deriving a capacitor's voltage

[Heidi]

Hello MrAl,

I'm not quite sure if you were telling me the way you used to measure the leakage current of a cap. Sometimes I can't understand the meaning of an English sentence! For example, I have completely no idea what " the pure part of the cap is out of picture with DC current" means.

I'm very interested in knowing the principle of your method though, so please allow me to ask a few more questions. : )

I used to have a very very old capacitance/inductance bridge that also had a way to test for leakage current.

Was that bridge used to charge a capacitor to a certain voltage?

But if you can measure the current with the required voltage then you can use Ohm's Law because the pure part of the cap is out of the picture with DC current, so we are left with just:
R=E/I

Did you mean that if we can measure the current of a cap which we already know its voltage, then the resistance of the parallel resistor that was used to model the leakage (let's call it the leakage resistance and indicate it by Rp) can be calculated by Rp=V/I?

It works because if we know the cap's voltage Vo, then it's going to decay due to leakage according to
v(t)=Vo*e^(-t/Rp*C),
and if we can measure the leakage current i(t)=-C*(dv/dt)=(Vo/Rp)*e^(-t/Rp*C) at any time t=T during it's decaying,
then Rp=v(T)/i(T)?

So far, did I misunderstand your descriptions?

 

[JoeJester]

Heidi,

As far as being practical, the current meter was in series with the capacitor and the voltage was s-l-o-w-l-y increased to the rated voltage as measured by the internal voltmeter, and keeping en eye on the current meter to keep it from pegging.

180 seconds is three minutes and it took all of that at times.

Today, the ESR dominates the failures in capacitors. There are schematics of ESR checkers on the internet.

Leakage could be checked with your DMM. You will see the "low ohmic value slowing increasing as the capacitor charges." The problem with that is the potential on the ohm scale probably doesn't approach the capacitors rated voltage. The same could be said with an analog ohmmeter using a d'Arsonval meter.

Look at that capacitor model, you have the shunt resistance, the series resistance (ESR) and the series inductance (typically the leads), The capacitor can fail from a capacitance out of specification, leakage, high ESR. The lead inductance probably is the least worry, unless your designs are at such a frequency they are troublesome. When the dielectric dries, the ESR increases.

These symptoms are more towards the electrolyte capacitors.

 

[MrAl]

Hello MrAl,

I'm not quite sure if you were telling me the way you used to measure the leakage current of a cap. Sometimes I can't understand the meaning of an English sentence! For example, I have completely no idea what " the pure part of the cap is out of picture with DC current" means.

I'm very interested in knowing the principle of your method though, so please allow me to ask a few more questions. : )


Was that bridge used to charge a capacitor to a certain voltage?


Did you mean that if we can measure the current of a cap which we already know its voltage, then the resistance of the parallel resistor that was used to model the leakage (let's call it the leakage resistance and indicate it by Rp) can be calculated by Rp=V/I?

It works because if we know the cap's voltage Vo, then it's going to decay due to leakage according to
v(t)=Vo*e^(-t/Rp*C),
and if we can measure the leakage current i(t)=-C*(dv/dt)=(Vo/Rp)*e^(-t/Rp*C) at any time t=T during it's decaying,
then Rp=v(T)/i(T)?

So far, did I misunderstand your descriptions?

Hi,

The "pure" part of the capacitance is the "C" in e^(-t/(R*C)). It's the capacitance ALONE.
So if we had a capacitor of say 10uf and it had an equivalent 'leakage resistance' of 1 megohm, the pure part of the capacitance would be just the 10uf part. It is the part with units of Farads.

When we charge up a capacitor with a voltage source E in series with a resistance R, the capacitor voltage charges up as:
Vc=E*(1-e^(-t/RC))

and the current is:
Ic=E/R*(e^(-t/RC))

After a long time though, the exponential part becomes equal to zero (t very very long like 1e32 seconds) so we get:
Vc=E*1=E
and
Ic=0

That is when the resistance is in series with the cap and that is the only resistance and notice that the current is zero. But if we have the second resistance that represents the leakage current, when the cap voltage stabilizes after a long time we end up with a small current that ONLY flows through the 'leakage' resistance and since that is then just a voltage across a resistance and we know the current we can calculate the resistance from R=E/I.
So with this kind of measurement we just need to measure the DC voltage and the DC current after a long time has passed.

The old piece of equipment i had actually had a fish eye tube (dont know if they make them anymore) which gave an indication of the current through the device under test and it was very sensitive because it did not require a physical movement of anything to show current, but i later used a highly sensitive current meter in series with the cap to get a numerical reading.
To set this up, as Joe pointed out, put the DC current meter in series with the capacitor, but shunt the meter with a shorting strip so that the meter is shorted out completely. Then apply some current and wait for the cap to charge up, then remove the shorting strip to take a reading. Measuring the voltage across the cap and the current, the resistance is then:
R=E/I

and E and I are then just DC values.

This works because once the cap charges up the capacitance no longer affects the readings because they are all DC.

This provides us with an equivalent resistance for the leakage at whatever voltage we are testing with.

 

[Heidi]

Hello MrAl,

Thank you very much for your detailed explanation. I think I finally "get the picture".

In your method, we don't have to know the capacitance, I think that is of great advantage over mine.

One more thing to be sure of, when the cap voltage stabilizes, we don't disconnect the DC voltage and the series resistor from the cap when we measure the leakage current, right?

The old piece of equipment i had actually had a fish eye tube (dont know if they make them anymore) which gave an indication of the current through the device under test and it was very sensitive because it did not require a physical movement of anything to show current,

I guess that piece of equipment must be a "graceful" one!

 

[MrAl]

Hi again,

Your diagram makes everything more clear.

Yes, everything stays connected so that you can read the current through R2 with the meter and the current will be coming from the source voltage. You can see from that diagram that when the cap charges up it does not draw current any longer so the only thing left to draw current is R2.

When you need to measure the leakage current at a specific voltage you would charge the cap until the cap voltage reached the target value. If R1 is small then that means it will be almost the same as the source voltage.

Your idea for the measurement is good too though. If you want a more average reading that might do it because the leakage may be a function of voltage:
Vc=Vs*(1-e^(-t/(R(Vc)*C)))

and so anything that depends on voltage might give a more average-like reading. If you really wanted a more exact average reading though you might have to use several test points and then average the results.

Also, if you dont have a current meter handy then measuring the time may be a good idea as in your method
Might need a scope though.

 

[JoeJester]

That is correct. The only time you disconnect the power source and meters, would be when your going to discharge the capacitor. There are special discharge sticks for the higher rated (kV) capacitors. You can get by with a resistor and an alligator clip jumper for lower voltage rated capacitors. Remember, this is a safety issue and the only one who has a keen interest in you IS you.