Electrical Engineering Chapter 14 Homework From the previous three equations, we obtain

CHAPTER 14

Exercises

E14.1

(a)

A

AR

v

i



B

BR

v

i



B

A

B

A

FR

v

R

v

iii



E14.2 (a)

1

(b)

E14.3

E14.4 (a)

in

(b)

(Note: We assume that

.

321

RRR



)

in

E14.5

E14.6 (a)

Using the above equations we eventually find that

E14.7 We have

2

R

A

vs



from which we conclude that

E14.8

Applying basic circuit principles, we obtain:

E14.9 Many correct answers exist. A good solution is the circuit of Figure 14.11

E14.11 Many correct selections of component values can be found that meet the

desired specifications. One possibility is the circuit of Figure 14.19 with:

E14.12

kHz 40

100

40105

0







CL

BOLOL

CL

t

BCL A

fA

A

f

The corresponding Bode plot is

E14.13 (a)

kHz 9.198

105

SR 6



 ππ

FP V

f

(c) With a load of 100  the current limit is reached when the output

(d) In deriving the full-power bandwidth we obtained the equation:

(e) Because the output,

E14.14 (a)

(b)

Applying basic circuit principles, algebra, and the summing-point

restraint, we have

B

bias

yx IRvv



BB

bias

xI

RR

R

I

R

v

i

2

1



(c)

(d)

off

I

off

2

off

I

R

I

R

v

x

E14.15 (a)

Because of the summing-point constraint, no current flows through

R

bias

so the voltage across it is zero. Because the currents through

R

1 and

R

2

E14.16

Because no current flows into the op-amp input terminals, we can use the

voltage division principle to write

E14.17 (a)

 

tt

odttvdttv

RC

tv

0

in

0

in )(1000)(

1

)(

(b) A peak-to-peak amplitude of 2 V implies a peak amplitude of 1 V. The

E14.18 The circuit with the input source set to zero and including the bias

current sources is:

E14.19

E14.20

E14.21 The transfer function in decibels is





E14.22 Three stages each like that of Figure 14.40 must be cascaded. From

Problems

P14.1 The most probable functions of the five op amp terminals are the

P14.2 An ideal operational amplifier has the following characteristics:

P14.3 The open-loop gain is the voltage gain of the op amp for the differential

P14.4 The differential voltage is:

P14.6* The steps in analysis of an amplifier containing an ideal op amp are:

P14.7 The inverting amplifier configuration is shown in Figure 14.4 in the text.

P14.8 According to the summing-point constraint, the output voltage of an op

P14.9* The circuit has negative feedback so we can employ the summing-point

v

P14.10 This is an inverting amplifier having a voltage gain given by

in(

P14.11 Because of the summing-point constraint, the voltages across the

resistors of values

R

and 2

R

are equal. Thus, the current in the resistor

of value

R

is twice that of the current in the 2

R

resistor as indicated:

in

vv

P14.12 Using the summing-point constraint, we have

P14.13 Using the summing-point constraint, we have

P14.14 This is an inverting amplifier with a voltage gain of 2. Thus, we have

P14.15 Using the summing-point constraint, we have

P14.16 This circuit has positive feedback and the output can be either +5 V or

P14.17* If the source has non-zero series impedance, loading (reduction in

P14.18* The circuit diagram of the voltage follower is:

P14.19* The circuit diagram is:

Writing a current equation at the noninverting input, we have

P14.20*

P14.21 The noninverting amplifier configuration is shown in Figure 14.11 in the

text.

P14.22 (a)

 

V 4mA 2k 2 

o

v

(b)

V 2503 

o

v

(d)

0

o

v

P14.23 Analysis of the circuit using the summing-point constraint yields

P14.24 (a) Working from left to right, we find the currents and voltages as

shown:

Then, we have