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 b) the value of the output filter capacitor, C, so that the output voltage ripple to be lower than 0.5 V. Use the inductance you calculated in part (a)
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 b) the value of the output filter capacitor, C, so that the output voltage ripple to be lower than 0.5 V. Use the inductance you calculated in part (a)
Solution
To solve this problem, we need to use the formulas for the boost converter in continuous mode and the formula for the output voltage ripple.
a) The smallest value of the inductor, L, in order to have continuous current can be calculated using the formula:
L = (Vin * D) / (2 * ΔIL * fs)
Where: Vin = input voltage = 36V D = duty cycle = Vin / Vo = 36V / 108V = 1/3 ΔIL = change in inductor current. For a boost converter in continuous mode, this can be approximated as (Vin * D) / fs fs = switching frequency = 51kHz
Substituting the values into the formula, we get:
L = (36V * 1/3) / (2 * (36V * 1/3) / 51kHz) = 0.5 mH
b) The value of the output filter capacitor, C, so that the output voltage ripple to be lower than 0.5 V can be calculated using the formula:
C = Iout / (8 * fs * Vripple)
Where: Iout = output current. This can be calculated as Po / Vo. Since Po varies from 107W to 200W, we will use the maximum value for the worst case scenario, so Iout = 200W / 108V = 1.85A fs = switching frequency = 51kHz Vripple = output voltage ripple = 0.5V
Substituting the values into the formula, we get:
C = 1.85A / (8 * 51kHz * 0.5V) = 90.2 μF
So, the smallest value of the inductor, L, in order to have continuous current is 0.5 mH and the value of the output filter capacitor, C, so that the output voltage ripple to be lower than 0.5 V is 90.2 μF.
Similar Questions
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
b) the value of the output filter capacitor, C, so that the output voltage ripple to be lower than 0.5 V. Use the inductance you calculated in part (a).
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
A Buck converter is charging a battery, with the following specifications: (Consider ideal switch and ideal diode) a) Vin = 25 V, Vb = 8 V, RL=20 Ω, b) fs= 100kHz (switching frequency), d=0.4 (duty ratio) c) Peak to peak inductor current ripple is 10% of the output current. d) Output voltage ripple across capacitor is 5% of the output voltage. What is the minimum inductor value L from the below Circuit
A Buck converter is charging a battery, with the following specifications: (Consider ideal switch and ideal diode) a) Vin = 25 V, Vb = 8 V, RL=20 Ω, b) fs= 100kHz (switching frequency), d=0.4 (duty ratio) c) Peak to peak inductor current ripple is 10% of the output current. d) Output voltage ripple across capacitor is 5% of the output voltage. What is the Capacitor C from the below Circuit in
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.