phases from one signal.

For the signal components to enhance each other at the output, we can invert one of the input

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Sec. 10.1 General Considerations 469

phases as shown in Fig. 10.3(b), obtaining

v

X

=

A

v

v

in

+

v

r

(10.8)

v

Y

=

,

A

v

v

in

+

v

r

(10.9)

and hence

v

X

,

v

Y

=2

A

v

v

in

:

(10.10)

Compared to the circuit of Fig. 10.2(a), this topology provides twice the output swing by exploit-

ing the ampliﬁcation capability of the duplicate stage.

The reader may wonder how

,

v

in

can be generated. Illustrated in Fig. 10.3(c), a simple

approach is to utilize a transformer to convert the microphone signal to two components bearing

a phase difference of

180

.

Our thought process has led us to the speciﬁc waveforms in Fig. 10.3(b): the circuit senses

two inputs that vary by equal and opposite amounts and generates two outputs that behave in a

similar fashion. These waveforms are examples of “differential” signals and stand in contrast to

“single-ended” signals—the type to which we are accustomed from basic circuits and previous

chapters of this book. More speciﬁcally, a single-ended signal is one measured with respect to

the common ground [Fig. 10.4(a)] and “carried by one line,” whereas a differential signal is

measured between two nodes that have equal and opposite swings [Fig. 10.4(b)] and is thus

“carried by two lines.”

V

out

V

in

Q

1

V

CC

Q

1

Q

2

CC

Input

Output

(a)

(b)

t

V

CM

V

1

V

2

(c)

V

Differential Signal

Differential Signal

V

2

0

Figure 10.4 (a) Single-ended signals, (b) differential signals, (c) illustration of common-mode level.

Figure 10.4(c) summarizes the foregoing development. Here,

V

1

and

V

2

vary by equal and

opposite amounts and have the same average (dc) level,

V

CM

, with respect to ground:

V

1

=

V

0

sin

!t

+

V

CM

(10.11)

V

2

=

,

V

0

sin

!t

+

V

CM

(10.12)

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BR Wiley/Razavi/Fundamentals of Microelectronics [Razavi.cls v. 2006] June 30, 2007 at 13:42 470 (1)

470 Chap. 10 Differential Ampliﬁers

Since each of

V

1

and

V

2

has a peak-to-peak swing of

2

V

0

, we say the “differential swing” is

4

V

0

. We may also say

V

1

and

V

2

are differential signals to emphasize that they vary by equal and

opposite amounts around a ﬁxed level,

V

CM

.

The dc voltage that is common to both

V

1

and

V

2

[

V

CM

in Fig. 10.4(c)]is calledthe “common-

mode (CM) level.” That is, in the absence of differential signals, the two nodes remain at a

potential equal to

V

CM

with respect to the global ground. For example, in the transformer of Fig.

10.3(c),

+

v

in

and

,

v

in

display a CM level of zero because the center tap of the transformer is

grounded.

Example 10.2

How can the transformer of Fig. 10.3(c) produce an output CM level equal to

+2

V.

Solution

The center tap can simply be tied to a voltage equal to

+2

V (Fig. 10.5).

t

2 V

v

in1

v

in2

+2 V

v

in1

v

in2

Figure 10.5

Exercise

Does the CM level change if the inputs of the ampliﬁer draw a bias current?

Example 10.3

Determine the common-mode level at the output of the circuit shown in Fig. 10.3(b).

Solution

In the absence of signals,

V

X

=

V

Y

=

V

CC

,

R

C

I

C

(with respect to ground), where

R

C

=

R

C

1

=

R

C

2

and

I

C

denotes the bias current of

Q

1

and

Q

2

. Thus,

V

CM

=

V

CC

,

R

C

I

C

.

Interestingly, the ripple affects

V

CM

but not the differential output.

Exercise

If a resistor of value

R

1

is inserted between

V

CC

and the top terminals of

R

C

1

and

R

C

2

,whatis

the output CM level?

Our observations regarding supply ripple and the use of the “duplicate stage” provide suf-

ﬁcient justiﬁcation for studying differential signals. But, how about the common-mode level?

What is the signiﬁcance of

V

CM

=

V

CC

,

R

C

I

C

in the above example? Why is it interesting

that the ripple appears in

V

CM

but not in the differential output? We will answer these important

questions in the following sections.

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