Series Caps in Power supply

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
Hi all,

I have a sub amp built that uses 2x 12000uF 50V caps in series to provide 6000uF 100V of filtering per rail.

I am aware that I should really have resistors across each of them to balance the voltage load across them so that any one of them doesn't exceed their voltage rating. The rails are at 56V unloaded, so even if the whole lot was across a single cap, they'd only be exceeding their rating by 6V or so. They're nice and balanced as they are (Due to ESR?), but if I were to add parallel resistors, what should I add?

Thanks,
Jonathan
 
diyAudio Senior Member
Joined 2002
SERIES CAPS.

Hi,

First check if there isn't a bleeder resistor already.
If so, check its' value and divide that by two than put those values across each cap.

It mustn't be the exact value, it's just a matter of not bleeding too much voltage off the rail.

If no bleeder is present you could try a pair of 50K Rs, just make sure to use equal values and that they have ample wattage for the current you're drawing..

Cheers,;)
 
diyAudio Senior Member
Joined 2002
'SPLANATION.

Hi,

I would advise derating any filter cap to 80% of its rated voltage.

Yup.
And Jonathan is already in the danger zone with these caps left to the elements.

The danger can be fatal to the caps at power-on where they take all the inrush current before the load is ready to soak it up.

(These are normally in parallel with the filter caps - right?), so I'll try the 50k resistors on each cap.

Yes, simply across each individual with the last one grounded as would be the bottom cap itself.

I hope this wasn't something you bought ready made?
If so check if no other "details" have been neglected.

OTOH, if you made it yourself count yourself lucky it survived so far.

Maybe some inrush current limiting may help, otherwise I'd consider using 63V caps iso the 50V rated ones to keep it running for some years.

Cheers,;)
 
Well, it's run well for 3 months or so...

I guess I've been lucky.

Will implement the 50k resistors across each cap ASAP.

Yes, I built it myself. Got the caps cheap (read free) and they're more than enough capacity - just a dodgy rating.

Just 1W types will be more than plenty right? (Transformer is 300VA 40-0-40 toroid, bridge is a standard 35A, caps total 6000uF per rail)

Thanks heaps for your help. Will grab the resistors today and solder them in tonight.

Cheers,
Jonathan
 
Not a filter cap, but...

I did pop a 63V 100uF Electro while "testing" this amp. (All due to dodgy 2SC3281/2SA1302 transistors - 2mm square dies don't take much to kill them).

Made a nice amount of goo and magic smoke - would hate to see what the filter caps would do as that's a lot of goo!

Would not having a load connected while powering up help the caps in their current state? I have a protection circuit (That's saved the woofer several times) in place that connects the speaker 3 seconds or so after power up. Just thought it may be one reason why the caps haven't had any problems yet.

Will definitely do those resistors before powering up again - the thought of all that goo is not good.

Cheers,
Jonathan
 
50k is wrong!

50k isn't even close to being low enough to do much good. I'm not sure where fdegrove came up with that value or where his concern about making sure they're rated for enough power comes from. 28 volts across 50k is 0.02 watts! Even a tiny 0603 surface mount resistor can handle that.

To do any good the resistors need to balance any differences in DC leakage which in old caps that big can be several milliamps or more. I'd say a couple of 2 watt or 5 watt 1.5K or 2.2K resistors would be a much better bet. 1.5K would give you 20ma of "protection" and only "drain" about 1 watt from your power supply which is very unlikely to be missed.

If your amp has a power on LED you can put it in series with one of the above resistors and then you're using the 20ma for a good cause :) (the LED may slightly offset the voltage across the caps by 2 volts but that's a non-issue)
 
Here's the info I have:

The power amp is Rod Elliotts P68 subwoofer amplifier. It delivers 300W or thereabouts into a 4 ohm load.

It is powered using a 300VA 40-0-40 Toroid to a standard 35A bridge rectifier.

The filter caps are NEW 12000uF 50V electros (Computer grade?) that are in series to give 6000uF 100V. This is per rail.

I want to add a resistor across each cap individually to balance the voltage across them.

Assuming the that 50k will be good, that's what I'll use. I don't see the point of using a smaller resistance (It'll draw more power) unless it is of particular advantage - please advise.

Thanks for the responses so far. I've mailed Rod as well to see his views on the matter.

Cheers,
Jonathan
 
Re: WRONG AGAIN.

fdegrove said:
Where did you come from with figures?

No way to calculate any values from what's given so far here, I reckon?

Why don't you reread the thread instead of contradicting yourself over and over again in a single post?
Cheers,;)
Look at a cap datasheet fdegrove. Most manufactures give the following forumula for their DC leakage specification:

DC Leakage = .01CV

So, we take .01 * 12000E-6 * 28v and we get 3.4ma. Your method of 100k total would allow0.56ma to flow which isn't enough to balance the caps if one has say 1ma leakage and the other 3ma. My values may have been overkill, but I'd rather be safe than sorry. I don't know the age of these caps and leakage current goes up with age.

As for the power specification, do I need to explain ohms law?

And where did I contradict myself?
 
Re: WRONG AGAIN.

fdegrove said:
If no bleeder is present you could try a pair of 50K Rs, just make sure to use equal values and that they have ample wattage for the current you're drawing..


fdegrove said:
Where did you come from with figures?

No way to calculate any values from what's given so far here, I reckon?
Well I reckon that P = V**2/R. So let's see... the voltage across each resistor will be 28 volts and they're 50k ohm resistors, so 28**2 = 784 divided by 50,000 = 0.016 watts which I rounded up to 0.02 watts.

So your advice to "make sure they have ample wattage" is really out of place as it's hard to buy a resistor that won't handle 0.02 watts. So please explain the "Wrong Again" part of your subject line?
 
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.