inding the Performances of a Dc–Dc Converter The dc converter in Figure 5.2a has a resistive load of R = 10 and the input voltage is Vs = 220 V. When the converter switch remains on, its voltage drop is v ch = 2 V and the chopping frequency is f = 1 kHz. If the duty cycle is 50%, determine

Question

inding the Performances of a Dc–Dc Converter

The dc converter in Figure 5.2a has a resistive load of R = 10 Ω and the input voltage is
[latex]V_s = 220 V[/latex]. When the converter switch remains on, its voltage drop is [latex]v-{ch} = 2 V[/latex] and the chopping frequency is f = 1 kHz. If the duty cycle is 50%, determine (a) the average output voltage [latex]V_a[/latex], (b) the rms output voltage [latex]V_o[/latex], (c) the converter efficiency, (d) the effective input resistance [latex]R_i[/latex] of the converter, (e) the ripple factor of the output voltage [latex]RF_o[/latex], and (f) the rms value of the fundamental component of output harmonic voltage.

Accepted Answer

[latex]V_s = 220 V, k = 0.5, R = 10[/latex] , and [latex]v_{ch }= 2 V[/latex].

a. From Eq. (5.8),

[latex]V_a=\frac{1}{T} \int_{0}^{t_1}{\upsilon _0dt}=\frac{t_1}{T}V_s=ft_1V_s=kV_s[/latex] (5.8)

[latex]V_a = 0.5 × (220 - 2) = 109 V[/latex].

b. From Eq. (5.9),

[latex]V_o=\left(\frac{1}{T}\int_{0}^{kt}{\upsilon _0^2dt} ight)^{1/2}=\sqrt{k}V_s[/latex] (5.9)

[latex]V_o =\sqrt{0.5}× (220 - 2) = 154.15 V[/latex].

c. The output power can be found from

[latex]P_0=\frac{1}{T}\int_{0}^{kT}{\frac{\upsilon _0^2}{R}dt }=\frac{1}{T}\int_{0}^{kt}\frac{(V_s -\upsilon _{ch}^2}{R}dt=k\frac{(V_s -\upsilon _{ch})^2}{R}[/latex] (5.12)

[latex]=0.5 imes \frac{(220 -2)^2}{10} =2376.2 W[/latex]

The input power to the converter can be found from

[latex]P_i=\frac{1}{T}\int_{0}^{kt}{V_sidt}=\frac{1}{T} \int_{0}^{kT}{\frac{V_s(V_s - \upsilon _{ch})}{R} } dt=k\frac{V_s(V_s -\upsilon _{ch})}{R}[/latex] (5.13)

[latex]=0.5 imes 220 imes \frac{220 -2}{10} =2398 W[/latex]

The converter efficiency is

[latex]\frac{P_0}{P_i}=\frac{2376.2}{2398}=99.09\%[/latex]

d. From Eq. (5.11),

[latex]R_i = V_s/I_a = V_s(V_a/R) = 220 × (109/10) = 20.18 Ω[/latex]

e. Substituting Va from Eq. (5.8) and Vo from Eq. (5.9) into Eq. (5.6) gives the ripple factor as

[latex]RF_0=\frac{V_r}{V_a} =\sqrt{\frac{1}{k}-1 } [/latex] (5.14)

[latex]=\sqrt{1/0.5 -1}=100\% [/latex]

f. The output voltage as shown in Figure 5.2b can be expressed in a Fourier series as

[latex]\upsilon _0(t)=kV_s +\sum\limits_{n=1}^{\infty }{\frac{V_s}{n\pi }sin 2nk cos 2nft }[/latex]

[latex]+ \frac{V_s}{n\pi }\sum\limits_{n=1}^{\infty }{(1 - cos 2nk)sin 2nft} [/latex] (5.15)

The fundamental component (for n = 1) of output voltage harmonic can be determined from Eq. (5.15) as

[[latex]\upsilon _1(t)=\frac{(V_s -\upsilon _{ch})}{\pi }[sin 2\pi k cos 2\pi ft + (1 - cos 2\pi k) sin 2\pi ft][/latex] (5.16)

[latex]=\frac{(220 -2) imes 2}{\pi }sin(2\pi imes 1000t)=138.78 sin(6283t)[/latex]

and its root-mean-square (rms) value is [latex]V_1 = 138.78/\sqrt{2} = 98.13 V[/latex].

Note: The efficiency calculation, which includes the conduction loss of the converter, does not take into account the switching loss due to turn-on and turn-off of practical converters. The efficiency of a practical converter varies between 92 and 99%.

Question 5.1: inding the Performances of a Dc–Dc Converter The dc converte...

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