Electrolytic Cap Leakage revised

Electrolytic Cap Leakage revised


As the result of previous discussions, and helpful
info supplied by RAT subscribers, I built a box to test
cap leakage and also reform some NOS electrolytics
that were unused but had been in storage for several
years.

I decided to re-form the 450V caps via a 10M resistor
chain, so that they charged *veeeery* slowly indeed.
After 3 hours, they were still only at 120V. After 24
hours they were at 158V so I added a switch to short
out two of the 3.19M resistors in the chain. This got me
to 250V in three hours.

Further info, leakage figs, and pics at:

* w w w .kolumbus.fi/iain.churches/Projects/ElectrolyticCapacitorTester.html



Iain

Re: Electrolytic Cap Leakage revised

On Fri, 28 Mar 2008 19:38:25 +0200, "Iain Churches"
<IainNG@kolumbus.fi> wrote:

>
>As the result of previous discussions, and helpful
>info supplied by RAT subscribers, I built a box to test
>cap leakage and also reform some NOS electrolytics
>that were unused but had been in storage for several
> years.
>
>I decided to re-form the 450V caps via a 10M resistor
>chain, so that they charged *veeeery* slowly indeed.
>After 3 hours, they were still only at 120V. After 24
>hours they were at 158V so I added a switch to short
>out two of the 3.19M resistors in the chain. This got me
>to 250V in three hours.
>
>Further info, leakage figs, and pics at:
>
> * w w w .kolumbus.fi/iain.churches/Projects/ElectrolyticCapacitorTester.html
>

Standard reforming current was conventionally limited to 5mA, with
4hrs or less being allowed to establish reformed leakage current
values below the initial part spec, at the rated voltage.There is no
concern for point-source power loss, as the leakage being corrected is
shared uniformly across the dielectric surface area.

When reforming higher voltage parts, you should reflect case size
temperature rise limits in the current limit imposed. No appreciable
rise in surface temperature (<delta20C)is expected. In most cases, IM
or greater limiting resistors have no place in the procedure

Unless the part is permitted to reform at the rated voltage, it will
not do so. Parts failing to reform are garbage. A signifigant
percentage of a batch failing to do so indicates that the whole batch
should probably be scrapped.

Any sudden reductions in terminal voltage, or audible sounds, indicate
non-recoverable defects.

RL

Re: Electrolytic Cap Leakage revised



Iain Churches wrote:

> As the result of previous discussions, and helpful
> info supplied by RAT subscribers, I built a box to test
> cap leakage and also reform some NOS electrolytics
> that were unused but had been in storage for several
> years.
>
> I decided to re-form the 450V caps via a 10M resistor
> chain, so that they charged *veeeery* slowly indeed.
> After 3 hours, they were still only at 120V. After 24
> hours they were at 158V so I added a switch to short
> out two of the 3.19M resistors in the chain. This got me
> to 250V in three hours.

You'll be quite safe with a 100k resistor.

Graham

Re: Electrolytic Cap Leakage revised


"Eeysore"

>> I decided to re-form the 450V caps via a 10M resistor
>> chain, so that they charged *veeeery* slowly indeed.
>> After 3 hours, they were still only at 120V. After 24
>> hours they were at 158V so I added a switch to short
>> out two of the 3.19M resistors in the chain. This got me
>> to 250V in three hours.
>
> You'll be quite safe with a 100k resistor.
>
> Graham the Charlatan


** 450 volts across 100kohms = 2 watts.

Single resistors of that value are not easy to get.

Better use a series string of say 3 x 33k, 1 or 2 watt resistors.



....... Phil

Re: Electrolytic Cap Leakage revised



Phil Allison wrote:

> "Eeysore"
>
> >> I decided to re-form the 450V caps via a 10M resistor
> >> chain, so that they charged *veeeery* slowly indeed.
> >> After 3 hours, they were still only at 120V. After 24
> >> hours they were at 158V so I added a switch to short
> >> out two of the 3.19M resistors in the chain. This got me
> >> to 250V in three hours.
> >
> > You'll be quite safe with a 100k resistor.
> >
> > Graham the Charlatan
>
> ** 450 volts across 100kohms = 2 watts.
>
> Single resistors of that value are not easy to get.
>
> Better use a series string of say 3 x 33k, 1 or 2 watt resistors.

Agreed.

