Neamen D. Microelectronics: Circuit Analysis and Design

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728 Part 2 Analog Electronics

The output voltage is

=−(g

π1

)(r

R

r

)

(10.93)

Since

π1

= V

, where V

is the ac input voltage, the small-signal voltage gain is

=−g

R

r

) =

−g





(10.94)

The small-signal voltage gain can also be written

−g

+ g

(10.95)

where g

and g

are the output conductances of Q

and Q

, and g

is the load con-

ductance. The transconductance is

= I

, the small-signal conductances are

= I

and

= I

, and the load conductance is

= 1/R

. There-

fore, Equation (10.95) becomes

−









(10.96)

If the passive load is an open circuit

→∞)

, the small-signal voltage gain is

identical to that determined from the dc analysis as given by Equation (10.86). If the

load resistance R

is not an open circuit, then the magnitude of the small-signal volt-

age gain is reduced.

EXAMPLE 10.12

Objective: Calculate the small-signal voltage gain of an ampliﬁer with an active

load and a load resistance R

For the circuit in Figure 10.37(a), the transistor parameters are

= 120

and

= 80

V. Let

= 0.026

V and

= 0.2

mA. Determine the small-signal

voltage gain for load resistances of

=∞

, 200 k



, and 20 k



Solution: For

=∞

, Equation (10.96) reduces to

−









−



0.026





120



=−1846

which is the same as that determined for the open-circuit conﬁguration in Exam-

ple 10.11.

For

= 200



, the small-signal voltage gain is

−



0.2

0.026





0.2

120

200

0.2



−7.692

0.001667 + 0.005 + 0.0025

=−839

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Chapter 10 Integrated Circuit Biasing and Active Loads 729

and for

= 20 k

, the voltage gain is

−



0.2

0.026





0.2

120

0.2



−7.692

0.001667 + 0.05 + 0.0025

=−142

Comment: The small-signal voltage gain is a strong function of the load resistance

. As the value of R

decreases, the loading effect becomes more severe.

Design Pointer: If an ampliﬁer with an active load is to drive another ampliﬁer

stage, the loading effect must be taken into account when the small-signal voltage

gain is determined. Also, the input resistance of the next stage must be large in order

to minimize the loading effect.

EXERCISE PROBLEM

Ex 10.12: For the circuit shown in Figure 10.37(a), the transistor parameters are

= 100

V and

= 60

V. Let

= 0.25

mA. (a) Determine the open-circuit

small-signal voltage gain. (b) Find the value of

such that the voltage gain is

60 percent of the open-circuit value. (Ans. (a)

=−1442

, (b)

= 225

)

The small-signal voltage gain of an active-load circuit can be increased by in-

creasing the effective resistance of the active load. Figure 10.38 shows the same type

of BJT ampliﬁer in which the active load is an “upside down” modiﬁed Widlar cur-

rent source. The small-signal voltage gain can be written in the same form as Equa-

tion (10.94), or

−g





(10.97)

Figure 10.38 BJT ampliﬁer with a modiﬁed Widlar current source as an active load

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730 Part 2 Analog Electronics

where R

is now the effective resistance looking into the collector of Q

. We ana-

lyzed the output resistance of a Widlar current source in the Section 10.1. In this case,

however, we are including a resistor in the emitter of both transistors Q

and Q

. We

can show that

= r



1 + g





(10.98a)

where



= R

[r

π2

+ R

(R

+ R

)]

(10.98b)

and where

= r

π1



(10.98c)

We may note that, if Q

and Q

are matched, then

= g

and

π1

= r

π2

since

the resistors in the emitters of Q

and Q

are the same value.

Problem-Solving Technique: Active Loads

1. Ensure that the active load devices are biased in the forward-active mode.

2. The small-signal analysis of the circuit with an active load then simply in-

volves considering the output resistance looking into the output of the active

load device as well as the equivalent circuit of the amplifying transistor.

Small-Signal Analysis: MOSFET Active Load Circuit

The small-signal voltage gain of a MOSFET ampliﬁer with an active load can also be

determined from the small-signal equivalent circuit. Figure 10.39 is the small-signal

equivalent circuit of the entire MOSFET active load in Figure 10.33. The signal volt-

ages

sg1

and

sg2

are zero, since there is no ac excitation in this part of the circuit.

This means that

sg2

= 0

and

= r

(10.99)

10.4.2

sg2

sg1

–

sg2

–

sg1

–

Figure 10.39 Small-signal equivalent circuit of the MOSFET active load circuit

A simple MOSFET ampliﬁer with an active load, and a load resistor R

capaci-

tively coupled to the output, is shown in Figure 10.40(a). Figure 10.40(b) shows the

small-signal equivalent circuit, in which the load R

, the active load resistance r

and the output resistance r

of transistor M

are included.

