libvolume3.xyzlibvolume3.xyz/.../feedbackamplifier/feedbackamplifiertutorial2.pdf · Microelectronics: Analysis and Design© February 26, 2004 Sundaram Natarajan 576 CHAPTER 8: FEEDBACK

Microelectronics: Analysis and Design© February 26, 2004 Sundaram Natarajan

576

CHAPTER 8: FEEDBACK AMPLIFIERS

8.0: INTRODUCTIONThe concept of feedback was originally introduced in 1934 by H. S. Black, an Electronics Engineer,

for building an amplifier with a gain that is insensitive to changes in the amplifier parameters. Since then, this

notion has played an important role in many areas of Engineering. We introduced the feedback concept in

Section 1.10 through the example of the noninverting amplifier of Fig. 1.10.6. Recall that the closed-loop gain

was almost insensitive to the change in the op-amp's gain in this circuit (see (1.10.25)). As pointed out in

Example 1.8 and in many other examples, the reverse transmission present in a circuit is the feedback, and

we used these two terms synonymously in the past. We have already seen several amplifiers in previous

chapters in which the feedback, both intentional and unintentional, has been present. As an example, to make

the gain insensitive to the variation in the β-value of a transistor in a CE-amplifier, we connected an emitter

resistance RE (see Fig. 4.6.6). This is the case of intentional feedback. Another example of unintentional

feedback is the current [ix/(βdc+1)] caused by the output port current ix in the voltage-follower of Fig. 4.6.10.

From these examples, we identify that the feedback (reverse transmission) exists whenever an output signal

is sensed and used to modify the effective input signal to the amplifier.

The desensitization of the gain to the variations in the active parameters of the BJTs and FETs is the

main reason for using negative feedback in amplifiers. There are also other advantages of using negative

feedback, such as controlling the input and output impedance levels (see Example 1.8) and extending the

bandwidth of an amplifier. Clearly, the designer can use feedback to exercise control over several important

characteristics, such as the gain, the bandwidth, the effect of noise etc. All the above issues provide the

motivation for our study of feedback amplifiers.

Although conventional circuit analysis methods can be used to analyze feedback amplifiers also, the

use of the feedback concept simplifies the process and provides better insight into the working of amplifier

circuits. Furthermore, the design of amplifiers becomes systematic and simplified. This is another important

motivation.

In this chapter, we first discuss the important properties of feedback amplifiers using the systems

approach. Four different feedback configurations are possible. We provide a unified analysis and obtain some

generalized expressions for the parameters of feedback amplifiers applicable to all four configurations. The

two-port concepts play a vital role in analyzing feedback amplifiers because one can easily represent the

feedback using the reverse transmission parameter. We also discuss some specific analysis/design examples

of these four configurations to illustrate their properties.


1Here F is called the feedback ratio (usually a fraction) and is a real constant just as A is; however, in ageneral situation, F can be a function of the complex variable s. Both A and F can also be impedances oradmittances.

2 The designation of Ve as the error signal arises from its significance in feedback control systems.

577

Fig. 8.1.1: An example of a feedback amplifier.

+

-Ve

RicIi

Vi Ri+-

Ro

AoVe

A-networkIo

Roc

+

-Vo = A Ve

RL

+

-Vo

+

-Vf

FVo

F-network

+-

+-

Fig. 8.1.2: The equivalent circuit of the closed-loop amplifier of Fig. 8.1.1 with feedback.

Ric

Ii

Vi Ric+-

Roc

AcoVi

Io

Roc

+

-

Vo = AcVi RL+-

8.1: FEEDBACK CONCEPT AND DEFINITIONSWe introduced the elementary ideas of feedback in Section 1.10 (see Fig. 1.10.6). Now we formally

introduce the concept of feedback and the related basic definitions using a specific circuit example shown in

Fig. 8.1.1. This circuit has an amplifier (A-network) with a finite input resistance Ri and a nonzero output

resistance Ro. Assume that the voltage gain of this amplifier with a load resistor of RL is A. This is known as

the open-loop or forward-path gain. Let Ao be the voltage gain of the amplifier as RL 6 4. The feedback

circuit (F-network) samples the output Vo and produces the feedback signal Vf = FVo, where F is the feedback

ratio 1. Vf is due to the reverse transmission. The so-called error signal Ve is the effective input to the

amplifier2. Vo is obtained by amplifying the error signal and not the original input Vi. The error signal is

influenced both by the input signal and the output signal through the feedback network. Whenever the output

has an influence on the effective input to an amplifier, the amplifier becomes a feedback amplifier as already


578

Ve ' Vi & Vf ' Vi & FVo . (8.1.1)

Vo 'RL

RL%Ro

Ao Ve / AVe . (8.1.2)

Vo 'RL

RL%Ro

Ao (Vi & FVo ) ' A (Vi & FVo ) .

Ac 'Vo

Vi

'A

1 % AF. (8.1.3)

/000Vo

Vi F ' 0

' Ac F ' 0' A , (8.1.4)

mentioned. It is also known as the closed-loop amplifier because the signal path through A- and F-networks

forms a loop.

The entire closed-loop amplifier can also be modeled by another VCVS with its own input resistance,

output resistance, and a gain as shown in Fig. 8.1.2 without paying any regard to its internal structure. The

gain of the feedback amplifier, Ac indicated in Fig. 8.1.2, is called the closed-loop gain. The dependent

voltage source has an voltage gain of Aco as RL 6 4. Similarly, the input and output impedances of the closed-

loop amplifier are denoted as Ric and Roc respectively. Clearly, the primary parameters of the closed-loop

amplifier are Ac, Ric, and Roc.

In Fig. 8.1.1, the feedback signal Vf opposes the input signal Vi in forming the effective input to the

A-network. This type of feedback is called the negative (degenerative) feedback. This is in contrast to the

situation in an oscillator circuit, where a positive (regenerative) feedback is used (see Section 10.3). Since

Vf = (FVo),

The output of the amplifier is

Assume that the amplifier’s gain Ao and hence A increases from its nominal value for some reason.

This will increase the output signal Vo for a given Vi. However, the feedback signal Vf will also increase thus

reducing the error signal Ve. This, in turn, does not permit the output Vo to increase as much as it would have

without feedback. Thus, an increase in the value of A does not automatically cause a proportionate increase

in the output. The automatic comparison between the input and the feedback signal serves to keep the output

signal at the desired value closely irrespective of the change in A. This is the most important property of the

negative feedback in an amplifier.

Let us next find the primary parameters, namely Ac, Ric and Roc, of the closed-loop amplifier.

Substituting (8.1.1) into (8.1.2),

Therefore, the closed-loop gain is (see also (1.10.24))

If the feedback is absent (i.e., F = 0), the input-output relationship reduces to

as it should. Therefore, without the feedback, the output signal Vo for a given input signal will increase by

the same percentage as the percentage increase in A. However, if F … 0, the percentage increase in Vo is


579

Ii 'Ve

Ri

'Vo

ARi

'Vi

(1 % AF )Ri

.

Ric /Vi

Ii

' (1 % AF )Ri / (A /Ac )Ri .

Vo ' Ac Vi 'RL

RL % Ro / (1 % Ao F )Ao

1 % Ao FVi /

RL

RL % Roc

Aco Vi ,

Ac (s ) ' A (s )1 % A (s ) F (s )

. (8.1.5)

reduced. This is an important advantage of using the negative feedback. We will study this property in depth

later in this section. Next, to find the input resistance Ric of the closed-loop amplifier, observe that

The input resistance Ric is therefore

If F = 0, the input resistance will simply be Ri. However, if F > 0, the input resistance of the circuit increases.

From (8.1.3), it is easy to show that

giving rise to the dependent source representation in Fig. 8.1.2 with an output resistance Roc. Hence, the

output resistance decreases from Ro to Roc = [ Ro/(1 + AoF)] for F > 0. Clearly, the input and output

impedances of the amplifiers can be controlled using the feedback.

In the specific example circuit of Fig. 8.1.1, we sampled the output voltage for comparison at the

input, and the feedback signal was also a voltage. However, we could have sampled the output current of the

amplifier. Similarly, the feedback signal could have been a current, which can be compared with the input

current by connecting the ports of the A- and F-networks at the input side in parallel. Thus, four different

feedback configurations (see Fig. 8.2.1) are possible. We discuss these four different configurations in detail

later in this chapter. For now, some important properties of the negative feedback applicable to all the four

configurations will be discussed. The closed-loop gain of all four feedback configurations have the same form

as that of (8.1.3).

The amplifier gain generally depends on frequency. If the A-network is a dc amplifier, the amplifier

gain can be approximated by a constant in the low frequency range. The F-network is usually a resistive

network in amplifiers, and therefore, F is usually a constant. Yet, F can also be a function of frequency. Let

us assume that both A and F are functions of the complex frequency s. If so, the closed-loop gain of an

amplifier of (8.1.3) becomes as follows:

The above equation is an important one and will be used often. If the forward-path gain A and the feedback

factor F are known or can be found in a circuit, we can use (8.1.3) or (8.1.5) to find the closed-loop gain Ac.

From the design point of view, for a given A and a desired closed-loop gain Ac, the designer can use (8.1.3)

to find the required feedback factor F.

If the feedback is absent or is removed intentionally, F = 0. If so, the closed-loop gain will be the

same as the forward-path gain A(s) as stated in (8.1.4). The equation (8.1.4) suggests a simple method of


580

1 % A (s )F (s ) ' 0. (8.1.6)

Ac (s ) . 1F (s )

. (8.1.7)

finding A(s) in a practical feedback amplifier. To find the forward-path gain A in a feedback amplifier, we

simply need to "kill" the feedback in the circuit, which will be used later in all four feedback topologies.

Amount of feedback and Loop GainThe denominator in (8.1.5), (1 + AF), is known as the amount of feedback. The amount of feedback

controls many important properties of feedback amplifiers. L = AF is called the loop-gain or the return ratio.

The loop-gain is a dimensionless quantity because the input signal and the feedback signal should have the

same dimension (voltage or current) for a proper comparison. Therefore, A and F have the inverse

dimensions. If A is a voltage or current ratio so is F. However, if A is a transfer impedance (admittance), F

must be a transfer admittance (impedance).

Characteristic RootsFinite poles of the closed-loop gain can be obtained from the denominator polynomial of (8.1.5). The

closed-loop poles are the roots of the characteristic equation

The values of s, which satisfy the above equation and known as the characteristic roots, are hence the poles

of the closed-loop gain. For the feedback amplifier to be strictly stable, the closed-loop poles (characteristic

roots) must be in the left half s-plane excluding the jω-axis. We discuss the stability of feedback amplifiers

in Chapter 10. For now, we bring out some important properties of feedback amplifiers to emphasize the

advantages of using negative feedback in amplifiers.

Properties of Negative feedback

1. Gain desensitizationThe desensitization of the closed-loop gain to the changes in the forward-path gain A was pointed

out earlier (see Example 1.9). Consider equation (8.1.5). If *A(jω)F(jω)* » 1 in the frequency range of interest,

the closed-loop gain can be approximated as

The above equation implies that the closed-loop gain is essentially controlled by the feedback factor F and

is insensitive to the changes in A if the loop-gain, L = AF, is large. This property of gain desensitization is

the most important reason for using negative feedback in amplifiers and other systems. Usually the forward-

path gain can change by a large amount due to the variations in the active parameters of the transistors. The

feedback network is usually realized by a passive network, whose gain (or more correctly its attenuation) can

be controlled and stabilized quite accurately by the designer, and one can thus control and stabilize the


581

∆Ac

Ac

. /001

1 % AF nominal values

∆AA

/ /000Ac

A nominal values

∆AA

, (8.1.8)

∆Ac

Ac

. 100100,000

(&10) ' &0.01 %.

Ac 2 ' 99.9 .

∆Ac

Ac

'Ac 2

Ac 1

& 1 ×100 . &0.1 %.

closed-loop gain quite accurately.

We next consider the quantitative assessment of the desensitization. Using the definition of (1.11.2),

the per-unit change in the closed-loop gain for small changes in A is

where (∆A/A) is the per unit change in the forward-path gain, the practical order of which depends on the

technology. The above equation clearly suggests that, if A changes for any reason, the per-unit change in Ac

will be smaller by the factor (1 + AF). Consequently, (1 + AF) is also known as the desensitivity factor.

The reason for the popularity of the high-gain operational amplifiers can be understood from the

equations (8.1.7) and (8.1.8). The op-amp (either a VOA or a CFA) is used as the A-network along with a

feedback network that is typically a passive network. Since the op-amp gain is very high, the loop-gain is also

large (at least in the low frequency range), and the closed-loop gain of such networks is mainly dependent

on F(s) as indicated in (8.1.7). Since F(s) can be adjusted externally by the designer, the closed-loop gain in

an op-amp circuit can be adjusted quite accurately. The very high value of A in an op-amp provides a large

desensitivity factor. Thus, even if the op-amp gain changes by as much as 50%, the change in the closed-loop

gain can be made small. Consider an example to illustrate this point.

