Question
Download Solution PDFAn R-L load is connected to a 250 V, 400 Hz step down dc converter. The average load current is 100 A. The load resistance is 0.5 Ohm. The allowed ripple limit is 15%. The minimum value of inductor to limit the maximum ripple current is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFRipple in inductor current
When the switch is ON:
\(V_L=V_S-V_O\)
\(L{dI_L\over dt}=V_S-DV_S\)
\({dI_L\over dt}={V_S(1-D)\over L}\)
\({Δ I_L\over DT}={V_S(1-D)\over L}\)
\({Δ I_L}={V_S(1-D)D\over Lf}\)
where, ΔIL = Ripple in inductor current
Vs = Source voltage
L = Inductor
f = Frequency
D = Duty ratio
The duty ratio is given by:
\(D={V_o\over V_S}\)
\(D={I_oR\over V_S}\)
Calculation
Given, Vs = 250 V
f = 400 Hz
ΔIL = 15 % of output current = 15 A
Io = 100 A
R = 0.5 Ω
\(D={I_oR\over V_S}\)
\(D={100× 0.5\over 250}=0.2\)
\({Δ I_L}={V_S(1-D)D\over Lf}\)
\({15}={250(1-0.2)0.2\over L× 400}\)
L = 6.67 × 10-3
L = 6.67 mH
Last updated on Jun 6, 2025
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