Neamen D. Microelectronics: Circuit Analysis and Design

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518 Part 1 Semiconductor Devices and Basic Applications

As previously stated, the equivalent circuit is the same for MOSFETs, JFETS,

and MESFETs. For JFETs, and MESFETS, capacitances

and

are depletion

capacitances rather than oxide capacitances. Typically, for JFETs,

and

are

larger than for MOSFETs, while the values for MESFETs are smaller. Also, for

MESFETs fabricated in gallium arsenide, the unity-gain bandwidths may be in the

range of tens of GHz. For this reason, gallium arsenide MESFETs are often used in

microwave ampliﬁers.

Miller Effect and Miller Capacitance

As for the bipolar transistor, the Miller effect and Miller capacitance are factors in

the high-frequency characteristics of FET circuits. Figure 7.53 is a simpliﬁed high-

frequency transistor model, with a load resistor

connected to the output. We will

determine the current gain in order to demonstrate the impact of the Miller effect.

7.5.3

–

Figure 7.53 Equivalent high-frequency small-signal circuit of a MOSFET with a load

resistance

Writing a Kirchhoff current law (KCL) equation at the input gate node, we have

= jωC

+ jωC

− V

)

(7.98)

where

is the input current. Likewise, summing currents at the output drain node,

we have

+ g

+ jωC

− V

) = 0

(7.99)

We can combine Equations (7.98) and (7.99) to eliminate voltage

. The input cur-

rent is then

= jω





1 + g

1 + jωR



(7.100)

Normally, (

ωR

) is much less than 1; therefore, we can neglect (

jωR

)

and Equation (7.100) becomes

= jω[C

(1 + g

)]V

(7.101)

Figure 7.54 shows the equivalent circuit described by Equation (7.101). The parame-

ter

is the Miller capacitance and is given by

= C

(1 + g

)

(7.102)

Equation (7.102) clearly shows the effect of the parasitic drain overlap capaci-

tance. When the transistor is biased in the saturation region, as in an ampliﬁer circuit,

the major contribution to the total gate-to-drain capacitance

is the overlap

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Chapter 7 Frequency Response 519

–

G D

–

Figure 7.54 MOSFET high-frequency circuit, including the equivalent Miller capacitance

capacitance. This overlap capacitance is multiplied because of the Miller effect and

may become a signiﬁcant factor in the bandwidth of an ampliﬁer. Minimizing the

overlap capacitance is one of the challenges of fabrication technology.

The cutoff frequency

of a MOSFET is deﬁned as the frequency at which the

short circuit current gain magnitude is 1, or the magnitude of the input current

equal to the ideal current

. From Figure 7.54, we see that

= jω(C

(7.103)

and the ideal short-circuit output current is

= g

(7.104)

The magnitude of the current gain is therefore



2π f (C

)

(7.105)

Setting Equation (7.105) equal to 1, we ﬁnd the cutoff frequency

2π(C

)

2πC

(7.106)

where

is the equivalent input gate capacitance.

EXAMPLE 7.12

Objective: Determine the Miller capacitance and cutoff frequency of an FET circuit.

The n-channel MOSFET described in Example 7.11 is biased at the same cur-

rent, and a 10 k



load is connected to the output.

Solution: From Example 7.11, the transconductance is

= 1.2

mA/V. The Miller

capacitance is therefore

= C

(1 + g

) = (10)[1 +(1.2)(10)] = 130 fF

From Equation (7.106), the cutoff frequency is

2π(C

)

1.2 × 10

−3

2π(50 +130) ×10

−15

= 1.06 ×10

= 1.06 MHz

Comment: The Miller effect and equivalent Miller capacitance reduce the cutoff

frequency of an FET circuit, just as they do in a bipolar circuit.

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520 Part 1 Semiconductor Devices and Basic Applications

EXERCISE PROBLEM

Ex 7.12: For the circuit in Figure 7.55, the transistor parameters are

= 0.8

mA/V

= 2

λ = 0

= 100

fF, and

= 20

fF. Determine

(a) the midband voltage gain, (b) the Miller capacitance, and (c) the upper 3 dB fre-

quency of the small-signal voltage gain. (Ans. (a)

=−6.69

, (b)

= 167.6

fF,

(c)

3dB

= 1.32

GHz)

