determining current from a discharging capacitor

Discussion in 'Power Electronics' started by Walks91, Apr 28, 2018.

  1. Walks91

    Thread Starter New Member

    Apr 28, 2018
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    So I know this is probably a simple question but I'm having a bit of trouble determining exactly how to calculate the max current generated by a capacitor bank. If I have a capacitor bank charged to say, 10kJ with 80000uf total capacitance and is charged by a 500V power supply. If I discharged the capacitor bank in parallel would my current still just be 500V divided by the total circuit resistance? (assuming all the components are in series)
     
  2. Walks91

    Thread Starter New Member

    Apr 28, 2018
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    I guess my time limit to edit the question expired but something I forgot to add was if my discharged current was simply I=V/R (capacitor bank charging voltage divided my total circuit resistance). What would be the difference between using a single 500V 80000uF capacitor vs. using a capacitor bank with ten 500V 8000uF capacitors? (These values are completely arbitrary)
     
  3. ebp

    Well-Known Member

    Feb 8, 2018
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    As long as total resistance includes the equivalent series resistance (ESR) of the capacitors, then Ohm's law applies to instantaneous conditions, except in circumstances where the total resistance is very low and equivalent series inductance (ESL) and other circuit inductance may be limiting. Be aware that capacitors suitable for very high discharge rate, especially if the the total energy through-put is high, will be large and very expensive. Ordinary electrolytic capacitors will not withstand this and even those made for high charge and discharge current are likely to be dubious if the rate is high. Power is dissipated in the ESR, so heating occurs and excessive heating can cause the capacitors to, if you are lucky, simply blow their safety vent, and if you aren't lucky, explode. Typically capacitors that will withstand very high energy throughput will be film-foil types. 80 millifarads at 500 volts will be a very very large bank making management of inductance and connection resistance an issue. I would expect the cost to be in the range of thousands of dollars.
     
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  4. Walks91

    Thread Starter New Member

    Apr 28, 2018
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    Thank you ebp that was a very helpful response. I'm researching magnetic pulse welding applications, so assuming I wanted to discharge say 10kA through a coil to produce a strong magnetic field inside and the capacitance is fairly high like around 4000uF could I assume that the instantaneous voltage is roughly the same or should I take the voltage from the equation Vcapacitor=Vsupply * e ^-(t/RC). The energy would be discharged through a high power switch and then through the aluminum coil so the total resistance would be very low.

    My simulation circuit is a 4000uF capacitor bank charged to 500VAC at a frequency of around 10kHz. I'm assuming the total resistance will be around 0.05 ohms and I want the discharge voltage after 20ns.

    So using the equation Vcapacitor=Vsupply * e ^-(t/RC) = 500*e^-(20ns/0.05*4000uF) I get about 450V, so that means I'd be getting an AC current of I=V/R = 450/0.05 = 9kA right?
     
  5. ebp

    Well-Known Member

    Feb 8, 2018
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    I'll give it some thought later, but my first response is eeeeee! That is huge current at very high rep rate which equates to huge power and extraordinary demands all round. The inductance of the coil may well be the limiting factor. The switch is about 11 orders of magnitude (any number base that suits your fancy) from trivial.

    "charged to 500VAC" - I assume you mean DC
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    7,653
    Just using a couple of numbers from your post gives some interesting results.

    Let's say that we want to apply 500 V to a coil and have the current rise from 0 to 10 kA in 20 ns. What is the most inductance that we can tolerate?

    V = L di/dt

    L = V / (di/dt)

    L = 500 V / (10 kA / 20 ns)

    L = 1 nH

    Keep in mind that this is the TOTAL allowed inductance of everything that the capacitor bank is driving -- including the coil, the lead wires, the switch, and its own internal inductance.

    You need to come at things from the other direction. First determine what the inductance is that you need to drive and then design your system to drive it.
     
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