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John Fields
Guest
Sun Mar 12, 2006 12:29 am
On Sun, 12 Mar 2006 09:35:33 +1100, Daniel
<dxmm_at_nospam.albury.net.au> wrote:
Quote:
John Fields wrote:
On Sat, 11 Mar 2006 14:36:05 -0600, "DB" <DLB613_at_charter.net> wrote:
"John Fields" <jfields_at_austininstruments.com> wrote in message
news:0aoo02lehaq7hgejn4vo6n1sp5t2oasvka_at_4ax.com...
On Sun, 5 Mar 2006 22:49:35 -0600, "DB" <DLB613_at_charter.net> wrote:
Unfortunately I don't have a schematic. There is only one capacitor, but
it
has three terminals. That and the notation on the side makes me think it
may be some kind of a dual capacitor in one big can. I had never heard of
such a thing before, but could that be what it is?
More than likely, that's precisely what it is. Back in the stony
ages it was quite common to have two or three electrolytics in the
same can. The way you describe yours though, makes it sound like
there's a separate terminal coming out of the base for the common
connection, when usually a tab which was part of the case was used
for the common connection. Is thare any other marking on the case?
Like "TYPE FP" or anything like that?
I picked up a couple of 5000 MFD 25VDC capacitors for $1 each, hoping that
they might be suitable replacements. The new capacitors are marked "Type
PFP." The old 5000 MFD 20VDC dual capacitor didn't have any markings to
indicate the type. The old capacitor case looks like it is made of
bakelite, by the way.
The new capacitors have three terminals that are part of the aluminum case,
and a fourth terminal slightly off-center from the base. The fourth
terminal has a little square marking next to it. There does not appear to
be any measureable resistance between any of the terminals, including the
central one.
I am wondering how I can determine if these capacitors are suitable
replacements. I'm also not sure how to figure out which terminal should be
connected to the positive output of the power supply, and which to the
negative. Any assistance would be greatly appreciated.
---
The capacitor you have is one of the Mallory PFP types which is a
twist-lock capacitor for PCB mounting. Twist-lock because usually
the tabs on the case were slipped through slots in the PCB (type
PFP) or through the chassis (type FP), and then the tabs were
twisted with a pair of pliers to lock the capacitor down.
The case is almost certainly negative, leaving the center pin
positive.
The reason you're measuring zero resistance between the center pin
and the case is because they're pretty good size capacitors and
it'll take a good while for them to charge up with the current from
your ohmmeter.
If the capacitors are discharged at the start, when connecting the
ohm-meter there should be maximum current flow, indicating zero ohms,
for an instant, then, as the capacitor charges, the amount of current
will decrease until, in the end, the capacitor is charged to the same
voltage as the ohm-meter is providing, so indicates zero current flow,
or as an open circuit, i.e. infinite resistance.
Depending on the internal resistance of the ohm-meter, this could take
almost zero time or substantial time
---
Most of my multimeters, on the 200 ohm full scale range, supply
about 1A into a short, so if we consider the ohmmeter a constant
current source it'll take:
dv C 1V * 5E-3F
t = ------ = ------------ = 5 seconds
I 1e-3A
to charge the cap up to a volt through the ohmmeter. Not exactly
what I'd call "almost zero time" ;)
But, to be fair, I put a Fluke 8060A on the 200 ohm range across a
4700µF 35V cap and it only had time to give me one reading before it
went "OL". OTOH, on the megohm range it took it 10s to get to
0.0020, so if the OP's using a cheap 3-1/2 digit meter on the high
resistance range, he may have seen 0.000 and not given it a chance
to rise any higher before disconnecting.
--
John Fields
Professional Circuit Designer
DB
Guest
Sun Mar 12, 2006 12:46 am
"Daniel" <dxmm_at_nospam.albury.net.au> wrote in message
news:441351d9$0$24335$6d36acad_at_titian.nntpserver.com...
Quote:
John Fields wrote:
On Sat, 11 Mar 2006 14:36:05 -0600, "DB" <DLB613_at_charter.net> wrote:
"John Fields" <jfields_at_austininstruments.com> wrote in message
news:0aoo02lehaq7hgejn4vo6n1sp5t2oasvka_at_4ax.com...
On Sun, 5 Mar 2006 22:49:35 -0600, "DB" <DLB613_at_charter.net> wrote:
Unfortunately I don't have a schematic. There is only one capacitor,
but
it
has three terminals. That and the notation on the side makes me think
it
may be some kind of a dual capacitor in one big can. I had never
heard of
such a thing before, but could that be what it is?
