0

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.

winny
  • 13,064
  • 6
  • 46
  • 63
zerobeat
  • 1
  • Why start with 3.3 V? Is that really the highest you have available? – Finbarr Aug 07 '23 at 13:57
  • 5v is available too, but 3.3v is already enough to make the circuit work. – zerobeat Aug 07 '23 at 14:36
  • 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. – rdtsc Aug 07 '23 at 14:39
  • If your Geiger tube is drawing 1 mA, your health is in danger. – John Doty Aug 07 '23 at 14:45
  • 1
    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 http://techlib.com/science/geiger.html). – GodJihyo Aug 07 '23 at 14:49

1 Answers1

0

Or is this even a issue?

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

Andy aka
  • 434,556
  • 28
  • 351
  • 777
  • 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. – zerobeat Aug 07 '23 at 14:39
  • It's still not a problem @zerobeat – Andy aka Aug 07 '23 at 17:39