In a Boost converter of Figure below, the inductor current has ΔiL = 2 A. It is operating in dc steady state under the following conditions: Vin = 5 V, Vo = 12 V, Po = 11 W, and fs = 200 kHz. Assuming ideal components, calculate the inductance value in Micro Henery
Question
In a Boost converter of Figure below, the inductor current has ΔiL = 2 A. It is operating in dc steady state under the following conditions: Vin = 5 V, Vo = 12 V, Po = 11 W, and fs = 200 kHz. Assuming ideal components, calculate the inductance value in Micro Henery
Solution
The question seems to be incomplete as there is no figure provided. However, assuming that it's a standard boost converter, we can use the following formula to calculate the inductance:
L = (Vin * (Vo - Vin)) / (fs * ΔiL * Vo)
Where: Vin = Input voltage = 5V Vo = Output voltage = 12V fs = Switching frequency = 200kHz = 200,000Hz ΔiL = Change in inductor current = 2A
Substituting the given values into the formula:
L = (5 * (12 - 5)) / (200,000 * 2 * 12) L = 35 / (4,800,000) L = 7.29e-6 H
To convert Henry to Micro Henry, we multiply by 1,000,000:
L = 7.29e-6 * 1,000,000 L = 7.29 uH
So, the inductance value is approximately 7.29 uH.
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