Chapter 12 Power Amplifiers
6. In every section of the field
7. Power-handling capability: The SCR can handle the most current, and the power FET the least current. Efficiency:
The SCR is the most efficient since the control signal can be removed once SCR is conducting, and the power FET is the next-most efficient since its control current is low.
Control input: The power FET and BJT are easier to control because they can be shut off using the control input.
Maximum frequency: The power FET switches the fastest.
PROBLEMS 15-1. Given:
VD = 0.7 V IH = 4 mA R = 1 kΩ Solution:
V = IH R + 0.7 V (Eq. 15-2) V = (4 mA)(1 kΩ) + 0.7 V V = 4.7 V
Answer: The power supply voltage will be 4.7 V at dropout.
15-2. Given:
VD = 0.7 V
VB = 12 V V = 19 V R = 5 kΩ
Solution: Just before breakover, the capacitor voltage is VB.
I = (V – VB)/R I = (19 V – 12 V)/5 kΩ I = 1.4 mA
While the diode is conducting, the voltage across it is 0.7 V.
I = (V – VD)/R
I = (19 V – 0.7 V)/5 kΩ I = 3.66 mA
Answer: The current through the resistor just before breakover is 1.4 mA, and during conduction is 3.66 mA.
15-3. Given:
VD = 0.7 V VB = 12V V = 19 V R = 5 kΩ C = 0.2 µF Solution:
RC = (5 kΩ)(0.02 µF) RC = 0.1 ms
T = 0.1 ms since the period equals the RC time f = 1/T
f = 1/10.1 ms f = 10 kHz
Answer: The RC time is 0.1 ms, and the frequency is 10 kHz.
15-4. Given:
VD = 0.7 V VB = 20 V IH = 3 mA R = 1 kΩ
Solution: Since the diode is open before breakover, no current flows before the device breaks over. Thus when the power supply reaches breakover voltage, the device will break over.
V = IH R + 0.7 V (Eq. 15-2) V = (3 mA)(1 kΩ) + 0.7 V V = 3.7 V
Answer: The power supply voltage will be 20 V at breakover and 3.7 V at dropout.
15-5. Given:
VD = 0.7 V VB = 12 V V = 19 V R = 10 kΩ C = 0.06 µF
Solution: The maximum voltage across the capacitor will be breakover voltage, because as soon as the device breaks over, the voltage drops to about 0.7 V.
RC = (10 kΩ)(0.06 µF) RC = 0.6 ms
Answer: The maximum voltage across the capacitor is 12 V, and the time constant is 0.6 ms.
15-6. Given:
VGT = 1.0 V output voltage when the SCR is off is the same as the power supply voltage.
Answer: The output voltage when the SCR is off is 12 V.
The input voltage required to turn on the SCR is 5.4 V, and the supply voltage required to turn the SCR off is 1.26 V.
Answer: The input voltage required to turn on the SCR is 7.3 V.
15-8. Answer: The highest output occurs when 0.8 V is across the 500-Ω resistor. The current through this resistor is 0.8 V divided by 500 Ω, which equals 1.6 mA. This 1.6 mA must flow through the 3.3-kΩ resistor. The 200 µA of gate current must also flow through the 3.3-kΩ resistor.
If we ignore the 200 µA on the grounds that it is much smaller than 1.6 mA, we get an approximate answer of:
V = 0.8 V + (1.6 mA)(3.3 kΩ) = 6.08 V
If we include the 200 µA, we get a slightly larger output voltage:
Answer: The input voltage required to turn on the SCR is 34.5 V, and the supply voltage required to turn the SCR off is 1.17 V.
Answer: The input voltage required to turn on the SCR is 54.8 V.
Answer: The charging time constant is 11.9 ms. and the Thevenin resistance is 611 Ω.
15-12. Given:
θcond° = 135
Answer: The firing angle is 45°, the conduction angle is 135°, and the voltage across the capacitor is 85 Vac.
15-13. Given:
R1= 1 kΩ R2 = 50 kΩ pot C = 0.47 μF
Solution: Perform the following calculations with an R value of 1 kΩ and 51 kΩ.
Answer: The minimum firing angle is 10°, and the maximum firing angle is 83.7°.
15-14. Given:
R1 = 1 kΩ R2 = 50 kΩ pot C = 0.47 μF
Solution: Perform the following calculations with an R value of 1 kΩ and 51 kΩ.
