Neamen D. Microelectronics: Circuit Analysis and Design

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248 Part 1 Semiconductor Devices and Basic Applications

EXAMPLE 4.13

Objective: Determine the small-signal voltage gain of the CMOS ampliﬁer.

For the circuit shown in Figure 4.45(a), assume transistor parameters of

=+0.8

=−0.8



= 80 μA/V



= 40 μA/V

(W/L)

= 15

(W/L)

= 30

, and

= λ

= 0.01 V

−1

. Also, assume

Bias

= 0.2

mA.

Solution: The transconductance of the NMOS driver is

= 2



= 2











Bias

= 2





0.08



(15)(0.2) = 0.693 mA/V

Since

= λ

, the output resistances are

= r

λI

(0.01)(0.2)

= 500 k

and M

are shown. Point A is the transition point for M

and point B is the transition

point for M

. The Q-point, to establish an ampliﬁer, should be approximately

halfway between points A and B, so that both transistors are biased in their saturation

regions. The voltage transfer characteristics are shown in Figure 4.45(d). Shown on

the curve are the same transition points A and B and the desired Q-point.

We again apply the small-signal equivalent circuits to ﬁnd the small-signal volt-

age gain. With v

SG2

held constant, the equivalent resistance looking into the drain of

is just

= r

. The small-signal equivalent circuit of the inverter is then as given

in Figure 4.46. The subscripts n and p refer to the n-channel and p-channel transis-

tors, respectively. We may note that the body terminal of M

will be tied to ground,

which is the same as the source of M

, and the body terminal of M

will be tied to

, which is the same as the source of M

. Hence, there is no body effect in this

circuit.

The small-signal voltage gain is

=−g

r

)

(4.52)

Again for this circuit, the small-signal voltage gain is directly proportional to the

output resistances of the two transistors.

–

Figure 4.46 Small-signal equivalent circuit of the CMOS common-source ampliﬁer

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Chapter 4 Basic FET Ampliﬁers 249

The small-signal voltage gain is then

=−g

r

) =−(0.693)(500500) =−173

Comment: The voltage gain of the CMOS ampliﬁer is on the same order of magni-

tude as the NMOS ampliﬁer with depletion load. However, the CMOS ampliﬁer does

not suffer from the body effect.

Discussion: In the circuit conﬁguration shown in Figure 4.45(a), we must again

apply a dc voltage to the gate of M

to achieve the “proper” Q-point. We will show in

later chapters using more sophisticated circuits how the Q-point is more easily

established with current-source biasing. However, this circuit demonstrates the basic

principles of the CMOS common-source ampliﬁer.

EXERCISE PROBLEM

Ex 4.13: For the circuit shown in Figure 4.45(a), assume transistor para-

meters of

=+0.5

=−0.5



= 80 μA/V



= 40 μA/V

, and

= λ

= 0.015 V

−1

. Assume

Bias

= 0.1

mA. Assume M

and M

are matched.

Find the width-to-length ratio of M

such that the small-signal voltage gain is

=−250

. (Ans.

(W/L)

= 35.2

)

CMOS Source-Follower Ampliﬁer

The same basic CMOS circuit conﬁguration can be used to form a CMOS source-

follower ampliﬁer. Figure 4.47(a) shows a source-follower circuit. We see that for

this source-follower circuit, the active load, which is M

, is an n-channel rather than

a p-channel device. The input signal is applied to the gate of M

and the output is at

the source of M

The small-signal equivalent circuit of this source-follower is shown in

Figure 4.47(b). This circuit, with two signal grounds, is redrawn as shown in

Figure 4.47(c) to combine the signal grounds.

EXAMPLE 4.14

Objective: Determine the small-signal voltage gain and output resistance of the

source-follower ampliﬁer shown in Figure 4.47(a).

