Hart Chapter 6 solutions

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CHAPTER 6 SOLUTIONS
5/17/10
6-1)
	
6-2)	
	
	
		
6-3)
	
6-4)
	
	
6-5)
	
6-6)
	
6-7)
	
6-8)	
	
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6-9)
	
6-10) 
	
Vo=5 V
	Vs, V
	D
	I, A.
	R, Ω
	Lmin, µH
	10
	0.5
	0.5
	10
	12.5
	10
	0.5
	1.0
	5
	6.25
	15
	1/3
	0.5
	10
	16.7 (worst case, D = 1/3, R = 10)
	15
	1/3
	1.0
	5
	8.33
	
6-11) Example design:
	
Other values of L and C are valid if the inductor current is continuous with margin.
6-12) (Based on the example design in 6-11)
Vmax, switch = Vs = 48 V
Vmax, diode = Vs = 48 V
Imax, switch = ILmax = 1.5 + 0.75/2 = 1.875 A
Iavg, switch = 
Irms, switch 
Imax,diode = ILmax = 1.875 A
Iavg,diode =IL- Iavg,switch = 1.875 – 0.586 = 1.289 A
Irms,diode 
6-13) Example design:
6-14) Example design:
 
Other values of L and C are valid if the inductor current is continuous with margin.
6-15)
	
 The output voltage is mainly the dc term and the first ac term.
6-16)
	
6-17)
	
6-18)
Inductor current: (see Example 2-8)
	
Capacitor current: (define t=0 at peak current)
	
6-19)
	
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6-20) Example design:
	
6-21)
	
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6-22) 
 6-23)
	
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6-24) 
 									
Inductor current: (see Example 2-8)
	
Capacitor current: For convenience, redefine t = 0 at the peak current. The current is then expressed as 
	
	
6-25)
	
	
6-26) Example design:
	
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6-27) Example design:
	
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6-28) Using the equations
	
 and using f = 100 kHz (designer’s choice), results are shown in the table. 
	Vs, (V)
	P (W)
	D
	R (Ω)
	Lmin (µH)
	IL (A)
	C (µF)
	10
	10
	0.545
	14.4
	14.9
	1.83
	37.9
	10
	15
	0.545
	9.6
	9.9
	2.75
	56.8
	14
	10
	0.462
	14.4
	20.9
	1.55
	32.1
	14
	15
	0.462
	9.6
	13.9
	2.32
	48.1
The value of L should be based on Vs = 14 V and P = 10 W, where Lmin = 20.9 µH. Select the value of L at least 25% larger than Lmin (26.1 µF). Using another common criterion of ΔiL = 40% of IL, again for 14 V and 10 W, L = 104 µH.
The value of C is 56.8 µF for the worst case of Vs = 10 V and P = 10 W.
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6-29)
	
6-30)
	
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6-31) Example design:
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6-32)
	
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6-33)
6-34) Equation (6-69) for the average voltage across the capacitor C1 applies:
	
	
When the switch is closed, the voltage across L2 for the interval DT is
	
	
Assuming that the voltage across C1 remains constant at its average value of Vs 
	
When the switch is open in the interval (1 - D)T,
	
Since the average voltage across an inductor is zero for periodic operation,
	
	
resulting in
	
6-35)
	
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6-36)
	
6-37)
	
6-38)
	
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6-39)
	
6-40)
	
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6-41) Discontinuous current for the buck-boost converter: Let DT be the time that the switch is closed and D1T be the time that the switch is open and the current in the inductor is positive. For a lossless converter, the output power is the same as the input power.
	
6-42) When switches “1” are closed, C1 and C2 are connected in series, each having Vs/2 volts. When the “1” switches are opened and the “2” switches are closed, Vo = Vs of the source plus Vs/2 of C1, making Vo = 1.5Vs.
6-43)
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6-44) Simulate the buck converter of Example 6-1 using PSpice. (a) Use an ideal switch and ideal diode. Determine the output ripple voltage. Compare your PSpice results with the analytic results in Example 6-1. (b) Determine the steady-state output voltage and voltage ripple using a switch with an on resistance of 2 Ω and the default diode model 
Using Ron =0.01 for the switch and n=0.01 for the diode, the p-p ripple voltage is 93.83 mV. 93.83/20 = 0.469%, agreeing precisely with the analytical results.
With Ron = 2 ohms, the p-p ripple is 90 mV, with a reduced average value.
6-45)
Note that for each converter topology, the average voltage across each inductor is zero, and the average current in each capacitor is zero.
Buck Converter:
Show from Eqs. (6-9) and (6-17) 
From the averaged circuit of Fig. 6.33b,
Boost Converter:
Show from Eqs. (6-27) and (6-28) that 
From the averaged circuit of Fig. 6.33c,
Buck-Boost Converter:
Show from Eqs. (6-47) and (6-49) and preceding equations that 
From the averaged circuit of Fig. 6.33d,
Ćuk Converter:
Show from Eqs. (6-59) and (6-61) that
From the averaged circuit,
	
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