Neamen D. Microelectronics: Circuit Analysis and Design

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508 Part 1 Semiconductor Devices and Basic Applications

Figure 7.41(a) is a common-emitter circuit with a signal current source at

the input. We will determine the small-signal current gain

= i

of the circuit.

Figure 7.41(b) is the small-signal equivalent circuit, assuming the frequency is

sufficiently high for the coupling and bypass capacitors to act as short circuits.

The transistor model is the simplified hybrid-

circuit shown in Figure 7.38

(assuming

=∞

). Capacitor C

is a feedback element that connects the output

back to the input. The output voltage and current will therefore influence the input

characteristics.

The presence of

complicates the analysis. Previously, we could write KCL

equations at the input and output nodes and derive an expression for the current gain.

Here, however, we will approach the problem differently. We will treat capacitor

as a two-port network and will develop an equivalent circuit, with elements between

the input base and ground and between the output collector and ground. This proce-

dure may appear more complicated, but it will demonstrate the effect of

clearly.

Consider the circuit segment between the two dotted lines in Figure 7.41(b). We

can treat this section as a two-port network, as shown in Figure 7.42. The input volt-

age is

and the output voltage is V

. Also, the input and output currents, I

and I

are deﬁned as shown in the ﬁgure.

Writing KVL equations at the input and output terminals, we now have

= I



jωC



+ V

(7.75(a))

and

= I



jωC



+ V

(7.75(b))

Using Equations (7.75(a)) and (7.75(b)), we can form a two-port equivalent circuit,

as shown in Figure 7.43(a). We then convert the Thevenin equivalent circuit on the

output to a Norton equivalent circuit, as shown in Figure 7.43(b).

(a)

(b)

⎪⎪ R

–

Figure 7.41 (a) Common-emitter circuit with current source input; (b) small-signal

equivalent circuit with simpliﬁed hybrid-

model

–

Figure 7.42 Two-terminal

network of capacitor

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Chapter 7 Frequency Response 509

(a)

–

(b)

–

(1/ jwC

)

Figure 7.43 (a) Two-port equivalent circuit of capacitor

with equivalent output circuits:

(a) Thevenin equivalent and (b) Norton equivalent

–

jwC

Figure 7.44 Small-signal equivalent circuit, including the two-port equivalent model of

capictor

The equivalent circuit in Figure 7.43(b) replaces the circuit segment between the

dotted lines in Figure 7.41(b), and the modiﬁed circuit is shown in Figure 7.44. To

evaluate this circuit, we will make some simplifying approximations.

Typical values of g

and

are

= 50 mA/V

and

= 0.05

pF. From these

values, we can calculate the frequency at which the magnitudes of the two dependent

current sources are equal. If

ωC

= g

(7.76(a))

then

f =

2πC

50 × 10

−3

2π(0.05 ×10

−12

)

= 1.59 ×10

⇒ 159 GHz

(7.76(b))

Since the frequency of operation of bipolar transistors is far less than 159 GHz, the

current source

= jωC

is negligible compared to the

source.

We can now calculate the frequency at which the magnitude of the impedance of

is equal to

R

. If

ωC

= R

R

(7.77(a))

then

f =

2πC

R

)

(7.77(b))

If we assume

= R

= 4k

, which are typical values for discrete bipolar

circuits, then

f =

2π(0.05 ×10

−12

)[(4 × 10

)(4 × 10

)]

= 1.59 ×10

(7.78)

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–

Figure 7.45 Small-signal equivalent circuit, including approximations

–

Figure 7.46 Small-signal equivalent circuit, including the equivalent Miller capacitance

510

Part 1 Semiconductor Devices and Basic Applications

If the frequency of operation of the bipolar transistor is very much smaller than

1.59 GHz, then the impedance of

will be much greater than

R

and

can

be considered an open circuit. Using these approximations, the circuit in Figure 7.44

reduces to that shown in Figure 7.45.

The I

versus V

characteristic of the circuit segment between the dotted lines is

− V

jωC

= jωC

− V

)

(7.79)

The output voltage is

=−g

R

)

(7.80)

Substituting Equation (7.80) into (7.79), we obtain

= jωC

[1 + g

R

)]V

(7.81)

In Figure 7.45, the circuit segment between the dotted lines can be replaced by

an equivalent capacitance given by

= C

[1 + g

R

)]

(7.82)

as shown in Figure 7.46. Capacitance C

is called the Miller capacitance, and the

multiplication effect of C

is the Miller effect.

For the equivalent circuit in Figure 7.46, the input capacitance is now

rather than just

had been ignored.