Graham

Re: Electrolytic Cap Leakage revised



Phil Allison wrote:
>
> "Eeysore"
>
> >> I decided to re-form the 450V caps via a 10M resistor
> >> chain, so that they charged *veeeery* slowly indeed.
> >> After 3 hours, they were still only at 120V. After 24
> >> hours they were at 158V so I added a switch to short
> >> out two of the 3.19M resistors in the chain. This got me
> >> to 250V in three hours.
> >
> > You'll be quite safe with a 100k resistor.
> >
> > Graham the Charlatan
>
> ** 450 volts across 100kohms = 2 watts.
>
> Single resistors of that value are not easy to get.
>
> Better use a series string of say 3 x 33k, 1 or 2 watt resistors.
>
> ....... Phil

WES supply a suitable range of 2W resistors of up to high values
and which probably might take 450V across them without converting
into a short, or going open.
But the voltage rating of the R isn't mentioned.
Some metal film R fail to a short if over voltaged, very
bad news if the short causes other bothers.

But you are dead right about series resistors of say
3 x 33k x 1W being better practice.
This is much better than say 3 x 330k in parallel, because each 330k
could be subject to
full 450V, but each 33k is subject to only 150V.


I have a number of dummy load R which consist of seriesed multiple
5W and 7W wire wound R slung between brass screws into 70mm wide timber
slats,
with a cover over them, so the wattage is quite high.

Patrick Turner.

Re: Electrolytic Cap Leakage revised


"Patrick Turner"

>
> WES supply a suitable range of 2W resistors of up to high values
> and which probably might take 450V across them without converting
> into a short, or going open.


** Errr - " probably might" ????

Sounds like dodgy lawyer speak.


> Some metal film R fail to a short if over voltaged,


** Sounds completely impossible.




..... Phil

Re: Electrolytic Cap Leakage revised



Phil Allison wrote:
>
> "Patrick Turner"
>
> >
> > WES supply a suitable range of 2W resistors of up to high values
> > and which probably might take 450V across them without converting
> > into a short, or going open.
>
> ** Errr - " probably might" ????
>
> Sounds like dodgy lawyer speak.

Weaving ex-builder spiel!

Ppl will know what I meant though.

>
> > Some metal film R fail to a short if over voltaged,
>
> ** Sounds completely impossible.
>
> ..... Phil

I once put in 3 x 470k 1W metal films in parallel in a
revised Quad-II amp where I made the two input EF86 act
as a real LTP, with input to one sideand FB signal to the other,
and with the 156k R from the commoned cathodes to a low current -400V
supply
easily created in the Quad-II.

In both amps, one of the 470k went to a short.

I didn't think its was possible either, but that shite did happen.

So I never like to see more than 200V across a resistance.

Patrick Turner.

Re: Electrolytic Cap Leakage revised


Iaim wrote
>>> I decided to re-form the 450V caps via a 10M resistor
>>> chain, so that they charged *veeeery* slowly indeed.
>>> After 3 hours, they were still only at 120V. After 24
>>> hours they were at 158V so I added a switch to short
>>> out two of the 3.19M resistors in the chain. This got me
>>> to 250V in three hours.
>>
>> You'll be quite safe with a 100k resistor.
>>
>> Graham the Charlatan
>
>
> ** 450 volts across 100kohms = 2 watts.
>
> Single resistors of that value are not easy to get.

I am using a 100K 2W resistors (it's a regular stock
item from RS)

> Better use a series string of say 3 x 33k, 1 or 2 watt resistors.

I see. I can make that change easily.
Thanks for the tip.

Iain




>
>
>
> ....... Phil
>
>
>

Re: Electrolytic Cap Leakage revised

Iain,

If your series resistor is 10 meg, and your meter is 10 meg-ohm input,
then you have a 2:1 voltage divider where the presence of the meter cuts
down the voltage available in half.

