What if a capacitor exposed to slightly higher voltage than rated?

Question

I'm building my 5 V circuitry for my bike's dynamo rated 3 W 6 V.

Today I went for testing peak voltages without load and capacitors, just with diode bridge 4 x 1N5819.

Unfortunately my multimeter doesn't have peak function, so I made peak detector from LM324N:

Capacitor used: 100 nF ceramic. Meter leads were connected to Vout and GND pin of LM324N. Maximum readings were 24 V at maximum speed I could do.

I tested this peak detector on my ATX PSU (bridge rectifier omitted), I got:

real 5V -> Vout = 3.9 + 0.1V (Vdrop of 1N5819) = 4V, delta = 1V
real 11.1V -> Vout = 9.8 + 0.1V = 9.9V, delta = 1.2.

Considering that real peaks are higher then 24V, real peak I'd say +2V for rough estimation = 26V

Can those peaks be damaging for 25V electrolytic capacitor with no load? And with load? I personally think if there is a load, the cap cannot be fully flooded (charged), hence cannot be damaged that way. But I'm not sure.

UPDATE: Just in case I'll use 24V transil

Answer 1

Your open circuit dynamo - actually it's an alternator since the output is AC - may be irrelevant to the real-world use. The unit will contain internal series resistance and inductance.

As desribed in my answer to Non-led simple bicycle dynamo light system, the impedance of an inductor is given by \$ Z=2ωL=2πfL \$. This shows that the impedance is proportional to the frequency which, of course, is directly related to the speed of the bike. If designed correctly the lamps will turn on to a reasonable brightness at quite low speed and will be noticeably brighter at high speed but without blowing the lamps - the reason being that the inductors and lamps form an L-R voltage divider. The inductance helps keep the voltage more constant over a wide range of speeds than if it was minimised.

Figure 1. Bicycle "dynamo" equivalent circuit.

If you place your regulator circuit after the switch shown in Figure 1 it will never run with a no-load situation and the alternator output voltage should be pulled down to a safe value for your capacitor.

Can those peaks be damaging for 25 V capacitor with no load?

You are right to be concerned but given the short duration of exposure to the higher voltages it's unlikely to fail. If going into production where warranty costs could become an issue you might take the cautious approach.

And with load? I personally think if there is a load, the cap cannot be fully flooded (charged), hence cannot be damaged that way.

Agreed.

Try your speed tests again with various loads connected to the output and I suspect the voltages obtained will be much less.

Answer 2

• For critical (product design) applications: check the datasheet and ensure in your design that the voltage across the capacitor doesn't ever exceed the absolute maximum rating.
• For non-critical applications (e.g. prototypes, testing): try it and see what happens. A bit overvoltage is usually going to be OK, since there will be a certain safety margin built into the design of any capacitor. When overvolting capacitors from sources that can potentially source a lot of current, be sure protect yourself and your environment against shrapnel from exploding caps.