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Charging Capacitor Bank

Discussion in 'General Electronics Discussion' started by LambdaTesseract, Dec 31, 2013.

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  1. LambdaTesseract


    Dec 31, 2013
    I want to charge a capacitor bank of 3 to 4 photo-flash capacitors in series. Each are 320v 110uF. I'm working on this with friends and one of them had the idea to create a modified "Joule Thief" where we would wrap the wire for the capacitor bank around the toroid of the joule thief in order to generate the high voltages we want.

    I was wondering a) will this idea work and b) does anyone have any better ideas on how we could charge this capacitor bank? We would like to use disposable batteries as our power supply but that is by no means set in stone; we would very strongly like to make this portable however. chaining our selves to a wall plug or something is rather last ditch.
    Last edited by a moderator: Dec 31, 2013
  2. davenn

    davenn Moderator

    Sep 5, 2009
    hi there
    welcome to the forums :)

    have you actually googled circuits for joule thief ?

    there's many ideas out there :)

  3. LambdaTesseract


    Dec 31, 2013
    Ya we have found designs for joule thievs but what my friend is proposing is to wrap another wire around the toroid to charge capacitor bank; I don't know as much about magnetism, electricity and such so I thought I'd ask someone else about the soundness of the idea. More over there may be better ways of charging this bank than what we have thus far thought of.
  4. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Four 110 µF 320V capacitors in series are equivalent to a 27.5 µF 1280V capacitor.

    You would need a winding with very large number of turns to generate this much voltage from the flyback energy from a 1.5V battery, and it would take a very long time to charge. You would need a high-voltage diode such as BY203-16.

    Here are some suitable diodes from Digikey:

    You'll also need to put some resistance in series with it, otherwise the oscillator probably won't start up because of the loading of the secondary.

    If you want to run the circuit from a 1.5V battery, let's assume a flyback voltage of around 10V at the collector. That requires a turns ratio of 1280/10 = 128 times, so the secondary winding to charge the capacitor needs to have 128 times as many turns as the collector winding. You'll need a pretty big inductor to fit all that wire! But you can use very thin wire... which breaks easily!

    You may also need some way of detecting when the target voltage has been reached and stopping the circuit. Otherwise the diode will fail.

    Charging a 27.5 µF capacitor from 0V to 1280V requires 22.53 Joules of energy. This is actually quite a lot of energy; if you want to charge the capacitor in ten seconds, for example, you'll need to provide an average (mean) power of 2.253 watts; far more than a joule thief can provide from a single 1.5V AA battery!

    My suggestion would be a current-mode flyback regulator using a UC3845 or similar, running at a relatively low frequency (e.g. 10~20 kHz), driving a MOSFET, powered from a supply voltage of at least 12V, preferably more like 24~36V, from a battery of C or D cells. That should have a drain flyback voltage around 150V so your turns ratio would only need to be about 8.5.

    A suitable core for the inductor would be a pot core, RM14 or ETD core. It needs to be pretty big because of the size of the secondary winding. You'll need to learn about winding high-voltage windings on transformers because the high voltages can cause breakdown.

    Another option would be a voltage multiplier. These can't supply much current, so your capacitor charge time will be even longer than with a proper flyback converter.

    Google some of these keywords for more information, or ask again here.
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