= 10 kΩ

→ ∞

= 4 kΩ

= 234 kΩ

= 166 kΩ

0.5 kΩ

→ ∞

20 kΩ

= 10 V

–

Figure 7.55 Figure for Exercise Ex 7.12

Test Your Understanding

TYU 7.9 An n-channel MOSFET has parameters

= 0.4

mA/V

= 1

V, and

λ = 0

. (a) Determine the maximum source resistance such that the transconduc-

tance is reduced by no more than 20 percent from its ideal value when

= 3

(b) Using the value of

calculated in part (a), determine how much

is reduced

from its ideal value when

= 5

V. (Ans. (a)

= 156 

, (b) 33.3%)

TYU 7.10 An n-channel MOSFET has a unity-gain bandwidth of

= 1.2

GHz. As-

sume overlap capacitances of

gsp

= C

gdp

3 fF, and assume



= 100 μ

A/V

W/L = 15

, and

= 0.4

V. If the transistor is biased at

= 100 μ

A, determine

. (Assume

is equal to zero.) (Ans.

= 66.6

fF)

TYU 7.11 For a MOSFET, assume that

= 1.2

mA/V. The basic gate capacitances

are

= 60

fF,

= 0

, and the overlap capacitances are

gsp

= C

gdp

. Determine

the minimum overlap capacitance for a unity-gain bandwidth of 2.5 GHz. (Ans.

gsp

= C

gdp

= 8.2

fF)

7.6 HIGH-FREQUENCY RESPONSE

OF TRANSISTOR CIRCUITS

Objective: • Determine the high-frequency response of basic transis-

tor circuit conﬁgurations including the cascode circuit.

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Chapter 7 Frequency Response 521

–

Figure 7.56 Common-emitter ampliﬁer

In the last sections, we developed the high-frequency equivalent circuits for the bipo-

lar and ﬁeld-effect transistors. We also discussed the Miller effect, which occurs

when transistors are operating in a circuit conﬁguration. In this section, we will ex-

pand our analysis of the high-frequency characteristics of transistor circuits.

Initially, we will look at the high-frequency response of the common-emitter and

common-source conﬁgurations. We will then examine common-base and common-

gate circuits, and a cascode circuit that is a combination of the common-emitter and

common-base circuits. Finally, we will analyze the high-frequency characteristics of

emitter-follower and source-follower circuits. In the following examples, we will use

the same basic bipolar transistor circuit so that a good comparison can be made be-

tween the three circuit conﬁgurations.

Common-Emitter and Common-Source Circuits

The transistor capacitances and the load capacitance in the common-emitter ampli-

ﬁer shown in Figure 7.56 affect the high-frequency response of the circuit. Initially,

we will use a hand analysis to determine the effects of the transistor on the high-

frequency response. In this analysis, we will assume that

and

are short cir-

cuits, and

is an open circuit. A computer analysis will then be used to determine

the effect of both the transistor and load capacitances.

The high-frequency small-signal equivalent circuit of the common-emitter

circuit is shown in Figure 7.57(a) in which

is assumed to be an open circuit. We

replace the capacitor

with the equivalent Miller capacitance

as shown in

Figure 7.57(b). From our previous analysis of the Miller capacitance, we can write

= C

(1 + g



)

(7.107)

where the output resistance



R

The upper 3 dB frequency can be determined by using the time constant tech-

nique. We can write

2πτ

(7.108)

7.6.1

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522 Part 1 Semiconductor Devices and Basic Applications

where

= R

. In this case, the equivalent capacitance is

= C

, and

the equivalent resistance is the effective resistance seen by the capacitance,

R

. The upper corner frequency is therefore

2π[r

R

](C

)

(7.109)

We determine the midband voltage gain magnitude by assuming

and

are

open circuits. We ﬁnd that



= g





r

+ R



(7.110)

The Bode plot of the high-frequency voltage gain magnitude is shown in Figure 7.58.

EXAMPLE 7.13

Objective: Determine the upper corner frequency and midband gain of a common-

emitter circuit.

For the circuit in Figure 7.56, the parameters are:

= 5

−

=−5

= 0.1



= 40



= 5.72



= 0.5



= 5



,and

10 k



. The transistor parameters are:

β = 150

(on) = 0.7

=∞

35 pF, and

= 4

pF.

