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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.

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2 Answers 2

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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.

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  • \$\begingroup\$ Thank you for the clarification and I see how you derived the final equation (CVoutdeltaV). However, what additional steps would I need to take to then Vout is not much larger than Vf-Vi, for example during the first few cycles? \$\endgroup\$
    – Pyrohaz
    Jan 5, 2016 at 1:05
  • \$\begingroup\$ Use the expressions on the left of the same equation with Vf and Vi (they do not have the assumption Vout >> Vf-Vi). \$\endgroup\$
    – rioraxe
    Jan 5, 2016 at 5:25
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See this for the energy delivered per cycle: SE Energy, and then the deltaV on the capacitor is calculated from deltaEnergy = C.VOUT.deltaV

You can operate at about twice the current if you operate in continuous conduction.

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  • \$\begingroup\$ Aha! I seem to be getting a much more reasonable answer in my Matlab simulations now. How did you derive the deltaEnergy equation? Also, from steady state, with an output capacitor charged to the supply voltage, how would one avoid a divide by zero from your energy equation? I'm currently avoiding that by setting the initial capacitor voltage to Vs+0.1. \$\endgroup\$
    – Pyrohaz
    Jan 3, 2016 at 15:00
  • \$\begingroup\$ The energy equation is only an approximation (assumes VOUT remains constant) -- so starting from VOUT=VIN, if VOUT remained at VIN (e.g. very very large COUT), then in fact it would take a long time for the inductor current to decay to 0, and there would actually be a large energy delivered. In reality, COUT is not infinite, and VOUT does change a small amount during the pulse; a more precise equation would be needed if you cared. @nd approximation -- deltaEnergy = C.VOUT.deltaV is only true if deltaV is small. Need a similar increase in detail if you cared for an exact expression. \$\endgroup\$
    – jp314
    Jan 4, 2016 at 0:38

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