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.
Ib = Ie(1 - α) = (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-baseconnection is 1.6 mA and the emitter current is 60 mA, what is α?
Ib = 1.6 mA; Ie = 60 mA; Ic = 60 mA – 1.6 mA = 58.4 mA
[pic] α = 0.973
41-3. For a common-base amplifier, the input resistance is 800 Ω and the output resistance is 600 kΩ. (a) Determine the voltage gain if the emitter current is 12 mA and α = 0.97. (b) What is the power gain?
(a) Vin = IeRin = (12 x 10-3A)(800 Ω); Vin = 9.60 V To find Vout, we need Ic.;
Ic = αIe = 0.97(12 x 10-3 A) = 0.0116 A; Vout = IcRout = (0.0116 A)(600,000 Ω)
Vout = 6980 V; [pic] Av = 727
(b) G = αAv = 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.[pic] Ic = αIe = (0.952)(1.2 mA); Ic = 1.14 mA
40-5. Determine the current gain in a common-emitter amplifier circuit when α = 0.98.
[pic] β = 49
40-6. In the previous problem, what is the collector current if the emitter current is 20 (A? What is the base current?
[pic] Ic = 0.98(20 (A); Ic = 19.6 (A
40-7. A transistor with an effectiveinput resistance of 400 Ω, an effective output resistance of 900 kΩ, and α = 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 Ω) = 3.2 mV; Ic = (0.96)(8 (A) = 7.68 (A
Vout = IcRout = (7.68 (A)(900,000 Ω) = 6.912 V; [pic] Av = 2160
G = αAv = (0.96)(2160); G = 207440-8. Calculate Ic and Ib for the conditions described in Problem 40-7.
Ic = αIe = 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 α = 0.97 connected with a common emitter, calculate β, Iin , and Iout.
[pic] β = 32.3
Ic = 0.97(8 (A) = 7.76 (A ; Ib = (8 (A – 7.76 (A) = 0.240 (A
40-10. For a transistor with α = 0.99connected with a common collector, calculate the current gain Ai.
40-11. A transistor has β = 99; what is the value of α?
[pic] α = 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 - α); Ib = 0.10 mA; Ie - Ieα = 0.10 mA;
[pic] Ie = 10 mA
[pic] Ic = αIe = 0.99(10 mA); Ic = 9.9 mACritical 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 isbest for temperature stability since the temperature effects are opposite and tend to cancel.
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, theconfiguration 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...
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