More than likely, that's precisely what it is. Back in the stony
ages it was quite common to have two or three electrolytics in the
same can. The way you describe yours though, makes it sound like
there's a separate terminal coming out of the base for the common
connection, when usually a tab which was part of the case was used
for the common connection. Is thare any other marking on the case?
Like "TYPE FP" or anything like that?
I picked up a couple of 5000 MFD 25VDC capacitors for $1 each, hoping
that they might be suitable replacements. The new capacitors are marked
"Type PFP." The old 5000 MFD 20VDC dual capacitor didn't have any
markings to indicate the type. The old capacitor case looks like it is
made of bakelite, by the way.
The new capacitors have three terminals that are part of the aluminum
case, and a fourth terminal slightly off-center from the base. The
fourth terminal has a little square marking next to it. There does not
appear to be any measureable resistance between any of the terminals,
including the central one.
I am wondering how I can determine if these capacitors are suitable
replacements. I'm also not sure how to figure out which terminal should
be connected to the positive output of the power supply, and which to
the negative. Any assistance would be greatly appreciated.
---
The capacitor you have is one of the Mallory PFP types which is a
twist-lock capacitor for PCB mounting. Twist-lock because usually
the tabs on the case were slipped through slots in the PCB (type
PFP) or through the chassis (type FP), and then the tabs were
twisted with a pair of pliers to lock the capacitor down.
The case is almost certainly negative, leaving the center pin
positive.
The reason you're measuring zero resistance between the center pin
and the case is because they're pretty good size capacitors and
it'll take a good while for them to charge up with the current from
your ohmmeter.
If the capacitors are discharged at the start, when connecting the
ohm-meter there should be maximum current flow, indicating zero ohms, for
an instant, then, as the capacitor charges, the amount of current will
decrease until, in the end, the capacitor is charged to the same voltage
as the ohm-meter is providing, so indicates zero current flow, or as an
open circuit, i.e. infinite resistance.
Depending on the internal resistance of the ohm-meter, this could take
almost zero time or substantial time
Daniel
Also, if they haven't been used in some time the
oxide on the plates might not be in the best condition, so what you
might want to do is put an ammeter in series with the cap and
measure the current going into it while it's charging from a low
voltage. When you first connect it the ammmeter will show a _LOT_
of current, but as time goes by you should see the current drop and
eventually settle on a very low value. When it gets stable,
increase the voltage and watch the current until it gets stable
again. Do that until the voltage across the capacitor is at 25V,
and then just let it sit there with power on it for a few hours. If
everything is OK the ammeter should read only a few milliamperes
after a few hours.
BTW, what kind of test equipment do you have access to?
The only test equipment I have is a $9.99 multi-meter from Harbor Freight.
The PFP capacitors have never been used, and appear to be brand new. The
ohmmeter reading is zero ohms across all terminals, for both the new and old
capacitors.
I tried connecting one of the new caps in series with a flashlight battery
and measuring the current. The current slowly but steadily declined. I did
the same thing with each side of the old dual capacitor, and in each case
there was a quick spike in current, followed by a steady current of approx.
0.9 mA.
- D. Boucher
John Fields
Guest
Sun Mar 12, 2006 1:35 am
On Sat, 11 Mar 2006 17:46:42 -0600, "DB" <DLB613_at_charter.net> wrote:
Quote:
The only test equipment I have is a $9.99 multi-meter from Harbor Freight.
The PFP capacitors have never been used, and appear to be brand new. The
ohmmeter reading is zero ohms across all terminals, for both the new and old
capacitors.
---
Quote:
I tried connecting one of the new caps in series with a flashlight battery
and measuring the current. The current slowly but steadily declined. I did
the same thing with each side of the old dual capacitor, and in each case
there was a quick spike in current, followed by a steady current of approx.
0.9 mA.
---
With a 1.5V flashlight battery supplying a steady 0.9mA into the
capS, that indicates that their internal resistances were:
E 1.5V
R = --- = ------ ~ 1.67K ohms,
I 09mA
so those caps are definitely bad
Why don't you cut them out of there, wire the new ones in where the
old ones were, just to test them, and see what happens as you crank
the voltage up _slowly_ from zero? If you put your meter in series
with the output filter:
RAW +DC>---[+METER-]--+--[CHOKE]--+---->+DC OUT
|+ |+
[5000µF] [5000µF]
| |
RAW -DC>--------------+-----------+---->-DC OUT
You'll be able to see what the total leakage current into the caps
is. With no load, I'd expect something substantially less than 5mA
with the full 25V across the caps.