Assume the reference bias current is

Bias

= 0.20

mA and the bias voltage is

= 3.3

V. Assume that all transistors are matched (identical) with parameters

= 0.4

= 0.20

mA/V

, and

λ = 0.01 V

−1

We may note that since M

and M

are matched transistors and have the same

gate-to-source voltages, the drain current in M

= I

Bias

= 0.2

mA.

Solution (voltage gain): From Figure 4.47(c), we ﬁnd the small-signal output volt-

age to be

= g

r

)

A KVL equation around the outside loop produces

= V

+ V

= V

+ g

r

)

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250 Part 1 Semiconductor Devices and Basic Applications

1 + g

r

)

Substituting this equation for V

into the output voltage expression, we obtain the

small-signal voltage gain as

r

)

1 + g

r

)

The small-signal equivalent circuit parameters are determined to be

= 2



= 2



(0.20)(0.20) = 0.40 mA/V

and

= r

λI

(0.01)(0.20)

= 500 k

The small-signal voltage gain is then

(0.40)(500500)

1 + (0.40)(500500)

Bias

Driver

Load

(a)

(c)

(b)

–

+–

–

(d)

+–

–

Figure 4.47 (a) All NMOS source-follower circuit, (b) small-signal equivalent circuit,

circuit for determining the output resistance

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Chapter 4 Basic FET Ampliﬁers 251

= 0.990

Solution (output resistance): The output resistance can be determined from the

equivalent circuit shown in Figure 4.47(d). The independent source V

is set equal to

zero and a test voltage V

is applied to the output.

Summing currents at the output node, we ﬁnd

+ g

From the circuit, we see that

=−V

. We then have

= V





The output resistance is then given as



We ﬁnd

0.40



500



500

= 2.48 k

Comment: A voltage gain of

= 0.99

is typical of a source-follower circuit. An

output resistance of

= 2.48

k is relatively small for a MOSFET circuit and is

also a characteristic of a source-follower circuit.

EXERCISE PROBLEM

Ex 4.14: The transconductance g

of the transistor in the circuit of Figure 4.47

is to be changed by changing the bias current such that the output resistance of

the circuit is

= 2

k. Assume all other parameters are as given in Exam-

ple 4.14. (a) What are the required values of g

and I

Bias

? (b) Using the results

of part (a), what is the small-signal voltage gain? (Ans. (a)

= 0.3125

mA;

(b)

= 0.988

)

COMPUTER ANALYSIS EXERCISE

PS 4.1: Using a PSpice analysis, investigate the small-signal voltage gain and

output resistance of the source-follower circuit shown in Figure 4.47 taking into

account the body effect.

CMOS Common-Gate Ampliﬁer

Figure 4.48(a) shows a common-gate circuit. We see that in this common-gate

circuit, the active load is the PMOS device M

. The input signal is applied to the

source of M

and the output is at the drain of M

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252 Part 1 Semiconductor Devices and Basic Applications

The small-signal equivalent circuit of the common-gate circuit is shown in

Figure 4.48(b).

EXAMPLE 4.15

Objective: Determine the small-signal voltage gain and output resistance of the

common-gate circuit shown in Figure 4.48(a).

Assume the reference bias current is

Bias

= 0.20

mA and the bias voltage is

= 3.3

V. Assume that the transistor parameters are

=+0.4

−0.4V

= 0.20

mA/V

= 0.20

mA/V

, and

= λ

= 0.01 V

−1

We may note that, since M

and M

are matched transistors and have the same

source-to-gate voltage, the bias current in M

= I

Bias

= 0.20

mA.

Solution (voltage gain): From Figure 4.48(b), we can sum currents at the output

node and obtain

+ g

−(−V

)

= 0





+ V





= 0

–

(b)

–

= 0

(c)

= 0

–

Bias

Load

Driver

(a)

Figure 4.48 (a) CMOS common-gate ampliﬁer, (b) small-signal equivalent circuit, and

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Chapter 4 Basic FET Ampliﬁers 253

From the circuit, we see that

=−V

. We then ﬁnd the small-signal voltage gain

to be









We ﬁnd the small-signal equivalent circuit parameters to be

= 2



= 2



(0.20)(0.20) = 0.40 mA/V

and

= r

λI

(0.01)(0.20)

= 500 k

We then ﬁnd



0.40 +

500





500



= 101

Solution (output resistance): The output resistance can be found from Figure 4.48(c).

