I need to know the current in a polarized electrolytic capacitor. What is the practical limit for this current and how do I calculate?
I want to know what is the maximum current I can get when I put the terminals in contact without considering the rest of the circuit. I know that if only the resistance of the terminals is considered, the current would be very high, but what prevents this current from reaching such high values? You can use a capacitor of 100uF and 250v as an example.
A half-decent polarized electrolytic capacitor model (in the forward regime, which is where you should be) is as follows:
simulate this circuit – Schematic created using CircuitLab
A decent capacitor datasheet will typically either directly give you the component values, or will give you the data to be able to calculate these.
From there, you can calculate the current across the capacitor, and power loss in the capacitor.
You can often - not always - neglect Ce, in which case a shorted capacitor is just a series RLC circuit. Don't forget to include the inductance (and resistance) of the wire completing the short - in some cases this can be significant.
The datasheet will then either tell you the max current, or max temperature, for a given lifespan.
Some datasheets are better than others.
The above being said, most electrolytics are not rated for a short circuit.
Here's an example. By playing around a bit, here's an approximate model for the ESK106M050AC3AA (this is very helpful when eyeballing this sort of thing...):
(You can kind of think of R1 as the lead+package resistance, L1 as the lead+package inductance, C1 as the capacitance of the oxide layer, R2 as the electrolyte resistance, and C2 as the electrolyte capacitance.)
At t=0, you magically short the 50v-charged capacitor across the leads with a zero-inductance zero-resistance crowbar. (Note you cant really do that with this simulation, so I'm faking it by adding a 50v step at t=0 instead). Current rises to ~40A in ~50ns while C2 charges, drops back down to ~12.5A, then decays from there as C1 charges. As a sanity check: total energy dissipated is ~12.5mJ, which matches the expected value (1/2 C V^2).
The capacitor has a total volume of ~0.22mL. If you make the (bad) assumption that the capacitor is made out of water with a heat capacity of 4.2 J/mLK, then this results in a temperature rise of ~0.01K. Negligible.
However. This overestimates the heat capacity of the capacitor, and also assumes that the power dissipation is uniformly distributed throughout the capacitor. Let's look at the datasheet to see what the manufacturer has to say...
Rated ripple current: 65mA @ 120Hz. With a compensation factor of 1.5 at high frequency, so ~98mA. We're out of spec by more than two orders of magnitude here. Whoops.
In practice this may be fine - ripple current rating is typically for continual operation whereas this is probably a one-shot - but it should certainly give you pause.
[ assumption: We are dealing with low frequency currents, or DC. If we were dealing with high frequencies, we wouldn't be dealing with electrolytic capacitors.
I have first-hand experience with discharging 10s of A from 2mF capacitors into LEDs.
Hundred times a second. On purpose. ]
The short circuit current will be limited by ESR (equivalent series resistance) of the capacitor.
\$ I = \cfrac {V} {R_{ESR}} \$ .
ESR is often listed in datasheets. Or, \$ \tan \delta \$ is listed in datasheets, and ESR can be calculated as \$ R_{ESR} = \cfrac {\tan \delta} {2 \pi f C} \$ . See this datasheet for an example.
Notice that ESR varies appreciably even within one family of capacitors. It also varies with temperature. Further reading about electrolytic capacitor ESR: here and here.
