| Prune |
Most small resistors nowadays are only rated to 250 V, beyond which the insulation starts conducting and changes the value.
My question is, in a position where exact value doesn't matter, will there be a problem if I run them at 400 V? In other words, besides the issue mentioned above, will the insulation degrade over time, perhaps failing eventually? |
|
|
| johnf |
| Any chance you could use two half resistance-value resistors in series? |
|
|
| Prune |
| Yes, but that means removing 32 and putting in 64 resistors. Otherwise I wouldn't have asked and just done it. |
|
|
| peranders |
Why have you in the first place used 250 V resistors if you have 400 Volts? You have two options as I see it:
1 Replace to high voltage types.
2 Insert resistors in series with the others.
3 Don't bother and take a chance (I wouldn't take a chance because off the very small cost to replace those resistors. It costs me 30-50 cents for all resistors.) |
|
|
| Prune |
| LOL, it's not cost, it's time. Soldering 64 resistors will take me two hours. |
|
|
| Geek |
| If they are Yaego or Xicon, they are extremely conservatively rated. I run 1/4 and 1/2 watt, 250V Yaego's at 400V for years in tube projects and have yet to have one fail. |
|
|
| fdegrove |
Hi,
| quote: | | If they are Yaego or Xicon, they are extremely conservatively rated. I run 1/4 and 1/2 watt, 250V Yaego's at 400V for years in tube projects and have yet to have one fail. |
Lucky you...
However it would be wise, very wise even, to do the resoldering now iso waiting for a fire....:att'n:
Two hours of your life is peanuts compared to no life at all....
Cheers,;) |
|
|
| johnnyx |
I would recommend using resistors well within their voltage ratings. I have had failures due to excessive voltage, not immediately, but over time, and in one instance a 1.8M startup resistor became o/c. The other resistors and leakage around the area was of a similar magnitude, with the result that it was very difficult to spot while probing with a dmm.
It will work - for a while, but in the interest of safety and reliability, spend a couple of hours changing them.
:) |
|
|
| jackinnj |
| quote: | Originally posted by Prune
Most small resistors nowadays are only rated to 250 V, beyond which the insulation starts conducting and changes the value.
My question is, in a position where exact value doesn't matter, will there be a problem if I run them at 400 V? In other words, besides the issue mentioned above, will the insulation degrade over time, perhaps failing eventually? |
1/2 Watt Carbon composition resistors can be used at 400 V -- you see them all the time in de-commissioned military equipment -- the problem is that they have to be de-rated as the resistance changes with voltage above. The resistance will also change as they absorb moisture -- and release it when they are heated.
There is a derating chart on page 373 of Horowitz and Hill. |
|
|
| rcavictim |
| quote: | Originally posted by Prune
LOL, it's not cost, it's time. Soldering 64 resistors will take me two hours. |
Prune,
By the looks of your avatar it isn't like you don't have two hours to spare.
:D |
|
|
| rcavictim |
| quote: | Originally posted by Prune
I don't get it. |
I guess being in a jail cell you are out of touch. ;)
I'm not sure if your avatar was supposed to be a UPC code made to look like bars on a jail cell but jail cell was what the image speaks to me at first glance. If you are in a jail cell you probably aren't too busy doing other stuff that you can't spend two hours soldering resistors. :D |
|
|
| jackinnj |
| quote: | Originally posted by rcavictim
I guess being in a jail cell you are out of touch. ;)
I'm not sure if your avatar was supposed to be a UPC code made to look like bars on a jail cell but jail cell was what the image speaks to me at first glance. If you are in a jail cell you probably aren't too busy doing other stuff that you can't spend two hours soldering resistors. :D |
actually, he probably has an RFID chip implanted behind his ear. |
|
|
| Prune |
| Attached is a larger version. As you can see, the 'prison' bars are a UPC (Universal Product Code). |
|
|
| Sch3mat1c |
So yeah, AFAIK:
1/4W = <300V
1/2W = up to 500V
1W = 600V
2W = 1kV I guess
Etc.
You get into 10kV ranges with long teens-watt types.
IIRC, the calculation goes where 330k on a 1/2W resistor is the maximum; any higher resistance is voltage limited (too much resistance to draw rated power), any lower is wattage-limited (draws power such that ratings are reached before voltage max). A rule of thumb around here should extend to all types.
Hmm, on second thought, no (unless you square or exponential-ize something). A quick calc shows a 10W 330k resistor at all of 2kV...