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Chapter 10 Integrated Circuit Biasing and Active Loads 731

(a) (b)

–

REF

Figure 10.40 (a) Simple MOSFET ampliﬁer with active load and load resistance and

(b) small-signal equivalent circuit

The output voltage is

=−g

R

r

)

(10.100)

and since

= V

, where V

is the ac voltage, the small-signal voltage gain is

=−g

(

R

r

)

−g

+ g

(10.101)

The parameters g

and g

are the output conductances of M

and M

, and g

is the

load conductance. This expression for the small-signal voltage gain of a MOSFET

ampliﬁer with active load is the same as that of the BJT ampliﬁer.

A load resistance R

tends to degrade the gain and to cause a loading effect, as it

did in the bipolar circuit with an active load. However, in MOSFET ampliﬁers, the

output may be connected to the gate of another MOSFET ampliﬁer in which the

effective R

is very large.

EXAMPLE 10.13

Objective: Calculate the small-signal voltage gain of an NMOS ampliﬁer with an

active load.

For the ampliﬁer shown in Figure 10.40(a) the transistor parameters are:

= λ

= 0.01 V

−1

= 1

V, and

= 1

mA/V

. Assume M

and M

are

matched and

REF

= 0.5

mA. Calculate the small-signal voltage gain for load resis-

tances of

=∞

and 100 k.

Solution: Since M

and M

are matched, then

= I

REF

, and the transconductance is

= 2



REF

= 2



(1)(0.5) = 1.41 mA/V

The small-signal transistor conductances are

= g

= λI

REF

= (0.01)(0.5) = 0.005 mA/V

For

=∞

, Equation (10.101) reduces to

−g

+ g

−1.41

0.005 + 0.005

=−141

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732 Part 2 Analog Electronics

For

= 100 k(g

= 0.01

mA/V), the voltage gain is

−g

+ g

−1.41

0.005 + 0.01 +0.005

=−70.5

Comment: The magnitude of the small-signal voltage gain of MOSFET ampliﬁers

with active loads is substantially larger than for those with resistive loads, but it is

still smaller than equivalent bipolar circuits, because of the smaller transconductance

for the MOSFET.

EXERCISE PROBLEM

Ex 10.13: For the circuit in Figure 10.40(a), the transistor parameters are:

= 0.015 V

−1

= K

= 0.10

mA/V

= 0.5

V, and

=−0.5

Let

= 5

V and

REF

= 0.20

mA. (a) Determine

. (b) Find the open-circuit

small-signal voltage gain. (c) Find the value of

that results in a voltage gain of

one-half the open-circuit value. (Ans. (a)

= 1.914

V, (b)

=−47.1

(c)

= 166

)

Small-Signal Analysis:

Advanced MOSFET Active Load

The active loads considered in the BJT (Figure 10.37) and MOSFET (Figure 10.40(a))

circuits correspond to the simple two-transistor current mirrors. We may use a more

advanced current mirror with a high output resistance as an active load to increase

the ampliﬁer gain. Figure 10.41(a) shows a MOSFET cascode amplifying stage with

a cascode active load. The small-signal equivalent circuit is shown in Figure 10.41(b),

10.4.3

REF

Bias

(a) (b)

–

gs2

–

gs1

–

gs1

gs2

Figure 10.41 (a) MOSFET cascode amplifying stage with cascode active load; (b) small-

signal equivalent circuit

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Chapter 10 Integrated Circuit Biasing and Active Loads 733

where R

is the effective resistance looking into the drain of M

. From our

discussion of the cascode current mirror, we found

= r

(1 + g

)

(Equation (10.57)).

We can assume all transistors are matched so that the currents in all transistors

are equal. Summing currents at D

, we have

gs1

(−V

gs2

)

= g

gs2

−(−V

gs2

)

(10.102)

Summing currents at the output node, we ﬁnd

−(−V

gs2

)

+ g

gs2

= 0

(10.103)

Eliminating

gs2

from the two equations, noting that

gs1

= V

, and assuming

 1/r

, we ﬁnd the small-signal voltage gain is

−g

(10.104)

The resistance R

is approximately

∼

, so the gain can be written as

−g

(10.105)

For the same transistor parameters given in Example 10.13, the small-signal

voltage gain of this circuit would be 39,762! However, a word of warning is in order.

As we mentioned previously, output resistances in the hundreds of megohm range

are ideal and will, in reality, be limited by leakage currents. For this reason, a voltage

gain of 39,000 in a one-stage ampliﬁer will probably not be achieved. However, the

voltage gain of this ampliﬁer should be substantially larger than the ampliﬁer using a

simple active load.