EXAMPLE 7.10

Objective: Determine the 3 dB frequency of the current gain for the circuit shown in

Figure 7.46, both with and without the effect of C

The circuit parameters are:

= R

= 4k

= 2.6k

= 200 k

= 0.8pF

= 0.05 pF

, and

= 38.5mA/V

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Chapter 7 Frequency Response 511

Solution: The output current can be written as

=−(g

)



+ R



Also, the input voltage is

= I



r



jωC



jωC



= I



r

1 + jω(R

r

)(C

)



Therefore, the current gain is

=−g



+ R



r

1 + jω(R

r

)(C

)



The 3 dB frequency is

3dB

2π(R

r

)(C

)

Neglecting the effect of

= 0)

, we ﬁnd that

2π[(200 × 10

)(2.6 × 10

)](0.8 × 10

−12

)

⇒ 77.5 MHz

The Miller capacitance is

= C

[1 + g

R

)] = (0.05)[1 +(38.5)(44)] = 3.9pF

Taking into account the Miller capacitance, the 3 dB frequency is

3 –dB

2π(R

r

)(C

)

2π[(200 × 10

)(2.6 × 10

)][0.8 + 3.9](10

−12

)

⇒ 13.2 MHz

Comment: The Miller effect, or multiplication factor of

, is 78, giving a Miller

capacitance of

= 3.9

pF. The Miller capacitance, in this case, is approximately

a factor of ﬁve larger than

. This means that the actual transistor bandwidth is

approximately six times less than the bandwidth expected if

is neglected.

The Miller capacitance, from Equation (7.82), can be written in the form

= C

(1 +|A

(7.83)

where

is the internal base-to-collector voltage gain. The physical origin of the

Miller effect is in the voltage gain factor appearing across the feedback element

A small input voltage

produces a large output voltage

=−|A

|·V

of the op-

posite polarity at the output of

. Thus the voltage across

(1 +|A

|)V

which induces a large current through

. For this reason, the effect of

on the

input portion of the circuit is signiﬁcant.

We can now see one of the trade-offs that can be made in an ampliﬁer design.

The tradeoff is between ampliﬁer gain and bandwidth. If the gain is reduced, then the

Miller capacitance will be reduced and the bandwidth will be increased. We will con-

sider this tradeoff again when we consider the cascode ampliﬁer later in the chapter.

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512 Part 1 Semiconductor Devices and Basic Applications

Discussion: In Equation (7.80), we assumed that

|jωC

|g

, which is valid even

for frequencies in the 100 MHz range. If

jωC

is not negligible, we can write

+ V



R

+ jωC



= 0

(7.84)

Equation (7.84) implies that a capacitance

should be in parallel with

and

in the output portion of the equivalent circuit in Figure 7.44. For

= R

= 4



and

= 0.05

pF, we indicated that this capacitance is negligible for

f  1.5

GHz.

However, in special circuits involving, for example, active loads, the equivalent

and

resistances may be on the order of 100 k



. This means that the

capaci-

tance in the output part of the circuit is not negligible for frequencies even in the low-

megahertz range. We will consider a few special cases in which

in the output

circuit is not negligible.

EXERCISE PROBLEM

Ex 7.10: For the circuit in Figure 7.41(a), the parameters are

= 200



= 220



= 2.2



= 4.7



= 1



= 100



, and

= 5

V. The transistor parameters are

= 100

(

)

= 0.7

=∞

and

= 1

pF. Consider the simpliﬁed hybrid-

model of the transistor. (a) De-

termine the midband current gain

= I

. (b) Find the Miller capacitance

for (i)

= 0

and (ii)

= 0.08

pF. (c) Determine the upper 3 dB frequency for

(i)

= 0

and (ii)

= 0.08

pF. (Ans. (a)

=−30.24

; (b) (i)

= 0

(ii)

= 4.38

pF; (c) (i)

3dB

= 60.2

MHz, (ii)

3dB

= 11.2

MHz)

Physical Origin of the Miller Effect

Figure 7.47(a) shows the hybrid-

equivalent circuit of the bipolar transistor with a

load resistor

connected at the output. Figure 7.47(b) shows the equivalent circuit

with the Miller capacitance. As a ﬁrst approximation, the output voltage is

=−g

. Assuming sinusoidal signals, Figure 7.48 shows the input signal

and the output signal

. As we have noted previously, the output signal is 180 de-

grees out of phase with respect to the input signal. In addition, because of the gain,

the magnitude of the output voltage is larger than the input voltage. The difference

between

and

is the voltage across the

capacitor as seen in Figure 7.47(a).

7.4.5

–

+–

–

(a)

(b)

Figure 7.47 (a) Hybrid-

equivalent circuit of a bipolar transistor with a load resistor

connected to the output. (b) Equivalent circuit with the Miller capacitance.

