ee321 ch9 solutions

G. Rizzoni, Principles and Applications of Electrical Engineering, 5th Edition Problem solutions, Chapter 9

Section 9.3: Circuit Models for the Semiconductor Diode

Problem 9.7

Solution:

Known quantities: The circuit of Figure P9.7.

Find:

Determine the diode is conducting or not.

Analysis: Assuming the diode is conducting, the current is found to be

y B - v -

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511 - v» +• 10 U -AWV M-

v. - 12 V V. z 10 V

]< V I t f

/ = -5+10

The voltage across the diode is

= - \>f f + 1 0 - 0 . 6 6 7 - 1 2 = - 3 . 9 9 V ]

This result contradicts the assumption, since the diode cannot conduct if V q is negative. Thus, the diode must be

off.


Problem 9.9

Solution: Copvr^«TlwMoG»»»w>WCoiTi(wiwMi mc P»mfeM>fi r»quk»d tor rtprodixaion or<fc*pt«y

Known quantities: The circuit of figure of P9.9.

Find: - V ' = , , ,V

Determine whether the diode is conducting or not.

Analysis: The Thevenin equivalent resistance seen by the diode is RT = 12.5Q . The Thevenin equivalent voltage is:

| x 5 - K r = 3.5F K-+ vc 10

Assuming the diode is conducting,

12.5

This contradicts the assumption. Thus the diode is off.

G. Rizzoni, Principles and Applications of Electrical Engineering, 5lh Edition Problem solutions, Chapter 9

Problem 9.13

Solution:

Known quantities: The circuit of f igure of P9.13.

Find: Sketch iD(i).

Analysis: (a) The diode c u r r e n t , is

i D = 100 mA for I < 10 ms

/ ' D = 0 m A for 10 < / < 20 ms

i j) = 100 mA for 20 < / < 30 ms

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100

0 10 20 30 , , l n S ,

(mA)

10 20 30 t (ms)

(b) The diode current /£>(/), is

1 0 - 0 . 6 <D 100

• 94 mA for / < 10 ms

' 0 = 0 mA for 10 < / < 20 ms

iD = 94 mA for 20 < / < 30 ms

'D (mA)'"

94

10 I

_J 20 30 t (ms)

(c) The iD(t) is

Vs - \\ >D=-

9.4

100 + id 1100 i c = 0 m A for 10 < f < 20 ms

iD= 8.55 mA for 20 < / < 30 ms

= 8.55 mA for / < 10 ms

<D (mAjf

8.55

10 20 30 t (ms)

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G. Rizzoni, Principles and Applications of Electrical Engineering, 5lh Edition Problem solutions, Chapter 9

Problem 9.14

Solution:


Find:

The range of Vjn for a forward-biased D| .

Analysis: Assume diode D\ is conducting; the diode current is

V - 2 1500

Since the current is positive, the initial assumption was

correct. Therefore, the range of Vm is: Vm > 7V .

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2 V

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> 0

Problem 9.15

Solution:


Find: Show iD is constant; Write an equation for

VD (T° ) based on figure P9.15(a).

Analysis:

(a) The current i p is

For VD = 0.6V, iD = 1.440mA

For VD = 0.66V , iD =1 .434mA

The percent change in iD is only 0.4%. Thus, iD

nearly constant.

(b) The diode voltage equation is

Vd{T) = -0.002T + 0.8

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Problem 9.16

Solution:

Known quantities: The circuit of Figure P9.16.

Find:

A plot of vL versus v<,-.

Analysis: For v s < 0 , the diode is reverse biased, and vL = 0 . For

f vs > 0 , the diode is forward biased, and vL = vs -—-— \RS + rL\\rI

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Where a = tan" - i f RL\\R\ Rc + R, II R>

VL

a

vs


Problem 9.21

Solution:

Known quantities: The configurations shown in Figure P9.21.

Find: Which diode are forward biased, and which are reverse biased.

Analysis: a) reverse-biased b) forward-biased c) reverse-biased d) forward-biased e) forward-biased

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+5 vo 44—VM-

c (a)

O—VM N--IOV k

6 - 5 V

I

<b)

© (4-12 V

i d )

- 5 V

-10 V

Problem 9.22

Solution:

Known quantities: The configuration shown in Figure P9.22.

Find: The range of Vin for which D, is forward-biased.

Analysis: The diode D, is clearly forward-biased for any Vm> 0.

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O VAV-F - = j r - 1.5 V

.SOOU

Problem 9.24

Solution:

Known quantities:

The circuit ofFigure P9.24; v s ( f )= 10sin(2,000sr).

Find: The output waveform and the voltage transfer characteristic. Assumptions: The diode is ideal.

