Finding the Values of LC Filter for the Buck Regulator The buck regulator in Figure 5.17a has an input voltage of Vs = 12 V. The required average output voltage is Va = 5 V at R = 500 Ω and the peak-to-peak output ripple voltage is 20 mV. The switching frequency is 25 kHz. If the peak-to-peak

Question

Finding the Values of LC Filter for the Buck Regulator

The buck regulator in Figure 5.17a has an input voltage of [latex]V_s = 12 V[/latex]. The required average output voltage is [latex]V_a = 5 V[/latex] at [latex]R = 500 Ω[/latex] and the peak-to-peak output ripple voltage is 20 mV. The switching frequency is 25 kHz. If the peak-to-peak ripple current of inductor is limited to 0.8 A, determine (a) the duty cycle k, (b) the filter inductance L, (c) the filter capacitor C, and (d) the critical values of L and C.

Accepted Answer

[latex]V_s = 12 V, ΔV_c = 20 mV, ΔI - 0.8 A, f = 25 kHz[/latex], and [latex]V_a = 5 V[/latex].

a. From Eq. (5.56),

[latex]V_a=V_s\frac{t_1}{T}=kV_s[/latex] (5.56)

[latex]V_a = kV_s[/latex] and [latex]k = V_a/V_s = 5/12 = 0.4167 = 41.67\%.[/latex]

b. From Eq. (5.59),

[latex]\Delta I=\frac{V_a(V_s - V_a)}{f LV_s}[/latex] (5.59)

[latex]L=\frac{5(12 -5)}{0.8 imes 25,000 imes 12} =145.83 \mu H[/latex]

c. From Eq. (5.61),

[latex]\Delta V_c=\upsilon _c -\upsilon _c(t=0)=\frac{1}{C}\int_{0}^{T/2}{\frac{\Delta I}{4}dt } =\frac{\Delta IT}{8C}=\frac{\Delta I}{8fC}[/latex] (5.61)

[latex]C=\frac{0.8}{8 imes 20 imes 10^{-3} imes 25,000}=200 \mu F[/latex]

d. From Eq. (5.64), we get [latex]L_c=\frac{(1-k)R}{2f}=\frac{(1 - 0.4167) imes 500}{2 imes 25 imes 10^3}=5.83 mH[/latex]

From Eq. (5.65), we get [latex]C_c=\frac{1 -k}{16Lf ^2} =\frac{1 - 0.4167}{16 imes 145.83 imes 10^{-6} imes (25 imes 10^3)^2}=0.4 \mu F [/latex]

Question 5.5: Finding the Values of LC Filter for the Buck Regulator The b...

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