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High-Voltage capacitor charge indicator.

Discussion in 'General Electronics Discussion' started by tip120, Dec 31, 2010.

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  1. tip120

    tip120

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    Sep 24, 2010
    So, I have a small railgun power supply I built for a 2 inch rail length railgun I'm building. The capcitor 'bay' consists of two 450v 560uf caps. I need to figure out a way to make a charge indicator for these caps. How can I make one using an LED or neon lamp that will light then the cap voltage reaches 450~?

    I've tried all manner of LEDs/Resistors and done a lot of research, I just can't seem to figure it out. Seems like no matter what ohm resistor I use, the LED still lights at low voltage. Using a 20k resistor, it lights at around 50v, normal brightness. Using a 550k resistor, it doesn't light until 100v, but it's VERY dim, and doesn't get any brighter past that point. LED is good, not blown, tested it afterward.

    Help? :(
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,269
    2,718
    Jan 21, 2010
    Your best bet is to get some small neon bulbs (which strike at between 45 and 90 volts) and power them from a voltage divider. When the voltage reaches the strike voltage, the neon lamp will light (and remain lit even if the voltage drops substantially).

    The "top" resistor of the voltage divider should be around 500k for your application. The lower resistor will then be between about 50k and 120k depending on the lamp.

    You should slowly charge the capacitor and note when the lamp lights. Modify the resistor if required.

    edit: Perhaps an easier way to describe this would be a 500k resistor in series with the lamp and a 50-120k resistor in parallel with the lamp.
     
  3. shrtrnd

    shrtrnd

    3,688
    457
    Jan 15, 2010
    Yep, *steve*'s right.
    You're going to zap just about anything else you use, besides a neon bulb.
    (If you don't zap yourself first)
    Get your car out of the way in your garage, before you fire that thing.
     
  4. tip120

    tip120

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    Sep 24, 2010
    Thanks steve, that worked perfectly!

    An un-intended but useful side effect I noticed:

    The neon bulb fires at 400v with one pole. At 450v, both poles fire, so it gets brighter. At first I thought it was being over-powered and I needed a bigger resistor, but I checked the voltage across the bulb and at 400v the bulb is getting about 60v. At 450, it's getting about 75. It's rated for 120v ac.

    I thought when powering neon bulbs with DC, only one pole would fire, and both would fire when powering them with AC. Am I mistaken or is there something else going on I should look into?

    BTW: Used a 470k ohm resistor in series, 100k in parallel. :)
     
  5. Resqueline

    Resqueline

    2,848
    2
    Jul 31, 2009
    I can't say why it does that, apart from that it obviously gets very excited.. ;)
    Notice that a 120V AC rating of neons includes a series resistor. Without that it would simply blow up. Did you use a complete bulb with a base?
     
  6. tip120

    tip120

    12
    0
    Sep 24, 2010

    Yep, it has a resistor inside the base.

    As for both poles firing...

    The only thing I can figure is that somehow when the caps are nearly charged and the load on the charging transformer decreases, the DC-DC converter starts switching faster and somehow a oscillating waveform is slipping past the rectifiers. But I'm not sure how that's happening. I'd do more in-depth testing with my scope but I don't have a HV probe for it. :/
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,269
    2,718
    Jan 21, 2010
    One thing that's worth noting. Resistors have voltage ratings.

    Unless you know otherwise, it's probably best to assume the voltage rating is around 150V. Calculate the voltage across the resistors (assume the neon bulb is a dead short). If any have more than 150V across them, consider using 2 or more resistors in series in its place.

    If you exceed the voltage rating on a resistor the resistance may fall significantly. This may be the cause for the sudden increase in brightness at the higher voltage.
     
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