I have literally never had a 1/2W resistor fail on me!! That doesn't count, of course, the occasions when the resistor was obviously overstressed (like... putting 20V across a 100 ohm resistor). I speak of RadioShank 5% carbon films. I have several in my collection which are slightly darkened around the midsection but still measure well within tolerance.
Tim |
|
|
| Prune |
I'm using 1/4 W 20M as equalizing resistors in the multi-mini-capacitor bank. I had one across each capacitor which gets charged about 400 V. I guess I'll replace each with two 1/4 W 10M.
I did fry a few ballast resistors due to overheating though, and they melted holes in my carpet. Besides resistors, my jumper cables keep exploding like fuses -- the gauge is too small. Last time I buy wire from this guy. |
|
|
| rcavictim |
Prune,
20 megohms is WAY too high a value to use across a power supply capacitor for equalizing. It may as well not be there. For electrolytics it is customary to use about 250 K ohms for 400 volts. I would not exxceed 500K for 500 volts in power supply cap equalization. That gives you 1000 ohms per volt. 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. A 1 watt or 2 watt resistor would be fine. |
|
|
| Prune |
Well, then I must have messed up my math here.
The capacitors are 330 uF 450 V Cornell Dubilier 380LX series. The datasheet gives leakage current as "<= 3root_sign(CV) uA at 5 min." The root_sign stands for the root sign which I cannot copy. I'm assuming that the three is a multiplier, not 3rd root. Now, I don't know what the "at 5 min." means, so maybe that's where my mistake is, but here's my calculation (each is run at under 400 V): 3 * sqrt(330E-6 F * 400) = 3 * sqrt(0.132) = 3 * 0.363318042 = 1.08995413 uA.
I've been told to use equalizing resistors that will pass about ten times the leakage current. So, 10 * 1.08995413E-6 A = 1.08995413E-5 A. 400 V / 1.08995413E-6A = 36.6987921E6 Ohms. Thus 20E6 Ohms should be just fine.
I hope someone comments soon so I know whether I'm mistaken and need to buy different resistors. |
|
|
| jane |
| quote: | Originally posted by Prune
Well, then I must have messed up my math here.
The capacitors are 330 uF 450 V Cornell Dubilier 380LX series. The datasheet gives leakage current as "<= 3root_sign(CV) uA at 5 min." |
That's the internal leakage current inside the cap, isn't it? |
|
|
| Prune |
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. |
|
|
| peranders |
| 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. |
|
|
| Prune |
| Huh? The manufacturer says "<=", i.e. it's AT MOST that much leakage. With 20M that's 20 times the maximum leakage current. |
|
|
| rcavictim |
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 |
| Allright then. It's just that now I've got 100x 10M resistors that I don't know what to do with... |
|
|
| rcavictim |
| quote: | Originally posted by Prune
Allright then. It's just that now I've got 100x 10M resistors that I don't know what to do with... |
Throw 'em in the kitchen drawer and use 'em as garbage bag twist ties. :D
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. |
|
|
| Sch3mat1c |
E-mail CD and ask them what maximum leakage current is in all circumstances, throughout the life of the cap.
Tim |
|
|
| Prune |
| The leakage DECREASES as the capacitor ages! The graph is in the datasheet link I posted above. |
|
|
| rcavictim |
| quote: | Originally posted by Prune
The leakage DECREASES as the capacitor ages! The graph is in the datasheet link I posted above. |
...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. |
|
|
| fdegrove |
Hi,
20M across a pair of series caps???:xeye:
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,;) |
|
|
| Prune |
| quote: | | n = number of caps in series |
Er, there's no 'n' in the formula you posted. |
|
|
| fdegrove |
Hi,
| quote: | | Er, there's no 'n' in the formula you posted. |
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,;) |
|
|
| Prune |
So I get (450 - 3000 / 8) / (0.00075 * 330E-6 *3000) ~= 101010.
You said R was in MOhms. That doesn't seem right. |
|
|
| fdegrove |
Hi,
| quote: | | You said R was in MOhms. That doesn't seem right. |
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,;) |
|
|
| Sch3mat1c |
| quote: | Originally posted by Prune
The leakage DECREASES as the capacitor ages! The graph is in the datasheet link I posted above. |
I mean, like, coming up on faliure. Long period at rated voltage and temperature. Etc.
Tim |
|
|
| Knarf |
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. |
|
|
| rcavictim |
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. |
|
|
| jackinnj |
| quote: | Originally posted by Knarf
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. |
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. |
|
|
| Prune |
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. |
|
|
|
