Finding the Load Inductance to Limit the Load Ripple Current
The converter in Figure 5.4 has a load resistance $R = 0.25 Ω$ , input voltage $V_s = 550 V$ , and battery voltage $E = 0 V$ . The average load current $I_a = 200 A$ and chopping frequency $f = 250 Hz$ . Use the average output voltage to calculate the load inductance L, which would limit the maximum load ripple current to 10% of $I_a$ .

Question

Finding the Load Inductance to Limit the Load Ripple Current
The converter in Figure 5.4 has a load resistance [latex]R = 0.25 Ω[/latex] , input voltage [latex]V_s = 550 V[/latex], and battery voltage [latex]E = 0 V[/latex]. The average load current [latex]I_a = 200 A[/latex] and chopping frequency [latex]f = 250 Hz[/latex]. Use the average output voltage to calculate the load inductance L, which would limit the maximum load ripple current to 10% of [latex]I_a[/latex].

Accepted Answer

[latex]V_s = 550 V, R = 0.25 Ω[/latex], [latex]E = 0 V, f = 250 Hz, T = 1/f = 0.004 s[/latex], and [latex]i = 200 ×0.1 = 20 A[/latex]. The average output voltage [latex]V_a = kV_s = RI_a[/latex]. The voltage across the inductor is given by

[latex]L\frac{di}{dt}=V_s-RI_a=V_s-kV_s=V_s(1 -k)[/latex]

If the load current is assumed to rise linearly, [latex]dt = t_1 = kT[/latex] and [latex]di = Δi[/latex]:

[latex]\Delta i=\frac{V_s(1 -k)}{L}kT[/latex]

For the worst-case ripple conditions,

[latex]\frac{d(\Delta i)}{dk}=0[/latex]

This gives k = 0.5 and

[latex]Δi L = 20 ×L = 55011 - 0.52 ×0.5 ×0.004[/latex]

and the required value of inductance is [latex]L = 27.5[/latex] mH.

Question 5.3: Finding the Load Inductance to Limit the Load Ripple Current...

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