Test Your Understanding

TYU 10.12 In the circuit shown in Figure 10.37(a), the transistor parameters are

= 120

V and

= 80

V. Let

= 0.5

mA and

= 50

k. (a) Determine

the small-signal parameters g

, r

, and r

. (b) Find the small-signal voltage gain.

(Ans. (a)

= 19.2

mA/V,

= 240

k,

= 160

k (b)

=−631

)

TYU 10.13 Repeat Example 10.12 for the case where a resistor

= 1

k is in-

cluded in the emitters of Q

and Q

as shown in Figure 10.38. (Ans. For

=∞

=−4404

; for

= 200



=−2800

; for

= 20



=−655

)

TYU 10.14 In the circuit in Figure 10.40(a), the transistor parameters are:

0.1 mA/V

= 0.2

mA/V

= 1

=−1

= 0.01 V

−1

, and

= 0.02 V

−1

. Let

= 10

REF

= 0.25

mA, and

= 100

k. (a) Determine

the small-signal parameters

(for M

, and

. (b) Find the small-signal volt-

age gain. (Ans. (a)

= 0.448

mA/V,

= 400

k,

= 200

k (b)

=−25.6

)

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734 Part 2 Analog Electronics

10.5 DESIGN APPLICATION: AN NMOS

CURRENT SOURCE

Objective: • Design an NMOS current source circuit to provide a

speciﬁed bias current and output resistance.

Speciﬁcations: Design an NMOS current source to provide a bias current of

= 100

μA and an output resistance greater than 20 M. The reference current is

to be

REF

= 150

μA. The circuit is to be biased at

±3.3

V and the voltage at the

drain of the current source transistor is to be no smaller than −2.2 V.

Design Approach: A simple two-transistor current source would yield an output

resistance of

= r

λI

(0.01)(0.1)

⇒ 1M

Therefore, to obtain a larger output resistance, a cascode current source is required.

The basic circuit is shown in Figure 10.42. The transistor M

may actually need to be

two or more transistors in series.

Choices: NMOS transistors are available with the following parameters:

0.5 V,



= 80 μA/V

, and

λ = 0.01 V

−1

. The minimum width-to-length ratio of

any transistor is to be unity.

Solution: The minimum voltage

is to be

−2.2V

. This voltage is given by

= V

GS1

+ V

DS3

(sat) + V

−

= V

GS1

+ V

GS3

− V

+ V

−

Assuming that M

and M

are matched, we ﬁnd

=−2.2 = 2V

GS1

−0.5 +(−3.3)

GS1

= V

GS3

= 0.8V

Now



(

GS1

− V

)

100 =





(0.8 − 0.5)

which yields





= 27.8

If we set









= 28

then we ﬁnd that

GS1

= V

GS3

= 0.799

V and

(min) =−2.202

REF

= 3.3 V

–

= –3.3 V

Figure 10.42 MOSFET

cascode current source circuit

for design application

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Chapter 10 Integrated Circuit Biasing and Active Loads 735

Assuming that M

and M

are matched, we have

REF

(W/L)

150

100

(W/L)

which yields









= 42

Now the equivalent V

GS5

is given by

GS5

= V

−2V

GS2

− V

−

= 3.3 −2(0.799) −(−3.3)

GS5

= 5.0V

The width-to-length ratio is found from

REF

= 150 =





(5.0 − 0.5)

which yields

(W/L)

= 0.185

. A width-to-length ratio less than unity is unaccept-

able. Putting two equivalent M

transistors in series yields a gate-to-source voltage of

GS5

= 5.0/2

V. Then

REF

= 150 =







5.0

−0.5



which yields

(W/L)

= 0.938

. This value is still less than unity. Putting three equiv-

alent M

transistors in series yields a gate-to-source voltage of

GS5

= 5.0/3

V. Then

REF

= 150 =







5.0

−0.5



which yields

(W/L)

= 2.76

. This is an acceptable solution.

The output resistance is given by

= r

(1 + g

)

We ﬁnd

= r

λI

(0.01)(0.1)

⇒ 1M

and

= 2









= 2







(28)(100) = 669 μA/V

We then ﬁnd

= 1 +1

[

1 + (669)(1)

]

= 671 M

This value certainly meets the design criteria.

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736 Part 2 Analog Electronics

Comment: The very large output resistance of 671 M assumes that we have ideal

MOS transistors. In fact there are leakage currents that, in reality, will lower the out-

put resistance. However, the cascode current source does provide a very high output

resistance that is useful in differential ampliﬁers as we will see in Chapter 11.

10.6 SUMMARY

• This chapter addressed the biasing of bipolar and FET circuits with constant-

current sources. The current source biasing technique eliminates the need for

resistor-intensive biasing used up to this point.

