Hart Chapter 3 solutions

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CHAPTER 3 SOLUTIONS
2/20/10
3-1)
	
 
3-2)
	
3-3)
	
3-4) Using Eq. 3-15,
	
3-5) Using Eq. 3-15,
	
3-6) Using Eq. 3-15,
	
	
	
3-7) Using an ideal diode model, R = 48 Ω for an average current of 2 A.
3-8) Using Eqs. 3-22 and 3-23,
	
�
3-9) Using Eqs. 3-22 and 3-23,
	
3-10) Using Eq. 3-33,
	
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3-11)
3-12) L ≈ 0.14 H for 50 W (51 W).
3-13) Using Eq. 3-34,
	a)
	
b)
	n
	Vn
	Zn
	In
	0
	54.02
	12.00
	4.50
	1
	84.85
	25.6
	3.31
	2
	36.01
	46.8
	0.77
	4
	7.20
	91.3
	0.08
The terms beyond n = 1 are insignificant.
3-14)
Run a transient response long enough to achieve steady-state results (e.g., 1000ms). The peak-to-peak load current is approximately 1.48 A, somewhat larger than the 1.35 A obtained using only the first harmonic. (The inductance should be slightly larger, about 0.7 H, to compensate for the approximation of the calculation.)
3-15)
a)
	
b) A PSpice simulation using an ideal diode model gives 0.443 A p-p in the steady state. This compares with 2(I1)=2(0.199)=0.398 A p-p.
3-16)
	
3-17) 	a) τ = RC = 10310-3=1 s; τ/T = 60. With τ >> T, the exponential decay is very small and the output voltage has little variation.
	b) Exact equations:
	
	c) Approximation of Eq. 3-51:
	
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3-18) 	a) R = 100 Ω: τ = RC (100)10-3 = 0.1 s; τ/T = 6.
	
	b) R = 10 Ω: τ = RC (10)10-3 = 0.01 s; τ/T = .6.
	
	In (a) with τ/T=6, the approximation is much more reasonable than (b) where τ/T=0.6.
3-19) 	a) With C = 4000 µF, RC = 4 s., and the approximation of Eq. 3-51 should be reasonable.
	
	b) With C = 20 µF, RC = 0.02, which is on the order of one source period. Therefore, the approximation will not be reasonable and exact equations must be used.
	
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3-20)	a) With C = 4000 µF, RC = 2 s., and the approximation of Eq. 3-51 should be reasonable.
	
	b) With C = 20 µF, RC = 0.01, which is on the order of one source period. Therefore, the approximation will not be reasonable and exact equations must be used.
	
3-21) From Eq. 3-51
	
3-22) Assuming Vo is constant and equal to Vm,
	
	From Eq. 3-51
	
3-23) Using the definition of power factor and Vrms from Eq. 3-53,
	
3-24)
	
3-25)
	
3-26)
	
3-27)
	
 
3-28) α ≈ 46°. Do a parametric sweep for alpha. Use the default (Dbreak) diode, and use Ron = 0.01 for the switch. Alpha of 46 degrees results in approximately 2 A in the load.
3-29) α ≈ 60.5°. Do a parametric sweep for alpha. Use the default (Dbreak) diode, and use Ron = 0.01 for the switch. Alpha of 60.5 degrees results in approximately 1.8 A in the load.
�
3-30) From Eq. 3-61,
	
3-31) From Eq. 3-61,
	
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3-32) α ≈ 75°. Alpha = 75 degrees gives 35 W in the dc voltage source. An Ron = 0.01 for the switch and n = 0.001 for the diode (ideal model).
�
3-33) From Eq. 3-61,
	
3-34) α ≈ 81°
3-35)
	
�
3-36)
	v0 = vs when S1 on, v0=0 when D2 on
	
3-37)
	
 
PSpice:
Use a current source for the constant load current:
�
D1 to D2
D2 to D1
3-38) u = 20°. Run the simulation long enough for steady-state results. From the Probe output, the commutation angle from D1 to D2 is about 20 degrees, and from D2 to D1 is about 18 degrees. Note that the time axis is changed to angle in degrees here. 
3-39) Run the simulation long enough for steady-state results. From the Probe output, the commutation angle from D1 to D2 is about 16.5 degrees, and from D2 to D1 is about 14.7 degrees. Note that the time axis is changed to angle in degrees here. 
 
3-40) At ωt = π, D2 turns on, D1 is on because of the current in LS (see Fig. 3-17).
	
3-41) At ωt = α,
	
3-42) A good solution is to use a controlled half-wave rectifier with an inductor in series with the 48-V source and resistance (Fig. 3-15). The switch will change the delay angle of the SCR to produce the two required power levels. The values of the delay angle depend on the value selected for the inductor. This solution avoids adding resistance, thereby avoiding introducing power losses.
3-43) Several circuit can accomplish this objective, including the half-wave rectifier of Fig. 3-2a and half-wave rectifier with a freewheeling diode of Fig. 3-7, each with resistance added. Another solution is to use the controlled half-wave rectifier of Fig. 3-14a but with no resistance. The analysis of that circuit is like that of Fig. 3-6 but without Vdc. The resulting value of α is 75°, obtain from a PSpice simulation. That solution is good because no resistance is needed, and losses are not introduced.
3-44 and 3-45) The controlled half-wave rectifier of Fig. 3-15 (without the resistance) can be used to satisfy the design specification. The value of the delay angle depends on the value selected for the inductor.
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