2
\$\begingroup\$

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 ?

\$\endgroup\$
  • 1
    \$\begingroup\$ As the Y axis is unlabelled, we can't tell. If the input voltage is 0V/5V you will be OK. If +/-2.5V, probably not. \$\endgroup\$ – Brian Drummond Dec 29 '14 at 11:39
  • \$\begingroup\$ You could use 2 electrolytes in series back-to-back. \$\endgroup\$ – Nazar Dec 29 '14 at 13:16
3
\$\begingroup\$

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.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Fantastic. Well explained and understood. \$\endgroup\$ – efox29 Dec 29 '14 at 19:37
1
\$\begingroup\$

This is not advisable, as it subjects the capacitor to reverse voltage. It's not "dv/dt" that hurts the cap, it's reverse voltage of more than a volt or so.

In actual practice, if the capacitor is an aluminum electrolytic type and is rated for a fairly high voltage, such as 63V or 100V, it may survive for some time.

Use a non-polarized or bipolar part for this kind of application.

|improve this answer|||||
\$\endgroup\$
1
\$\begingroup\$

An electrolytic capacitor will generally leak much more when reverse-biased than when forward biased. Any current which leaks through the cap in the reverse direction will cause the dielectric to become thinner; this effect will be greatest at the point where it is already thinnest. Current which leaks in the forward direction will cause the dielectric to become thicker, but because forward leakage is much below reverse leakage this effect will generally be much less significant.

What is important is not so much the duration for which the cap sees reverse voltage across it, but rather the total number of electrons that leak through the cap in each direction. If one charges some caps once to their "normal" voltage level, but in reverse, and then disconnects them with the charge sitting there, the caps which would be most quickly damaged by a reverse applied voltage will also be the ones which self-discharge the fastest; those which would self-discharge the most slowly would also be the ones least damaged by the reverse voltage. In either case, one "capacitor-load" of reverse charge flow should not cause significant damage.

This sort of principle is most important to understand when connecting polarized caps in series. If one connects e.g. two 10uF 6V caps in back-to-back series (with the - terminals together, e.g.) and drives the combination with 5 volts, then initially one cap will have +2.5 volts on it and the other will have -2.5; charge leaking through the -2.5V cap will cause that to shift toward one cap having 5 volts and the other zero. If polarity of the applied voltage were then reversed, the cap which had been charged to 5 volts would be discharged, and the one which had been discharged would be charged to 5 volts. Note that the total charge flowing through the series combination would be that required to charge one of the caps from 0 to 5 volts, but the voltage change would be twice that. The series combination of caps would thus pass half as much charge per volt as each cap individually, which is to say it would have half the capacitance.

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.