A buck converter is shown in Figure 5.40. The input voltage is Vs = 110 V, the average load voltage is Va = 60 V, and the average load current is Ia = 20 A. The chopping frequency is f = 20 kHz. The peak-to-peak ripples are 2.5% for load voltage, 5% for load current, and 10% for filter Le current.

Question

A buck converter is shown in Figure 5.40. The input voltage is [latex]V_s = 110 V[/latex], the average load voltage is [latex]V_a = 60 V[/latex], and the average load current is [latex]I_a = 20 A[/latex]. The chopping frequency is [latex]f = 20 kHz[/latex]. The peak-to-peak ripples are 2.5% for load voltage, 5% for load current, and 10% for filter [latex]L_e[/latex] current. (a) Determine the values of [latex]L_e, L[/latex], and [latex]C_e[/latex]. Use PSpice (b) to verify the results by plotting the instantaneous capacitor voltage [latex]v_c[/latex], and instantaneous load current [latex]i_L[/latex]; and (c) to calculate the Fourier coefficients and the input current [latex]i_S[/latex]. The SPICE model parameters of the transistor are [latex]IS = 6.734f, BF = 416.4, BR = 0.7371, CJC = 3.638P, CJE = 4.493P, TR = 239.5N, TF = 301.2P[/latex], and that of the diode are [latex]IS = 2.2E - 15, BV = 1800V, TT = 0[/latex].

Accepted Answer

[latex]V_s = 110 V, V_a = 60 V, I_a = 20 A[/latex]

[latex]ΔV_c = 0.025 × V_a = 0.025 × 60 = 1.5 V[/latex]

[latex]R=\frac{V_a}{I_a}=\frac{60}{20}=3 \Omega[/latex]

From Eq. (5.56),

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

[latex]k=\frac{V_a}{V_s}=\frac{60}{110}=0.5455[/latex]

From Eq. (5.57),

[latex]I_s = kI_a=0.5455 × 20 = 10.9 A[/latex]

[latex]ΔI_L = 0.05 × I_a = 0.05 × 20 = 1 A[/latex]

[latex]ΔI=0.1 × I_a = 0.1 × 20 = 2 A[/latex]

a. From Eq. (5.59),we get the value of [latex]L_e[/latex]:

[latex]\Delta I=\frac{V_a(V_s -V_a)}{f LV_s}=\frac{60 imes (110 - 60)}{2 imes 20 kHz imes 110}=681.82 \mu H[/latex]

From Eq. (5.61) we get the value of [latex]C_e[/latex]:

[latex]C_e=\frac{\Delta I}{\Delta V_c imes 8f}=\frac{2}{1.5 imes 8 imes 20 kHz}=8.33 \mu F[/latex]

Assuming a linear rise of load current iL during the time from [latex]t = 0[/latex] to [latex]t_1 = kT[/latex], we can write approximately

[latex]L\frac{\Delta I_L}{t_1}=L\frac{\Delta I_L}{kT}=\Delta V_C[/latex]

which gives the approximate value of L:

[latex]L=\frac{kT\Delta V_c}{\Delta I_L}=\frac{k\Delta V_c}{\Delta I_Lf}[/latex] (5.160)

[latex]=\frac{0.5454 imes 1.5}{1 imes 20 kHz}=40.91 \mu H[/latex]

b. [latex]k = 0.5455, f = 20 kHz, T = 1/f = 50 µs[/latex], and [latex]t_{on} = k ×T = 27.28 s[/latex]. The buck chopper for PSpice simulation is shown in Figure 5.41a. The control voltage [latex]V_g[/latex] is shown in Figure 5.41b. The list of the circuit file is as follows:

Question 5.10: A buck converter is shown in Figure 5.40. The input voltage ...

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