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Chapter 7 Frequency Response 513

We may write the sinusoidal signals as

= V

jωt

= V

jωt

,and

jωt

. The current

through the capacitor

can be written as

= C

(7.85(a))

Using phasor notation, we ﬁnd

= jωC

(7.85(b))

This current inﬂuences the input impedance looking into the base terminal of the

transistor.

For the two circuits shown in Figures 7.47(a) and 7.47(b) to be equivalent, the

current

in the two circuits must be the same. From Figure 7.47(b), we can write

= C

(7.86(a))

or using phasors, we have

= jωC

(7.86(b))

For the two capacitor currents in Equations (7.85(b)) and (7.86(b)) to be equal, we

must have

= C

(7.87)

From the signals shown in Figure 7.48, we see that

< V

so that we must

have

> C

. Because of the 180 degree phase shift and voltage gain, the voltage

across

is quite large leading to a relatively signiﬁcant value of capacitor current

. In order to have the current in the Miller capacitor

be the same with a smaller

voltage across

, the value of

must be relatively large. This, then, is the physi-

cal origin of the Miller multiplication effect.

Test Your Understanding

TYU 7.6 A bipolar transistor is biased at

= 120 μ

A and its parameters are

= 120

= 0.08

pF, and

= 15

MHz. Determine the capacitance

. (Ans.

= 0.328

pF)

Figure 7.48 Input signal voltage

and output signal voltage

for the circuits in

Figure 7.47

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514 Part 1 Semiconductor Devices and Basic Applications

TYU 7.7 For the transistor described in Example 7.9 and biased at the same

-point,

determine



and the phase at (a)

f = 150

MHz, (b)

f = 500

MHz, and (c)

f = 4

GHz. (Ans. (a)



= 89.3

φ =−41.9°

; (b)



= 38.0

φ =−71.5°

;

(c)



= 5.0

φ =−87.6°)

TYU 7.8 The parameters of a bipolar transistor are:

= 150

= 1

GHz,

= 12



, and

= 0.15

pF. (a) Determine

and

. (b) What is the bias cur-

rent in the transistor? (Ans. (a)

= 1.84

pF,

= 6.67

MHz; (b)

= 0.325

mA)

7.5 FREQUENCY RESPONSE: THE FET

Objective: • Determine the frequency response of the MOS transis-

tor, and determine the Miller effect and Miller capacitance.

We have considered the expanded hybrid-

equivalent circuit of the bipolar transis-

tor that models the high-frequency response of the transistor. We will now develop

the high-frequency equivalent circuit of the FET that takes into account various

capacitances in the device. We will develop the model for a MOSFET, but it also

applies to JFETs and MESFETs.

High-Frequency Equivalent Circuit

We will construct the small-signal equivalent circuit of a MOSFET from the basic

MOSFET geometry, as described in Chapter 3. Figure 7.49 shows a model based on

the inherent capacitances and resistances in an n-channel MOSFET, as well as the

elements representing the basic device equations. We make one simplifying assump-

tion in the equivalent circuit: The source and substrate are both tied to ground.

Two capacitances connected to the gate are inherent in the transistor. These ca-

pacitances,

and

, represent the interaction between the gate and the channel

7.5.1

gsp

gdp

Figure 7.49 Inherent resistances and capacitances in the n-channel MOSFET structure

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Chapter 7 Frequency Response 515

inversion charge near the source and drain terminals, respectively. If the device is

biased in the nonsaturation region and

is small, the channel inversion charge is

approximately uniform, which means that

∼





WLC

where

(F/cm

) = 

. The parameter



is the oxide permittivity, which for

silicon MOSFETs is



= 3.9

, where



= 8.85 ×10

−14

F/cm is the permittivity

of free space. The parameter

is the oxide thickness in cm.

However, when the transistor is biased in the saturation region, the channel is

pinched off at the drain and the inversion charge is no longer uniform. The value of

essentially goes to zero, and

approximately equals

(2/3)WLC

. As an ex-

ample, if a device has an oxide thickness of 100 Å, a channel length of

L = 0.18 μ

and a channel width of

W = 20 μ

m, the value of

∼

8.3

fF. The value of

changes as the device size changes, but typical values are in the tens of femto-

farads range.

The remaining two gate capacitances,

gsp

and

gdp

, are parasitic or overlap

capacitances, so called because, in actual devices, the gate oxide overlaps the source

and drain contacts, because of tolerances or other fabrication factors. As we will see,

the drain overlap capacitance

gdp

lowers the bandwidth of the FET. The parameter

is the drain-to-substrate pn junction capacitance, and

and

are the series re-

sistances of the source and drain terminals. The internal gate-to-source voltage con-

trols the small-signal channel current through the transconductance.