Answer: The minimum conduction angle is 170°, and the maximum conduction angle is 96.3°.
15-15. Given:
Answer: The voltage needed to trigger the crowbar is 10.8 V.
Answer: The voltage needed to trigger the crowbar is 12.5 V.
Answer: The SCR will trigger at 12.8 V.
15-18. Given:
Answer: The voltage needed to trigger the crowbar is 12.8 V.
15-19. Given:
VB = 20 V VGT = 2.5 V
Solution: Ignore the gate current in the triac. Then VC = VB + VGT
VC = 20 V + 2.5 V VC = 22.5 V
Answer: The capacitor voltage required to turn on the triac is 22.5 V.
15-20. Given:
VCC = 100 V R = 15 Ω
Solution: Ideally, when the triac is conducting, the voltage drop across it is 0 V.
I = VCC/R I = 100 V/15 Ω I = 6.67 A
Answer: The load current is 6.67 A.
15-21. Given:
VB = 28 V VGT = 2.5 V
Solution: Ignore the current through the diac and triac.
Then VC = VB + VGT
VC = 28 V + 2.5 V VC = 30.5 V
Answer: The capacitor voltage required to turn on the triac is 30.5 V.
15-22. Given:
VCC = 15 V R2 = 1 kΩ R3 = 2V Solution:
Vgate = [R3/(R2 + R3)]VCC (Voltage divider formula) Vgate = [2 kΩ/(1 kΩ + 2 kΩ)]15 V
Vgate = 10 V Vanode = VTrig + 0.7 V Vanode = 10 V + 0.7 Vanode = 10.7 V
Answer: The gate trigger voltage is 10 V and the anode is 10.7 V.
15-23. Given:
VCC = 15 V Vgate = 10 V Vanode = 10.7 V Solution:
VR4 = Vanode – 0.7 V VR4 = 10.7 V – 0.7 VR4 = 10 V
Answer: The peak voltage across R4 = 10 V.
15-24. Answer: The output waveform will be a sawtooth waveform from 0 V to 10.7 V.
CRITICAL THINKING
15-25. Answer: The breakover voltage of the diode, which is 10 V.
15-26. Answer: The breakover voltage of the diode, which is 10 V.
15-27. Given:
R1 = 0 to 50 kΩ R2 = 1 kΩ C = 0.1 µF T = 20%(RC)
Solution:
Rmax = R2 + R1 (max) Rmax = 1 kΩ + 50 kΩ Rmax = 51 kΩ Rmin = R2 + R1 (min)
Rmin = 1 kΩ + 0 kΩ Rmin = 1 kΩ RCmax = RmaxC
RCmax = (51 kΩ)(0.1 µF) RCmax = 5.1 ms
RCmin = RmaxC RCmin = (1 kΩ)(0.1 µF) RCmin = 0.1 msec Tmax = 0.2(RCmax) Tmax = 0.2(5.1 ms) Tmax = 1.02 ms Tmin = 0.2(RCmin) Tmin = 0.2(0.1 ms) Tmin = 0.02 ms fmax = 1/Tmin fmax = 1/0.02 ms fmax = 50 kHz fmin = 1/Tmax
fmin = 1/1.02 ms fmin = 980 Hz
Answer: The maximum frequency is 50 kHz, and the minimum is 980 Hz.
15-28. Given:
R = 100 Ω VCC = 15 V
Solution: In a dark room the SCR is off and the output voltage is 15 V. Once the SCR fires, its voltage drops to 0.7 V.
I = (VCC – 0.7 V)/R I = (15 V – 0.7 V)/100 Ω I = 143 mA
Answer: The output voltage when it is dark is 20 V and when it is light is 0.7 V, and the current through the resistor is 143 mA when it is light.
15-29. Answer:
Trouble 1: Since there is voltage at D and not at E, the wire connecting the two is open.
Trouble 2: No supply voltage.
Trouble 3: Since there is voltage at B and not at C, the transformer is the problem.
Trouble 4: Since there is voltage at A and not at B, the fuse is open.
15-30. Answer:
Trouble 5: Since there is an overvoltage and the crowbar is off, the problem is the crowbar.
Trouble 6: Since there is voltage at C and not at D and the load resistor is not shorted, the rectifier is the problem.
Trouble 7: Since there is voltage at E and not at F, the wire connecting the two is open.
Trouble 8: Since there is voltage at A and not at B, the fuse is open.