Electrolytic caps in series

What you do is calculate the bleeder resistor value based on the following worst case conditions:

- The maximum voltage rating of one one of the caps

- Assuming one cap has maximum leakage and the other has zero leakage

i.e. The capacitor with the leakage with drag the mid rail voltage away from it's nominal half rail, creating a greater voltage across the cap with no leakage. You don't want to have more than the maximum capacitor voltage across the non-leaky cap.

Once you assume these worse case conditions then the circuit is easy to analyse.

Each resistor is: R=(CapLeakRes * (MaxCapVolt-(Vrail/2)) / (Vrail-MaxCapVolt)) * 2

The leakage will be voltage dependant, but this a simple way to look at it.

Hope that helps.

Regards Dave :)

"David L. Jones"

** The scenario concerns *identical, new electro* caps being used as post inductor filters. ** Why not assume the earth is flat while you are at it?? ** You forget the caps are in series - so any current MUST be identical in both caps at all times !!! ** I got news for you David - if an electro shows little or no leakage, then it is well able to stand the applied voltage. ** Shame all your assumptions are false.

........... Phil

Sorry I cant hear you very well at all perhaps removing roddles dick from your mouth might help you articulate toaster boy ?

"Uncle-Fester" = another anencephalic prick

** Fuck off - you rote learning moron.

.......... Phil

It's called calculating for worst case conditions, it's a perfectly valid way to calculate bleeder resistors.

Not when you put bleed resistors across them! That's what we are talking about. If the leakage of a capacitor changes then the mid rail voltage will change also, thus increasing the voltage across the other cap. It ain't a simple series circuit any more when you put bleeder resistors in PARALLEL.

Bleeder resistor values are always calculated using ball-park figures and typical expected worst case conditions. In this case one cap could be leaky and the other cap may not have any leakage, how is that a false way to view this circuit? The question was how to calculate bleeder resistor values for series caps, not if they are needed or not. I gave an answer for calculating bleeder resistor values, how would you calculate it Phil?

Dave :)

** No electro has zero leakage and the "maximum" cannot be found except by testing a huge number of caps. ** Not with your totally mad assumptions. ** Learn to read David, the current flowing in series connected caps must be identical. ** Go back - you have jumped a crucial step.

** Correct - the problem is your assumptions about these matters are wacky. ** Zero leakage electros do not exist.

You are pulling wild assumptions out of mid air.

** The two questions are not separable.

You need realistic figures for the leakage performance and leakage v voltage curve of the ACTUAL caps in question BEFORE any calc can be done.

** You gave a ** bleeding stupid ** one - I doubt a *digital* person like you has ever worked on gear with more than a 15 volt supply in your life.

Do you claim to have any engineering experience with high voltage electros ??

** My position is that the OP does not need any in **his** app - if done as I suggested with new, identical caps that have a 30 % or more margin of voltage.

" ** Forget it - just use caps that have a large margin in excess of the needed voltage.

Eg - two 350 volt types applied to a 500 volt supply.

The caps will very soon reach a mutual, acceptable agreement on what precise voltage suits their individual taste !! "

The reason I said this is that I have done it at least 100 times with 350 and 400 volt caps from WES and Farnell and in every case the resulting centre voltage was within 5% of half supply.

There are OTHER situations where bleed resistors might be very worthwhile or even essential - ie on the first stage after the rectifier where the caps may undergo significant ripple current and hence self heating.

........... Phil

Yep, that's why the question is relevant and I gave a way to calculate the values, as asked. How you get the "worst case" or "best case" leakage values doesn't change the formula presented, or the way you calculate it.

You still haven't told us how you would calculate the values Phil.

Dave :)

"David L. Jones"

( snip lots of good stuff that DLJ rudely ignored)

** The OP has a specific case in mind - but being a novice he asked an overly general question thinking it would contain the answer he needed.

NG posters do that over and over and over - then wind up with a totally useless answers from pedantic fuckheads like David L Jones who must

**insanely** imagine he has been presented with an some problem to solve !!! ** So you deny posting this ?

" What you do is calculate the bleeder resistor value based on the following worst case conditions:

- The maximum voltage rating of one one of the caps

- Assuming one cap has maximum leakage and the other has zero leakage. "

** The info you supplied is utterly useless to the OP as he has no idea what leakage figures to use.

Apparently - since you are a digital tech using 5 volt supplies - neither do you.

