Finding the Peak Load Current Ripple of Two Multiphase Dc–dc Converters

Two dc–dc converters control a dc separately excited motor and they are phase shifted in operation by π/2. The supply voltage to the dc–dc converter drive is $V_s = 220 V$ , the total armature circuit resistance is $R_m = 4 Ω$ , the total armature circuit inductance is $L_m = 15 mH$ , and the frequency of each dc–dc converter is $f = 350 Hz$ . Calculate the maximum peak-to-peak load ripple current.

Question

Finding the Peak Load Current Ripple of Two Multiphase Dc–dc Converters

Two dc–dc converters control a dc separately excited motor and they are phase shifted in operation by π/2. The supply voltage to the dc–dc converter drive is [latex]V_s = 220 V[/latex], the total armature circuit resistance is [latex]R_m = 4 Ω[/latex], the total armature circuit inductance is [latex]L_m = 15 mH[/latex], and the frequency of each dc–dc converter is [latex]f = 350 Hz[/latex]. Calculate the maximum peak-to-peak load ripple current.

Accepted Answer

The effective chopping frequency is [latex]f_e = 2 × 350 = 700 Hz, R_m = 4 Ω, L_m = 15 mH, u = 2[/latex], and [latex]V_s = 220 V. 4ufL_m = 4 × 2 × 350 × 15 × 10^{-3} = 42[/latex]. Because [latex]42 \gg 4[/latex], approximate Eq. (14.49)

[latex]\Delta I_{\max}=\frac{V_s}{4ufL_m}[/latex] (14.49)

can be used, and this gives the maximum peak-to-peak load ripple current, [latex]ΔI_{\max}= 220/42 = 5.24 A[/latex].

Question 14.11: Finding the Peak Load Current Ripple of Two Multiphase Dc–dc...

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