Voltage (or power)-current relation of a small supercapacitor

Question

Is size a factor that limits the maximal current that a capacitor can take?

I got this question when seeing Figure 7 of this supplementary material of the research of the smallest super-capacitor now. Why the voltage drops to zero as the current goes high? Similarly the power increases as current increases at first yet it suddenly drops after a point of a current and drops to zero as well as current is getting higher.

Should I interpret (based on the description of device studied and general principle of supercapacitor) the result that the supercapacitor, due to its small size, cannot withstand higher currents (relatively, the current displayed in the diagram are not high in themselves)? Is it generally true that the smaller the size of a (super)capacitor, the lower the maximal current it can withstand?

Answer 1

These capacitors have a significant internal resistance -- basically a resistance in series with the terminals.

As you draw current, the voltage drop across this internal resistance increases, thereby reducing the voltage at the terminals. At maximum current, the voltage drop equals the voltage stored on the capacitor, so you get 0 V at the terminals.

Generally, the size of the capacitor (and construction/technology) defines the internal resistance. Larger capacitors will have lower internal resistance.

There is a theorem in electronic circuits that you get maximum power when the load resistance equals the internal resistance. In this condition however, your efficiency is only 50 %, and you get only 50 % of the capacitor's stored voltage.