Hart Chapter 9 solutions

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CHAPTER 9 SOLUTIONS
3/13/10
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9-6)
	
9-7) a) The circuit is shown with diode D2 added to make the switch unidirectional.
(a) The average output (capacitor) voltage is 5.16 V, agreeing with the 5.17 V computed analytically. (b) Peak capacitor voltage= 20 V.; (c) Inductor currents: peak = 10.5 A.; average = 2.59 A.; rms = 4.54 A.
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9-11)
	
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9-14) A suitable circuit is shown. The values of the output filter components L1 and C2 are not critical. The load resistor is chosen to give 10 A. The switch must be open for an interval between t2 and t3; 50 ns is chosen.
Results from Probe for steady-state output: a) Vo ≈ 14.6 V., b) VC,peak= 120 V., c) Integrate instantaneous power, giving 72.7 μJ per period (supplied).
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9-16)
	
The output file shows that the THD is 10.7%. Increase Q by increasing L, and adjust C accordingly. L=1.4 mH and C=12.6 µF gives THD=9.8%. Switching takes place when load (and switch) current is approximately 3.6 A.
9-17)
	
From the output file, THD = 10.8%. From Probe: VC,peak=149 V.; IL,peak=8.12 A.
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9-20)
	
A PSpice simulation using the circuit of Fig. 9-6(a) gives an output voltage of approximately 5.1 V.
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9-22)
A PSpice simulation using the circuit of Fig. 9-6a shows a steady-state output voltage of approximately 14.4 V, slightly less than the target value of 15 V. Note that the current in Lr and Cr is not quite sinusoidal.
9-23)
	
A PSpice simulation using the circuit of Fig. 9-6a shows a steady-state output voltage of approximately 53 V, slightly less than the target value of 55 V. Note that the current in Lr and Cr is not quite sinusoidal.
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9-32)
Using a circuit based on Fig. 9-11a but with a square-wave source implemented with Vpulse (see Fig. 9-6a), the result is approximately 9.4 V.
9-33)
 (a) A PSpice simulation using the circuit shown reveals that the capacitor voltage returns to zero at 15.32 μs and the switch must remain closed for 5.58 μs for the inductor current to return to 12 A. Initial conditions for the inductor (12 A) and for the capacitor (0 V) must be applied. Ideal models for the switch and diode are used.
b) Using the expression S(W(V1)) for energy (S is integration), 15.7 mJ are supplied by the voltage source in one period.
c) Average power is 754 W, obtained by entering AVG(W(V1)).
d) Average resistor power is 104 W.
e) With R = 0, the capacitor voltage returns to zero at 15.44 μs and the switch must remain closed for 5.45 μs. The source power and energy are not changed significantly.
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