solucionario capitulo 40 - paul e. tippens

8/2/2019 Solucionario Capitulo 40 - Paul E. Tippens

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Chapter 40. Electronics Physics, 6th Editon.

Chapter 40. Electronics

Transistors and Applications

41-1. Given an NPN transistor with a common-base connection and α = 0.98, determine the base

current and the collector current when the emitter current is 40 mA.

I b = I e(1 - α ) = (40 mA)(1 – 0.98); I b = 0.800 mA

I c = I e – I b = 40 mA – 0.800 mA; I e = 39.2 mA

41-2 If the base current of a transistor with a common-base connection is 1.6 mA and the emitter

current is 60 mA, what is α?

I b = 1.6 mA; I e = 60 mA; I c = 60 mA – 1.6 mA = 58.4 mA

58.4 mA;

60 mA

c

e

I

I α = = α = 0.973

41-3. For a common-base amplifier, the input resistance is 800 Ω and the output resistance is 600

k Ω. (a) Determine the voltage gain if the emitter current is 12 mA and α = 0.97. (b)

What is the power gain?

(a) V in = I e Rin = (12 x 10-3 A)(800 Ω); V in = 9.60 V To find V out , we need I c.;

I c = α I e = 0.97(12 x 10-3 A) = 0.0116 A; V out = I c Rout = (0.0116 A)(600,000 Ω)

V out = 6980 V;6980 V

;9.6 V

out

in

V A

V = = Av = 727

(b) G = α Av = 0.97(727); G = 705

41-4. The power gain is 800 for a common-base amplifier, and the voltage amplification factor is

840. Determine the collector current when the base current is 1.2 mA.

8000.952;

840v

G

Aα = = = I c = α I e = (0.952)(1.2 mA); I c = 1.14 mA

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40-5. Determine the current gain in a common-emitter amplifier circuit when α = 0.98.

0.98;

1 1 0.98

α β

α = =

− − β = 49

Supplementary Problems

40-6. In the previous problem, what is the collector current if the emitter current is 20 µA? What

is the base current?

;c

e

I

I α = I c = 0.98(20 µA); I c = 19.6 µA

40-7. A transistor with an effective input resistance of 400 Ω, an effective output resistance of 900

k Ω, and α = 0.96 is connected in a common-base circuit. (a) What is the voltage gain when

the input current I e is 8 µA? (b) What is the power gain?

V in = I e Rin = (8 µA)(400 Ω) = 3.2 mV; I c = (0.96)(8 µA) = 7.68 µA

V out = I c Rout = (7.68 µA)(900,000 Ω) = 6.912 V; -3

6.912 V;

3.2 x 10 Vv A = Av = 2160

G = α Av = (0.96)(2160); G = 2074

40-8. Calculate I c and I b for the conditions described in Problem 40-7.

I c = α I e = 0.96(8 µA) = 7.68 mA; I b = 8 µA – 7.68 µA = 0.320 µA

40-9. For a transistor with I e = 8 µA and α = 0.97 connected with a common emitter, calculate β ,

I in , and I out .

0.97;

1 1 0.97

α β

α = =

− − β = 32.3

I c = 0.97(8 µA) = 7.76 µA ; I b = (8 µA – 7.76 µA) = 0.240 µA

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40-10. For a transistor with α = 0.99 connected with a common collector, calculate the current

gain Ai.

1 1;

(1 ) 1 1 0.99

out ci i

in c

I I A A

I I α α = = = =

− − −= 100

40-11. A transistor has β = 99; what is the value of α?

; ;1

α β β βα α α αβ β

α = − = + =

−

990.99;

1 1 99

β α

β = = =

+ + α = 0.99

40-12. For the transistor of Problem 40-11, if I b = 0.10 mA, what are the values of I e and I c?

I b = I e(1 - α ); I b = 0.10 mA; I e - I eα = 0.10 mA;

0.10 mA;

1 1 0.99

be

I I

α = =

− −I e = 10 mA

;c

e

I

I α = I c = α I e = 0.99(10 mA); I c = 9.9 mA

Critical Thinking Problems

40-13. It is known that zener breakdown and avalanche breakdown are affected oppositely by

temperature. From this fact, can you determine a good voltage rating for a zener diode

when temperature stability is important?

There is a transition region between 5 and 6 volts where the breakdown mechanism is a

combination of both zener and avalanche. This voltage range is best for temperature

stability since the temperature effects are opposite and tend to cancel.

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40-14. A lab worker is directed to build the circuit shown in Fig. 40-38, but connects the diode as

shown below. Can you predict how the circuit will work?

This configuration is often used for voltage clipping or

bilateral voltage regulation. Properly connected, the

configuration will turn on at the zener voltage plus 0.7 volts. Thus, with two 10 volt

zeners, we would have turn-on voltages of +10.7 volts and -10.7 volts. If one diode is

reversed, then the symmetry is lost. Using the same zeners, the turn-on thresholds would

be 20 volts and 1.4 volts.

40-15. Incandescent lamps have a large current flow when first turned on and then the current

drops as they heat up. Thus, they often burn out at the moment they are turned on. Is there

a solution based on the information covered in this chapter?

Conductors have a positive temperature coefficient and semiconductors have a negative

temperature coefficient. Thus, one solution is to place a semiconductor in series with the

lamp. At turn-on, the semiconductor will be cool, have a higher resistance, and limit the

surge current. After a time, the temperature of the semiconductor will go up, it's resistance

will drop and allow normal lamp operation.

40-16. Incandescent lamps produce a lot of heat and are thus rather inefficient. Why can't they be

replaced with LEDs that are more efficient?

LEDs are monochromatic (they emit one color). Incandescent lamps produce mostly

white light which is a mixture of all the visible colors of the spectrum. It might be

feasible (economically) to combine red, green and blue LED light to make white light

at some time in the future.

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40-17. Write an equation that describes the current gain from base to emitter in terms of B and

identify the amplifier configuration where it would be most useful.

1 e

b

I

I

β + =

The common collector configuration.

40-18. Suppose there is an application where the common-emitter amplifier would serve best but

the phase reversal would not be acceptable. Can you think of a solution?

A second common-emitter amplifier will invert the phase again and the overall result is no

phase change.

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solucionario capitulo 40 - paul e. tippens

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