模电双语(英语教材)习题及解答[第三版]

模电英语教材习题及解答[第三版]

Chapter 1

Problem 1.43

The cut-in voltage of the diode shown in the circuit is V =0.7V. The diode is to remain biased “on” for a power VPS

D

supply voltage in the range 5V VPS 10V. The minimum diode current is to be ID(min)=2mA. The maximum power dissipation in the diode is to be no more than 10mW. Determine approximate values of R1 and R2. (Fig. P1.43)

Solution:

Current relations in the circuit can be written as follows as long as the diode is “on”

I1 I2 ID

V R ID

2

;

The maximum diode current is

IPDmax10mW

Dmax

V 7V

14.3mA 0.

Using Kirchhoff’s Current Law, we obtain

V I1max I2 IDmax

R 14.3mA 0.7V

14.3mA(1.43.1) 2R2 IIV

.7V1.43.2)

1min I2 Dmin R 2mA 0 2mA(2R2

Using Kirchhoff’s Voltage Law and express the equation in terms of I1, we obtain

VPSmax V 10V 0 I1max

R .7V 9.3V(1.43.3) 1R1R V V1 5V 0.7V4.3V

Imin 1min PSR (1.43.4)

1R1R1

Equating I1max and I1min from the above expressions, we obtain

9.3V0. 7V

14.3mA(1.43.5) R 41R.3V02

R .7V(1.43.6)

1

R 2mA2

Solving from (1.43.5) and (1.43.6) for R1 and R2, we have

R1 0.407k

模电英语教材习题及解答[第三版]

R2 0.0819k

Problem 1.53

Consider the Zener diode circuit shown in Fig, P1.41. The Zener breakdown voltage is Vz=5.6V at Iz=0.1mA, and the incremental Zener resistance is rZ=10 .

(a) Determine Vo with no load (RL= ). (b) Find the change in the output voltage

if VPS changes by 1V.

(c) Find Vo if VPS=10V and RL=2k .

VPS=10V

o

(Fig. P1.53)

Solution:

(a) From the circuit we can see that the zener diode in the circuit is reverse biased, so it operates in the “reverse breakdown” region. R=0.5k

We can use the equivalent circuit to replace the zener diode in the circuit as shown in Fig.S1.53(a) Vo

Since VZ

Iz 0.1mA

IZ rZ V'Z 5.6V

VPS=10Vso, we have

V 5.6V 0.1mA 10 5.599V Applying KVL around the loop for the circuit with RL= , we get

'

Z

Fig.S1.53(a)

VPS VZ'10V 5.599V

Vo IZ rZ V rZ VZ' 10 5.599V 5.685V

R rZ0.5k 10

'

Z

(b) Vomax

VPS(max) VZ'

R rZVPS(min) VZ'

R rZ

rZ VZ

11V 5.599V

10 5.599V 5.706V

0.5k 10

9V 5.599V

10 5.599V 5.667V

0.5k 10

RTH=9.8

Vomin rZ VZ

The output voltage change is between Vomax and Vomin.

Vo Vomax Vomin 5.706V 5.667V 0.039V

(c) Using Thevenin equivalent circuit, we have

VTH=5.686V

o

模电英语教材习题及解答[第三版]

the circuit and parameters as shown in Fig.S1.53(b), where

RTH R//rZ 0.5k//10 9.8

VPS VZ

TH

VR rrZ VZ

Z

10V 5.6V

0.5k 10

10 5.6V 5.686V

The output voltage Vo is

VVTH5.686V

o

R RRL 2k 5.658V

THL0.5k//10 2k

模电英语教材习题及解答[第三版]

Chapter 2

Problem 2.26

For the circuit shown, let V =0.7V and assume the input voltage varies over the range -10V vI +10V, Plot (a) vO versus vI

(b) iD versus vI over the input voltage range indicated

vvO

Solution:

(a) When iD=0, vO

10V ( 10V)

R2 10V 3.3V

R1 R2

when vI < 4V, the diode D is “off”,

Fig. P2.26 vO

10V ( 10V)

R2 10V 3.3V

R1 R2

When vI > 3.3V+V =4V, the diode D is “on”,

vO vI

(b) When vI<3.3V, that is vO=4V, iD=0. When vO>4V, then

Fig. S2.26 (a)

iD i2 i1

vO 10V10V vO

R2R1

R1vO R2vO10V10V

R1R2R2R1(R1 R2)vO10V10V

R1R2R2R1

v3 vO 0.5mA20k 3 (vI V ) 0.5mA20k 33 0.7V vI 0.5mA20k 20k

vO

Fig. S2.26 (b)

Fig. S2.26 (c)

模电英语教材习题及解答[第三版]

The diode in the circuit has piecewise linear parameters V =0.7V, rf =10 , Plot

(a) vO versus vI for -30V vI 30V

(b) If the triangular wave is applied, plot output versus time.

v

O

DCSolution:

Fig. P2.27(a)

(a) for vO < VDC+V =10.7V, the diode in the circuit is “off”, so vO=vI.

for vO > VDC+V =10.7V, the diode in the circuit is “on”, Using KVL for the equivalent circuit in Fig.S2.27(a), we get

vO

vI V VDC

R rf

rf V VDC

110.7VvI 10.7V 1111

30V 10.7V

10.7V 12.45V

11

The output transfer characteristic is plotted in Fig.S2.27 (b)

if vI = 30V, then vO

R=100

vO vI

Fig.S2.27 (a)

(b) The output waveform versus time is plotted in Fig. S2.27(C)

模电英语教材习题及解答[第三版]

Sketch vO versus time for each circuit with the input shown in Fig. 2.32 Assume the capacitors in the circuit are initially uncharged, V =0 and the RC time constant is large.

+ +

vI

Fig.P2.32(a)

+ +

v I

Fig.P2.32(b)

Solution:

Fig.S2.32

Suppose the capacitor is initially uncharged, that is VC = 0V. For 0 t t1,

The diodes in Fig.P2.32(a) and in Fig.P2.32(b) are all reverse biased and the effective RC time constant in both circuit are large, the voltage across the capacitors do not change, and vo1 = vI, vo2 = vI.

For t1 t t2,

Both the diodes become forward biased; so vo1 = v =0V, vo2 = VB+V = 5V. At the same time, since the effective time constants are zero for both the circuits, the capacitors are charged instantaneously, the voltage across the capacitor in Fig.P2.32 (a) remains constant at VC = vI = 20V, and the voltage across the capacitor in Fig.P2.32 (b) remains constant at VC = V …… 此处隐藏：9856字，全部文档内容请下载后查看。喜欢就下载吧 ……