Jim
WD5JKO


"Iain Churches" <IainNG@kolumbus.fi> wrote in message
news:o2aHj.317015$cL7.6790@reader1.news.saunalahti.fi...
>
> As the result of previous discussions, and helpful
> info supplied by RAT subscribers, I built a box to test
> cap leakage and also reform some NOS electrolytics
> that were unused but had been in storage for several
> years.
>
> I decided to re-form the 450V caps via a 10M resistor
> chain, so that they charged *veeeery* slowly indeed.
> After 3 hours, they were still only at 120V. After 24
> hours they were at 158V so I added a switch to short
> out two of the 3.19M resistors in the chain. This got me
> to 250V in three hours.
>
> Further info, leakage figs, and pics at:
>
> * w w w .kolumbus.fi/iain.churches/Projects/ElectrolyticCapacitorTester.html
>
>
>
> Iain
>
>
>
>
>
>

Re: Electrolytic Cap Leakage revised



"JC" <jcandela@prodigy . net > wrote in message
news:vetHj.20652$xq2.9715@newssvr21.news.prodigy . net ...
> Iain,
>
> If your series resistor is 10 meg, and your meter is 10 meg-ohm input,
> then you have a 2:1 voltage divider where the presence of the meter cuts
> down the voltage available in half.
>
> Jim
> WD5JKO


Hi Jim,

I don't quite follow you.

The 10M is in series with the cap to reduce the charging current.
The meter is across the cap to read the voltage on it.

The second meter is across the 100k resistor to measure the
voltage across that (with the 10M switched out) to allow the
voltage across the 100k to be read and the leakage current
calculated.

* w w w .kolumbus.fi/iain.churches/Projects/ElectrolyticCapacitorTester.html

Regards
Iain

Re: Electrolytic Cap Leakage revised


"Iain Churches" <IainNG@kolumbus.fi> wrote in message
news:hntHj.317560$wd5.295311@reader1.news.saunalahti.fi...
>
> Hi Jim,
>
> I don't quite follow you.
>
> The 10M is in series with the cap to reduce the charging current.
> The meter is across the cap to read the voltage on it.
>
> The second meter is across the 100k resistor to measure the
> voltage across that (with the 10M switched out) to allow the
> voltage across the 100k to be read and the leakage current
> calculated.
>
> * w w w .kolumbus.fi/iain.churches/Projects/ElectrolyticCapacitorTester.html
>
> Regards
> Iain
>


Iain,

Draw your circuit in the charge mode with the 10 meg resistor between the
power supply and the capacitor. Now draw in another 10 meg resistor across
the capacitor. The capacitor cannot charge to anything beyond 50% of the
power supply voltage, and will be less than that because of the capacitor
leakage. The second 10m resistor is your DVM. You are right about the 3
hours since 5 R-C time constants work out to 183 seconds. In cases like
yours I remove the voltmeter to allow for a higher capacitor charge voltage,
and then I connect the DVM later to measure the voltage even though the
voltage will start dropping as soon as the dvm starts loading the capacitor.
Remenber probing a circuit changes the circuit.

I think a series resistance of 100K is more appropriate than either 3m or
10m. You can still measure the voltage across that 100K to measure leakage
current.

Having a variable power supply is handy too so you can charge the
capacitor gradually to a given voltage. The old Eico capacitor bridges with
a 1629 magic eye tube are ideal for forming electrolytics in a non
destructive way. Those old Eico's go up to around 450 volts with the turn of
a knob.

Jim

Re: Electrolytic Cap Leakage revised



"JC" <jcandela@prodigy . net > wrote in message
news:bSyHj.6388$6H.2327@newssvr22.news.prodigy . net ...
>
> "Iain Churches" <IainNG@kolumbus.fi> wrote in message
> news:hntHj.317560$wd5.295311@reader1.news.saunalahti.fi...
>>
>> Hi Jim,
>>
>> I don't quite follow you.
>>
>> The 10M is in series with the cap to reduce the charging current.
>> The meter is across the cap to read the voltage on it.
>>
>> The second meter is across the 100k resistor to measure the
>> voltage across that (with the 10M switched out) to allow the
>> voltage across the 100k to be read and the leakage current
>> calculated.
>>
>> * w w w .kolumbus.fi/iain.churches/Projects/ElectrolyticCapacitorTester.html
>>
> Draw your circuit in the charge mode with the 10 meg resistor between the
> power supply and the capacitor. Now draw in another 10 meg resistor across
> the capacitor. The capacitor cannot charge to anything beyond 50% of the
> power supply voltage, and will be less than that because of the capacitor
> leakage. The second 10m resistor is your DVM.

Hi Jim. OK. Now I see your point:-)

> You are right about the 3 hours since 5 R-C time constants work out to 183
> seconds.

I had not thought about being able to calculate the
charging time.Please elaborate.

