# Electrolytic caps in series

Discussion in 'Hobby Electronics' started by Rheilly Phoull, Jun 22, 2005.

1. ### Mark HarrissGuest

You haven't seen my cat then: a very hairy date!!

2. ### David L. JonesGuest

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

3. ### Phil AllisonGuest

"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

4. ### Phil AllisonGuest

"Uncle-Fester" = another anencephalic prick

** **** off - you rote learning moron.

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

5. ### David L. JonesGuest

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
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

6. ### Phil AllisonGuest

"David L. Jones"

** No electro has zero leakage and the "maximum" cannot be found except
by testing a huge number of caps.

** 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

7. ### David L. JonesGuest

Yep, that's why the question is relevant and I gave a way to calculate
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

8. ### Phil AllisonGuest

"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
**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

9. ### Phil AllisonGuest

"Phil Allison"

**correction:

10. ### Franc ZabkarGuest

Series caps are often used in dual voltage PSUs. Here is a typical
circuit with balancing resistors:
http://www.pavouk.comp.cz/hw/atxps.png

- Franc Zabkar

11. ### dmmGuest

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.

12. ### Phil AllisonGuest

"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

13. ### Rheilly PhoullGuest

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

14. ### Phil AllisonGuest

"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

15. ### dmmGuest

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.

16. ### PoxyGuest

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

17. ### Terry GivenGuest

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

18. ### Phil AllisonGuest

"dmm"
"Phil Allison"

** More fool you - dickhead.

(snip irrelevances)

** Hmmm - so I see you actually have no fucking idea what particular
failure mode the voltage rating refers to.

( So you pulled in all those extraneous ones to cover that fact. )

** Obviously a Sea Food lover here - red herrings galore is a favourite.

** Hmmm - so I see you actually have no fucking idea what particular
failure mode the voltage rating refers to.

** Just for fun - how about YOU tell all of US what failure happens
( and after what time span ) when a resistor is run somewhat beyond its
rated voltage BUT well within its power rating.

Then explain how a high voltage cap can fail ( either short or open) BUT
and the gear keeps working for a significant time so is left switched on.

Then explain why this matters to anyone ??

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

19. ### Phil AllisonGuest

"Terry Given"

** 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

20. ### KLRGuest

If you have 2 caps in series, or if you have a single cap of the same
value (as the series combination) across the rail why is the failure
of ONE cap in a series pair some major problem, or somehow worse than
the single capacitor failing in the same way and deserving of a lot of
fuss and bother ?

Either way, depending on the way it fails, (eventually either goes
open circuit or short) you are going to have either an almighty bang,
or a hell of a lot of hum on the rail that the cap is filtering.
Either way the cap(s) is fucked

I would just use caps in series, as long as they are same value and
voltage rating everything should be fine. I wouldnt go mixing values
and WOULD have a rating on EACH cap significantly higher than 50% of
the rail voltage.

I have done it before too on a few occasions, with no problems at all.
Both on high and low voltage rails.