I have been for the past 3 months trying to design and experiment with topologies to charge a high voltage capacitor up to 1 kV. I have settled on a classic flyback topology:

Flyback topology used and high voltage capacitor on output

I have tried several transformers and understand how important elements such as the turn ratio, switching frequency, core type and leakage inductance impact my application. I struggle however to understand the following formula: Vo = N . Vi . D / (1-D), where D is the duty-cycle (D = ton. fSW). This would imply that to get my high voltage capacitor up to 1 kV from a 12V source I would need an 83 turn ratio. However, I have seen multiple designs where a low turn ratio transformer is used to generate high voltage such as the following:

350V from a 12V input

I found contradicting information left and right on the forum about a high turn ratio. I have read explanation to view the flyback topology from the perspective of energy transfer and that what I am looking at is delivering pulses of current on my secondary to gradually charge my capacitor to the desired voltage. Is there an assumption on the formula that I am missing? How is it possible that we can observe a high voltage on the output with low turn ratios?


1 Answer 1


A flyback design whether step up or step down can operate in two modes of operation: -

  • DCM (discontinuous conduction mode)
  • CCM (continuous conduction mode)

In CCM your formula is correct but, in DCM, what you can transfer to the secondary is governed by this formula: -

enter image description here

Image from my basic website, flyback section.

In other words, in DCM, whatever energy you transfer to the secondary gets used by the secondary and, can therefore continue to charge up a capacitor to whatever voltage you require.

This cannot happen in CCM because of the constraints imposed by switching waveforms. In other words, you cannot arbitrarily produce a larger output voltage without increasing the duty cycle, D.

In DCM, if you keep increasing \$R_L\$, in order to burn the energy transferred from the primary, a higher secondary DC voltage must be reached. It's all in the formulas.

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    \$\begingroup\$ @TimWilliams and Andy aka (you'll get notified as the post owner even without an @) - We would prefer site members to resolve minor disagreements themselves. However, the comments here been flagged and therefore we have to get involved. I'll move the previous comments into a chatroom and I plan to post some site usage information in that chatroom shortly, but not immediately (real life takes priority). It should be later tonight EU (UTC+2) time. I'll share the chatroom link & invite you both then. Please don't post further comments here. Thanks. \$\endgroup\$
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    Commented Jul 13, 2023 at 19:02

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