• The basic bipolar current source is the simple two-transistor circuit with a resis-

tor to establish the reference current. The basic FET current source is also a sim-

ple two-transistor circuit but includes additional transistors in the reference

portion of the circuit. The relation between the bias current and reference current

was determined.

• One parameter of interest in the current source circuit is the output resistance,

which determines the stability of the bias current. More sophisticated current-

source circuits, such as the Widlar and Wilson circuits in the BJT conﬁguration

and the Wilson and cascode circuits in the FET conﬁguration, have larger output

resistance parameters and increased bias-current stability.

• Multitransistor output stages, in both bipolar and FET constant-current circuits,

are used to bias multiple ampliﬁer stages with a single reference current. These

circuits, called current mirrors, reduce the number of elements required to bias

ampliﬁer stages throughout an IC.

• Both bipolar and MOSFET active load circuits were analyzed. Active loads are

essentially “upside down” current-source circuits that replace the discrete col-

lector and drain resistors. The active loads produce a much larger small-signal

voltage gain compared to discrete resistor circuits.

• As an application, a MOSFET current source circuit was designed to provide a

speciﬁed bias current and output resistance.

CHECKPOINT

After studying this chapter, the reader should have the ability to:

✓ Analyze and design a simple two-transistor BJT current-source circuit to pro-

duce a given bias current.

✓ Analyze and design more sophisticated BJT current-source circuits, such as the

three-transistor circuit, cascode circuit, Wilson circuit, and Widlar circuit.

✓ Design a BJT current-source circuit to yield a speciﬁed output resistance.

✓ Analyze and design a basic two-transistor MOSFET current-source circuit with

additional MOSFET devices in the reference portion of the circuit to yield a

given bias current.

✓ Analyze and design more sophisticated MOSFET current-source circuits, such

as the cascode circuit, Wilson circuit, and wide-swing cascode circuit.

✓ Design a MOSFET current-source circuit to yield a speciﬁed output resistance.

✓ Describe the operation and characteristics of a BJT and MOSFET active load

circuit.

✓ Discuss the reason for the increased small-signal voltage gain when an active

load is used.

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Chapter 10 Integrated Circuit Biasing and Active Loads 737

REVIEW QUESTIONS

1. Sketch the basic BJT two-transistor current source and explain the operation.

2. Explain the signiﬁcance of the output resistance of the current-source circuit.

3. Discuss the effect of mismatched transistors on the characteristics of the BJT

two-transistor current source.

4. Sketch the BJT three-transistor current source and discuss the advantages of this

circuit.

5. What is the primary advantage of a BJT cascode current source?

6. Sketch a Widlar current source and explain the operation.

7. Can a piecewise linear model of the transistor be used in the analysis of the

Widlar current source? Why or why not?

8. Discuss the operation and signiﬁcance of a multiple-output transistor current

mirror.

9. Sketch the basic MOSFET two-transistor current source and explain the operation.

10. Discuss the effect of mismatched transistors on the characteristics of the

MOSFET two-transistor current source.

11. Discuss how the reference portion of a MOSFET current source can be designed

with MOSFETs only.

12. Sketch a MOSFET cascode current source circuit and discuss the advantages of

this design.

13. Discuss the operation of an active load.

14. What is the primary advantage of using an active load?

15. Sketch the voltage transfer characteristics of a simple ampliﬁer with an active

load. Where should the Q-point be placed?

16. What is the impedance seen looking into a simple active load?

17. What is the advantage of using a cascode active load?

PROBLEMS

Section 10.1 Bipolar Transistor Current Sources

10.1 Figure P10.1 shows another form of a bipolar current source. (a) Neglecting

base currents, derive the expression for I

in terms of the circuit, transistor,

and diode parameters. (b) If the transistor B–E and diode voltages are equal,

show that, for

= R

, the expression for I

reduces to

(−V

−

)

−

=−10

V and

(on) = V

= 0.7

V, design the circuit such

that

= I

= 2

mA.

10.2 The matched transistors

and

in Figure 10.2(a) have parameters

= 10

−16

A. (a) For

β =∞

, determine

and

BE1

for (i)

REF

= 50 μ

(ii)

REF

= 150 μ

A, and (iii)

REF

= 1.5

mA. (b) Repeat part (a) for

β = 50

10.3 Consider the circuit in Figure 10.2(a). Let

REF

= 200 μ

A. Assume transis-

tor parameters of

β = 80

= 5 ×10

−15

A, and

= 2 ×10

−15

A. Find

BE1

BE2

, and

10.4 Reconsider the circuit in Figure 10.2(a). Let

REF

= 150 μ

A. Assume tran-

sistor parameters of

β = 120

= 10

−16

A, and

= 3 ×10

−16

A. Find

BE1

BE2

, and

–

Figure P10.1

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