Example 8.1 (Design and Analysis)In a feedback amplifier, assume that A and F are both real constants for simplicity. The nominal value

of A = 100,000. It is needed to have a nominal closed-loop gain of Ac = 100. Find the required value of F.

Find the expected change in the value of Ac if A decreases by (a) 10 % and (b) 50 %.

SOLUTIONUsing A = 100,000 and Ac = 100 in (8.1.5), we find that F = 0.00999. If the decrease in A is only 10%,

which is small, we can use (8.1.8) to find the percent change in Ac. Thus,

When the change in A is 50%, the change is not small, and we cannot use (8.1.8) in a strict sense. Therefore,

the closed-loop gain must be evaluated with each A to find the required change. Let Ac1 and Ac2 be the closed-

loop gains, if A is 100,000 and 50000 respectively. Ac1 = 100. Using (8.1.5) and the value of F,

Therefore, the percentage change in Ac is


582

Ac (s ) ' A (s )1 % A (s )F

'Am s

(s%ωL ) (1%s /ωH )%Am Fs.

Amc 'Am

1 % Am F. (8.1.11)

ωHc ' (1 % Am F )ωH / (Am /Amc )ωH , and ωLc ' ωL / (1 % Am F ) / (Amc /Am )ωL . (8.1.12)

A (s ) 'Am s

(s % ωL ) (1 % s /ωH ). (8.1.9)

Ac (s ) .Am

(1%Am F )s

s%ωL / (1%Am F ) 1%s / [ωH (1%Am F ) ]. (8.1.10)

Clearly, the relative changes in the closed-loop gain are very small in comparison to the relative changes in

the value of the forward-path gain because of the negative feedback. One can also find the effects of the

tolerance in open-loop gain on the distribution of the closed-loop gain using PSPICE simulation (see Example

1.9).

2. Bandwidth extensionFor simplicity, assume that A(s) has only one low frequency pole at s = -ωL and one high-frequency

pole at s = -ωH. This could be the case of retaining only dominant poles both in the low- and high-frequency

ranges. Incidently, for the particular case of a dc amplifier, ωL may be taken as zero. If the midband gain is

Am, the forward-path gain is

Typically ωL « ωH, and we can assume that the bandwidth of the amplifier is ωH (see 1.8.16)). Let the feedback

factor F be a constant as is typically the case. Then, using (8.1.5),

Using the fact that ωL « ωH, the closed-loop gain can be written as

From the above expression, we identify the midband gain of the closed-loop amplifier as

Also, the upper and lower 3-dB frequencies of the closed-loop amplifier are:

It is clear form (8.1.11) that the mid-band gain of the closed-loop gain has the same form as that of

(8.1.5), if F is a constant. Besides, we find that the upper 3-dB frequency has increased by the amount of

feedback in the midband and so has the bandwidth. Besides, the lower 3-dB frequency is also reduced by the

factor of (1+AmF). The asymptotic plots of the forward-path and the closed-loop gains are shown in Fig. 8.1.3.

Exercise

E8.1. In a feedback amplifier, the nominal value of A = 1000. When the open-loop gain changes by

10%, it is required for the closed-loop to remain within 2%. Find the required value of F and

the corresponding closed-loop gain. Find the expected change in the value of Ac if A decreases

by (a) 10 % and (b) 50 %. Answers: F = 0.004, Ac = 200, (a) -2.17%, (b) -16.67%


583

Gain&Bandwidth product (GB&product ) ' Mid&band or dc Gain (asapplicable ) × Bandwidth . (8.1.13)

A (s ) ' 104

(1 % 10&4 s ) (1 % 10&5 s ) (1 % 10&6 s ),

Ac (s ) / A (s )1 % A (s ) F

'100

1 % 1.11×10&6 s % 11.1×10&12 s 2 % 10×10&18 s 3.

Fig. 8.1.3: Illustration of the bandwidth extension of an amplifier with a constant negative feedback (F = constant).

ω (log scale)ωLc ωL ωH ωHc

Gain in dB A

Ac

20log(Amc)

20log(Am)

Observe that the bandwidth extension is achieved with a reduction in the gain.

The Gain-bandwidth (GB) product of an amplifier is

The GB-product of an amplifier without feedback is (AmωH). With a constant F, we find that the closed-loop

gain in the midband and the bandwidth are given by (8.1.11) and (8.1.13). The GB-product (AmcωHc) of the

closed-loop amplifier with constant feedback also equals (AmωH). Thus, the GB-product is ideally a constant

independent of F and serves as a figure of merit for the basic amplifier. To increase the bandwidth of an

amplifier using the negative feedback, the gain must be correspondingly sacrificed.

Example 8.2A dc amplifier having a gain of

is used with a feedback network that has F = 0.0099. Find the dc gain and the bandwidth of the closed-loop

amplifier.

SOLUTIONSince there is a dominant-pole at 10 kr/s in the forward-path gain, the bandwidth of the dc amplifier

is 10 kr/s. Using (8.1.5), the closed-loop gain can be found to be

The dc gain of the closed-loop amplifier is 100. From the denominator polynomial of Ac(s), we find that the


584

s ' &8.901 ± j302.46 kr / s , and s ' &1.092 Mr/s .

Xo 'A1 A2 Xi

1 % A1 A2 F%

A2 Xn

1 % A1 A2 F. (8.1.14)

closed-loop poles are:

It is interesting to note that, although the poles of the forward-path gain are real and negative, two poles of

the closed-loop gain are complex-conjugate poles. Since a real dominant pole is not present in the closed-loop

gain, we find using (1.8.8) that ωHc = 459 kr/s.

Clearly, the bandwidth has increased with negative feedback. However, the increase in the bandwidth

is not by the amount of feedback in the midband (100) in this example. The reason is that the poles of the

closed-loop gain nearest to the origin are not even real, and therefore, the previous theoretical analysis does

not hold in this example.

3. Reduction of noise and distortionWe discussed a scheme in Fig. 6.3.6 to reduce the distortion in the class-AB power amplifier. This

example is a clear illustration of how the negative feedback can be used to reduce the distortion. Let us now

see how the negative feedback can be used to combat the effects of extraneous signals, such as noise,

introduced at an arbitrary point in the forward path of the loop. Consider the block diagram of an amplifier

shown in Fig. 8.1.4. Assume that a noise signal Xn is added at the input of the amplifier A2 as shown in this

figure. In terms of the input and noise signals, the output signal Xo is

The first component of the output signal is due to the input signal to the amplifier whereas the second

component is due to the noise injected in the amplifier. A figure of merit, which is used to assess the noise

performance of an amplifier, is the signal-to-noise ratio (S/N). The larger this ratio is at the output, the better

is the amplifier performance. Now, the signal-to-noise ratio of the amplifier output is Hence the(A1 Xi /Xn ) .

feedback has no direct impact on improving the output (S/N) ratio because this is independent of the feedback

factor F. However, since the amplified signal has a higher magnitude, the (S/N) ratio increases by theA1 Xi

same amount as that of the amplifier gain A1. In the absence of feedback, the signal at the input of the

amplifier A2, viz. cannot be increased beyond a certain level to avoid saturation and distortion in itsA1 Xi ,

amplification process. However, with feedback, the strength of signal at the input of amplifier A2 being

A (s ) ' 103 s(s % 100)(1 % 10&3 s )2

.

Exercise

E8.2. An amplifier has a gain of

If the feedback factor F = 0.5 is employed, what are the values of the mid-band gain and the

lower and upper 3-dB frequencies of the closed-loop gain?

Answers: ωLc . 0.1995 r/s, ωHc . 34.73 k r/s.


585

Xo 'A ( Xi % Xn )

1 % AF. (8.1.15)

Xo'AXi

1 % AF%

Xn

1 % AF' Ac Xi % (Ac /A )Xn . (8.1.16)

SN output

' AXi

Xn

. (8.1.17)

can be increased by times (increasing Xi or A1 or both) still keepingA1 Xi / (1%A1 A2 F ) , A1 Xi (1%A1 A2 F )

the input to the amplifier A2 within permissible limit.

Let us now consider a feedback amplifier of a given forward path gain of A = A1A2 and see how the

location of the entry of the noise signal affects the output (S/N) ratio in comparison with the ratio of the input

signal to the noise signal. Consider two particular cases:

(a) The noise is additive at the input of the forward path. A1 = 1, and A2 = A.

(b) The noise is additive at the output of the forward path. A1 = A, and A2 = 1.

In case (a), the output Xo is

If the noise (or distortion) signal is additive at the input, the noise signal is amplified by the same amount as

the input signal. The (S/N) ratio at the input is (Xi/Xn). At the output also, it has the same ratio. There is no

improvement. In case (b), however, we find that

If the additive noise signal occurs in the forward path near the output, the noise component in the output is

reduced by the amount of feedback. The (S/N) ratio at the output is A-times larger than the one in case (a);

i.e.,

Similar considerations are true for the distortion also. If the distortion occurs at the output (such as

in the power amplifier), the harmonic distortion at the output can be reduced by a factor of (A/Ac) with the

use of negative feedback.

4. Control over Input and Output impedancesRecall that the input impedance increased in the feedback amplifier of Fig. 8.1.1. In Chapter 4, it was

shown that an unbypassed emitter resistance increases the input and output impedances of the CE-amplifier.

Clearly, we can control both the input and output impedances using feedback. We discuss several examples

to illustrate these properties later in this chapter.

Potential Problems of FeedbackThere are also some potential problems with feedback. It is clear from the previous two examples that

the negative feedback decreases the overall gain. However, this is not a serious disadvantage. We can add as

many stages as we want to increase the gain because the active devices are inexpensive these days. The


586

property of insensitivity of the closed-loop gain outweighs this disadvantage.

The most serious problem in feedback amplifiers is the possibility of undesired oscillations although

the forward-path amplifier may be absolutely stable. As seen from Example 8.2, the closed-loop amplifier

may have complex-conjugate poles. For some higher value of F, the poles may move to the jω-axis causing

unforced oscillations to occur in the output. In an amplifier with negative feedback, the feedback signal

generally opposes the original input signal. This may be true at low frequencies. However, although F may

be a constant, A(s) is frequency dependent. Therefore, the phase of the feedback signal at some high

frequency may be such that the feedback signal adds to the original input signal. If so, an output signal at this

frequency may be sustained without the input resulting in oscillations at this frequency. Once the oscillations

start in a circuit, the input will not have any control on the output. This results in an unstable amplifier. We

address the stability problem in Chapter 10.

8.2: FOUR BASIC FEEDBACK TOPOLOGIESThe A- and F-networks are two-port networks, and a feedback amplifier is an interconnection of two

two-port networks. Let their ports on the input signal side have the label , and the ports on the output1

signal side carry the label . If the input signal is a voltage, the feedback and error signals must also be2

voltages (see (8.1.6)). To mix these voltages, the ports with the label of the A-network and F-network1

must be connected in series as shown in Fig. 8.1.1. However, if the currents are to be mixed at the input side,

we have to connect the ports with the label of the two networks in parallel. At the output side, depending1

on the desired output, we can either sample the output voltage as shown in Fig. 8.1.1 or the output current

to generate the feedback signal. To sample the output voltage, we have to connect the ports with the label 2

of the two networks in parallel. However, to sample the output current, the ports with the label of the A-2

and F-networks must be connected in series. Therefore, depending on how the input signal and the feedback

signal are combined at the input side and how the output is sampled, the feedback amplifiers are classified

into four different topologies. These four basic configurations, shown in Fig. 8.2.1, are:

(a) Series-Shunt Configuration,

(b) Series-Series Configuration,

(c) Shunt-Shunt Configuration,

and

(d) Shunt-Series Configuration.

In the Series-Shunt configuration, to mix the feedback voltage Vf with the input source voltage Vs,

the ports of the amplifier and feedback networks are connected in series. This is the reason for the first1


587

Fig. 8.2.1: Four feedback configurations: (a) Series-Shunt (VCVS), (b) Series-Series (VCCS), (c) Shunt-Shunt (CCVS), and (d) Shunt-Series (CCCS).

(a)

+-

A-network

F-network

ZocIo

ZL

+

-Vo

+

-Vo

+

-Vf

Ii

IiZs

Zic

Vs

+

-Ve 1 2

1 2

(b)

+-

A-network

F-network

Zoc

Io

ZL

+

-Vo

+

-Vf

Ii

IiZs

Zic

Vs

+

-Ve

Io

1 2

1 2

(c)

A-network

F-network

ZocIo

ZL

+

-Vo

+

-Vo

If

Ie

Zs

Zic

Is 21

1 2

(d)

A-network

F-network

Zoc

Io

ZL

+

-Vo

IoIf

Ie

Zs

Zic

Is 21

1 2

term "Series" in this name. Port of the F-network is connected in shunt (parallel) with port of the2 2

A-network to sample the output voltage. This is the reason for the second term "Shunt." The example circuit

of Fig. 8.1.1 belongs to this category. In this configuration, with negative feedback, we found that the input

impedance of the closed-loop amplifier will be the input impedance of the original amplifier Ri multiplied by

the amount of feedback and will therefore be higher. The output impedance of the closed-loop amplifier will

be the output impedance Ro of the original amplifier divided by a factor equal to the amount of feedback.