The small-signal equivalent circuit for the n-channel common-source MOSFET

is shown in Figure 7.50. Voltage



is the internal gate-to-source voltage that con-

trols the channel current. We will assume that the gate-to-source and gate-to-drain

capacitances,

and

, contain the parasitic overlap capacitances. One parame-

ter,

, shown in Figure 7.50 is not shown in Figure 7.49. This resistance is associated

with the slope of

versus

. In the ideal MOSFET biased in the saturation re-

gion,

is independent of

, which means that

is inﬁnite. However,

is ﬁnite

in short-channel-length devices, because of channel-length modulation, and is there-

fore included in the equivalent circuit.

Source resistance r

can have a signiﬁcant effect on the transistor characteristics.

To illustrate, Figure 7.51 shows a simpliﬁed low-frequency equivalent circuit in-

cluding r

but not r

. For this circuit, the drain current is

= g



(7.88)

–

Figure 7.50 Equivalent circuit of the n-channel

common-source MOSFET

–

G D

Figure 7.51 Simpliﬁed low-frequency

equivalent circuit of the n-channel

common-source MOSFET including

source resistance

but not resistance

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516 Part 1 Semiconductor Devices and Basic Applications

and the relationship between

and



= V



+(g



= (1 + g



(7.89)

From Equation (7.88), the drain current can now be written as



1 + g



= g



(7.90)

Equation (7.90) shows that the source resistance reduces the effective transconduc-

tance, or the transistor gain.

The equivalent circuit of a p-channel MOSFET is exactly the same as that of an

n-channel device, except that all voltage polarities and current directions are re-

versed. The capacitances and resistances are the same for both models.

Unity-Gain Bandwidth

As for the bipolar transistor, the unity-gain frequency or bandwidth is a ﬁgure of

merit for the FETs. If we neglect

,and

, and connect the drain to signal

ground, the resulting equivalent small-signal circuit is shown in Figure 7.52. Since

the input gate impedance is no longer inﬁnite at high frequency, we can deﬁne

the short-circuit current gain. From that we can deﬁne and calculate the unity-gain

bandwidth.

Writing a KCL equation at the input node, we ﬁnd that

jωC

= V

[ jω(C

)]

(7.91)

From a KCL equation at the output node, we obtain

jωC

+ I

= g

(7.92(a))

= V

− jωC

)

(7.92(b))

Solving Equation (7.92(b)) for

produces

− jωC

)

(7.93)

7.5.2

–

Figure 7.52 Equivalent high-frequency small-signal circuit of a MOSFET, for calculating

short-circuit current gain

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Chapter 7 Frequency Response 517

Substituting Equation (7.93) into (7.91) yields

= I

[ jω(C

)]

− jωC

)

(7.94)

Therefore, the small-signal current gain is

− jωC

jω(C

)

(7.95)

If we assume typical values of

= 10

fF and

= 1

mA/V, and a maximum

frequency of

f = 1

GHz, we ﬁnd that

ωC

 g

. The small-signal current gain, to

a good approximation, is then

∼

jω(C

)

(7.96)

The unity-gain frequency

is deﬁned as the frequency at which the magnitude

of the short-circuit current gain goes to 1. From Equation (7.96) we ﬁnd that

2π(C

)

(7.97)

The unity-gain frequency or bandwidth is a parameter of the transistor and is inde-

pendent of the circuit.

EXAMPLE 7.11

Objective: Determine the unity-gain bandwidth of an FET.

Consider an n-channel MOSFET with parameters

= 1.5

mA/V

= 0.4

λ = 0

= 10

fF, and

= 50

fF. Assume the transistor is biased

= 0.8

Solution: The transconductance is

= 2K

(

− V

)

= 2(1.5)

(

0.8 − 0.4

)

= 1.2

mA/V

From Equation (7.97), the unity-gain bandwidth, or frequency, is

2π(C

)

1.2 × 10

−3

2π(50 +10) × 10

−15

= 3.18 ×10

= 3.18 GHz

Comment: As with bipolar transistors, high-frequency FETs require small capaci-

tances and a small device size.

EXERCISE PROBLEM

Ex 7.11: The parameters of an n-channel MOSFET are

= 1.2

mA/V

= 0.5

λ = 0

= 8

fF, and

= 60

fF. The unity-gain frequency is

found to be

= 3

GHz. Determine the transconductance and the bias current of

the MOSFET. (Ans.

= 1.282

mA/V,

= 0.342

mA)

Typically, values of

for MOSFETs are in the range of 10 to 50 fF and values

are typically from 0.1 to 0.5 fF.

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