I'm designing a Geiger counter power supply that boosts 3.3 V to 500 V.

The circuit is a basic voltage booster.

3.3 V power line is directly connected to inductor where high voltage is generated. How can i protect the 3.3 V line = other devices connected to it, from high voltage transients generated in the inductor? Or is this even a issue?

My initial thoughts were using TVS diode before the voltage booster.

  • \$\begingroup\$ Why start with 3.3 V? Is that really the highest you have available? \$\endgroup\$
    – Finbarr
    Aug 7, 2023 at 13:57
  • \$\begingroup\$ 5v is available too, but 3.3v is already enough to make the circuit work. \$\endgroup\$
    – zerobeat
    Aug 7, 2023 at 14:36
  • \$\begingroup\$ What Finbarr's getting at is, 500:3.3 = 151.515 times as much voltage. So to get 1mA out at 500V, 151.515mA would be drawn from 3.3V (ignoring about 20% loss.) At 5V this current drops to 100mA plus losses - so higher voltages are easier to boost. \$\endgroup\$
    – rdtsc
    Aug 7, 2023 at 14:39
  • \$\begingroup\$ If your Geiger tube is drawing 1 mA, your health is in danger. \$\endgroup\$
    – John Doty
    Aug 7, 2023 at 14:45
  • 1
    \$\begingroup\$ You don't need 1mA with a Geiger tube, they need maybe 100uA when there's a pulse, average current is much less. There are circuits online using 3V (see techlib.com/science/geiger.html). \$\endgroup\$
    – GodJihyo
    Aug 7, 2023 at 14:49

1 Answer 1


Or is this even a issue?

It's not an issue with any boost converter circuit that I've ever seen.

  • \$\begingroup\$ This is propably the case, but most of the geiger counters i know are standalone units powered from batteries and are not connected directly to other circuits than their own. This one is going to be added to a bigger system with other circuits powered from the same source. \$\endgroup\$
    – zerobeat
    Aug 7, 2023 at 14:39
  • \$\begingroup\$ It's still not a problem @zerobeat \$\endgroup\$
    – Andy aka
    Aug 7, 2023 at 17:39

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