Thus, this type of feedback makes the original amplifier A closer to an ideal voltage-controlled voltage source

(VCVS), or simply, a voltage amplifier. In a Series-Shunt feedback configuration, finding the output voltage

is very convenient. Even if we are interested in finding the output current, we first find the output voltage in

this configuration and then find the output current. Similarly, even if the input source is a non ideal current

source, the current source is converted to a voltage source before carrying out the analysis of a Series-Shunt

configuration.

In the Series-Series topology of Fig. 8.2.1(b), both sets of ports of the A- and F-networks are

connected in series. At the input side, the voltages are mixed. The output current is sampled. Both the input

and output impedances of the closed-loop amplifier will be higher than those of the A-network in a series-

series feedback configuration. Thus, this type of feedback makes the amplifier A closer to an ideal

voltage-controlled current source (VCCS). Therefore, even if we need to find the output voltage across the


588

load, we find it convenient to find the output current first in this configuration. The CE-amplifier with an

unbypassed emitter-lead resistor is an example of this type of feedback.

In a Shunt-Shunt configuration, the ports of the A- and F-networks are connected in parallel.1

There is a current mixing at the input side. Ports are also connected in parallel for the F-network to2

sample the output voltage as shown in Fig. 8.2.1(c). The Shunt-Shunt configuration can be used to make an

amplifier closer to an ideal current-controlled voltage source (CCVS). With this type of feedback, the input

and output impedances of the closed-loop amplifier will be lower than those of the original amplifier A. For

a convenient analysis of the Shunt-Shunt configuration, the input should be a current, and the output should

be a voltage.

The Shunt-Series feedback, shown in Fig. 8.2.1(d), makes an amplifier closer to an ideal current-

controlled current source (CCCS). The input impedance decreases, and the output impedance increases with

this type of feedback connection. Both the input and output signals are taken as currents in this configuration.

We provide a unified analysis, applicable to all four configurations, in the next section. We also

consider specific examples of feedback amplifiers in the next few sections. Practical amplifiers have finite

and nonzero input and output impedances. In the example circuit of Fig. 8.1.1, the F-network was an ideal

VCVS. However, in a practical amplifier, the F-network is usually a passive network and may load the

amplifier. Therefore, we have to include the loading effects of both A- and F-networks in a practical circuit.

The analysis of these four configurations can be accomplished quite easily with the use of two-port

parameters for the forward-path amplifier as well as the feedback network. An understanding of the two-port

parameters greatly enhances this analysis, and we therefore provide a brief account of their properties and

notations. The port-voltages and port-currents are the variables of interest in a two-port network. The vector-

matrix equations of the four pertinent sets of two-port parameters are listed in Table 8.1. Depending on the

parameter set under use (h, z, y, or g), two of the port variables, one from each port, are considered to be the

inputs or excitations (port currents for example in the z-parametric set) and the other two variables, called

the outputs or responses (port voltages in the z-parametric set) are expressed in terms of the inputs. Each

individual parameter is defined as the ratio of a specific response to a specific input while keeping the other

input zero. Furthermore, each parameter is distinguished by two subscripts first subscript giving the response

location and the second the input location. For example z12 is the ratio of the voltage response at port to1

the current excitation in port while the current in port is kept at zero. As pointed out earlier, this2 1

parameter describes the feedback (reverse transmission) if the transmission from port to port is1 2

assumed to be a forward transmission. The parameter with a "11" subscript describes the input immittance

(the word immittance is used to mean either an impedance or an admittance) seen at port of a two-port1

network. The parameter with the "22" subscript describes the input immittance seen at port , and that with2


589

Table 8.2: Properties of the Feedback configurations

Feedbackconnection

AppropriateTwo-port parameter

representation

Input Variable (source form)

Outputvariable

Transfer functionstabilized

ZicInput

Impedance

ZocOutput

Impedance

Series-Shunt

h-parameters Voltage, Vs(Thevenin)

Voltage, Vo (Vo/Vs) Voltagetransfer function

Increases Decreases

Series-Series

z-parameters Voltage, Vs(Thevenin)

Current, Io (Io/Vs) Transferadmittance

Increases Increases

Shunt-Shunt

y-parameters Current, Is(Norton)

Voltage, Vo (Vo/Is) TransferImpedance

Decreases Decreases

Shunt-Series

g-parameters Current, Is(Norton)

Current, Io (Io/Is) Currenttransfer function

Decreases Increases

Input variables of A&network 'X1A

X2A

, Output variables of A&network 'Y1A

Y2A

, (8.3.1a)

Table 8.1: Table of Two-port parameters used in analysis of feedback amplifiers.

.

h-parametricrepresentation

,=h11 h12

h21 h22

V1

I2

I1

V2

y-parametricrepresentation

, and=y11 y12

y21 y22

I1

I2

V1

V2

=g11 g12

g21 g22

I1

V2

V1

I2

g-parametricrepresentation

=z11 z12

z21 z22

V1

V2

I1

I2

z-parametricrepresentation

,

a "21" subscript describes the "forward transmission" or the "forward-path gain." These meanings hold for

all the sets of two-port parameters considered here. Now, any set of two-port parameters can be conceivably

used to analyze feedback amplifiers. However, the analysis of a given configuration becomes simplified if

a particular set of two-port parameters suitable for the configuration is used. For example, to analyze a

feedback amplifier with a Series-Shunt configuration, the h-parametric representation of the two networks

is most convenient. Besides, the use of the appropriate set of parameters permits a unified analysis of all the

four configurations. A summary of the properties of the four configurations discussed so far in this section

is given in Table 8.2.

8.3: A UNIFIED ANALYSIS OF ALL FOUR FEEDBACK CONFIGURATIONSWe present a unified analysis of feedback amplifiers in this section applicable to all four

configurations and obtain the expressions for the closed-loop gain, and the input and output immittances of

the feedback amplifiers. Using the vector-matrix equations to describe the A- and F-networks in terms of their

two-port parameters is very convenient. Let the input and output variables of A- and F-networks be


590

Ri ' /000V1

I1 I2 ' 0

, Av ' /000V2

V1 I2 ' 0

and Ro ' /000V2

I2 V1 ' 0

.

y11 '476.2×10&6 (1 % 110×10&9 s )

(1%5.328×10&9 s )S , y12 '

&4.762×10&12 s(1%5.328×10&9 s )

S ,

y21 '(38.1×10&3&4.762×10&12 s )

(1%5.328×10&9 s )S , and y22 '

(200×10&6%25.1×10&12 s%23.81×10&21 s 2 )(1%5.328×10&9 s )

S .

Fig. E8.3.

5 kΩ20 kΩ104×VX

+

-

V2

+ -VX

I2

+

-

V1

100 kΩI1

+

-

180 kΩ

1 kΩ

RiRo

Fig. E8.5.

5 kΩ2 kΩ 40ms×VX

+

-

V2

5 pF

50 pF+

-

VX

I2

+

-

V1

100 ΩI1

Exercise

E8.3. Find the h-parameters of the two-port network shown in Fig. E8.3.

Answers: h11 = 118 kΩ, h12 = 0.1 V/V, h21 = -106 A/A, h22 = 1.205 mS.

E8.4. Use the h-parameters of the two-port network of Fig. E8.3 and find the input resistance, overall

gain, and the output resistance defined as follows:

Answers: Ri = 83.1 MΩ, Av = 9.986 V/V, Ro = 1.178 Ω.

E8.5. Show that the y-parameters of the two-port network shown in Fig. E8.5 are

E8.6. The g-parameters of a two-port amplifier are: g11 = 10 µS, g12 = -0.01 A/A, g21 = 1 V/V, g22 =

25 Ω. The output port is terminated 1- kΩ resistor. Find the input resistance and the overall

voltage gain. Answers: Ri = 50.62 kΩ, Av = 0.9756 V/V


591

Input variables of F&network 'X1F

X2F

, and Output variables of F&network 'Y1F

Y2F

. (8.3.1b)

p11A p12A

p21A p22A

, andp11F p12F

p21F p22F

, (8.3.2)

Y1A

Y2A

'

p11A p12A

p21A p22A

X1A

X2A

, andY1F

Y2F

'

p11F p12F

p21F p22F

X1F

X2F

. (8.3.3)

In the above equations, X and Y represent the port-voltages and port-currents (see Table 8.1). The subscripts

1 and 2 in the above variables refer to ports and respectively. Using the generalized symbol p for the1 2

two-port parameters, we can describe the two-port parameters of the two networks as follows:

where p’s represents h-, z-, y-, or g-parameters as appropriate to the feedback configuration. Using the

definitions of (8.3.1) and (8.3.2), the input-output descriptions of the A- and F-networks are

An ideal amplifier is one in which only the forward transmission parameter p21A is nonzero. Thus, all

other parameters of the A-network should ideally be zero, i.e., p21A …0, and p11A = p22A = p12A = 0. However,

in a practical amplifier, the other parameters may be nonzero. While the immittances p11A and p22A represent

the loading effects, p12A represents the reverse transmission (feedback) that is internal in the amplifier.

Similarly, in an ideal feedback network, only the reverse transmission parameter p12F that represents the

feedback should be nonzero (p12F …0), and the other parameters should be zero. The feedback network,

usually being a passive network, causes additional loading through the immittances p11F and p22F at the input

and output ports respectively, and these loading effects should be included in the analysis. The reverse

transmission (feedback) through the amplifier is usually absent or negligible in comparison to the one through

the feedback network, i.e., usually *p12A* « *p12F*. Therefore, for now, we will ignore the reverse transmission

through the amplifier by setting p12A = 0. We suggest a simple modification later to account for p12A, if it is

not negligible.

For a feedback amplifier configuration, the choice of the parametric representation is determined by

p12F appropriate to the type of feedback employed. For example, in a Series-Shunt configuration, the reverse

transmission should be a voltage ratio. Since h12 is the only two-port parameter that stands for this voltage

ratio, we need to use the h-parametric representation to analyze this feedback configuration. With this

preliminary background on the two-port parameters of the A- and F-networks, we proceed to develop the

required equations for the closed-loop parameters in terms of the generalized parameters that will be

applicable to all four configurations.

We use the specific example of the Series-Shunt configuration to illustrate this general development

in this section. The interconnection of the A- and F-networks in a Series-Shunt configuration is shown Fig.

8.3.1 using the h-parametric representation of the two networks, where the internal feedback (i.e., in the A-

network) is assumed to be zero. In this configuration, we identify that


592

X1A ' X1F ' Ii , X2A ' X2F ' Vo , Y1A'Ve , Y1F'Vf , Y2A' IoA , and Y2F' IoF . (8.3.4)

Vi ' Ve % Vf ' Y1A%Y1F / Y1 , and Io ' IoA % IoF ' Y2A%Y2F / Y2 . (8.3.5)

Y1

Y2

'

( p11A%p11F ) p12F

( p21A%p21A ) (p22A%p22F )

X1

X2

/p11T p12F

p21T p22T

X1

X2

, (8.3.6)

pijT ' pijA % pijF , i , j ' 1,2 . (8.3.7)

Ys ' Y1 % X1 WS , and Y2 ' &X2 WL , (8.3.8)

Fig. 8.3.1: The equivalent circuit of Series-Shunt configuration using the h-parameters of the A- and F-networks.

h11A

h21A Ii h22A

1

+-

h12FVo

h11F

h21F Ii

F-network

h22F

1

Zic

Zs

+-

Ii

IiVs

Zoc

Io

+

-

Vo

ZL

+

-

Ve

A-network

IoA

IoF

+

-

Vi

+

-

Vf

+

-Vo

While the inputs of the two networks are equal, the responses of the two networks add because

This property is not only true in this configuration but in the other three configurations as well, if the

appropriate two-port representation is used. Defining X1 / X1A = X1F, and X2 / X2A = X2F, we have

where

The equation (8.3.6) implies that the feedback amplifier can be represented with a single two-port network,

where the two-port parameters can be obtained by simply adding the respective two-port parameters of the

A- and F-networks. That the parameters of the A- and F-networks become additive in the combined network

underlines the appropriateness of choosing a specific set of two-port parameters to analyze a particular

feedback configuration. The implementation of (8.3.6) for the Series-Shunt configuration results in a

simplified equivalent circuit of Fig. 8.3.2. Now, the input-output relationships can be obtained quite easily

by adding two more constraints. Let WS and WL represent the source and load immittances (for example: WS

= ZS and WL = YL in the Series-Shunt configuration). Then, we have the following constraints:

where Ys is the input signal to the feedback amplifier, and X2 is the output signal (for example: Ys = Vs, and

X2 = Vo, and Y2 = Io in the Series-Shunt configuration). Using (8.3.8) in (8.3.6) and rearranging,


593

Ys

0'

(p11T%WS ) p12F

p21T (p22T%WL )

X1

X2

. (8.3.9)

X1

X2

'1

[(p11T%WS ) (p22T%WL ) & p12F p21T ]

(p22T%WL ) &p12F

&p21T (p11T%WS )

Ys

0, (8.3.10)

Ac 'X2

Ys

'&p21T

[ (p11T%WS ) (p22T%WL ) & p12F p21T ]. (8.3.11)

F / p12F , (8.3.12)

A ' Ac F ' 0' /000

X2

Ys F ' 0

'&p21T

( p11T%WS ) ( p22T%WL ). (8.3.13)

F ' h12F , (8.3.14)

A / /000Vo

Vs F'0

'&h21T

(h11T % Zs ) (h22T % YL ). (8.3.15)

Fig. 8.3.2: A simplified equivalent circuit for the circuit shown in Fig. 8.3.1.

