Voltage increase per cycle across capacitor for a boost converter

boostcapacitorchargeconverterinductor

So I'm looking to make a high voltage capacitor charger using a boost converter and I'm wondering how much voltage the capacitor will increase per cycle of the boost converter. In reality, I would most likely design around a high powered inductor, charging to just below the maximum saturation current, then allowing enough time for the inductor to fully discharge into the capacitor (just on the brink of discontinuous operation).

I know that the discharge time will be dependent on the output voltage thanks to a previous answer. The voltage however will be changing with time as the current from the inductor decreases. The rate of decrease will also be dependent on this output voltage meaning for a fixed duty cycle and frequency, the output voltage will not rise linearly and the rate of increase will decrease as time increases.

How would I go about finding an equation for voltage increase per cycle? I'm assuming it will be likely iterative, depending on the output voltage of the last cycle.

Best Answer

Energy stored in a capacitor is \$ \frac{1}{2}CV^2 \$.

Energy stored in a inductor is \$ \frac{1}{2}LI^2 \$.

So the conservation of energy equation would be: $$ \frac{1}{2}LI_i^2 - \frac{1}{2}LI_f^2 = \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2 $$ (subscript i means initial, f means final)

Using an assumption that \$V_{out}\$ is near constant over one cycle, that is \$ V_{out} \gg V_f-V_i \$, then \$ 2 V_{out} \approx V_{f} + V_{i} \$. $$ \frac{1}{2}CV_f^2 - \frac{1}{2}CV_i^2 = \frac{1}{2}C(V_f^2-V_i^2) = \frac{1}{2}C(V_f+V_i)(V_f-V_i) = C V_{out} \Delta V$$ This is the answer given by jp314.