> In cases like yours I remove the voltmeter to allow for a higher capacitor
> charge voltage, and then I connect the DVM later to measure the voltage
> even though the voltage will start dropping as soon as the dvm starts
> loading the capacitor. Remenber probing a circuit changes the circuit.


Good. I will try that.
>
> I think a series resistance of 100K is more appropriate than either 3m or
> 10m. You can still measure the voltage across that 100K to measure leakage
> current.

It seems to work well.
>
> Having a variable power supply is handy too so you can charge the
> capacitor gradually to a given voltage. The old Eico capacitor bridges
> with a 1629 magic eye tube are ideal for forming electrolytics in a non
> destructive way. Those old Eico's go up to around 450 volts with the turn
> of a knob.

I have a Farnell bench supply and so could charge the cap
by switching first to say 50V and leaving it there for an hour
or two, and then to 100V and so on.

I have always understood that a cap, if it is to be reformed,
needs to be charged very slowly indeed, over the course of
many hours. That's how I decided on the 10M resistor for
reforming and a 100k for leakage testing,

Jim. It's nice to discuss in a rational way with someone who
doesn't rant and rave and call one a f*ckw*t. Thanks.

Regards
Iain

Re: Electrolytic Cap Leakage revised


"Iain Churches" <IainNG@kolumbus.fi> wrote in message
news:wRGHj.317880$rF5.220153@reader1.news.saunalahti.fi...

>
>> You are right about the 3 hours since 5 R-C time constants work out to
>> 183 seconds.
>
> I had not thought about being able to calculate the
> charging time.Please elaborate.
> Regards
> Iain
>
>

Reply by Jim,

Iain, look at the following link. They explain it way better than I can.

* w w w .allaboutcircuits . com /vol_1/chpt_16/4.html


So your 220E-6 Farad capacitor and 10E6 ohm resistor when multiplied make
T=RC= 2200 seconds, or 36 2/3 minutes for one RC time constant. The units
need to be Farads and Ohms.

The formula at the link says that for 1, 2, 3, 4, 5, 10 time constants, the
capacitor will charge to 63.2%, 86.5%, 95.0%, 98.2%, 99.3%, 99.995% of the
power supply voltage. I draw the line at 5 time constants where I declare
the capacitor fully charged. Of course theoreticially with infinite time we
still never quite get there. Since I am not a purist, I draw the line at 5
RC time constants. So 36.66 * 5 = 183.33 minutes. That is pretty close to
the 3 hours you mentioned. Maybe this is a coincidence since the capacitor
leakage was constantly decreasing, and not modeled into the RC time
constant.

Regards,
Jim

Re: Electrolytic Cap Leakage revised



"JC" <jcandela@prodigy . net > wrote in message
news:PSTHj.5710$qT6.667@nlpi070.nbdc.sbc . com ...
>
> "Iain Churches" <IainNG@kolumbus.fi> wrote in message
> news:wRGHj.317880$rF5.220153@reader1.news.saunalahti.fi...
>
>>
>>> You are right about the 3 hours since 5 R-C time constants work out to
>>> 183 seconds.
>>
>> I had not thought about being able to calculate the
>> charging time.Please elaborate.
>> Regards
>> Iain
>>
>>
>
> Reply by Jim,
>
> Iain, look at the following link. They explain it way better than I can.
>
> * w w w .allaboutcircuits . com /vol_1/chpt_16/4.html
>
>
> So your 220E-6 Farad capacitor and 10E6 ohm resistor when multiplied make
> T=RC= 2200 seconds, or 36 2/3 minutes for one RC time constant. The units
> need to be Farads and Ohms.
>
> The formula at the link says that for 1, 2, 3, 4, 5, 10 time constants,
> the capacitor will charge to 63.2%, 86.5%, 95.0%, 98.2%, 99.3%, 99.995% of
> the power supply voltage. I draw the line at 5 time constants where I
> declare the capacitor fully charged. Of course theoreticially with
> infinite time we still never quite get there. Since I am not a purist, I
> draw the line at 5 RC time constants. So 36.66 * 5 = 183.33 minutes. That
> is pretty close to the 3 hours you mentioned. Maybe this is a coincidence
> since the capacitor leakage was constantly decreasing, and not modeled
> into the RC time constant.
>
I will take a look at the link.
Thanks Jim, for your time and patience:-)

Iain