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
Question
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
Solution 1
To find the value of the capacitor C, we first need to find the output current (Io) using Ohm's law (V = I*R).
Given: Vb (output voltage) = 8V RL (load resistance) = 20Ω
So, Io = Vb / RL = 8V / 20Ω = 0.4A
The output voltage ripple across the capacitor is given as 5% of the output voltage. So, the voltage ripple (ΔV) is 0.05 * Vb = 0.05 * 8V = 0.4V
The switching frequency (fs) is given as 100kHz or 100,000 Hz.
The value of the capacitor C can be found using the formula for the ripple voltage of a buck converter:
ΔV = Io / (fs * C)
Rearranging the formula to solve for C gives:
C = Io / (fs * ΔV) = 0.4A / (100,000Hz * 0.4V) = 0.4A / 40,000V = 10^-5 F or 10 μF
So, the value of the capacitor C is 10 μF.
Solution 2
The question seems to be incomplete. Could you please provide the complete question or the missing information about the circuit?
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