Okay, all you doubting Thomases, here's a little experiment. Two carefully matched electrolytics: one is a fairly new Xicon 10µF 450V; the other is a Sprague 10µF 50V (that's fifty Volts) from the 1960s. What do you suppose happens when I put them in series across about 530 Volts?

Nothing.


They share the voltage quite nicely. All by themselves. No "equalizing" resistors. They're drawing 12 microamps, and the meter itself takes 5µA so the 50V cap is passing 7µA. I could put a 100 meg resistor in series with the meter so it would draw less but you get the idea.

Cheers

Thanks Alan!

WOW !

Thanks, Alan.

I gotta think about this for a while. Speechless.

Rich

What I would take away from this (amazing!) demonstration is that as long as you follow the sound longstanding advice of series-connecting identical value caps, you are perfectly OK to leave off the equalizing resistors.

Wow thank you Alan, as they say a picture is worth a thousand words .
with all the reading about this I have done- your demo helped me understand best of all.
take care Gary

Doesn't this mean that the ESR of the 50v cap is 7 times less than the ESR of the 350v cap?? Hence the voltage dividing safely. Wonder if this is true in general with all caps?? Hard to believe.

I tried two 47 uF caps - one rated at 35v and the other at 450v. I applied 70vdc and they shared voltages equally - 35vdc ech.

I have the feeling that your capacitors are sharing the voltage nicely because the 50 volt one is breaking down near that value so it is acting somewhat like a zener. the 450v one hasn't broken down and is supplying the few microamps as a current limited source for the 50v one. That's why it doesn't get hot.

With resistors only being a few cents each and not a bad idea to have a bleeder, I would include them especially in a power supply application.

Maybe I'm missing the point of Alan's post.

OK, so you have 2 capacitively matched units that charge to their rated voltage value. They use what they need from the power supply. That is hardly a useful circuit, in that it will not amplify, count, oscillate or do anything else.

If you were to place those caps in a circuit in which the loading forced too much current/voltage to the lower voltage cap, you would quickly see a problem. Of course you can put an electrolytic across a high voltage power supply source with a series resistance and it will charge up to a certain voltage and won't go higher; but what happens when you eliminate the series resistor?

[quote="MarkA"]I have the feeling that your capacitors are sharing the voltage nicely because the 50 volt one is breaking down near that value so it is acting somewhat like a zener. the 450v one hasn't broken down and is supplying the few microamps as a current limited source for the 50v one. That's why it doesn't get hot.

quote]

Pretty close. The leakage current will increase rapidly as the operating voltage is reached or exceeded. In a series circuit the voltages across the caps will reach equilibrium when both capacitors, in series, share the voltage in a manner that results in drawing the least possible amount of leakage current. Balancing resistors will not improve the situation.

Thanks for the demo Alan.

Pete

Yes, it's a stable configuration.

One minor point:

That's so, but the term "breakdown" can be misleading. As with a zener* diode, "breakdown" is its normal operating point and is not damaging the device.

*"Zener" diodes above about 5V are actually avalanche diodes; they don't use the Zener effect.

There's a little smoke and a couple of mirrors involved here, since I selected capacitors that had about equal leakage currents. And this would not be a practical idea, because if you discharge the two capacitors, the voltage goes negative across the 50-Volt one. That goes back to the equal energy storage mentioned previously. A large AC/ripple component would cause the same thing.

But the demo shows that the system is stable, tending toward minimum leakage current.


No, as long as the other capacitor could stand that extra voltage. That's exactly what is happening in my demo, where the 450V cap actually has 480V on it.


ESR would only matter for impressed AC/ripple, and yes, this would be a bad idea for a practical filter.

Alan... could you humor me?

One of the tech notes that I read mentioned that capacitor leakage current radically increases at max voltage and hot temperatures.

If you slightly warmed that lower voltage cap, say with the heat from a nearby soldering iron, how would the voltages split?

Rich

If the leakage increased, its voltage would go down. The 450V cap would have to take up the slack (which would increase its leakage). The final system would be stable at the new, higher leakage.

Two capacitors of the same value in Farads will have
an equal DC voltage across each, the sum of which will
equal the appied voltage.

To make electrolytics behave as capacitors, equalizing
resistors must be used. This thread breaks new ground
in proving a debate can be started without a diode.

And to measure what is happening, with honest
capacitors, you need one of these.

I have one.

Cute little tyke, but not much use below 1000V.

Is there any reason most of the time not to put the resistors in?

I would see where some consideration would be needed is at the output of a resistive Pi filter, if you used too low of equalizing resistors you could disturb (lower) the output voltage of it.

I like that Ferranti meter alot, however this is one time I would follow the warning to disconnect the test leads from the source before handling the meter.

Mark

Resistors add heat, reduce available power and could fail over time.

I understand Alan's point. As an electrolytic comes up to its rating leakage increases. Higher leakage causes the other electrolytic to take voltage until leakage equalizes between units.

Here is a situation. Over time electrolytics may age differently? One has little leakage even past it's voltage rating. The other draws some current but is still good. Under this condition the one with little leakage could arc over?

I've only used balancing resistors a few times. This was using 470K or 1 meg in high voltage, transmitter power supplies.

Best to use 450 volt electrolytic caps for AC radios. This voltage covers just about everything needed. Don't mess with 300 volt caps.





.

Of course the leakage current will be the same in both caps! IT'S A SERIES CIRCUIT FOR HEAVEN'S SAKE! No magic involved.

John

Ah, you'd think that would be obvious, but it's hard getting the concept through, at times.



It's bad engineering, and I'm an electrical engineer. YVMV.

I'd like to see someone post the innards of a high voltage cap to see the construction techique.

Alan, is there much variation in the leakage current below the operating voltage; my only question is whether it is best to use caps that at near their rated voltage to ensure the voltage division is nearly equal.

Pete