Solution: From a dc analysis, we ﬁnd

= 1.03

mA. The small-signal parameters

are therefore

= 39.6

mA/V and

= 3.79



The Miller capacitance is then

= C

(1 + g



) = C

[1 + g

R

)]

= (4)[1 +(39.6)(510)] = 532 pF

and the upper 3 dB frequency is therefore

2π[r

R

](C

)

2π[3.79405.720.1](10

)(35 + 532)(10

−12

)

⇒ 2.94 MHz

(a)

–

⎪⎪ R

–

(b)

–

Figure 7.57 (a) High-frequency equivalent circuit of common-emitter ampliﬁer; (b) high-

frequency equivalent circuit of common-emitter ampliﬁer, including the Miller capacitance

Figure 7.58 Bode plot of the

high-frequency voltage gain

magnitude for the common-

emitter ampliﬁer

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Chapter 7 Frequency Response 523

Finally, the midband gain is

= g





r

+ R



= (39.6)(510)



405.723.79

405.723.79 + 0.1



= 126

Comments: This example demonstrates the importance of the Miller effect. The

feedback capacitance

is multiplied by a factor of 133 (from 4 pF to 532 pF), and

the resulting Miller capacitance

is approximately 15 times larger than

. The

actual corner frequency is therefore approximately 15 times smaller than it would be

were neglected.

PSpice Veriﬁcation: Figure 7.59 shows the results of a PSpice analysis of this

common-emitter circuit. The computer values are:

= 35.5

pF and

= 3.89

pF.

The curve marked “

only” is the circuit frequency response if

is neglected; the

curve marked “

and

only” is the response due to

, and the Miller effect.

These curves illustrate that the bandwidth of this circuit is drastically reduced by the

Miller effect.

The corner frequency is approximately 2.5 MHz and the midband gain is 125,

which agree very well with the hand analysis results.

f (Hz)

100

200

= 150 pF

= 5 pF

only

and C

only

Figure 7.59 PSpice analysis results for common-emitter ampliﬁer

The curves marked “

= 5

pF” and “

= 150

pF” show the circuit response

if the transistor is ideal, with zero

and

capacitances and a load capacitance

connected to the output. These results show that, for

= 5

pF, the circuit re-

sponse is dominated by the

and

capacitances of the transistor. However, if a

large load capacitance, such as

= 150

pF, is connected to the output, the circuit

response is dominated by the

capacitance.

EXERCISE PROBLEM

*Ex 7.13: The transistor in the circuit in Figure 7.60 has parameters

β = 125

(on) = 0.7

= 200

= 24

pF, and

= 3

pF. (a) Calculate the

Miller capacitance. (b) Determine the upper 3 dB frequency. (c) Determine the

small-signal midband voltage gain. (Ans. (a)

= 155

pF, (b)

= 1.21

MHz,

(c)

|=37.3

)

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524 Part 1 Semiconductor Devices and Basic Applications

= 1 kΩ

→ ∞

= 20 kΩ

= 2.3 kΩ

20 kΩ

→ ∞

+5 V

–5 V

→ ∞

5 kΩ

= 5 kΩ

–

Figure 7.60 Figure for Exercise Ex 7.13

The high-frequency response of the common-source circuit is similar to that of

the common-emitter circuit, and the discussion and conclusions are the same.

Capacitance

is replaced by

, and

is replaced by

. The high-frequency

small-signal equivalent circuit of the FET is then essentially identical to that of the

bipolar transistor.

Common-Base, Common-Gate,

and Cascode Circuits

As we have just seen, the bandwidth of the common-emitter and common-source cir-

cuits is reduced by the Miller effect. To increase the bandwidth, the Miller effect, or

the

multiplication factor, must be minimized or eliminated. The common-base

and common-gate ampliﬁer conﬁgurations achieve this result. We will analyze a

common-base circuit; the analysis is the same for the common-gate circuit.

Common-Base Circuit

Figure 7.61 shows a common-base circuit. The circuit conﬁguration is the same as

the common-emitter circuit considered previously, except a bypass capacitor is

added to the base and the input is capacitively coupled to the emitter.

Figure 7.62(a) shows the high-frequency equivalent circuit, with the coupling

and bypass capacitors replaced by short circuits. Resistors

and

are then effec-

tively short circuited. Also, resistance r

is assumed to be inﬁnite. Capacitance

which led to the multiplication effect, is no longer between the input and output ter-

minals. One side of capacitor

is tied to signal ground.

