Hart Chapter 7 solutions

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CHAPTER 7 SOLUTIONS
4/03/10
7-1)
	
	 
7-2)
	
	
7-3)
	
	
7-4) Example design
	
	
7-5) For continuous current
	
	
7-6) Switch is closed for DT, current returns to zero at t = tx:
	
	
7-7)
	
	
	
7-8)
	
 b) 
The currents in the converter are shown below. The winding currents are for the windings in the ideal transformer model, not the physical windings. The physical primary winding current is the sum of winding #1 and Lm currents.
 
 
 
	
	
7-9)
	
7-10)
	
 b) 
The currents in the converter are shown below. The winding currents are for the windings in the ideal transformer model, not the physical windings. The physical primary winding current is the sum of winding #1 and Lm currents.
	
7-11)
	
7-12) 
	
7-13) 
	
7-14)
The current in the physical primary winding is the sum of iL1 and iLm in the model. The physical currents in windings 2 and 3 are the same as in the model.
7-15)
	
7-16) 
7-17)
	
7-18) The input voltage vx to the filter is Vs(Ns / Np) when either switch is on, and vx is zero when both switches are off. (See Fig. 7-8.) The voltage across Lx is therefore
	
7-19)
	
7-20)
	
		
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7-21)
	
7-22)
The simulation is run using a Transient Analysis with a restricted time of 3 to 3.02 ms, representing two periods of steady-state operation. The steady-state output voltage has an average value of approximately 30 V and peak-to-peak ripple of approximately 600 mV, ignoring the negative spike. The average transformer primary and secondary currents are 912 mA and 83.5 mA, respectively. The output voltage is lower than the predicted value of 36 V because of the nonideal switch and diode, mostly from the switch. The output voltage ripple is 2%, matching the predicted value. The converter would operate much better with a switch that has a lower on resistance.
7-23)
Using a nonideal switch and diode produces lower values for the currents. For iLx, the maximum, minimum, and average values in PSpice are 1.446 A, 0.900 A, and 1.17 A, compared to 1.56 A, 1.01 A, and 1.28 A, respectively. However, the peak-to-peak variation in iLx in PSpice matches that of the ideal circuit (0.55 A).
7-24) 
Design for θco= -210° and a gain of 20 dB for a cross over frequency of 12 kHz.
	
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7-25)
	
7-26)
 Using Vs = 6 V as in Example 7-8, the frequency response of the open-loop system shows that the crossover frequency is approximately 16.8 kHz. The phase angle at the crossover frequency is 17°, which is much less than the desired value of at least 45°. Therefore, the system does not have the desired degree of stability.
7-27)
a) A frequency response of the circuit yields Vo ≈ -2.5 dB and θ ≈ 103° at 10 kHz.
b) With Vp = 3, the gain of the PWM function is 20log10(1/3) = -9.54 dB. The required gain of the compensated error amplifier is then 2.5 + 9.54 = 12.06 dB, corresponding to a gain magnitude of 4.0. The phase angle of the compensated error amplifier at the crossover frequency to give a phase margin of 45° is
	
From 7-75, 7-85, 7-86, and 7-87,
	
c) Referring to Example 7-9, the PSpice simulation results are shown indicating a stable control system. The switching frequency was not specified, and 50 kHz was used here. Use initial conditions for the capacitor voltage at 8 V and the inductor current at 2 A.
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7-28) 
a) The gain at 8 kHz is approximately -2.44 dB, and the phase angle is -100°.
b) This design is for fco = 8 kHz. With Vp = 3, the gain of the PWM function is 20log10(1/3) = -9.54 dB. The required gain of the compensated error amplifier is then 2.44 + 9.54 = 11.98 dB, corresponding to a gain magnitude of 3.97. The phase angle of the compensated error amplifier at the crossover frequency to give a phase margin of 45° is
	
From 7-75, 7-85, 7-86, and 7-87,
	
c) Referring to Example 7-9, the PSpice simulation results are shown indicating a stable control system. The switching frequency was not specified, and 50 kHz was used here. Use initial conditions for the capacitor voltage at 8 V and the inductor current at 1.6 A.
 If designing for fco = 10 kHz, the gain of the converter is -4.38 dB, and θco = -98°. R1 = 1k, R2 = 4.97k, C1 = 9.58 nF, and C2 = 1.07 nF. 
7-29)
	
7-30)
	
7-31)
	
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