A boost converter has the following specifications: Vin = 36 V, Vo= constant = 108 V, Fs = 51 kHz, and the output power may vary with a range of Po(range) = 107 W to 200 W. Assuming that the converter is ideal, calculate: a) the smallest value of the inductor, L, in order to have continuous current
Question
A boost converter has the following specifications:
Vin = 36 V, Vo= constant = 108 V, Fs = 51 kHz, and the output power may vary with a range of Po(range) = 107 W to 200 W. Assuming that the converter is ideal, calculate:
a) the smallest value of the inductor, L, in order to have continuous current
Solution
To calculate the smallest value of the inductor, L, for continuous current in a boost converter, we can use the following formula:
L = (Vin * (Vo - Vin)) / (2 * ΔI * fs * Vo)
Where: Vin = Input voltage = 36V Vo = Output voltage = 108V fs = Switching frequency = 51kHz ΔI = Change in current
The change in current, ΔI, can be calculated using the formula:
ΔI = Po_min / (Vin * D)
Where: Po_min = Minimum output power = 107W D = Duty cycle
The duty cycle, D, for a boost converter is given by:
D = 1 - (Vin / Vo)
Substituting the given values:
D = 1 - (36V / 108V) = 0.67
Substituting D back into the ΔI equation:
ΔI = 107W / (36V * 0.67) = 4.46A
Finally, substituting all the values back into the L equation:
L = (36V * (108V - 36V)) / (2 * 4.46A * 51000Hz * 108V) = 0.000014 H or 14 μH
So, the smallest value of the inductor, L, in order to have continuous current is 14 μH.
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