I am making a 500 V voltage supply using DC-DC boost converter from 5 V.

Apparently inductors have maximum transient voltage rating to prevent the breakdown of the wire insulation material or the ferrite core. But it seems most inductors don't list this voltage rating.

In practice, is the max voltage rating a parameter of concern in designing high voltage boost converters?

  • 2
    \$\begingroup\$ As a practical matter you'll probably wind up needing a transformer using a flyback topology instead of an inductor. 5 V -500 V in a simple boost stage is going to be difficult. \$\endgroup\$
    – John D
    Commented Sep 16, 2021 at 22:43
  • \$\begingroup\$ You should choose an inductor that does have the breakdown voltage rating of the wire insulation. \$\endgroup\$
    – Andy aka
    Commented Sep 17, 2021 at 7:10
  • \$\begingroup\$ I would assume 500 V winding to winding isolation unless anything else is specified, but I would not trust that to be put into mass production. As Andy says, find one with stated voltage withstand. \$\endgroup\$
    – winny
    Commented Sep 17, 2021 at 8:57

1 Answer 1


So I was possible to get 500 V from 5 V supply using the boost converter topology. Be careful that many commercial chips have maximum duty cycle around 80 - 90% and thus would not work in this case. 500 V can be achieved either by using PWM of a microtroncoller / function generator, or instead operate it in the discontinuous mode.

I grabbed a off-the-shelf inductor with no voltage rating, and it worked fine. Most inductors seem not to come with this rating, and the actual functioning depends on how the wires are wound. If one is generating 500 V using an inductor with 100 turns, then the voltage between each turn is about 5 V and poses no problem, unless the wire is wound on top of each other and the last turn is on top of or near the first turn. In this case, the voltage will be close to the full high voltage.


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