Not exponentially, because the cap leakage is nonlinear.
If voltage is applied quickly, the voltage across each cap is inverse on C.
As the leakage time constants lick in, the voltage redistributes into the minimum-current situation. It might take hours to mostly settle down, or maybe days.
Of course they don't equalize to equal voltage. They do magically find the voltage distribution that minimizes the leakage current.
--
John Larkin Highland Technology, Inc
lunatic fringe electronics
The capacitance leakage current is a non-linear function of what?
What Steve Wilson was observing was that it was a non-linear function of time at constant voltage - specifically an exponential function of time.
John Larkin seems to be saying that the time dependence wouldn't be exponential because the leakage current is a non-linear function of something else.
Since he hasn't specified what the something else might be - he probably had voltage in mind - he's just being confused rather than saying anything that might have been useful.
Voltage stress is what ruptures the layer of aluminium oxide that forms the dielectric in an electrolytic capacitor.
Once this has broken down, all bets are off.
There's no requirement that current damages each capacitor in the same way.
Allowing the capacitor to exceed the rated working voltage is not a suitable engineering goal.
You can use your method. I'll stick with the industry standard bleeder resistors where the bleed current is 10 times the maximum capacitor leakage spec
A typical spec is
I = 0.01CV or 3uA, whichever is greater
Where I : Max. leakage current (uA) at 20C after 2 minutes C : Nominal capacitance (uF) V : Rated voltage (V)
The leakage current is measured at 20C by applying the rated voltage to the capacitor through a series resistor of 1000 Ohms. The leakage current is the value 2 minutes after the capacitor has reached the rated voltage.
This test requires the capacitor to be already properly formed.
Example for a 1000uf, 450V capacitor
I = 0.01 CV = 0.01 * 1000 * 450 = 4,500 uA
So the bleeder current should be 45mA, or
R = 450 / 45e-3 = 10,000 Ohms
The bleeder dissipation is
P = 45e-3 * 450 = 20.25 Watts for each resistor.
In addition to supplying a known voltage across the capacitor, this also provides a means of discharging the capacitors, which your method ignores.
The bleeder time constant is
RC = 10,000 * 1000e-6 = 10 seconds.
So the capacitor voltage should be reduced to a safe level one minute after power off. Your method would allow a dangerous or lethal voltage to remain on the capacitor for an unknown amount of time. That is hardly an example of good engineering.
** With most tube gear, the tubes themselves drain all the electros after switch off - thy do keep on working for quite a while with no supply to the heaters. It is only those amplifiers with standby by switches that allow this to happen - however the correct procedure is to turn off the AC first.
That's a common, but horrible formula, especially if the units aren't stated. Should be farads and amps.
Wasting 40W in a pair of bleeder resistors is not an attractive situation, leading one to change the "rule".
I routinely make HV supplies, with big caps for high pulse output currents, but with a maximum dissipation of under 10W. Wouldn't want to waste more than 4W.
I designed a fairly simple HV discharge circuit, that works automatically when the operating voltages are removed, fully discharges the HV in under 5 seconds.
Now you're getting prissy. When people can't explain things, they fall back on "good engineering practice" as their reason for doing, or usually not doing, things.
Fine. Add fans as needed.
Make that LOTS of fans!
Presumably something would discharge the entire string; that's a separate issue. Caps leak like sieves in the reverse direction.
--
John Larkin Highland Technology, Inc
lunatic fringe electronics
I like depletion fets to discharge caps linearly. May as well throw in an LED in series.
On my big (as in BIG) gradient amps, under the cover we had a bright LED and a warning sign and a discharge pushbutton. The depletion fet insures that the LED is bright almost all the way to zero volts.
--
John Larkin Highland Technology, Inc
lunatic fringe electronics
40w of bleeder is excessive. Your approach may work but it's a prodigious e nergy waste. There are assorted other ways to do things. Eg smaller resisto rs, fet constant current leaks, checking there's always some load on the ci rcuit etc. Maybe permit a longer discharge time & justify it based on acces s time if that's permitted where you are.
You might have looked at the sym more closely. The lower capacitor was
100uF instead of 1000uf. Typo - missing a zero.
This shows a problem if the capacitors are not equal. You need to add a dump diode across the lower capacitor to prevent reverse bias on the electrolytic.
The sym assumes a constant load. You still have no protection if the load is intermittent or shuts down when power is lost.
Here's the result when the capacitors are not matched.
