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Ludic science light dimmer driver

I was watching this video made by ludic science and at 3:14 he said that at the maximum output of the high voltage ignition coil, the distance was 10mm/1cm which he claimed to be 10-12kv. Isn't the gold standard for the breakdown voltage of air 30kv/cm? Can someone explain?

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    \$\begingroup\$ No standard. Breakout voltage depends on air pressure, humidity, amount of dust, ionization. That value just for reference for some normalized condition. \$\endgroup\$
    – user263983
    Jan 25 '21 at 13:01
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There is a big difference in dielectric Breakdown Voltage BDV for smooth parallel surfaces and point sources from sharp wire tips due to the E-field gradient force effects on ionization.

Yes 30kV/cm or 3kV/mm is true for smooth area surfaces.

But for wire tips it is 10kV/cm or 1kV/mm This also raises the resistance and this begins arcing with a lower current, which is less visible.

This of course varies with dust , humidity and pressure.

This BDV has nothing to do with the "holding current" gap which may be increased after conduction occurs.

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The guy says that at maximum power, the distance of the arc is around 10 mm but, in order to initiate that arc he moves the electrodes to a much smaller distance i.e. he lowers the distance to "start" breakdown (about 5mm). Once breakover has occurred, the arc can be drawn to a greater distance. That is what you are seeing.

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  • \$\begingroup\$ That's probably correct but to quote the guy "the arc is around 10mm, which indicates an output voltage of 10-12kv" Isn't he saying 10kv means 10mm? \$\endgroup\$ Jan 25 '21 at 13:07
  • \$\begingroup\$ I'm not going to try and interpret what he says or what he meant to say. I can only point out that arcing initiates at about half the distance that it may be drawn to and that's because the air becomes ionized. Once ionized, the arc can be extended. You interpret this one way and he may have meant another thing @import_hill \$\endgroup\$
    – Andy aka
    Jan 25 '21 at 13:13
  • \$\begingroup\$ You make a valid point, I just wasn't sure as he didn't really specify in the video. He said in a comment though(that I have just linked to the post) that it's roughly 1000 v per mm. I'm not sure if that's quite accurate. \$\endgroup\$ Jan 25 '21 at 13:19
  • \$\begingroup\$ Air breaks over with pointed conductors at a lower voltage too. Go do some research on it is my advice. What I see in the video seems about right for the voltage mentioned. \$\endgroup\$
    – Andy aka
    Jan 25 '21 at 13:26
  • \$\begingroup\$ Yup it seems another answer has reached the same consensus, thanks \$\endgroup\$ Jan 25 '21 at 13:28

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