Electronic – What happens to a polarity sensitive capacitor when you feed a square wave through it

capacitor

Consider the following circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

The output wave form will look like this

enter image description here
Orange – input signal
Blue – output signal

The Wikipedia entry on Electrolytic capacitors says

Nevertheless, electrolytic capacitors can withstand for short instants
a reverse voltage for a limited number of cycles. In detail aluminum
electrolytic capacitors with non-solid electrolyte can withstand a
reverse voltage of about 1 V to 1.5 V. This reverse voltage should
never be used to determine the maximum reverse voltage under which a
capacitor can be used permanently

It does not specify what a short duration is.

Since polarized capacitors are meant to be used one way, will the falling edge's dv/dt damage the capacitor ?

Best Answer

I disagree with the answer given. I think the answer confuses a negative voltage (with respect to zero or ground) with potential difference between two points in a circuit (i.e. the plates of the capacitor).

The voltage ACROSS a capacitor cannot be changed instantly, so the positive plate is still MORE POSITIVE than the negative plate at the transition, thus it always maintains the normal polarisation.

enter image description here

Case 1: Suppose the input was 0V to 5V (or 0 to +V).

On the negative edge of the input pulse the positive plate falls to 0V BUT the negative plate falls to -5V at the same time maintaining the correct polarisation direction (positive plate more positive than the negative). The negative plate then discharges through R1 back to zero.

On the positive edge the voltage across the capacitor is zero. The positive plate rises to +5V as does the negative plate. We still have zero volts across the capacitor. The negative plate then discharges through R1 to zero BUT the positive plate remains at +5 volts. The capacitor is polarised normally.

At no time is the negative plate more positive than the positive plate.

Case 2: (As suggested in comments) Suppose the input was +5v to -5V (or +V , -V)

This simply cannot be the circuit as drawn above. There is a common (ground) connection clearly shown between the pulse signal and R1 with a positive end shown connected to the capacitor.