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I have a large capacitor made of two smaller 200V rated 560uF capacitors. If I charge this capacitor to 30V with 4.5A from a bench power supply and disconnect it after around 7 seconds, how do I calculate the short circuit current between the positive and negative terminals?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ The short circuit current of any charged ideal capacitor is infinity. \$\endgroup\$
    – Andy aka
    Oct 1, 2021 at 7:14
  • \$\begingroup\$ best not try to determine this by experiment as well. \$\endgroup\$
    – danmcb
    Oct 1, 2021 at 7:39
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    \$\begingroup\$ 2 * 560uF = 1.12 mF and not 1.02 mF. \$\endgroup\$
    – Elec1
    Oct 1, 2021 at 9:04

4 Answers 4

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What you are asking is very complicated. Looks easy enough until you realise that you are connecting a 0Ω resistor across a voltage source. The current you would get if that 0Ω was ideal, and the capacitor is ideal:

$$ I = \frac{V}{R} = \frac{V}{0\Omega} = \infty A $$

The time it would take to discharge the capacitor is zero seconds, since the time constant is also zero:

$$ \tau = R \times C = 0\Omega \times C = 0 $$

So you have inifinite current passing for no time, which doesn't make sense at all.

In reality the capacitor is not ideal, and the wire you use to short circuit the capacitor is not 0Ω. So the first thing you need to find out is what is the actual resistance between the capacitor's terminals when you "short circuit" it.

The other thing you need to know is the equivalent series resistance (ESR) of the capacitor, and that will not be trivial. It is highly dependent on rate of discharge, temperature, and other factors.

Lastly, the physical loop you form (around which current will flow) has inductance, which will depend on loop area, and whatever else is present inside the loop. Inductance will have the effect of preventing current from rising instantly, further complicating your calculations.

In short (pun intended) you are asking for something way beyond a simple formula, and you can easily damage the capacitor by discharging it in this way.

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The short circuit current decreases rapidly with time and depends critically on the internal resistance of the cap. Also, to a charged up capacitor, even the wire you use to short the terminals may contribute resistance that appreciably reduces the peak amplitude of the short circuit current. So the test setup matters, too.

So you can estimate it based on the resistance of the capacitors, or you can short the cap with a precision shunt and measure the voltage vs time with an oscilloscope. But ideally the shunt should be smaller than the series resistance of the cap.

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You don't yet have enough information to calculate it.

You need to know the ESR and ESL (equivalent series resistance and inductance respectively) of the capacitor, and the same for the external circuit.

The current will be 'large'.

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A simplified model of a capacitor is shown below.
It consists of:

  • Cp = parallel capacitance, or the bulk capacitance. In your case, the 1020uF capacitance.
  • Rp = parallel resistance, or, the leakage resistance. Typically in the tens of meg-ohm range for a properly formed aluminum electrolytic capacitor.
  • Ls = series inductance. This includes the inductance of internal plates, internal wiring, and leads. This could be hundreds of nH to tens of uH depending on the construction of the capacitor and lead length.
  • Rs = series resistance. Commonly called Equivalent Series Resistance (ESR) which has a frequency dependency. This includes the resistance of the capacitor structure, internal wiring, and external leads. This can be in the low milli-ohms to a few ohms depending on construction.
  • Rshort and Lshort are the resistance and inductance of your shorting conductor. These items are added to Rs and Ls.

The series inductances, Ls + Lshort, affects the rise time of short-circuit current.

The series resistance, Rs + Rshort, determines the short-circuit current (use Ohm's law). Rs & Rshort will change when you short the capacitor due to heating which is hard to quantify. Repeated high current shorts can damage the internal structure of the capacitor due to spot heating of the internal structure causing parametric changes.

The parallel resistance, Rp, will determine the discharge of the capacitor when not connected to the charging source. With a 1000uF capacitor, you won't see any significant change in voltage over 7 seconds after you remove the charging source.

Trying to calculate the short circuit current with any accuracy is a bit dicey since the loss parameters are dynamic during the short event. You also have changes in loss parameters with age and usage for aluminum electrolytics.

Shorting capacitors is fun to do, but err on the side of caution and wear eye protection. Good quality capacitors charged to 30V could generate 300+ Amps which can cause molten bits of metal to be expelled. I've had a 150V, 2 Farad, capacitor bank vaporize a PCB copper ground plane which caused a plasma arc which melted the head of a stainless steel bolt a few cm away. Took a few hours for hearing to return to normal.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ That's an immense amount of stored energy. I imagine it didn't all escape, but ... ! || LASER? Other? \$\endgroup\$
    – Russell McMahon
    Oct 1, 2021 at 23:21
  • \$\begingroup\$ @RussellMcMahon For a high power sonar. I have a picture of the affected area hanging on my cork-board. The bang was probably equivalent to a .38 handgun going off in the room, i.e. very loud and debilitating. And then it recharged and arced a second time before someone got enough sense to kill the power. \$\endgroup\$
    – qrk
    Oct 2, 2021 at 1:29
  • \$\begingroup\$ A 1.5 litre PET softdrink bottle at 150 psi makes an awesomely loud explosion Stand next to it and you can feel the pressure pulse all over - but nowhere the energy of your discharge. Hearing slowly trickles back in after a few minutes. \$\endgroup\$
    – Russell McMahon
    Oct 2, 2021 at 7:53

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