An R-L load is connected to a 230V, 400 Hz step down dc converter. The average load current is 100 A. The load resistance is 0.5 Ohm...
35) An R-L load is connected to a 230V, 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 :
Solution
This is a standard single-phase step-down (buck) chopper design problem.
Given
- Supply voltage, Vs = 230 V
- Chopper frequency, f = 400 Hz
- Average load current, Io = 100 A
- Load resistance, R = 0.5 Ω
- Ripple current limit = 15%
First find the average output voltage:
Vo = Io × R = 100 × 0.5 = 50 V
For a buck converter,
Vo = DVs
So the duty ratio is
D = 50 / 230 = 0.2174
Ripple Current
Allowed peak-to-peak ripple current:
ΔI = 0.15 × 100 = 15 A
For a buck converter operating in continuous conduction mode (CCM),
ΔI = Vo(1 − D) / (Lf)
Substituting the values:
15 = [50 × (1 − 0.2174)] / (L × 400)
15 = 39.13 / (400L)
L = 39.13 / 6000
L = 0.00652 H
L ≈ 6.5 mH
The nearest option is:
Answer: (B) 6.67 mH
The key formula used is:
ΔI = Vo(1 − D) / (Lf)
This is the inductor ripple-current expression for a buck converter operating in continuous conduction mode.