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Resistor Suggestion for Cap Bank

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Sir Fixalot

Tue Nov 16 2010, 02:53AM

Registered Member #2810 Joined: Sat Apr 17 2010, 07:17PM
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Posts: 22

Hokay, I have 8 350 v 2900 uf lytics, and I want to arrange them in 4 sub-banks of 2 in series (each sub-bank will be 700v, and 1450 Uf). Of course I wanna include some safety so none of the caps overcharge the others, so im going to put a resistor across each cap. Any of these Oughta be sufficient, right?

or is the resistance too high? Any help appreciated.

ScotchTapeLord

Tue Nov 16 2010, 04:39AM

Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
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Why not 1/8 Watt, 1 Meg resistors? They're all you need and won't waste power.

Unless your capacitors have terrible, terrible leakage... in which case they are probably not safe to use at all...

klugesmith

Tue Nov 16 2010, 08:09AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1716

ScotchTapeLord wrote ...

Why not 1/8 Watt, 1 Meg resistors? They're all you need and won't waste power.
Unless your capacitors have terrible, terrible leakage... in which case they are probably not safe to use at all...

Aside from the safety bleeder function, I think the goal of the resistors is to balance the voltages in a series string of C's (since you guys think it's silly to charge to anything less than 100% of rated voltage).

Without R's, the steady state voltage ratio is determined by the matching of leakage currents -- but 1 megohm R's are too weak to matter much with datasheet-tolerable leakage currents in 2900 uF 350 V Al-can capacitors, which are in the milliamp range.
More important than leakage matching, the voltage ratio during rapid charging depends on the matching of capacitance values (remember the individual units are probably rated -10%, + 50% tolerance).
I think a conservative design wants balancing resistors that waste a fair fraction (10% ?) of the charger output power -- unless you selectively match the capacitors.

My recommendation, which also saves resistors: Form all 8 capacitors in parallel for a while, then measure them individually & make two groups of four to dividing total C in half, as closely as possible.
Each group of 4 is wired in parallel, with one bleed resistor. The two groups are connected in series.

A secondary benefit of matching C values: equal voltage change during a pulse discharge shot, which happens too fast for bleeder R's to matter. To avoid reversing the voltage on the half-bank with smaller C.

ScotchTapeLord

Tue Nov 16 2010, 02:07PM

Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
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Oh, thanks for that. I never realized the leakage was so bad in those capacitors.

Sir Fixalot

Tue Nov 16 2010, 02:10PM

Registered Member #2810 Joined: Sat Apr 17 2010, 07:17PM
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Posts: 22

Klugesmith wrote ...

Aside from the safety bleeder function, I think the goal of the resistors is to balance the voltages in a series string of C's (since you guys think it's silly to charge to anything less than 100% of rated voltage).

That's the main reason I wanted the resistors; so the caps are as balanced as possible when charged.

Jasonr

Wed Nov 17 2010, 02:19AM

Registered Member #167 Joined: Mon Feb 13 2006, 06:41PM
Location: Waterloo, WI
Posts: 54

BE CAREFUL!!!! it will not really matter the leakage of the caps. it is more the body style of the resistors

i=v/r and then p=iv those little 1/8 guys will be smoking hot!

plus if memory serves 1/8 watt is only good to 150volts?

Jason

Saz43

Wed Nov 17 2010, 03:12AM

Registered Member #1525 Joined: Mon Jun 09 2008, 12:16AM
Location: America
Posts: 294

Jasonr wrote ...

BE CAREFUL!!!! it will not really matter the leakage of the caps. it is more the body style of the resistors

i=v/r and then p=iv those little 1/8 guys will be smoking hot!

plus if memory serves 1/8 watt is only good to 150volts?

Jason

I've used 1/8 watts up to 400V with no problem.

ScotchTapeLord

Wed Nov 17 2010, 03:50AM

Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
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Hm? Obviously a 1 M ohm, 1/8 watt resistor can tolerate 350V.

P = V^2/R or I^2*R

using V^2/R

.125 Watts = V^2/(10^6 ohms)

V^2 = 125000 Volts^2

V = 353.553 Volts

having 350V across it is the only way to get the wattage that high! Just do math to determine what wattage you need for your application.

But if your capacitors have leakage resistances that are lower than a few M ohms, then you'll have to use a lower rated resistor which will require a higher wattage rating.

Sir Fixalot

Wed Nov 17 2010, 06:21AM

Registered Member #2810 Joined: Sat Apr 17 2010, 07:17PM
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Klugesmith wrote ...

My recommendation, which also saves resistors: Form all 8 capacitors in parallel for a while, then measure them individually & make two groups of four to dividing total C in half, as closely as possible.
Each group of 4 is wired in parallel, with one bleed resistor. The two groups are connected in series.

The reason I don't want to do that is to keep the ESR Low-I would much rather do sub-banks of 2 in series.

Sir Fixalot

Wed Nov 17 2010, 07:50AM

Registered Member #2810 Joined: Sat Apr 17 2010, 07:17PM
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Posts: 22

I really really appreciate all this help, but this still doesn't answer my question: what do yall think about the resistors I suggested? are they fitting? overkill?

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