Writing a KCL equation at the emitter, we ﬁnd that

+ g

(1/sC

)

= 0

(7.111)

Since

=−V

, Equation (7.111) becomes

+ g

+sC

(7.112)

7.6.2

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Chapter 7 Frequency Response 525

–

Figure 7.61 Common-base ampliﬁer

(b)

(c)

= –V

–

1 +

–

(a)

–

Figure 7.62 (a) High-frequency common-base equivalent circuit, (b) equivalent input

circuit, and (c) equivalent output circuit

where

is the impedance looking into the emitter. Rearranging terms, we have

1 + r

+sC

1 + β

+sC

(7.113)

The equivalent input portion of the circuit is shown in Figure 7.62(b).

Figure 7.62(c) shows the equivalent output portion of the circuit. Again, one side

is tied to ground, which eliminates the feedback or Miller multiplication effect.

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526 Part 1 Semiconductor Devices and Basic Applications

We then expect the upper 3 dB frequency to be larger than that observed in the

common-emitter conﬁguration.

For the input portion of the circuit, the upper 3 dB frequency is given by

Hπ

2πτ

pπ

(7.114(a))

where the time constant is

Pπ



1 + β





·C

(7.114(b))

In the hand analysis, we assume that

is an open circuit. Capacitance

will

also produce an upper 3 dB frequency, given by

Hμ

2πτ

Pμ

(7.115(a))

where the time constant is

Pμ

= [R

R

] · C

(7.115(b))

is much smaller than

, we would expect the 3 dB frequency

Hπ

due to

to dominate the high-frequency response. However, the factor

/(1 + β)

in the

time constant

Pπ

is small; therefore, the two time constants may be the same order

of magnitude.

EXAMPLE 7.14

Objective: Determine the upper corner frequencies and midband gain of a common-

base circuit.

Consider the circuit shown in Figure 7.61 with circuit parameters

= 5

−

=−5

= 0.1



= 40



= 5.72



= 0.5



= 5



and

= 10



. (These are the same values as those used for the common-emitter

circuit in Example 7.13.) The transistor parameters are:

β = 150

(on) = 0.7

=∞

= 35

pF, and

= 4

pF.

Solution: The dc analysis is the same as in Example 7.13; therefore,

= 1.03

mA,

= 39.6

mA/V, and

= 3.79



. The time constant associated with

Pπ



1 + β





·C



3.79

151



(0.5)



(0.1)



×10

(35 × 10

−12)

⇒ 0.675 ns

The upper 3 dB frequency corresponding to

is therefore

Hπ

2πτ

Pπ

2π(0.675 ×10

−9

)

⇒ 236 MHz

The time constant associated with

in the output portion of the circuit is

Pμ

= [R

R

] · C

= [510] ×10

(4 × 10

−12

) ⇒ 13.33 ns

The upper 3 dB frequency corresponding to

is therefore

Hμ

2πτ

Pμ

2π(13.3 ×10

−9

)

⇒ 11.9 MHz

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Chapter 7 Frequency Response 527

So in this case,

Hμ

is the dominant pole frequency.

The magnitude of the midband voltage gain is

= g

R

)

⎡

⎢

⎣





1 + β







1 + β



+ R

⎤

⎥

⎦

= (39.6)(510)

⎡

⎢

⎣

0.5





3.79

151



0.5





3.79

151



+0.1

⎤

⎥

⎦

= 25.5

Comment: The results of this example show that the bandwidth of the common-base

circuit is limited by the capacitance

in the output portion of the circuit. The band-

width of this particular circuit is 12 MHz, which is approximately a factor of four

greater than the bandwidth of the common-emitter circuit in Example 7.14.

Computer Veriﬁcation: Figure 7.63 shows the results of a PSpice analysis of the

common-base circuit. The computer values are

= 35.5

pF and

= 3.89

pF,

which are the same as those in Example 7.13. The curve marked “

only” is the cir-

cuit frequency response if

is neglected. The curve marked “

and

only” in-

cludes the effect of both

and

. As the hand analysis predicted,

dominates

the circuit high-frequency response.

The corner frequency is approximately 13.5 MHz and the midband gain is 25.5,

both of which agree very well with the hand analysis results.

f (Hz)

0.1

1.0

= 150 pF

= 5 pF

only

and C

only

Figure 7.63 PSpice analysis results for common-base circuit

The curves marked “

= 5

pF” and “

= 150

pF” are the circuit response if

the transistor is ideal and only a load capacitance is included. These results again

show that if a load capacitance of

= 150

pF were connected to the output, the cir-

cuit response would be dominated by this capacitance. However, if a 5 pF load

capacitor were connected to the output, the circuit response would be a function of

both the

and

capacitances, since the two response characteristics are almost

identical.

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