rcavictim wrote: "That will draw only 1 mA of equalizing current (which is not a lot compared to the leakage current of the electrolytics) so dropping 500 volts will dissipate exactly 1/2 watt of power."
I don't know of any other leakage, so I'm assuming the datasheet is referring to the same one that rcavictim was. In that case his statement is way off, as 1 mA is very much a lot compared to the 1.1 uA leakage of the capacitors.
I don't know of any other leakage, so I'm assuming the datasheet is referring to the same one that rcavictim was. In that case his statement is way off, as 1 mA is very much a lot compared to the 1.1 uA leakage of the capacitors.
1 uA seems little and those formulas you have used give you only a teorethical value. I'll doubt that 20 Mohms is common praxis.
Huh? The manufacturer says "<=", i.e. it's AT MOST that much leakage. With 20M that's 20 times the maximum leakage current.
I only tend to remember what is successful and works. Here is a classical case of how theory, or too much knowledge can get in the way. Like I said 20 megohms is not the way to go. A new electrolytic, after 5 minutes, may indeed under ideal circumstances have it's leakage current drop to such low values, at an ideal laboratory controlled temperature, but maybe not in the first 5 minutes where current imbalance could cause one cap to receive long term fatal overvoltage. This is after all what the eq R is supposed to prevent. As caps age this situation will get worse. There is another advantage to using lower R in the eq chain in that it doubles as a bleeder to slowly discharge the power supply to save the life or nerves of an unsuspecting, poorly trained technician when he services the unit. A good tech will disharge the caps expecting a possible hazard.
Prune was asking for good advice. I gave him the best advice for the real world that I know. End of story.
Prune was asking for good advice. I gave him the best advice for the real world that I know. End of story.
Allright then. It's just that now I've got 100x 10M resistors that I don't know what to do with...
Prune said:
Throw 'em in the kitchen drawer and use 'em as garbage bag twist ties. 😀
That high value could be a good static drain resistor on the input end of sensitive cmos circuits, etc., or TC resistors in long time constant integrators, but at a penny a piece (if you weren't ripped too badly when you bought them) who cares.
E-mail CD and ask them what maximum leakage current is in all circumstances, throughout the life of the cap.
Tim
Tim
Prune said:
...and cars improve mechanically with age. I've told you what to do based on my experience. if you do what I suggest the worst that can happen is nothing. If you rely on where you are headed, if the data you rely on is not 100% accurate you can suffer a meltdown later down the road.
If electrolytic capacitors got better with age that would save those of us who restore vintage electronics a lot of time and money.
Hi,
20M across a pair of series caps???
If the caps in series are identical in insulation voltage and capacitance only a pair of identical voltage dividing resistors are required and Ct = (C1 + C2)/2
To determine the divider resistors you can use the following formula:
For 2 capacitors in series: R = (2Vr - Vb) / (0.0015 C Vb)
R = resistance in Megohms
Vr = max rated surge voltage
Vb = max voltage across entire bank of caps
n = number of caps in series
C = capacitance in F
Remember no two caps are absolutely identical, moreover they tend to change with usage so balancing resistors are no luxuary and will force equal voltage drop across the series of caps.
Cheers,😉
20M across a pair of series caps???
If the caps in series are identical in insulation voltage and capacitance only a pair of identical voltage dividing resistors are required and Ct = (C1 + C2)/2
To determine the divider resistors you can use the following formula:
For 2 capacitors in series: R = (2Vr - Vb) / (0.0015 C Vb)
R = resistance in Megohms
Vr = max rated surge voltage
Vb = max voltage across entire bank of caps
n = number of caps in series
C = capacitance in F
Remember no two caps are absolutely identical, moreover they tend to change with usage so balancing resistors are no luxuary and will force equal voltage drop across the series of caps.
Cheers,😉
Hi,
No need for n when only two caps are involved.
Otherwise the formula would go like this:
R = (Vr - Vb/n) / (0.00075 C Vb)
Cheers,😉
No need for n when only two caps are involved.
Otherwise the formula would go like this:
R = (Vr - Vb/n) / (0.00075 C Vb)
Cheers,😉
So I get (450 - 3000 / 8) / (0.00075 * 330E-6 *3000) ~= 101010.
You said R was in MOhms. That doesn't seem right.
You said R was in MOhms. That doesn't seem right.
Hi,
You're quite right...
That should be plain Ohms.
I guess they never really doublechecked that formula, they had the capacitance in µF as well....
Sorry about that.
Cheers,😉
You're quite right...
That should be plain Ohms.
I guess they never really doublechecked that formula, they had the capacitance in µF as well....
Sorry about that.
Cheers,😉
Prune said:
I mean, like, coming up on faliure. Long period at rated voltage and temperature. Etc.
Tim
When dimensioning power supply capacitors, you customarily overrate the voltage rating at least 40% to account for AC mains voltage fluctuations and just general bad karma.
So in this case Prune should use 10x 330uF/450V.
If I'm not quite mistaken, then fdegrove's formula is a hard limit on the maximum size of the equalizing resistors, which only works when everything is lined up 100%. Meaning that that value should be decreased by around 40% as well when determining real world values. So for the 10 capacitors in this example the value is 200K * 0.6 = 120Kohm.