Summing currents at the output node, we ﬁnd

+ g

−(−V

)

However,

= 0

so that

= 0

. We then ﬁnd

= r

r

= 500500

= 250 k

Comment: A voltage gain of

=+101

is typical of a common-gate ampliﬁer. The

output signal is in phase with respect to the input signal and the gain is relatively

large. Also, a large output resistance of

= 250

k is typical of a common-gate

ampliﬁer in that the circuit acts like a current source.

EXERCISE PROBLEM

Ex 4.15: The transconductance g

of the transistor in the circuit of Figure 4.48 is

to be changed by changing the bias current such that the small-signal voltage gain

= 120

. Assume all other parameters are as given in Example 4.15. (a) What

are the required values of g

and I

Bias

? (b) Using the results of part (a), what is the

output resistance? (Ans. (a)

= 0.14

mA,

= 0.335

mA/V; (b)

= 357 k

)

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254 Part 1 Semiconductor Devices and Basic Applications

COMPUTER ANALYSIS EXERCISE

PS 4.2: Using a PSpice analysis, investigate the small-signal voltage gain and

output resistance of the common-gate ampliﬁer shown in Figure 4.48 taking into

account the body effect.

Test Your Understanding

TYU 4.11 For the enhancement load ampliﬁer shown in Figure 4.39(a), the parame-

ters are:

TND

= V

TNL

= 0.8



= 40 μA/V

(W/L)

= 80,(W/L)

= 1

, and

= 5

V. Determine the small-signal voltage gain. Determine V

such that the

Q-point is in the middle of the saturation region. (Ans.

=−8.94

= 1.01

4.8 MULTISTAGE AMPLIFIERS

Objective: • Analyze multitransistor or multistage ampliﬁers and

understand the advantages of these circuits over single-transistor

ampliﬁers.

In most applications, a single-transistor ampliﬁer will not be able to meet the

combined speciﬁcations of a given ampliﬁcation factor, input resistance, and output

resistance. For example, the required voltage gain may exceed that which can be

obtained in a single-transistor circuit. We will consider, here, the ac analysis of the

two multitransistor circuits investigated in Chapter 3.

Multistage Ampliﬁer: Cascade Circuit

The circuit shown in Figure 4.49 is a cascade of a common-source ampliﬁer followed by

a source-follower ampliﬁer. As shown previously, the common-source ampliﬁer

4.8.1

383 kΩ

16.1 kΩ

3.9 kΩ

8 kΩ

4 kΩ

= 5 V

–

= –5 V

4 kΩ

–

135 kΩ

Figure 4.49 Common-source ampliﬁer in cascade with source follower

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Chapter 4 Basic FET Ampliﬁers 255

provides a small-signal voltage gain and the source follower has a low output impedance

and provides the required output current. The resistor values are those determined in

Section 3.5.1 of the previous chapter.

The midband small-signal voltage gain of the multistage ampliﬁer is determined

by assuming that all external coupling capacitors act as short circuits and inserting

the small-signal equivalent circuits for the transistors.

EXAMPLE 4.16

Objective: Determine the small-signal voltage gain of a multistage cascade circuit.

Consider the circuit shown in Figure 4.49. The transistor parameters are

= 0.5

mA/V

= 0.2

mA/V

TN1

= V

TN2

= 1.2

V, and

= λ

= 0

. The

quiescent drain currents are

= 0.2

mA and

= 0.5

mA.

