Hart Chapter 4 solutions

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CHAPTER 4 SOLUTIONS
2/17/10
4-1) Load:
 Each diode:
	
4-2)
	
4-3)
	
4-4)
4-5)
	
			 b) Power is determined from the Fourier series. Using Eq. 4-4 and 4-5.
	n
	Vn, V.
	Zn. Ω
	In, A.
	2
	72.0
	27.1
	2.65
	4
	14.4
	47.7
	0.302
	
4-6
	
			 b) Power is determined from the Fourier series. Using Eq. 4-4 and 4-5.
	n
	Vn, V.
	Zn. Ω
	In, A.
	2
	72.0
	19.3
	3.74
	4
	14.4
	32.5
	0.444
	
4-7)
	
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4-8) 	Load:
	
4-9)
	
4-10)
	
�
4-11)
	
 b) Fourier Series
	n
	Vn, V.
	Zn. Ω
	In, A.
	2
	72.2
	11.7
	6.16
	4
	14.4
	22.8
	0.631
	
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4-12
	
 b) Fourier Series
	n
	Vn, V.
	Zn. Ω
	In, A.
	2
	144.3
	30.6
	4.72
	4
	28.9
	60.5
	0.477
	
4-13)	
4-14) 	a) Continuous current; P=474 W.
	b) Discontinuous current; P=805 W.
4-15
	
 b) Fourier Series
	n
	Vn, V.
	Zn. Ω
	In, A.
	2
	72.0
	30.4
	2.37
	4
	14.4
	60.5
	0.238
	
�
4-16 
	
 b) Fourier Series
	n
	Vn, V.
	Zn. Ω
	In, A.
	2
	72.0
	45.5
	1.58
	4
	14.4
	90.6
	0.159
	
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4-17)
The current with the 100 μH inductor is discontinuous.
4-18)
	
4-19)
	
4-20) C ≈ 3333/2 = 1667 µF. Peak diode currents are the same. Fullwave circuit has advantages of zero average source current, smaller capacitor, and average diode current ½ that for the halfwave. The halfwave circuit has fewer diodes, and has only one diode voltage drop rather than two.
4-21)
	
> 1 ( continuous current
			
	
	 
 < 1 ( 
Calculated Vo is slightly larger than initial estimate. Try Vo=120 V.:
	
Therefore, 119 < Vo < 120 V. (Vo=119.6 with more iterations.)
c) PSpice results:
R = 7 results in continuous current with Vo = 108 V. R = 20 results in discontinuous current with Vo = 120 V. The simulation was done with C = 10,000 μF.
4-22) PSpice results with a 0.5 Ω resistance in series with the inductance: For Rload = 5 Ω, Vo=56.6 V. (compared to 63.7 volts with an ideal inductor); for Rload = 50 Ω, Vo=82.7 V. (compared to 84.1 volts with an ideal inductor).
4-23)
	
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4-24)
	
4-25) a) α = 15° : Check for continuous current. First period:
	
 b) α = 75° Check for continuous current. First period:
	
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4-26)a) α = 20°: Check for continuous current. First period:	
	
 b) α = 80°: Check for continuous current. First period:
	
4-27) The source current is a square wave of ±Io.
	
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4-28)
	
Note that the turns ratio could be lower (higher secondary voltage) and α adjusted accordingly.
4-29)
	
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4-30)
	
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4-31)
	
Choose L somewhat larger, say 120 mH, to allow for approximations.
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4-32) In Fig. 4-14, Pac = Pbridge = -VoIo = 1000 W. Using Vdc = -96 V gives this solution:
	
	
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4-33)
	
4-34)
	
4-35) 
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4-36)
	
4-37)
There are no differences between the calculations in Problem 4.36 and the PSpice results. The power absorbed by each diode ia approximately 1.9 W.
4-38)Equation (4-46) gives values of of I1 = 28.6 A, I5 = 5.71 A, I7 = 4.08 A, I11 = 2.60 A, and I13 = 2.20 A. All compare well with the PSpice results. The total harmonic distortion (THD) is 27.2% when including harmonics through n = 13.
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4-39)
	
		c) 
4-40)
c)
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4-41)
	
4-42)
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4-43)
	
4-44)
	
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4-45)
	
4-46)
	
	
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4-47)
	
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4-48)
	
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4-49)
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4-50)
	
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4-51)
	
Uncontrolled rectifiers with additional resistances added can also satisfy the specifications. However, adding resistance would increase power loss and decrease efficiency.
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