+-

h12FVo

h11T

h21T Ii h22T

1

Zic

Zs

+-

Ii

Vs

ZocIo

+

-

Vo ZL

+

-

Vi

Solving the above matrix-equation,

The output being X2, the closed-loop gain is

Since the feedback factor is

the forward-path gain is then found by setting F = p12F = 0, and thus

For the Series-Shunt configuration, the feedback factor is

and the forward-path gain is

Since the A-network is unilateral, the forward-path gain can also be found by directly evaluating (VoA/Vs)

using the circuit of Fig. 8.3.3 in which F = h12F has been set to zero. This means that we do not need to

evaluate the h-parameters of the amplifier. This is one of the powerful advantages of using the feedback

concept. As the currents and voltages in this network are different from those in the circuit of Fig. 8.3.1, we

use IiA, VoA, and IoA in this circuit.


594

Ac 'A

1 % A×F. (8.3.16)

RLeff ' (RL2RE2ro ) / WL .

F'g12' /000ii

ix vi'0

'&1

βdc % 1.

Fig. 8.3.3: The equivalent circuit to find the parameters A, ZiA, and ZoA of a Series-Shunt feedback amplifier.

A-network

h11F

Zs

+-Vs

+

-

ZL

IoA

h22F

1 VoA

IiA

h21F IiA

ZiAYoA

+

-

ViA

Using the definitions of (8.3.13) and (8.3.12) for the forward-path gain and the feedback factor, the

closed-loop gain Ac of (8.3.11) becomes the following familiar equation:

Consider an example illustrating the process of finding the overall gain of an amplifier using the feedback

concept.

Example 8.3Consider the small-signal equivalent circuit of the emitter-follower of Fig. 4.8.5 that has been

reproduced here in Fig. 8.3.4(a). Find the overall gain Avs = (vo/vs).

SOLUTION

The controlled current source samples the current ix and is connected in parallel at the input side. If

βdc 6 4 as in an ideal BJT, this feedback will be absent. Since the feedback factor is a current ratio, the g-

parametric representation is appropriate in this case. Therefore, it is convenient to consider this amplifier as

a current-controlled current source (CCCS) for the purpose of analysis. Of course, we can obtain the voltage

gain after finding the current gain (ix/is). The previous feedback analysis can be used to find current gain.

Since ix is the output, we define an effective load of

Also, we identify that WS = (1/RS) = GS.

The feedback parameter is

Suppressing the feedback, the network to find A is shown in Fig. 8.3.4(b). Once the feedback (reverse

transmission) is suppressed, the circuit becomes a unilateral circuit. By the inspection,


595

viA 'is

(Gs % GB ), and ixA '

&viA

(re % RLeff )'

& is

(Gs % GB ) (re % RLeff ).

A 'ixA

is

'&1

(GB%GS ) (RLeff%re ).

Ac 'ix

is

'A

1 % A×F'

& (βdc%1)1 % (βdc%1)(GB%GS ) (RLeff%re )

.

Avs 'vo

vs

'&RLeff

RS

×Ac '(βdc%1) RLeff

RS % (βdc%1)(1 %RS /RB ) (RLeff%re ).

Fig. 8.3.4: (a) The SS-equivalent circuit of the voltage-follower of Fig. 3.8.5 using the CC-model, and (b) the circuit with the feedback suppressed (modified A-network).

(a)

e

c

+

-1×vi

re

REviRs RB

Ri iib

RL

+vo

Ro

ro

ix

vsRs

is =

io

ix

βdc+1

+

-

Roe

(b)

c

+

-1×viA

re

viARs RB

iiA RLeffixA

vsRs

is =+

-

Gs+GiA RoA+RLeff

Therefore, the forward-path gain is

Using (8.3.16), the overall (closed-loop) gain of the amplifier of Fig. 8.3.4 is

Next, using the facts that vs = (isRS) and vo = -(ixRLeff) in the above equation,

We can show that the voltage gain Avs would be the same as that obtained in Section 4.8 for this amplifier

once we express the above in terms of the input resistance of the amplifier. Observe the simplicity of the

present analysis.

We obtain the expressions for the input and output immittances next. The input immittance is


596

Wic 'Y1

X1

'Ys

X1

& WS '[ (p11T%WS ) (p22T%WL ) & p12F p21T ]

(p22T%WL )& WS ' (p11T%WS ) (1 % A F ) & WS .

Wic ' (p11T % WS ) (A /Ac ) & WS . (8.3.17)

Woc ' /000Y2

X2 Ys ' 0

' /000p21T X1 % p22T X2

X2 Ys ' 0

' p22T % /000p21T X1

X2 Ys ' 0

(p11T%WS ) X1 % p12F X2 ' 0, YX1

X2

'&p12F

(p11T%WS ).

Woc ' p22T &p12F p21T

(p11T % WS )' (p22T % WL ) (1 % A F ) & WL ' (p22T % WL ) (A /Ac ) & WL (8.3.18)

Zic ' (h11T%ZS ) (A /Ac ) & ZS , and Yoc ' (h22T%YL ) (A /Ac ) & YL .

Zic ' (ZiA%ZS ) (A /Ac ) & ZS , and Yoc ' (YoA%YL ) (A /Ac ) & YL . (8.3.19)

Wic ' (WiA%WS ) (A /Ac ) & WS , and Woc ' (WoA%WL ) (A /Ac ) & WL . (8.3.20)

ZiA % Zs ' /000ViA

IiA VoA ' 0

, and YoA % YL ' /000IoA

VoA IiA ' 0(8.3.21)

Since (1+A×F) = (A/Ac),

If the ports of the A- and F-networks are connected in series at the input, Wic, WS, and p11T are impedances,

and in a parallel connection, all these parameters are admittances.

Next, the output immittance is

If Ys = 0,

Using the above result in the previous equation,

If the ports of the A- and F-networks are connected in series (parallel) at the output side, WL, p22T, and Woc are

all impedances (admittances). For example, in the Series-Shunt configuration, the input immittance is an

impedance while the output immittance is an admittance. Applying (8.3.17) and (8.3.18) to the Series-Shunt

topology,

Observe that h11T and h22T are nothing but the input impedance ZiA and the output admittance YoA of the A-

network shown in Fig. 8.3.3. Using these familiar parameters for the Series-Shunt topology,

Reverting again to the general case, defining the input and output immittances of p11T = WiA and p22T

= WoA,

Again, one does not need the individual two-port parameters of the amplifier because, in a unilateral

amplifier, both WiA and WoA can be found almost by inspection.

We can also find (ZiA + Zs) and (YoA + YL) by applying the following the formal definitions to the

circuit shown in Fig. 8.3.5:

The network of Fig. 8.3.5 is the same as that of Fig. 8.3.3 except for the introduction of the independent


597

shunt (voltage) port Y voltage Y short&circuit Y admittance ,and

series (current) port Y current Y open&circuit Y impedance .(8.3.22)

Gi ' Gic ' (GB % GS )[1 % (βdc%1)(GB%GS ) (RLeff%re ) ]

(βdc%1)(GB%GS ) (RLeff%re )& GS ' GB %

1(βdc%1)(RLeff%re )

.

Ri ' Ric ' [RB 2 (βdc%1)(RLeff%re ) ] ,

Fig. 8.3.5: The circuit to find ZiA + Zs and YoA + YL in a Series-Shunt feedback amplifier.

A-networkh11F

Zs

+-IiA ZL

IoA

ViA

+

-

VoAh22F

1 h21F I1

sources IiA and VoA to conceptualize the definitions of the driving point immittances (h11T+Zs) and (h22T+YL).

Recall that the modified A-network is a unilateral network. Therefore, (ZiA+Zs) can be found by inspection

of the circuit provided in Fig. 8.3.3. When the input current IiA is set to zero to find (YoA+YL), the controlled

sources representing the forward transmissions on the output side become ‘dead.’ Therefore, only passive

elements will exist on the output side also. Hence, one can obtain the admittance (YoA+YL) also by inspection

directly from the modified A-network without having to analyze the network of Fig. 8.3.5.

The foregoing conclusions are true in all four configurations. It is however important to use proper

terminations while the driving-point immittances of the modified A-network. If we designate a port of this

network as a shunted port or a series port depending on the mode of its connection with a port of the feedback

network, then we can frame the following rules:

The driving-point immittance of the modified A-network is an admittance if the connection at the port

is a shunt connection and is an impedance if the ports are connected in series. Furthermore, for determining

these driving-point immittances, the other port is shorted if the other port is a shunt port and is open-circuited

(severed) if it is a series port. So, in summary, it is worth remembering the following associated words:

Consider the voltage-follower of Fig. 8.3.4(a) earlier in Example 8.3 again. Since this is a Shunt-

Series configuration, to find the input admittance Gi and output impedance Ro in this amplifier, we need (GiA

+GS) and (RoA +RLeff) from the modified A-network of Fig. 8.3.4(b). They can be easily identified from the

circuit as (Gs+GB) and (re+RLeff) respectively. We can use these parameters, and the forward-path and closed-

loop gains in (8.3.20) and find the input conductance and output resistances. First,

Therefore,

which is identical to that given by (4.8.13). Next,


3 See Section 17.9 of Richard C. Dorf and James A. Svoboda, “Introduction to Electric Circuits,” 5th

Edition, John Wiley & Sons, Inc., 2001.

598

Ro ' (re % RLeff )[1 % (βdc%1)(GB%GS ) (RLeff%re ) ]

(βdc%1)(GB%GS ) (RLeff%re )& RLeff ' re %

1(βdc%1)(GB%GS )

' re % (RB2RS ) / (βdc%1) ,

F / p12A . (8.3.23)

A , WiA , and WoA . (8.3.24)

A 'A

1%A×F, WiA ' (WiA%Ws ) A

A&Ws , and WoA ' (WoA%WL ) A

A&WL . (8.3.25)

A ' A , WiA ' WiA , and WoA ' WoA . (8.3.26)

which is identical to one obtained in (4.8.12). Observe that, although we did not use the two-port parameters

explicitly, we did use the two-port concepts developed earlier.

We observed earlier that the appropriate two-port parameters of the A- and F-networks add to give

the parameters of the overall network. A word of caution is necessary at this point. This result is valid only

if the port principles continue to be valid in the combined network. In a series-connection, for instance, the

currents flowing into the port of the A-network must be the same as the current flowing into the port of the

F-network. If this is not the case, the terminal relations defined in terms of the pertinent parameters for a two-

port network may not be valid when it is interconnected3. Fortunately, in most amplifier networks, it is either

true or approximately true. Nevertheless, it is important and useful to keep this in mind and examine the

topology of an interconnection of the A- and F-networks.

We assumed so far that the internal feedback is absent or negligible in comparison to the external

feedback. If the A-network has an internal feedback that cannot be ignored, one can use a two-step process

and determine A, WiA, and WoA that includes the influence of the internal feedback. Since the formulas

obtained earlier are independent of whether the feedback is internal or external, the same set of formulas can

be used to account for the internal feedback. Assume that the internal feedback is

We can suppress internal feedback in the A-network and find the following auxiliary forward-path gain, and

the input and output immittances

using the same process that we discussed earlier. Then, the parameters A, WiA, and WoA that include the

influence of the internal feedback can be found first using

Then, we can use the equations of (8.3.16) and (8.3.20) and find the closed-loop parameters that include the

external feedback. Observe that, if p12A =F ' 0,

In such a case, we need only one step process to find the closed-loop parameters.

Steps for the analysis of Feedback AmplifiersWe have now completed the discussion of all the steps required to find the gain, the input and output


599

F ' p12F .

Ad (s ) ' 104

1 % 0.01s,

Fig. 8.3.6: A noninverting amplifier using an op amp considered in Example 8.4.