Analysis: For v.v<8 V, v„=4 V. For v4 > 8 V, v0= vs 12. The voltage transfer characteristic is

i

8

k i

8

• y

n r

8 16

n n nnr rT i m i Trrir i i t-«. . \ 4 . ...... 11:11 p .

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V 0

(V)

-5 4 I T v I T Y t \

0.5 1 1.5 t(ms)

9.15


Problem 9.35

Solution:

Known quantities: A half-wave rectifier is to provide an average voltage of 50 V at its output.

Find: a) Draw a schematic diagram of the circuit. b) Sketch the output voltage waveshape. c) Determine the peak value of the input voltage. d) Sketch the input voltage waveshape. e) The rms voltage at the input.

Analysis:

a)

+ " v;„ ( •'V ) R vou,

<

b)

/ /

c) vme = 0 .318v p e a t = 50 =J> vpeak = 157.2 V

v„

_ 157.2 = r i

= 111.2 V

9.24 P R O P R I E T A R Y MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


Problem 9.36

Solution:

Known quantities: A rectifier circuit similar to that of Figure 9.25 in the text. The load resistance is 100 Q, the AC source voltage is 30 V (rms).

Find: The peak and average current in the load.

Assumptions: Ideal diode.

Analysis: The peak voltage is

J/p i n VR = 3 0 V 2 V = > / « = — ^ i - = i - ^ - = 0 .424A

*°k R 100 The average current in the load is

/ IR = - ^ = 0.135A

M 71

Problem 9.37

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Section 9.5: DC Power Supplies, Zener Diodes and Voltage Regulations

Problem 9.49

Solution:

Known quantities: Figure P9.49. The piecewise characteristic that passes through the points (-10 V, -5 |iA), (0, 0), (0.5 V, 5 raA) and (1 V, 50 mA).

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2 V

Find: Determine the piecewise linear model, and, using that model, solve for i and v.

Analysis: Assume that the diode is forward-biased, and operating in the region between (0.5V, 5mA) and (IV, 50mA). If this is true, then the diode can be modeled by the resistance

_ 1 - 0 . 5 0.5 = 11.11 Q in series with a battery having value

I !Lo « — | — — AiD ( 50 -5 )10" 3 4510 ' 3

= 0.5-510 _ 3 (11.11)= 0.444 V. 9 _ 0 444

Then, i= =14 mA and v = 0.444+ 0.014(111 0 = 0.6 V. 111.11 v '

This solution is within the range initially assumed, justifying the assumption.

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1.800 a - W A -

18 V

Problem 9.50

Solution:

Known quantities: Figure P9.50. The output voltage at 5.6 V.

Find: Determine the minimum value of RL for which the output voltage remains at just 5.6 V.

Analysis: r l , X =22 ( 1 8 ) = 5 . 6 = > 12.4K, = 10080 => K, = 8 1 2 . 9 O

R, +1800 ™

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A minimum and maximum Zener current is caused by [respectively]: Minimum source voltage Maximum source voltage Maximum load current Minimum load current Maximum Zener voltage Minimum Zener voltage

Substituting with these extreme values and solving for the load current: ; _ min ~ ^ Z - m a x ~ ^ Z - m i n ^ Z r — 1 ft 7Q m A ' L - m a x ' Z - m i n ^ m A

/ — ^S-max ~~ ^Z-miu " 'I-inin

Problem 9.59

Solution:

Known quantities: The circuit of Figure P9.59; the desired load voltage is 14V; The Zener diode is rated at 14V, 5W.

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Find:

The range of load resistance for regulation.

Analysis:

ys = 50 V , Zener is rated Vz = 14 V, 5 W

The minimum load resistance can be calculated by assuming

that all the source current goes to the load, and that the load voltage is regulated at the nominal value:

.W v

j — A W . °—VAV

R, = 14

is (ys~vz)l30 36 /30 = 11.7 n

The maximum current through the Zener diode that does not exceed the diode power rating is P? 5

i z = — = 0.357 A 14

The source current is

30 30

So the maximum load resistance is V, 14

R = 16.6 n 'Smax —'Zmax 0 8 4 3

Thus, the range of allowable load resistance:

11.7 O Z RL <16 .6 n

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Problem 9.60

Solution:

Known quantities: The circuit of Figure P9.60(b); Zener diode i-v character; VZ=0.77V;

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Find:

The output voltage Voul.

Analysis: (a) The voltage across the diode is

50 + 50 VZ=VS- = 6 < 7.7 V

100 + 50 + 50

Therefore, the Zener diode is off. Thus, the output voltage is:

Kut= 6 — = 3 V out 100

(b) The voltage across the diode is

K7 = 2 0 — = 10 > 7.7 V 100

In this case, the Zener diode is on and the output voltage is:

Kui = 7 . 7 — = 3.85 V out 100

I

V?

0 VD

(a)

10012 -WvV-

Vv

H

( b )

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