balancing resister for series capacitor in a power supply
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 6th April 2012, 01:10 PM #1 tim614   diyAudio Member     Join Date: Jun 2008 balancing resister for series capacitor in a power supply just wanted to see im doing this right. i got a pair of 40uf 440vac ASC oil cap in series for cap input power supply, i know that you need a balancing resisters to help them share voltage equally, so i've done some reading and found the formula to calculate for the resister. For 2 capacitors in series: R = (2Vm - Vb) / (0.0015 C Vb) R = resistance in megohms Vm = max voltage you'll permit on either capacitor Vb = max voltage across the entire bank of two (or N) capacitors N = number of caps in series C = capacitance in microfarads so here what i got 440vac X 1.41 = 620vac (620x2-600)= 640 (0.0015x40ufx600)=36 36/640=.056m----56Kohms so 56k is what i need for the balancing resisters but i dont have any 56k on hand what i have is 150k, so if i put 150k in parallel i got 70k, would this work ok? __________________ Tim " perfect imperfection "
 6th April 2012, 03:16 PM #2 Stefkorn   diyAudio Member   Join Date: Mar 2012 Location: Gosport, Hampshire It will work, even with 100K on each capacitor. I have never calculate them, always been uning 100K and never had any problems with it.
 6th April 2012, 03:34 PM #3 MelB   diyAudio Member   Join Date: Feb 2005 Location: BC Canada I'm not an expert but I thought the resistors needed to be based on the leakage current of the capacitors?? That said 100K = more than fine
 6th April 2012, 03:44 PM #4 Stefkorn   diyAudio Member   Join Date: Mar 2012 Location: Gosport, Hampshire I'm not an expert either. I never considered the values as critical and the amps I built are still working and have never caused problems. Sure there is a science behind it but in this case I thik the 'over the thumb' rule works ok.
 6th April 2012, 03:45 PM #5 Simon B   diyAudio Member   Join Date: Dec 2011 Location: North-East England If you want to find a solution to this analytically, you'll need to consider rather more variables than you currently are doing - like leakage current of each of the capacitors - I can't see the basis for your formula at all. Stefkorn's suggestion of 100k sounds a good starting point to me, make sure you use resistors of sufficiently high power and voltage rating. Myself, I usually use electrolytics for reservoir caps, which are leakier than your oil-filleds - leakage is the cause of imbalance in this situation. If you want to do it analytically, either measure the leakage of your caps at full voltage or assume that you have worst case leakage on one, and none on the other, then calculate the resistor required to stop the good cap from suffering over-voltage. I don't usually bother with anything so proper, I just put 220k in! The thing you want to get right here is that the resistors are up to the job, don't fret overly about the value. For oil caps 56k sounds unnecessarily wasteful of power - 3.5 watts each at 440vdc per cap for instance, 100k resistors would each dissipate just under 2 watts and 220ks less than 1 watt. Thse things are sitting there steadily being cooked, so choose a power rating about twice what they'll actually be burning if you want them to last.
 6th April 2012, 06:24 PM #6 tim614   diyAudio Member     Join Date: Jun 2008 Thanks guys! i was researching for my project and i came accross this Capacitors - Multiple Capacitors i like to wing it too but i dont have the proper experience. __________________ Tim " perfect imperfection "
 6th April 2012, 11:08 PM #7 MelB   diyAudio Member   Join Date: Feb 2005 Location: BC Canada I just built a stack. 820v with 3 x 400v @ 470uF. I used 120K on each one and the voltage divided pretty much exactly 1/3 (273v) across each cap. The resistors are 3 watt rated 700v.
 6th April 2012, 11:22 PM #8 tim614   diyAudio Member     Join Date: Jun 2008 hey thanks! i guess i'll be using the 150k i have and call it good. this why i love this forum, i rather learn from people who have done it than trying to read and decide whats right and wrong on the net. thanks again! __________________ Tim " perfect imperfection "
 6th April 2012, 11:46 PM #9 DigitalJunkie   diyAudio Member   Join Date: Apr 2003 Location: Portland,Oregon From what I understand,you basically just need to 'swamp out' the capacitors leakage current,which should be fairly low,so it shouldn't take much. I have never done the math,I always just guesstimate. I've used anything from 100K to 470K,and never had any problems.I do like to double-check that the voltage is balanced across the capacitors pretty equally. As a bonus,the resistors can double as bleeders.

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