+-

+

-

R2R1

Rs = 10 k

RL = 2 kVs

Ad

+

-Vf

+

-

Vo

immittances of feedback amplifiers and illustrated their use for the analysis of the Series-Shunt feedback

amplifier as a specific example. The steps are summarized below.

1. Find the two-port parameters p11F, p12F, p21F, and p22F of the F-network that are appropriate to the

feedback configuration. Identify the reverse transmission parameter as the feedback factor, i.e., set

2. Connect the two-port equivalent of the F-network to the A-network and suppress the feedback.

Although it does not add any complexity to the analysis process, the forward transmission of the F-

network can also be ignored as an approximation. Find the forward-path gain A of this modified A-

network with feedback suppressed (see (8.3.11) and (8.3.13)).

3. Find the input and output immittances of the modified A-network. Including the source and load

immittances in this step is usually convenient as illustrated for the case of the Series-Shunt

configuration in Fig. 8.3.5.

4. Find the closed-loop gain and the input and output immittances of the feedback amplifiers using

(8.3.16) and (8.3.20). If the internal feedback cannot be neglected, use the two-step process suggested

earlier.

Example 8.4 (Design and Analysis)The circuit shown in Fig. 8.3.6 is the same noninverting amplifier circuit designed in Example 1.7

and analyzed in Example 1.8. However, the source and load resistors have been added in this circuit. Using

the feedback concept and assuming that the VOA is ideal, design the amplifier with a closed-loop dc gain of

10. A practical op-amp (VOA) has finite gain, finite input impedance, and nonzero output impedance. If the

op-amp has a differential-mode gain of

a finite differential-mode input resistance of Rid = 100 kΩ, and an output resistance of Ro = 100 Ω, find the

gain and the input and output impedances of the designed closed-loop amplifier.


600

h11F' /000Vf

I1 Vo'0

' (R12R2 ) , F'h12F' /000Vf

Vo I1'0

'R1

R2%R1

, h21F' /000I2

I1 Vo ' 0

'&R1

R2%R1

,

h22F ' /000I2

Vo I1'0

'1

R2 % R1

.

Fig. 8.3.8: The circuit to find A, ZiA + Rs, and YoA + GL in Example 8.4.

+-

Rs = 10 k

RL = 2 kVs Ad Vd

+

-

+-

Ro

+

-

Vd Rid VoA

h11F = (R12R2)

= (R1+R2)h22F

1

IiA

h21FIiA

YoA + GLZiA + RsIoA

Fig. 8.3.7: (a) The equivalent circuit of the feedback amplifier of Fig. 8.3.6 and (b) the feedback network of the amplifier.

(b)

R2R1

+

-Vf

+

-Vo

I1 I2

(a)

+-

Rs = 10 k

RL = 2kVs

Ad Vd

+

-

Vo

+-

Ro

+

-

VdRid

R2R1

+

-Vf

SOLUTIONThe equivalent circuit of the closed-loop amplifier is shown in Fig. 8.3.7(a), where the op-amp has

been replaced with its model having a finite gain, finite input resistance, and the nonzero output resistance.

On the output side, the ground terminal is the common terminal between the A- and F-networks. Therefore,

this circuit has a Series-Shunt feedback configuration. The F-network is shown separately in Fig. 8.3.7(b).

Following the step (1), we first find the h-parameters for the F-network of Fig. 8.3.7(b). They are:

and

Connecting the two-port equivalent of the feedback network and setting the feedback F to zero, the

network of Fig. 8.3.8 results. No internal feedback exists in this amplifier. From this network, by inspection,

we first find that


601

IiA 'Vs

(Rs % Rid % h11F ), and Vd '

Rid Vs

(Rs % Rid % h11F ),

VoA '(Go Ad Vd & h21F IiA )

(Go % GL % h22F ).

A (s ) 'VoA

Vs

'1

(Rs % Rid % h11F )(Go Rid Ad & h21F )(Go % GL % h22F )

. (8.3.27)

Ac ' (1 / F ) ' 1 % (R2 / R1 ) . (8.3.28)

(R2 / R1 ) ' 9 .

A (s ) ' 8478.2 (1 % 100×10&12 s )(1 % 0.01s )

.

A (s ) . 8478.2(1 % 0.01s )

.

Ac 'A

1 % AF. 9.988

1 % 11.78×10&6 s.

Using KCL at the output node,

Substituting for IiA and Vd in the above equation and solving for VoA in terms of Vs, we obtain the forward-path

gain of

If the op-amp is ideal, Ad 6 4, (AF) 6 4, and the closed-loop gain is

This equation is identical to (1.10.20). If the closed-loop gain is required to be 10, then the resistance ratio

must satisfy that

We can choose the standard resistors of which are the same values chosenR1 ' 2 kΩ , and R2 ' 18 kΩ ,

in Example 1.7. This completes the design. At this point, we should emphasize the power of the feedback

concept. The design process became simplified if (AF) is assumed to be very high. We will use this fact very

often in the future designs also.

Let us now find the parameters of the designed feedback amplifier including the characteristics of

the practical op-amp. Using the selected resistance values and the op-amp's parameters in (8.3.27),

There is a dominant pole at s = -100 r/s. The high frequency zero at s = -10 Gr/s is due to the forward-

transmission h21F of the feedback network and can be ignored for all practical purposes. This implies that the

forward transmission through the feedback network could have been ignored. Ignoring this zero,

Using the above A and the feedback factor F in (8.3.16),

The dc gain is 9.988 that is very close to the design specification of 10 even with loading effects. The

bandwidth of the closed-loop amplifier is This is indeed equal to [(1 + AoF)ωH]. TheωHc ' 84.89 kr / s .

frequency responses of the op-amp Ad and the closed-loop gain Ac are shown in Fig. 8.3.9. Clearly, with the


602

ZiA%Zs' /000Vs

IiA VoA'0

'111.8 kΩ , and YoA%GL' /000IoA

VoA IiA'0

'10.55 mS.

Zic ' (ZiA % Zs ) (A /Ac ) & Zs '94.9(1 % 10.73×10&6 s )

1 % 0.01sMΩ ,

Yoc ' (YoA % YL ) (A /Ac ) & YL '8.955(1 % 11.22×10&6 s )

1 % 0.01sS.

Zoc '1

Yoc

'0.112(1 % 0.01s )

(1 % 11.22×10&6 s )Ω .

Fig. 8.3.9: Magnitude responses of Ad and Ac in the Example 8.4

Frequency in r/s (log scale)

Gai

n in

dB

10 100 1.0K 10K 100K 1.0M-20

0

20

40

60

80

100Op-amp'sMagnituderesponse

(100.000,76.989)

(84.658K,16.991)

Magnituderesponse of

the Amplifier

negative feedback, the bandwidth has increased along with the reduction in the dc gain.

By the inspection of the circuit of Fig. 8.3.8, we find that

Using the calculated values of the various parameters in (8.3.20) (or (8.3.19) for this topology),

and

The output impedance is

Although the input and output impedances of the op-amp are real, the closed-loop amplifier has

frequency dependent impedances because of the op-amp’s frequency dependent gain. The magnitude plots

of the input and output impedances are shown in Fig. 8.3.10. We find that the input (output) impedance of

the closed-loop amplifier is high (low) at low frequencies as expected. With a series-shunt feedback, the

closed-loop amplifier behaves nearly as an ideal VCVS; however, as the operating frequency increases, the

input impedance decreases and the output impedance increases. Indeed, the input impedance can be modeled

by a parallel combination of a resistance and a capacitance in the useful frequency range from dc to ωHc.

Similarly, the output impedance can be modeled by a series combination of a resistance and an inductance

in the same frequency range.


603

F ' z12F . (8.4.1)

Fig. 8.3.10: Magnitude responses of input and output impedances of the noninverting amplifier in Example 8.4.

Inpu

t im

peda

nce

in Ω

(log

scal

e)

Frequency in r/s (log scale)

Out

put i

mpe

danc

e in

Ω (l

og sc

ale)

10 100 1.0 K 10 K 100 K 1.0 M100 K

1.0 M

10 M

100 M

0.1

1.0

10

100

101.8 kΩ

94.9 MΩ

0.112 Ω

99.5 Ω

8.4: SERIES-SERIES FEEDBACK AMPLIFIERWe consider the Series-Series feedback configuration in this section. In a Series-Series configuration,

the feedback network samples the output current Io and produces a feedback voltage Vf. The feedback voltage

modifies the input voltage. F should be clearly a transfer impedance, and thus

For A and F to have inverse dimensions, A should be a transfer admittance. The closed-loop gain Ac is the

transfer admittance (Io/Vs). Therefore, the input must be a voltage, and the output must be a current in this

configuration. We should use the z-parametric representation. Therefore, the Series-Series configuration using

the z-parametric representation of the A- and F- networks is shown in Fig. 8.4.1.

Suppressing the feedback, the modified A-network is shown in Fig. 8.4.2. One can find the forward-

path gain using the definition of

Exercise

E8.7. In the amplifier of Fig 8.3.6, the VOA is replaced with a CFA (see the model of Fig. 1.10.12

reproduced here in Fig. 8.5.3). Ignore the input capacitor Ci for now. The parameters of the

CFA are: Rx = 50 Ω, Ri = 1 MΩ, Zt = 900 kΩ, and Ro = 15 Ω. Also, R1 = R2 = 820 Ω. Find the

amplifier parameters of the closed-loop amplifier using the feedback concept. Observe that

input resistance of the closed-loop amplifier is not affected by the feedback in this case.

Answers: Ac = 1.9979 V/V, Ro = 0.0155 Ω, Ri = 1 MΩ,.


604

A 'IoA

Vs

. (8.4.2)

ZiA % Zs ' /000ViA

IiA IoA'0

, and ZoA % ZL ' /000VoA

IoA IiA'0(8.4.3)

Fig. 8.4.1: The equivalent circuit of Series-Series configuration using the z-parameters of the A- and F-networks.

+-

z12FIo

z11F

F-network

Zic

Zs

+-

Ii

IiVs

ZocIo

+

-

Vo

+

-

VoF

ZL

+

-

Vf

+

-

Ve

IoA

IoF

+

-

Vi

+-

z21FIi

+

-

VoA

z22F

A-network

Fig. 8.4.2: The circuit to find the forward-path gain A, (ZiA+Zs), and (ZoA+ZL) of a Series-Series feedback amplifier.

Zs

+-

Vs

ZLIoA

z11F z22F

IiA

+ -

z21FIiAA-network

ZiA+Zs ZoA+ZL

from this network. The open-circuit input and output impedances (ZiA+Zs) and (ZoA+ZL) can be found by

applying the following formal definitions

to the circuit of Fig. 8.4.2. We should sever the output (current) port to find (ZoA+ZL).

One should note that (ZiA+Zs) is not necessarily the same (ZiA+Zs) found in the Series-Shunt topology

because the terminating conditions of the output ports are different. However, since the feedback is absent

and the modified A-network is a unilateral network, they should be the same because the terminating

condition of the output port has no effect at the input. If there is internal feedback, one should use the two-

step process. When we suppress the internal feedback, the situation is the same, and the terminating condition

at the output port has no influence at the input port. Once we find A, F, and the impedance parameters, we

can find the parameters of the closed-loop amplifier using


605

Ac 'Io

Vs

'A

1 % AF, Zic ' (ZiA%Zs ) (A /Ac )&Zs , and Zoc ' (ZoA%ZL ) (A /Ac )&ZL . (8.4.4)

z11F ' RE , z22F ' RE , z21F ' RE , and F ' z12F ' RE . (8.4.5)

Ac 'Io

Vs

'(&Vo /RL )

Vs

'104.7

' 2.128 mS. (8.4.6)

Fig. 8.4.3: A CE-amplifier with an unbypassed emitter resistor designed in Example 8.5, and itssmall-signal equivalent circuit.

+-

4.7 k

Q+Vo

Rs = 1k

RE

Vs

-12 V

1 mA

+12 V

+- Ii Io

RE

F-network

rπ

Rs = 1 kΩ

Vs

βdcIi

+

-

Vo

RL =4.7 kΩ

Ii = Ib Io = Ic

ro

Example 8.5 (Design and Analysis)In the CE-amplifier shown in Fig. 8.4.3, the β-value of the BJT may range from 100 to 300 with

Gaussian distribution and VA = 100 V. Assume that the nominal value of β is the average of the extreme

values of β. It is required to obtain a midband voltage gain of (Vo/Vs) = -10 V/V ± 10%. Find the value of RE

and verify if the specifications are met. Find the input and output impedances of the design.

The resistors have 5% tolerance with Gaussian distribution. Find the distribution of the midband

voltage gain using PSPICE simulation and the ‘expected’ minimum, maximum, and mean values of the open-

loop voltage gain without feedback (RE = 0) and the closed-loop gain with feedback.

SOLUTIONAssume that the transistor operates in the active mode. Otherwise, the circuit cannot be an amplifier.

