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Physics, 6th Editon.
Chapter 40. Electronics
Chapter 40. Electronics
Transistors and Applications
41-1. Given an NPN transistor with a common-base connection and  = 0.98, determine the
base current and the collector current when the emitter current is 40 mA.
Ib = Ie(1 - ) = (40 mA)(1 – 0.98);
Ib = 0.800 mA
Ic = Ie – Ib = 40 mA – 0.800 mA;
Ie = 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 ?
Ib = 1.6 mA;
Ic = 60 mA – 1.6 mA = 58.4 mA
Ie = 60 mA;

I c 58.4 mA

;
Ie
60 mA
 = 0.973
41-3. For a common-base amplifier, the input resistance is 800  and the output resistance is
600 k. (a) Determine the voltage gain if the emitter current is 12 mA and  = 0.97.
(b) What is the power gain?
(a) Vin = IeRin = (12 x 10-3 A)(800 );
Vin = 9.60 V To find Vout, we need Ic.;
Ic = Ie = 0.97(12 x 10-3 A) = 0.0116 A;
Vout = 6980 V;
(b)
G = Av = 0.97(727);
A
Vout = IcRout = (0.0116 A)(600,000 )
Vout 6980 V

;
Vin
9.6 V
Av = 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.

G 800

 0.952;
Av 840
Ic = Ie = (0.952)(1.2 mA);
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Ic = 1.14 mA
Physics, 6th Editon.
Chapter 40. Electronics
40-5. Determine the current gain in a common-emitter amplifier circuit when  = 0.98.


0.98

;
1   1  0.98
 = 49
Supplementary Problems
40-6. In the previous problem, what is the collector current if the emitter current is 20 A? What
is the base current?

Ic
;
Ie
Ic = 0.98(20 A);
Ic = 19.6 A
40-7. A transistor with an effective input resistance of 400 , an effective output resistance of
900 k, and  = 0.96 is connected in a common-base circuit. (a) What is the voltage gain
when the input current Ie is 8 A? (b) What is the power gain?
Vin = IeRin = (8 A)(400 ) = 3.2 mV;
Vout = IcRout = (7.68 A)(900,000 ) = 6.912 V;
G = Av = (0.96)(2160);
Ic = (0.96)(8 A) = 7.68 A
Av 
6.912 V
;
3.2 x 10-3 V
Av = 2160
G = 2074
40-8. Calculate Ic and Ib for the conditions described in Problem 40-7.
Ic = Ie = 0.96(8 A) = 7.68 mA;
Ib = 8 A – 7.68 A = 0.320 A
40-9. For a transistor with Ie = 8 A and  = 0.97 connected with a common emitter, calculate
, Iin , and Iout.


0.97

;
1   1  0.97
Ic = 0.97(8 A) = 7.76 A ;
 = 32.3
Ib = (8 A – 7.76 A) = 0.240 A
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Physics, 6th Editon.
Chapter 40. Electronics
40-10. For a transistor with  = 0.99 connected with a common collector, calculate the current
gain Ai.
Ai 
I out
Ic
1


;
I in I c (1   ) 1  
Ai 
1
= 100
1  0.99
40-11. A transistor has  = 99; what is the value of ?


;
1


1 

     ;
    
99
 0.99;
1  99
 = 0.99
40-12. For the transistor of Problem 40-11, if Ib = 0.10 mA, what are the values of Ie and Ic?
Ib = Ie(1 - );
Ie 

Ic
;
Ie
Ib = 0.10 mA;
Ib
0.10 mA

;
1   1  0.99
Ie - Ie = 0.10 mA;
Ie = 10 mA
Ic = Ie = 0.99(10 mA);
Ic = 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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Physics, 6th Editon.
Chapter 40. Electronics
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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Physics, 6th Editon.
Chapter 40. Electronics
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 
Ie
Ib
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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