Version 4 SHEET 1 1648 680 WIRE 720 -256 656 -256 WIRE 768 -256 720 -256 WIRE 864 -256 832 -256 WIRE 928 -256 864 -256 WIRE 1168 -256 928 -256 WIRE 656 -240 656 -256 WIRE 928 -240 928 -256 WIRE 1168 -208 1168 -256 WIRE 656 -144 656 -160 WIRE 928 -144 928 -176 WIRE 1008 -144 928 -144 WIRE 1072 -144 1008 -144 WIRE 928 -112 928 -144 WIRE 1072 -112 1072 -144 WIRE 1168 -112 1168 -128 WIRE 928 -32 928 -48 WIRE 1072 -32 1072 -48 FLAG 656 -144 0 FLAG 928 -32 0 FLAG 1168 -112 0 FLAG 864 -256 D1C1 FLAG 1008 -144 C1C2 FLAG 720 -256 V2 FLAG 1072 -32 0 SYMBOL voltage 656 -256 R0 WINDOW 123 0 0 Left 2 WINDOW 3 -61 161 Invisible 2 SYMATTR Value PULSE(800 0 1 10m 10m 20 100 1) SYMATTR InstName V2 SYMBOL cap 912 -240 R0 SYMATTR InstName C1 SYMATTR Value 1000uf SYMBOL cap 912 -112 R0 SYMATTR InstName C2 SYMATTR Value 900uf SYMBOL res 1152 -224 R0 SYMATTR InstName R1 SYMATTR Value 2k SYMBOL diode 768 -240 R270 WINDOW 0 32 32 VTop 2 WINDOW 3 0 32 VBottom 2 SYMATTR InstName D1 SYMATTR Value 1N4007 SYMBOL diode 1056 -48 M180 WINDOW 0 24 64 Left 2 WINDOW 3 24 0 Left 2 SYMATTR InstName D2 SYMATTR Value 1N4007 TEXT 720 -352 Left 2 !.tran 0 10 0 1m TEXT 720 -384 Left 2 ;'Capacitor Discharge TEXT 1016 -352 Left 2 !.ic V(c1c2) =400
Why? If the caps are about equal, capacitive division works for the transient cases. You are proposing adding a lot of parts that nobody actually uses, to solve a problem that doesn't exist.
In a series string of similar electrolytic caps, they automatically voltage-equalize in both directions.
This is not a problem. Sure, discharge the entire series string for safety, as you would if there were one big cap. But no electrolytic cap in a series string is ever going to have much negative voltage across it.
I did a leakage test on a real cap, beyond the positive rated voltage. You might volunteer to do a similar leakage test, but for negative voltage. I should have done that when I had it set up.
Other people could measure some alum caps too, to get a better sampling.
--
John Larkin Highland Technology, Inc
picosecond timing precision measurement
jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com
The typical electrolytic capacitance tolerance is -50% to +100%. You are not going to get matched caps often.
The problem is near then end of the discharge. With unbalanced caps, the lower cap would be reverse biased. The diode prevents that from happening.
We have been through this. You previously said the capacitor leakage current minimizes itself automatically after a few hours or days.
How do they equalize voltages? You need bleed resistors to do that.
The reverse bias occurs near the end of the discharge. Run the sym and disconnect the dump diode. Then change C1 to 2000uF. Look at the voltage at the junction.
Yes, we all remember your test. You also stated the reading was very noisy, which meant the measurement was bad.
There is no point. You can damage electrolytics by applying reverse bias. That's what the dump diode is for.
When I was a kid and in the Boy Scouts, at summer camp, I and a few others were known to have put an un-opened can of beans in the 'enemy's' campfire. It didn't take too long for a big bang to occur and the hot coals scattered about. Thankfully no one was ever hurt but a few tents developed small holes in the top.
Another trick I learned in the air force was taking a small camera flash bulb, the kind that had the small wires bent up the side of the flattened base and used for contacts could be reformed to connect the leads to the filament pins of a
7 or 9 pin miniature tube and placed in the target's radio.
When turned on, there was a nice flash an a small stream of smoke coming from the melted plastic film on the bulb. All those present would shout "WTF was that." Some of the guys who did this also worked on the B-52s weapons systems. Apparently they never thought to take this to another level. (:-0)
-------------------------------------------------------------------- From: John Larkin Newsgroups: sci.electronics.design Subject: Re: Film capacitor as power-supply filter Date: Wed, 16 Oct 2019 16:48:02 -0700
Here's possibly the only curve like this ever posted online:
formatting link
The data is crude, because there is reforming and dielectric absorption going on, and I don't have a month to play with this. But the curve is clearly radically upward. Not a zener, more like an MOV.
The current is very noisy above maybe 70 volts. I see what looks like brief high current spikes.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.