The small-signal equivalent circuit in the midband is also shown in Fig. 8.4.3. As far as the amplifier is

concerned, the input-port current is Ii = Ib, and the output-port current is Io = Ic. However, the current through

RE is Ie. Since Ie = Ib + Ic, assume that Ie splits into Ib and Ic again at the emitter node, and the F-network is a

two-port formed by the single element RE. Clearly, this is a Series-Series feedback configuration. Separating

the F-network, we find that

The nominal value of the closed-loop gain should be the transfer admittance of


606

rπ ' 5.67 kΩ , and ro ' 108.5 kΩ .

IiA 'Vs

Rs % rπ % RE

.

(RL%RE ) IoA %RE IiA % ( IoA & βdc IiA )ro ' 0.

A 'IoA

Vs

'(βdc ro & RE )

(Rs % rπ % RE ) (ro % RE % RL ).

βdc ro

(Rs % rπ % RE ) (ro % RE % RL ) (8.4.7)

F ' RE '1Ac

&1A'

12.128 mS

&(Rs % rπ % RE ) (ro % RE % RL )

βdc ro

.

Fig. 8.4.4: The equivalent circuit to find the forward-path gain A, (ZiA+Rs), and (ZoA+RL)in Example 8.5.

Rs = 1 kΩ

+-Vs

RL = 4.7 kΩ IoA

rπ βdcIiAz11F = RE

IiA

z22F = RE

ro

+ -

REIiA

ZiA + Rs ZoA + RL

If *AF* » 1, then Ac . (1/F). This means that RE = z12F = F should be close to (1/Ac) = 0.47 kΩ.

We will now proceed to find the accurate value of RE. To do this, we need the forward-path gain.

With the nominal value of β = 200, the collector bias current is approximately 0.995 mA. Then, we find that

VCB . 7.3 V > 0, and VCE . 8 V. βdc . 216. The BJT operates in the active mode. Also,

The small-signal equivalent circuit to find the forward-path gain and the impedance parameters is obtained

by suppressing the feedback as shown in Fig. 8.4.4. By inspection of the input side,

Using KVL on the outer loop containing RL, ro, and RE, we have

Solving for IoA in terms of IiA and then substituting for IiA in terms of Vs, we find that

The approximation in the above equation is valid because (βdcro) = 23.44 MΩ » RE. This approximation is

equivalent to neglecting the forward-path gain (z21F) through the feedback network while determining the

forward-path gain of the feedback amplifier. Had we neglected the forward transmission through the feedback

network, we could have obtained the output current using a simple current division. In any case, proceeding

further, the value of RE can be selected using the above equation for A and the required value of Ac in the

following equation,


607

Ac /A

1 % AF' 2.15 mS.

Vo

Vs

' &2.15×4.7 ' &10.11 V/V.

Ac 'Io

Vs

'A

1 % AF'

βdc ro

(Rs % rπ % RE ) (ro % RE % RL ) % βdc ro RE(8.4.8)

ZiA % Rs ' (Rs % rπ % RE ) , and ZoA % RL ' (ro % RE % RL ) .

Zic ' (ZiA % Rs ) (A /Ac ) & Rs 'βdc ro

ro % RL % RE

% 1 RE % rπ , (8.4.9a)

Zoc ' (ZoA % RL ) (A /Ac ) & RL 'βdc RE

Rs%rπ % RE

% 1 ro % RE . (8.4.9b)

Zic ' 94.79 kΩ , and Zoc ' 1.528 MΩ .

Substituting the various known parameters in the previous equation, the required value of RE can be found.

This value is 0.435 kΩ. Observe that this value is close to the approximate value determined earlier. We select

a standard resistor of RE ' 0.43 kΩ .

To verify if the specifications are met, using (8.4.7), we find that A = 29.05 mS. Using this value of

A and F = 0.43 kΩ, the closed-loop gain is

The midband voltage gain is

This value is in error only by 1.05% from the specification. The PSPICE simulation shows that the midband

voltage gain is Acm = -10.13 V/V.

We next find the input and output impedances of the amplifier and compare these results with those

obtained in Section 3.6 for the same amplifier configuration. The closed-loop gain is

By inspection of the circuit of Fig. 8.4.4 (observe when the input current is set to zero, the controlled sources

representing both forward transmissions become ‘dead’), we find that

The input and output resistances of the feedback amplifier are

and

All the dominant terms present in (8.4.9a) and (8.4.9b) are the same as those found in (3.6.19). However,

observe the ease with which the amplifier parameters have been obtained now without having to find the

individual z-parameters as we did in Section 3.6. It is clear from the above two equations that the negative

feedback caused by RE increases both the input and output impedances. If RE = 0, the input and output

impedances would reduce to rπ and ro respectively as they should. Substituting the various parameters in the

above two equations,

Monte-Carlo simulation was performed and the distributions of the voltage gain with and without


608

Fig. 8.4.5: The distribution of the gains with and without feedback in Example 8.5.

(a) Without feedback

median90th %ilemaximum

n samples = 5000n divisions = 10mean = 145.221

sigma = 7.76556minimum = 31.41910th %ile = 136.168

= 146.643= 152.974= 160.953

Voltage Gain

Perc

enta

ge o

f Am

plifi

ers

40 80 120 16020 1800

20

40

60

(b) With feedback

Voltage Gain

Perc

enta

ge o

f Am

plifi

ers

9.0 9.5 10.0 10.5 11.00

10

20

30

n samples = 5000n divisions = 10mean = 10.1178

sigma = 0.180531minimum = 9.3606610th %ile = 9.89124

median = 10.113990th %ile = 10.351maximum = 10.8218

feedback were obtained. The model statements for the BJT and the resistors are as follows:

.model QbreakN-X NPN IS=1.5e-15 VA = 100 BF = 200 DEV/gauss 25%

.model Rbreak-X RES R=1 dev/gauss 1.25%

The distributions are shown in Fig. 8.4.5. All the values, maximum, minimum and ‘expected’ mean

values of the gains are listed in the tables below each distribution. The ‘expected’ mean value of the midband

voltage gain is 10.12 V/V, which is very close to the analytical value of 10.11 V/V. Besides, with feedback,

almost 99.8% of the amplifiers have gains within ±10% of the nominal value of the gain. Indeed, 97% of the

amplifiers have errors within about ±5%. More important, the variance indicates how sensitive the amplifier

is to the changes in the parameter values. To compare the performance characteristics of the amplifier with

and without feedback, we can use the normalized value of variance. Without feedback, the normalized value


609

F ' y12F . (8.5.1)

A ' (VoA / Is ) . (8.5.2)

YiA % Ys ' /000Is

ViA VoA'0

, and YoA % YL ' /000IoA

VoA ViA'0(8.5.3)

is (7.7656/145.22) . 5.35 %, and with feedback, the same is (0.1805/10.12) . 1.8 %. This variance is mainly

due to the 5% tolerance in the resistors in RL and RE (especially RE because Ac . (1/RE)). Clearly, the feedback

reduces the variation of the gain to (1/3) of the original variation without feedback.

8.5: SHUNT-SHUNT CONFIGURATIONIn a shunt-shunt feedback configuration, the feedback network samples the output voltage and

provides the feedback current If, F being the transfer admittance

Therefore, y-parametric representation should be used to analyze a Shunt-Shunt feedback configuration as

shown in Fig. 8.5.1. Both Ac and A should be transfer impedances. Let Is and Vo be input and the output

respectively. Suppressing the feedback, the network shown in Fig. 8.5.2 can be used to find the forward-path

gain and the admittance parameters. From this network, we can find

One can find the input and output admittances using the formal definitions of

to the circuit of Fig. 8.5.2. However, just as in other configurations, the above admittances can usually be

Fig. E8.8.

+-

+

-Vs

Rs

Zic

Ad

M

RF

Zoc Io +Vo

ZL

IBIAS

Exercise

E8.8. In the transconductance

amplifier of Fig. E8.8, the

op-amp has a finite gain of

Ad = 103 with an input

resistance of 100 kΩ and an

output resistance of 100 Ω.

T h e M O S F E T h a s

K ' 2 mA/V 2 , Vt'0.8 V,

and λ = 0.03 V-1. Also, RS =

10 kΩ, IBIAS = 1 mA, and ZL

= 1 kΩ, and RF = 1 kΩ. Find

the values of A, Ac, Zic, and Zoc. Also, find (Vo/Vs).

Answers: A = 2.644 S, Ac = 0.9996 mS, Zic = 293.6 MΩ, Zoc = 93.45 MΩ, (Vo/Vs) = -0.9996 V/V.


610

Ac 'Vo

Is

'A

1 % AF, Yic ' (YiA%Ys ) (A /Ac )&Ys , and Yoc ' (YoA%YL ) (A /Ac )&YL . (8.5.4)

F ' g12F ' /000If

Io V1 ' 0

. (8.6.1)

Fig. 8.5.1: An amplifier with the Shunt-Shunt feedback using the y-parameteric representation for the feedback network.

Yic

Zs

Ie

If

Is

Yoc

ZL

+

-

V1 y21FV1

Io

+

-Vo

+

-Vo

1y22F

y12FVo1

y11F

A-network

F-network

Fig. 8.5.2: The circuit to find the forward-path gain A, (YiA+Ys), and (YoA+YL) of a Shunt-Shunt feedback amplifier.

ZsIs ZLy21FViA

+

-VoA

1y22F

1y11F

A-network

+

-ViA

YiA + Ys YoA + YLIoA

obtained by inspection. Once we find A, F, and the admittance parameters, we can find the parameters of the

closed-loop amplifier using

8.6: SHUNT-SERIES CONFIGURATIONThe Shunt-Series configuration is the dual of the Series-Shunt configuration. For the two-port

parameters of the A- and F-networks to add in this configuration, it is necessary to use the g-parametric

representation. The current Io is the output. At the input side, the feedback current is mixed with the input

current. The closed-loop gain is (Io/Is). The closed-loop amplifier is closer to an ideal CCCS with this

feedback. The voltage-follower circuit of Example 8.3 is an example of this configuration. The general form

of the Shunt-Series configuration is shown in Fig. 8.6.1, where the F-network is represented with the

g-parameters. In this configuration,


611

A 'IoA

Is

. (8.6.2)

Suppressing the external feedback, the modified A-network to find the forward-path gain is shown in Fig.

8.6.2. The forward-path gain is

Of course, one can also find the input admittance (YiA + Ys) and the output impedance of (ZoA + ZL) from this

network. The parameters of the closed-loop gain can be found using

Rx ' 50 Ω , Ri 64 , Ci ' 0, Zt (s ) ' (3×106 ) / (1 % 16.5×10&6 s ) Ω , and Ro ' 15 Ω .

Fig. E8.9.(b)

+VoCFA

+

-

IxR1

R2

+Vs

Zic Zoc

Ii

If

+Vx

RL

+Vy

+Vw

(a)

Ix

Rx

1 Zt(s)×Ix

+Vx

+VyIy

Ci

Ro

Iw

+Vw

Ri

+

-

-

+

Exercise

E8.9. The macro-model of a current feedback amplifier (CFA) in the low frequency range was

developed in the previous chapter and is shown in Fig. 7.6.8. Its frequency-domain macro-model

is shown in Fig. E8.9(a). Some model parameters are assumed to be zero for simplicity. Only the

dominant model parameters have been included here. The unity gain amplifier represents a voltage

buffer. Assume that the model parameters of the CFA are:

This CFA is used to build an inverting amplifier shown in Fig. E8.9(b). Assume that R1 = 100 Ω,

R2 = 1 kΩ, and RL = 1 kΩ. Find the overall voltage gain and the bandwidth of the closed-loop

amplifier. Also, find the input and output impedances.

Answers:Vo

Vs

'Ac

R1

. &9.9951 % 8.772×10&9 s

V/V, ωHc'114 Mr/s ( 18.14 MHz) ,

Zic '1

Yic

'16.91 (1 % 16.5×10&6 s )

(1 % 5.855×10&9 s )mΩ , and Zoc '

1Yoc

'7.746 (1 % 16.5×10&6 s )

(1 % 8.644×10&9 s )mΩ .


612

Ac 'Io

Is

'A

1 % AF, Yic ' (YiA%Ys ) (A /Ac )&Ys , and Zoc ' (ZoA%ZL ) (A /Ac )&ZL . (8.6.3)

KP'51.169 µA/V 2 , Vto'0.7339 V, λ'0.03122 V &1 , γ'0.4823 V 1/2 , and (2φf )'0.7 V.

KP'16.528 µA/V 2 , Vto'&0.7776 V, λ'0.04349 V &1 , γ'0.6727 V 1/2 , and (2φf )'0.7 V.

WL 1

'605

, and WL 2

'WL 3

'WL 4

'140

5.

Rs ' 10 kΩ , RL ' 10 kΩ , and Rf ' 50 kΩ .

Fig. 8.6.1: An amplifier with the Shunt-Series feedback using the g-parameteric representation for the feedback network.