** You rudely ignored what I posted on that question:

" You need realistic figures for the leakage performance and leakage v voltage curve of the ACTUAL caps in question BEFORE any calc can be done. "

You also need to know all about the application and determine if the DC load current of the bleed resistors is acceptable at all - in the OPs one, the original caps had no bleed resistors and additional DC current would significantly disturb supply voltage values and increase supply ripple so that it became audible as hum.

........... Phil

"Phil Allison"

**correction:

On Wed, 22 Jun 2005 23:10:57 +1000, Mark Harriss put finger to keyboard and composed:

Series caps are often used in dual voltage PSUs. Here is a typical circuit with balancing resistors:

- Franc Zabkar

--
Please remove one 's' from my address when replying by email.

Whilst others have given their ideas as to the values (and how they are determined) to be used, ensure that the voltage rating of the resistors exceeds the voltages expected across the capacitors, especially for the worst possible scenario, ie if a capacitor fails for some reason.

"dmm"

** You can't be serious ???

Worry about some 10 cent resistor failing AFTER a high voltage electro has exploded ??

In any case, it would only do if its power ratings were exceeded.

............ Phil

Sorry roddles , what's that you say ? hard to hear over your spanking the monkey so furiously .

One day Phil Allison got dressed and committed to text

Thanks folks for all the info, I think I have the gist of it. Whilst I have not a great deal of theory I have a lot of exploded devices behind me :-) I'm going along with Phil, that master of gentle explanation. Cheers........... Rheilly

"Rheilly Phoull"

** Hey, I'm doing real well this week - that is the third "back handed" compliment in a row !!!

BTW

Are you working on an old Fender guitar amp ??

They are full of 20 or 22 uF @ 500 volt caps - sort of reddish brown coloured ??

.............. Phil

Perfectly serious.

Caps can fail for may reasons, not necessarily by exploding, but that is one of their more spectacular results. I remember many years ago having to clean the guts of a Radford valve power amplifier whose main filter caps had let go. What a mess.

A resistor could conceivably cause a fire if it isn't specified properly and the correct value, voltage, and power ratings and deratings aren't correctly calculated and defined. When playing with high voltages it would be prudent to spec the resistor to be flameproof, or at least to have a flame retardant coating.

If the leakage of one capacitor changed, the voltage across both bleed resistors would change as well, possibly exceeding their voltage rating.

A standard MRS16 330K ohm 400mW metal film resistor across a 300 volt supply would not exceed its power rating (273mW), but would exceed its voltage rating of 200V.

I think being favourably compared to Rod Speed probably doesn't quite count as a compliment, forward- or back-handed.

Here is a front-handed complement: you are right.

To design the balancing resistors, simply choose resistors that draw more than the leakage current, to swamp out the variations. how much more depends on how well the voltage needs to be balanced.

Thats all well and good, but leakage current increases with increasing temperature, and the variation is extremely wide. So an effective balancing resistor is a fairly low value, and gets hot.

Last time I did the calcs (Hitachi AIC caps), it was around 30k-40k. So I thought "?!" and looked at some existing product. which used 470k. which draw far less current than the *measured* leakage current of the caps at room temperature. voltage measurements showed DC balance was not great, and did not change when balancing resistors were removed. hardly surprising really.

In addition, the balancing resistors affect only the DC voltage. AC voltage sharing is governed entirely by capacitance ratios. This is very important at power-on, when Vdc ramps from 0 to 100%, perhaps quickly. Its easy to calculate the dV/dt needed to draw more current than the balancing resistors, or the frequency at which Xcap < Rbalance. Any faster than that, and Rbalance does nothing.

Cheers Terry

** Why even bother ????

The two caps will sort it out between themselves - as long as there is a

30%+ voltage margin.

** Correct - in many cases there is no need and no benefit at all. ** Some 56 uF, 400 volt caps ( WES HSW types) I tested today showed a leakage of < 2 uA at 250 volts, at room temp.

This increased to a mere 10 uA when quite hot.

The appropriate ballast resistor value is therefore about 20 Mohms.

Forget it !!!!

With 2 in series across 500 volts, the middle point read 265 volts.

IME - this is how most modern electro caps behave.

............... Phil