Solution: The small-signal equivalent circuit is shown in Figure 4.50. The small-

signal transconductance parameters are

= 2



= 2



(0.5)(0.2) = 0.632 mA/V

and

= 2



= 2



(0.2)(0.5) = 0.632 mA/V

The output voltage is

= g

gs2

R

)

Also,

gs2

+ V

=−g

gs1

where

gs1



+ R



· V

Then

gs2

=−g



+ R



· V

− V

Therefore

= g



−g



+ R



· V

− V



R

)

gs1

gs2

–

gs1

–

gs2

⎪⎪ R

–

Figure 4.50 Small-signal equivalent circuit of NMOS cascade circuit

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256 Part 1 Semiconductor Devices and Basic Applications

The small-signal voltage gain is then

−g

R

)

1 + g

R

)



+ R



−(0.632)(0.632)(16.1)(84)

1 + (0.632)(84)



100

100 + 4



=−6.14

Comment: Since the small-signal voltage gain of the source follower is slightly

less than 1, the overall gain is due essentially to the common-source input stage.

Also, as shown previously, the output resistance of the source follower is small,

which is desirable in many applications.

EXERCISE PROBLEM

Ex 4.16: For the cascade circuit shown in Figure 4.49, the transistor and circuit

parameters are given in Example 4.16. Calculate the small-signal output resis-

tance R

. (The small-signal equivalent circuit is shown in Figure 4.50.) (Ans.

= 1.32 k

)

Multistage Ampliﬁer: Cascode Circuit

Figure 4.51 shows a cascode circuit with n-channel MOSFETs. Transistor M

connected in a common-source conﬁguration and M

is connected in a common-gate

conﬁguration. The advantage of this type of circuit is a higher frequency response,

which will be discussed in Chapter 7. The resistor values are those determined in

Section 3.5.2 of the previous chapter.

We will consider additional multistage and multitransistor circuits in Chapters 11

and 13.

4.8.2

54.6 kΩ

150 kΩ

95.4 kΩ

= 2.5 kΩ

= 5 V

–

= –5 V

10 kΩ

–

Figure 4.51 NMOS cascode circuit

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Chapter 4 Basic FET Ampliﬁers 257

gs1

–

gs2

= g

gs1

⎪⎪ R

–

Figure 4.52 Small-signal equivalent circuit of NMOS cascode circuit

Since

gs1

= V

, the small-signal voltage gain is

=−g

=−(1.13)(2.5) =−2.83

Comment: The small-signal voltage gain is essentially the same as that of a single

common-source ampliﬁer stage. The addition of a common-gate transistor will

increase the frequency bandwidth, as we will see in a later chapter.

EXERCISE PROBLEM

Ex 4.17: The transistor parameters of the NMOS cascode circuit in Figure 4.51

are

TN1

= V

TN2

= 0.8

= K

= 3

mA/V

, and

= λ

= 0

. (a) Deter-

mine

DSQ1

, and

DSQ2

. (b) Determine the small-signal voltage gain. (Ans.

(a)

= 0.471

mA,

DSQ1

= 2.5

DSQ2

= 1.61

V; (b)

=−5.94)

EXAMPLE 4.17

Objective: Determine the small-signal voltage gain of a cascode circuit.

Consider the cascode circuit shown in Figure 4.51. The transistor parameters are

= K

= 0.8

mA/V

TN1

= V

TN2

= 1.2

V, and

= λ

= 0

. The quiescent

drain current is

= 0.4

mA in each transistor. The input signal to the circuit is

assumed to be an ideal voltage source.

Solution: Since the transistors are identical and since the current in the two transis-

tors is the same, the small-signal transconductance parameters are

= g

= 2



= 2



(0.8)(0.4) = 1.13 mA/V

The small-signal equivalent circuit is shown in Figure 4.52. Transistor M

supplies

the source current of M

with the signal current (g

). Transistor M

acts as a

current follower and passes this current on to its drain terminal. The output voltage is

therefore

=−g

gs1

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