A-network

g12F Io

Yic

Zs

Ie

If

Is

Zoc

ZL

+

-

V1g21FV1

Io

g11F

1

Io

+

-

Vo

+-

g22F

F-network

+

-

V2

Fig. 8.6.2: The circuit to find A, (YiA+Ys), and (ZoA+ZL) of a Shunt-Series feedback amplifier.

ZsIs

ZL

g11F

1

g22F

IoA

A-network

+

-

ViA

+ -

g21FViA

YiA + Ys ZoA + ZL

Example 8.6Consider the two-stage MOSFET amplifier shown in Fig. 8.6.3. The feedback is provided by the

combination of RL and Rf. The process parameters of the NMOS FET are

The process parameters of the PMOS FETs are

The aspect ratios of the MOSFETs are

The circuit parameters are


613

K1'0.051169

2605

'0.307 mA/V 2 , and K2'K3'K4'0.016528

2140

5'0.2314 mA/V 2 .

F' /000If

Io V1'0

'RL

RL % Rf

'&16

, g11F' /000If

V1 Io'0

'1

Rf%RL

'1

60 kΩ,

g21F' /000V2

V1 Io'0

'RL

Rf%RL

'16

, and g22F' /000V2

Io V1'0

' (Rf 2 RL )' 506

kΩ .

Fig. 8.6.3: A MOSFET amplifier analyzed in Example 8.6 and the feedback network used in the amplifier.

F-network

RfRL

IoIf

+

-

V1

+

-

V2

Ric

RfRLRsIs

IoVB

IREF

M4M3M1

M2

F-network

Roc

+5 V

The dc bias values are IREF = 100 µA, and VB = 2.7 V. Is and Io are the input and output currents of the

amplifier. Use the feedback concept to find the closed-loop gain and the input and output resistances. Also,

verify the analytical evaluations using the PSPICE simulations.

SOLUTIONThe conductivity parameters of the MOSFETs are

The first stage is a common-gate amplifier, while the second stage is a common-source amplifier. The

feedback not only affects the small-signal analysis but also the dc bias currents and voltages. It is very

interesting that the bias currents, practically dependent on the current mirror and the feedback network, would

be stable. The feedback network can be separated as shown in Fig. 8.6.3, and we can find the feedback factor

and its loading effect. They are

Using the current mirror properties, we can replace the current mirror with a bias current and an output

resistance. The feedback network can also be replaced by its two-port equivalent. In this process, we can


614

VS1 ' VB%γ 2φf % (γ2 /2 )& Vt1o& ID1 /K1 &γ VB%2φf%γ 2φ %γ2 /4&Vt1o& ID1 /K1 , (8.6.4)

1.144 ' ( ID1% ID2 /6 )×8.571 , Y ID2 . 0.2008 mA, and VD2.1.673 V.

VSG2 ' Vt2%ID2

K2 (1%λp VSD2 )'1.648 V, and VD1 . 3.352 V.

Fig. 8.6.4: The MOSFET amplifier of Example 8.6 for dc analysis.

(1/6)ID2

ID2VB

IBIAS =0.1147 mA

M1

M2

+5 V

ro3 =244.1 kΩ

8.571 kΩ 8.333 kΩ

+VS1 +VD2

+VD1

ignore the forward transmission through the feedback network. Doing so, the amplifier network for dc

analysis is shown in Fig. 8.6.4.

Neglecting the secondary effect of the channel-length modulation for now, the source-node potential

of M1 in terms of its bias current can be obtained by adapting (3.7.11) and (3.7.12). Therefore, substituting

vI = VB, VSS = 0 in (3.7.12) and using (3.7.11),

where ID1 = (IBIAS - VD1/rO3). The value of VD1 = (VDD - VSD3) is not known now. For now, we assume that VSD3

. VSG3 = 1.415 V. Using this approximate value, we have VD1 . 3.585 V, and ID1 . IREF = 0.1 mA. Using this

current and other known values in (8.6.4), we get an approximate value of VS1 . 1.144 V. With this value,

we also find that Vt1 . 0.9853 V. Using the above approximate value of VS1 in

Using the above approximate values of ID2 and VD2 in the following equation:

Since we have a better estimate of VD1, we iterate the process again (needed anyway to verify the answers),

and find that ID1 = 0.101 mA. Observe that this current is very close to the value used earlier. The value of

VS1 is also expected to be approximately the same as before. Therefore, VDS1 . 2.208 V. Besides, the

previously estimated value Vt1 can be used to find the better estimate of VS1 using the following simple

equation:


615

VS1 ' 2.7 & Vt1&ID1

K1 (1%λn VSD1 )'1.16 V, ID2 . 0.206 mA, and VD2 ' 1.717 V.

ID1 ' 0.101 mA, ID2 ' 0.206 mA, VDS1 ' 2.18 V, VS1'1.16 V, and VSD2 ' 3.283 V.

gm1'0.364 mS, ro1'338.7 kΩ , gmb1'0.0644 mS, gm2'0.4668 mS, and ro2'127.6 kΩ .

rs1'1

gm1%gmb1%1/ro1

'2.318 kΩ , and (ro12ro3 )'141.9 kΩ .

R1 '1

g11F%1/ (rs12Rs )'1.825 kΩ , and R2 ' (ro12ro3 )'141.9 kΩ .

Vdg1 'ViA

rs1

R2 ' 61.22ViA , and Is 'ViA

R1

&Vdg1

ro1

' (0.3672 mS) ViA .

ViA ' (2.723 kΩ ) Is , YiA % Gs ' 0.3672 mS, and Vdg1 ' (166.7 kΩ ) Is .

g21F ViA % g22F IoA % ( IoA % gm2 Vdg1 ) ro2 ' 0.

IoA '& (gm2 ro2Vdg1 % g21F ViA )

ro2 % (RL2Rf )Y IoA ' &73.05 Is Y A '

IoA

Is

' &73.05 A/A.

Repeating the calculations with this new current, we find that VSG2 = 1.66 V and VD1 = 3.34 V. Since all the

values calculated during the second iteration are close to the values calculated during the first iteration, we

take the values obtained during the second iteration as the correct values. Thus,

One can verify that all the MOSFETs operate in the pinch-off mode. The small-signal parameters of M1 and

M2 can be found to be

The small-signal equivalent circuit of the whole amplifier is shown in Fig. 8.6.5, where

After connecting the feedback network and suppressing the feedback, the modified small-signal equivalent

circuit is shown in Fig. 8.6.6, where

To find the forward-path gain, observe that (Vdg1/ro1) is an internal feedback in the CG-amplifier. This

Shunt-Shunt feedback is not used by the output variable but the amplifier’s internal node voltage. Although,

one can use the two-step feedback analysis, it is not necessary now because of the simplicity of the circuit.

One can find the parameters A, (YiA + Gs), and (ZoA + RL) directly from the circuit. Using Ohm’s law and KCL,

we get

Solving the above,

Using KVL at the right most loop,

Solving the above equation, the output current is

After suppressing the input, we find that


616

ZoA%ZL ' ro2%g22F ' 135.9 kΩ .

Ac'A

1 % AF' &5.545 A/A, Yic ' (YiA%Gs ) (A /Ac )&Gs ' 4.738 mS, Ric ' 211.1 Ω ,

Zoc ' (ZoA%ZL ) (A /Ac )&ZL ' 1.791 MΩ .

ID1 ' ID3 ' 101 µA, and ID2 ' 204.17 µA,VSG3 ' 1.416 V, VSD3 ' VSG2 ' 1.659 V, VDG1 ' 0.641 V > & Vt1 ' 0.988 V,

andVGD2 ' 1.447 V > & Vt2 ' 0.7776 V.

Fig. 8.6.5: The complete small-signal equivalent circuit of the amplifier in Example 8.6.

Is Rs

Zic

vdg1

ro1

Ii

(ro12ro3)+

vdg1

-

gm2vdg1 ro2

RL

Io

RfIf

Zoc

rs1

vsg1

rs1

Fig. 8.6.6: The complete small-signal equivalent circuit of the A-network in Example 8.6.

Is

Vdg1

ro1

+Vdg1

-gm2Vdg1 ro2

IoA

+

-ViA

+-

g21FViAg22F

ViA

rs1

YiA + Gs ZoA + RL

R1 R2

The contribution of (g21FViA) (the contribution of the feedback network) in the above forward-path gain is

negligible again proving the point that *g21A* » *g21F*. Clearly then, we could neglect the influence of g21F.

In any case, proceeding further, the closed-loop gain is

and

The PSPICE simulation provided the following dc bias values.

Although the value of ID1 is the same as the one estimated earlier, the value of ID2 slightly differs. This is due

to the reason that we neglected the forward transmission during the dc analysis. PSPICE simulation also

showed that Ac = -5.541 A/A, which is very close to the value calculated earlier.


617

8.7: CONCLUSIONSWe have completed the study of all four feedback configurations along with analysis/design examples

in each type. The feedback concept simplifies the analysis and design processes of these amplifiers. In the

practical situations, one can start with F . (1/Ac), design the feedback network accordingly for a given closed-

loop gain, and then adjust the components in the feedback circuit to obtain the desired gain. The various

formulas to find the different quantities in all four configurations can be obtained from the unified analysis

of Section 8.3. The analysis process of the amplifiers is also simplified with the use of feedback concept. By

suppressing the feedback, the circuit to find the forward-path gain usually becomes a unilateral one, and that

makes it easy to find the forward-path gain. This is an enormous advantage in analyzing amplifier circuits

to find the transfer functions in the next chapter where the capacitances couple the input and the output

causing reverse transmission.

PROBLEMS

SECTION 8.18.1. A feedback amplifier has an open-loop gain of A = 1000 and a feedback factor of F = 0.1. Find the

KP'51.169 µA/V 2 , Vto'0.7339 V, λ'0.03122V &1 .

KP'16.528 µA/V 2 , Vto'&0.7776 V, λ'0.04349 V &1 .

WL 1

'882

, WL 2

'92

, WL 3

'272

2, W

L 4

'272

.

ID1'599.6 µA, ID3'316.5 µA, ID2'60.67 µA, ID4'24.41 µA,VD1'&3.193 V, VG1'&3.556 V, and VG3'3.767 V.

Fig. E8.10.

io

RL

+5 V

M3

M1

iD3

iD1

M4

M2

RF

-5 V

ii

RBIAS

Exercise

E8.10. In the MOSFET amplifier of Fig. E8.10,

process parameters of the NMOS FET are


The aspect ratios of the MOSFETs are

The circuit parameters are RF = 10 kΩ, RBIAS = 300 kΩ,

and RL = 10 kΩ. The dc bias currents and voltages are:

Find the values of A, Ac, Ric, and Roc.

Answers: A = -6.45 A/A, Ac = -1.527 A/A, Ric = 1.056 kΩ, and Zoc = 167.6 kΩ.


618

Fig. P8.3.

RicIi

IsRi

+

-

Ro

AIi

A-networkIo

Roc

+

-Vo

10k

+

-VoFVo

F-network

Fig. P8.4.

+

F

Xi +

-

+

F

Xo+

-

(a)

A1 A2

(b)

+

F

Xi +

-

XoA1 A2

closed-loop gain Ac. If the open-loop gain changes to 900, what will be the percentage change in the

closed-loop gain?

8.2. If a feedback amplifier has an open-loop gain of A = 10,000 and a closed-loop gain of Ac = 10, what

is the feedback factor F? If the feedback factor decreases by 10%, find the percentage change in the

closed-loop gain.

8.3. In the network shown in Fig. P8.3, Ri = Ro= 100 Ω, A = 1 kΩ, and F = 10 mS. Find the values of Ac

= (Vo/Is), Ric, and Roc.


619

A (s ) ' 103 s(s % 100)(1 % 10&3 s )

.

Fig. P8.8.

Output Voltage

Input Voltage

5 V

10 V12 V

0.5 V 1.0 V

(0.8 V,12 V)

(0.7 V,1 V)-0.5 V-1.0 V

(-0.8 V,-12 V)

(-0.7 V,-1 V)

D8.4. It is required to design an amplifier with a closed-loop gain of 100.Two alternative schemes are

shown in Fig. P8.4. Assume that A1 = A2 = 1000.

(a) Find the value of F required in each case.

(b) If both A1 and A2 decrease by 10%, find the percentage change in the closed-loop gain in each

case. Comment on your results.

D8.5. It is required to design an amplifier represented by the block diagram of Fig. 8.1.3 with a closed-loop

gain of 100. If A changes by 10%, the closed-loop gain should not change by more than 0.01%.

Determine the values of the amplifier gain and the feedback factor. If the amplifier gain decreases

by 50%, what is the expected change in the closed-loop gain?

8.6. An amplifier has a gain of

If the feedback factor F = 0.1 is employed, what are the values of the mid-band gain and the lower

and upper 3-dB frequencies of the closed-loop gain?