Also they must be overrated by a factor of four in wattage, ie. use a 5W rated wirewound in this case.
The executive answer is that if you get a wattage below 5W in the steps above, then use a 5W wirewound anyway. Do not use those inky, dinky 1-2W metal film resistors. Also if the equalizing resistors doesn't consume substantial power, then something is not right as well.
The reason why any sane person want to overrate the capacitor bank in this way, is that if even one of the resistors ever opens up, then the outcome is in all likelyhood a catastrophic cascade failure of the whole capacitor bank! Do the math if you don't believe me. So you want to make really, really sure this never happens.
Coincidentally I'm sitting here doing the initial work on a 1300V/150mA PSU, and here each 220uF/350V capacitor, all six of 'em, gets a 51K/7W resistor to keep them company. That value might be a bit too conservative, but that is what the junk box provided...
Frank.
So in this case Prune should use 10x 330uF/450V.
If I'm not quite mistaken, then fdegrove's formula is a hard limit on the maximum size of the equalizing resistors, which only works when everything is lined up 100%. Meaning that that value should be decreased by around 40% as well when determining real world values. So for the 10 capacitors in this example the value is 200K * 0.6 = 120Kohm.
Also they must be overrated by a factor of four in wattage, ie. use a 5W rated wirewound in this case.
The executive answer is that if you get a wattage below 5W in the steps above, then use a 5W wirewound anyway. Do not use those inky, dinky 1-2W metal film resistors. Also if the equalizing resistors doesn't consume substantial power, then something is not right as well.
The reason why any sane person want to overrate the capacitor bank in this way, is that if even one of the resistors ever opens up, then the outcome is in all likelyhood a catastrophic cascade failure of the whole capacitor bank! Do the math if you don't believe me. So you want to make really, really sure this never happens.
Coincidentally I'm sitting here doing the initial work on a 1300V/150mA PSU, and here each 220uF/350V capacitor, all six of 'em, gets a 51K/7W resistor to keep them company. That value might be a bit too conservative, but that is what the junk box provided...
Frank.
Frank,
Glad to see another sees this as I do. Do you happen to know how much voltage can you safely put across those small rectangular 5 watt ceramic type resistors? According to you it sounds like 400 VDC is OK.
Glad to see another sees this as I do. Do you happen to know how much voltage can you safely put across those small rectangular 5 watt ceramic type resistors? According to you it sounds like 400 VDC is OK.
Knarf said:
I know it's a little expensive, but the TO-220 size thick film power resistors are just great space savers and you can heat sink them very easily. I believe that Caddock, Huntington and Ohmite make them.
You have to watch out for is that the heat doesn't migrate from the leads onto the board (and soften it up.) I am using them in a high voltage linear supply.
I now have two multi-mini capacitors, each 2 * 330uF@450V in parallel * 8 in series. That's 32 total, and I have two spare ones (I had 3 spares but one got squished in shipping and some electrolyte leaked out). With a resistor betweeen the two MMC banks, I have a CRC filter. Then a handwound choke into 2 * 40uF@3kV tin can caps (under load the voltage drops to below 2.7kV by the time it reaches these caps). That leaves me two of the tin cans left to put one each at the load end amp/speaker after the HV distribution cable.
With two spare of the electrolytics, the most I can do is put them as a 9th level in the first C, as that's what sees the highest voltage.
I'd like a suggestion for a circuit that will turn off the power supply when there is insufficient load connected, for two reasons: prevent voltage rising near the maximum rating of the tin can capacitors, and to prevent a shock hazard if the HV cable is unplugged with the power supply on.
Another question: how do I figure out how much current I can draw without overheating the transformer? From a discussion elsewhere I've been told the large MOT I used (core is over 7 kilograms) would be good for 1500 W, but I've completely rebuilt the core so it should be a bit better than that; also, I'm going to have a fan in the enclosure.
After the bridge rectifier it's about 3 kV and looses about 300 V in the CRCLC. I'm having trouble figuring out how much current I can draw because the complication is that the rectifiers only conduct near the peaks of the voltage waveform, and so I can't just multiply the RMS figures.
With two spare of the electrolytics, the most I can do is put them as a 9th level in the first C, as that's what sees the highest voltage.
I'd like a suggestion for a circuit that will turn off the power supply when there is insufficient load connected, for two reasons: prevent voltage rising near the maximum rating of the tin can capacitors, and to prevent a shock hazard if the HV cable is unplugged with the power supply on.
Another question: how do I figure out how much current I can draw without overheating the transformer? From a discussion elsewhere I've been told the large MOT I used (core is over 7 kilograms) would be good for 1500 W, but I've completely rebuilt the core so it should be a bit better than that; also, I'm going to have a fan in the enclosure.
After the bridge rectifier it's about 3 kV and looses about 300 V in the CRCLC. I'm having trouble figuring out how much current I can draw because the complication is that the rectifiers only conduct near the peaks of the voltage waveform, and so I can't just multiply the RMS figures.
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