D8.7. A feedback amplifier is built with an amplifier having a dc gain of 104 to obtain a closed-loop dc gain

of 10. What is the required F? Assume that A(s) has a one-pole roll-off. If the closed-loop gain should

have a bandwidth of 100 kHz, what should be the minimum bandwidth of the basic amplifier?

8.8. An amplifier A has a transfer characteristic shown in Fig. P8.8. If F = 0.1, find the transfer

characteristic of the closed-loop gain. Also, find the closed-loop gain for small signals in each

operating region.

D8.9. Consider the feedback amplifier configuration of Fig. P8.9. The output stage is a power amplifier


620

Fig. P8.13.(a)

R1 R2

R3

V1

+

-V2

+

-

I1 I2

(b)

R1 R2

R3V1

+

-V2

+

-

I1 I2

Fig. P8.9.

Vi+

F

+

-

VoA 1

with a voltage gain of 1, and A is a pre-amplifier. When a 10-kHz signal is applied at the input of the

power amplifier, its output has 1-V 10-kHz signal contaminated by a power supply hum at 120-Hz

having an amplitude of 1 V. Design the closed-loop amplifier using the pre-amplifier A and the

feedback network so that the closed-loop gain is approximately unity while the hum is reduced to

at the output.10 mV

SECTION 8.28.10. In the series-shunt feedback amplifier of Fig. 8.2.1(a), the basic amplifier is ideal (an ideal VCVS).

If (Vo/Ve) = 100 V/V and (Vf/Vo) = 0.09 V/V, find the closed-loop gain. If Vs = 100 mV, what are the

values of the output signal, feedback signal, and error signal (the input to the amplifier)?

8.11. In a series-series feedback amplifier of Fig. 8.2.1(b) using an ideal amplifier (ideal VCCS) in its

forward-path, Vs = 1 mV, Vf = 0.9 mV, and Io = 1 mA. What are the values of A, F, and Ac? What are

their units?

8.12. A shunt-series feedback amplifier, represented by Fig. 8.2.1(d), uses an ideal current amplifier (ideal

CCCS). If Is = 1 mA, If = and Io = 100 mA, what are the values of A and F?0.9 mA,

8.13. The most general resistive feedback circuits that can be used in any of the four feedback

configurations are shown in Fig. P8.13. Find the z-, y-, h-, and g-parameters of both networks.

SECTION 8.3


621

A (s ) ' 10001%10&4 s

.

KP'51.169 µA/V 2 , Vto'0.7339 V, λ'0.03122 V &1 , γ'0.4823 V 1/2 , and (2φf )'0.7 V.

KP'16.528 µA/V 2 , Vto'&0.7776 V, λ'0.04349 V &1 , γ'0.6727 V 1/2 , and (2φf )'0.7 V.

VD1 ' 3.59 V, VD2 ' 2.47 V, VS ' &1.804 V, VS5 ' 60.2 mV,ID1 ' 48.9 µA, ID2 ' 51.1 µA, and ID5 ' 1.002 mA.

Fig. P8.17.

+-

+Vo

+15 V

Roc∞

1 mA

Ric

2 kΩ

Vs

Q1

20 kΩ

Q2 9 kΩ

1 kΩ 2 mA

Q3

50 kΩ

Fig. P8.16.

+

-Vo

+

-Vf 1 kΩ

9 kΩ

8.14. In the series-shunt amplifier shown in Fig. 8.2.1(a), the forward-path amplifier has a gain of

with an input resistance of 10 kΩ and an output resistance of 100 Ω. The F-networkA ' 1000 V/V

is an ideal VCVS with F = 0.1. If Rs = RL = 1 kΩ, find the voltage gain, the input impedance, and

output impedance of the closed-loop amplifier.

8.15. Repeat Problem 8.14, if the amplifier has a frequency dependent gain of

Also, find the dc gain and the bandwidth of the closed-loop amplifier.

8.16. Repeat Problem 8.15, if the F-network is as shown in Fig. P8.16.

8.17. Consider the amplifier shown in Fig. P8.17. Assume that β = 200 for the transistors. It uses a series-

shunt feedback formed by the 9-kΩ and 1-kΩ resistors. Use the feedback concept and find the values

of A, F, Ac, Ric, and Roc in the mid-frequency range. Neglect the effects of the base-bias currents.

8.18. The amplifier shown in Fig. P8.18 uses a series-shunt feedback. The process parameters of the

NMOS FET are


The dc analysis provided

Verify if the MOSFETs operate in the pinch-off mode. If they do, use the feedback concept and find

the closed-loop voltage gain (Vo/Vs) and the output resistance Roc.


622

Io

Vs

. 1RE

.

Fig. P8.18.

-

+Vo

+5 V

-5 V

0.1 mA

M1 M2

M3 M4

Vs+-

+VS

Roc

10 kΩ

10 kΩ 20 kΩ

1 mA

M5

(30/5) (30/5)

(70/5) (70/5)(200/5)

SECTION 8.48.19. A series-series feedback amplifier shown in Fig. 8.4.1 uses a basic amplifier with a gain of (Io/Ve) =

100 mS (transconductance). The basic amplifier has a finite input resistance of 10 kΩ and an output

resistance of 100 kΩ. Rs = RL = 1 kΩ. Assume that the feedback network is an ideal CCVS (i.e., z11F

= z22F = z21F = 0) with F = 1 kΩ. Find the values of Ac, Zic, and Zoc. Also, find the voltage gain (Vo/Vs).

8.20. An amplifier circuit, called the feedback-triple, with a series-series feedback is shown in Fig. P8.20.

The biasing details have been suppressed. The β for the transistors is 100. The dc bias currents are

IC1 = 0.6 mA, IC2 = 1 mA, and IC3 = 4 mA. Assume that ro 6 4 for both Q1 and Q2 but ro = 25 kΩ for

Q3. Obtain the values in Ac = (Io/Vs), Ric, and Roc.

D8.21. A transconductance amplifier is shown in Fig. P8.21 without the biasing details. Show that, if the op

amp is ideal,

(a) Design the circuit for a closed-loop gain of 1 mS.

(b) The op-amp has a finite gain of Ad = 105 with an input resistance of 100 kΩ and an output

resistance of 100 Ω. The BJT has β = 100 and VA = 100 V. The collector-bias current is 1 mA.

If RE = 1 kΩ, Rs = 1 kΩ, and ZL is a resistance of 10 kΩ, find the values of A, Ac, Zic, and Zoc.

Also, find (Vo/Vs).


623

Ad (s ) ' 105

1 % 10&2 s.

VD1 ' 3.59 V, VD2 ' 2.0 V, VS1 ' &1.802 V, VS5 ' 28.7 mV,ID1 ' 48.4 µA, ID2 ' 51.6 µA, and ID5 ' 128.7 µA.

Fig. P8.20.

+-

Io

Roc

Ric

Vs

Q1

100 Ω640 Ω

100 Ω

1 kΩ

Q15 kΩ

1 kΩ

Fig. P8.21.

+-

+

-Vs

Rs

Zic

Ad

Q

RE

Zoc Io +Vo

ZL

Fig. P8.23.

-

+Vo

+5 V

-5 V

0.1 mA

M1 M2

M3 M4

Vs+-

+VS1

Roc

1 kΩ

10 kΩ

10 kΩ

M5

(30/5) (30/5)

(70/5) (70/5)(100/5)

0.1 mA

+VS5

8.22. Repeat Problem 8.21, if all the parameters remain the same except that

Find the bandwidth of the closed-loop gain. If ZL is a capacitor having a capacitance value of 10 nF,

find the voltage gain (Vo/Vs).

8.23. The process parameters of the MOSFETs in the amplifier of Fig. P8.23 are the same as those in

Problem 8.18. The dc bias voltages and currents are


the closed-loop voltage gain (Vo/Vs) and the output resistance Roc.


624

s 2 % (ωn /Q )s % ω2n .

Fig. P8.27.

+-

+12 V

4.7 kΩ

RF

10 kΩ

Vs

+VoZic

Zoc

∞

Fig. P8.25.

1 kΩ

V1

+

-

I1 I2

V2

+

-

Fig. P8.26.

+-

+12 V

6.2 kΩ

10 kΩ

Vs

+VoZic

Zoc

∞

100 kΩ

SECTION 8.58.24. In the shunt-shunt feedback amplifier of Fig. 8.5.1, the feedback network is ideal with F = -1 mS.

That is, y11F = y22F = y21F = 0, and y12F = -1 mS. The A-network is a Transimpedance amplifier with

a gain of -100 kΩ under no-load condition. It has an input resistance of 1 kΩ and an output

resistance of Rs = 10 kΩ and RL = 1 kΩ. Find the values of Ac = VO/Is, Zic, and Zoc.100 Ω .

8.25. Repeat Problem 8.24, if the F-network is as shown in Fig. P8.25.

8.26. The MOSFET in the circuit of Fig. P8.26 has Vt = 2 V and K = 1 mA/V2. The amplifier uses the

Shunt-Shunt feedback formed by the 100-kΩ resistor. Find the gain (Vo/Vs), Zic, and Zoc in the

midband using the feedback concept.

D8.27. Consider the amplifier shown in Fig. P8.27 that uses a shunt-shunt feedback. It is required to design

this amplifier for (Vo/Vs) = -5 ± 5% V/V in the midband. Using the feedback concept, select a

standard value for RF. The BJT has a nominal value of β = 200. Verify analytically if your design

meets the specification. Determine the input and output resistances of the closed-loop amplifier in

the midband. Assume that resistors have 10% tolerance. The β value of the BJT can vary between

100 and 300. All the variable parameters have Gaussian distribution. Use Monte-Carlo simulation

of 5000 circuits and find the ‘expected’ mean, maximum, and minimum values for the voltage gain.

D8.28. The circuit shown in Fig. P8.28 uses the Shunt-Shunt feedback configuration. Assume that the op-

amps are ideal. The feedback resistor Rf samples the output voltage and provides the feedback current

If. The boxed-part is the forward path. The closed-loop trans-impedance (Vo/Is) has a second order

denominator polynomial in the form of

A particular application requires that Q = 10 and ωn = (20π) kr/s. Use the feedback concept and

determine the values of R1 and Rf to meet this requirement.


625

Fig. P8.30.

If Io

R2V1

+

-V2

+

-

R1

Fig. P8.28.

R12.461 kΩ

0.01 µF

Is

+Vo

Rf

A1

+

-

A2

+

- 10 kΩ

10 kΩ

0.01 µF

If

Fig. P8.33.

+-Vs

10 kΩRic

0.5 mARoc

RE

+12 V

+Vo

Q1

10 kΩ

1 kΩ

Q2

Fig. P8.32.

+

-

Is Rs

Ric

Io

r

Ad

Rf

Roc

RL

SECTION 8.68.29. The shunt-series feedback amplifier of Fig. 8.6.1 has its A-network (a current amplifier) with a short-

circuit current gain of (Io/Ie) = 1000 A/A, an input resistance of 1 kΩ and an output resistance of 10

kΩ. (Note: The output of the A-network should have Norton's form.) Rs = RL = 1 kΩ. If the feedback

network is ideal with F = (If/Io) = 0.1 A/A, find the values of Ac, Zic, and Zoc.

D8.30. Design a F-network to implement F = (If/Io) = -0.1 A/A using the network shown in Fig. P8.30.

8.31. Assume that the F-network in Problem 8.30 has R1 = 18 kΩ and R2 = 2 kΩ. It is used in the feedback

amplifier of Fig. 8.6.1. The forward-path amplifier has a short-circuit current gain of

The A-network has an input resistance of 1 kΩ and an output resistance of 10 kΩ.&1000 A/A.

Assume that Rs = RL = 10 kΩ. Find the values of Ac, Zic, and Zoc.

D8.32. Design the circuit shown in Fig. P8.32 to meet the specification of (Io/Is) = -10 A/A. Assume that the

op-amp is ideal and Rs = RL = 10 kΩ. If the op-amp is non-ideal having Ad = 103 V/V, Rid = 100 kΩ,

and Ro = 1 kΩ, find the values of Ac = Io/Is, Ric, and Roc.

D8.33. The circuit shown in Fig. P8.33 is to be designed for a voltage gain of (Vo/Vs) = 3 V/V. The BJTs

have β = 100 and VA = 100 V. Select a standard value for RE.


626

VD1 ' 4.742 V, VG1 ' 1.072 V, and VS2 ' 2.144 V.

Fig. P8.35.

+-Vs

10 kΩRic

20 µA

+5 V

M1

10 kΩ2 kΩ

M2

(30/5)(100/5)

Io

2 MΩ

8.34. RE = 330 Ω in the circuit of Fig. P8.33. Find the values of (Vo/Vs), Ric, and Roc.

8.35. The process parameters of the MOSFETs in the amplifier of Fig. P8.35 are the same as those in

Problem 8.18. The dc bias voltages are


the closed-loop voltage gain (Io/Vs) and the input resistance Ric.

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