--
John Fields
Professional Circuit Designer
Daniel
Guest
Sun Mar 12, 2006 10:50 am
John Fields wrote:
Quote:
On Sun, 12 Mar 2006 09:35:33 +1100, Daniel
dxmm_at_nospam.albury.net.au> wrote:
John Fields wrote:
On Sat, 11 Mar 2006 14:36:05 -0600, "DB" <DLB613_at_charter.net> wrote:
"John Fields" <jfields_at_austininstruments.com> wrote in message
news:0aoo02lehaq7hgejn4vo6n1sp5t2oasvka_at_4ax.com...
On Sun, 5 Mar 2006 22:49:35 -0600, "DB" <DLB613_at_charter.net> wrote:
Unfortunately I don't have a schematic. There is only one capacitor, but
it
has three terminals. That and the notation on the side makes me think it
may be some kind of a dual capacitor in one big can. I had never heard of
such a thing before, but could that be what it is?
More than likely, that's precisely what it is. Back in the stony
ages it was quite common to have two or three electrolytics in the
same can. The way you describe yours though, makes it sound like
there's a separate terminal coming out of the base for the common
connection, when usually a tab which was part of the case was used
for the common connection. Is thare any other marking on the case?
Like "TYPE FP" or anything like that?
I picked up a couple of 5000 MFD 25VDC capacitors for $1 each, hoping that
they might be suitable replacements. The new capacitors are marked "Type
PFP." The old 5000 MFD 20VDC dual capacitor didn't have any markings to
indicate the type. The old capacitor case looks like it is made of
bakelite, by the way.
The new capacitors have three terminals that are part of the aluminum case,
and a fourth terminal slightly off-center from the base. The fourth
terminal has a little square marking next to it. There does not appear to
be any measureable resistance between any of the terminals, including the
central one.
I am wondering how I can determine if these capacitors are suitable
replacements. I'm also not sure how to figure out which terminal should be
connected to the positive output of the power supply, and which to the
negative. Any assistance would be greatly appreciated.
---
The capacitor you have is one of the Mallory PFP types which is a
twist-lock capacitor for PCB mounting. Twist-lock because usually
the tabs on the case were slipped through slots in the PCB (type
PFP) or through the chassis (type FP), and then the tabs were
twisted with a pair of pliers to lock the capacitor down.
The case is almost certainly negative, leaving the center pin
positive.
The reason you're measuring zero resistance between the center pin
and the case is because they're pretty good size capacitors and
it'll take a good while for them to charge up with the current from
your ohmmeter.
If the capacitors are discharged at the start, when connecting the
ohm-meter there should be maximum current flow, indicating zero ohms,
for an instant, then, as the capacitor charges, the amount of current
will decrease until, in the end, the capacitor is charged to the same
voltage as the ohm-meter is providing, so indicates zero current flow,
or as an open circuit, i.e. infinite resistance.
Depending on the internal resistance of the ohm-meter, this could take
almost zero time or substantial time
---
Most of my multimeters, on the 200 ohm full scale range, supply
about 1A into a short, so if we consider the ohmmeter a constant
current source it'll take:
dv C 1V * 5E-3F
t = ------ = ------------ = 5 seconds
I 1e-3A
to charge the cap up to a volt through the ohmmeter. Not exactly
what I'd call "almost zero time"
All fine and dandy if the Ohm-meter were a constant current source and
the capacitor could just keep absorbing those electrons, but neither is
the case so your logic fails.
(As a side note, I'll give you 5E-3F is the same as 5000 microfarads,
but 1e-3A is not the one amp you say your multimeter can supply into a
short. If you fix this in the workings, t = 5 milli-seconds, which I
would call "almost zero time" when talking about digital multi-meter
response time.)
Quote:
But, to be fair, I put a Fluke 8060A on the 200 ohm range across a
4700µF 35V cap and it only had time to give me one reading before it
went "OL". OTOH, on the megohm range it took it 10s to get to
0.0020, so if the OP's using a cheap 3-1/2 digit meter on the high
resistance range, he may have seen 0.000 and not given it a chance
to rise any higher before disconnecting.
Daniel
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John Fields
Guest
Sun Mar 12, 2006 12:22 pm
On Sun, 12 Mar 2006 20:50:41 +1100, Daniel
<dxmm_at_nospam.albury.net.au> wrote:
Quote:
John Fields wrote:
---
Most of my multimeters, on the 200 ohm full scale range, supply
about 1A into a short, so if we consider the ohmmeter a constant
current source it'll take:
dv C 1V * 5E-3F
t = ------ = ------------ = 5 seconds
I 1e-3A
to charge the cap up to a volt through the ohmmeter. Not exactly
what I'd call "almost zero time" ;)
All fine and dandy if the Ohm-meter were a constant current source and
the capacitor could just keep absorbing those electrons, but neither is
the case so your logic fails.
---
Not at all. On the higher resistance ranges the ohmmeter _does_
approximate a constant current source; at least for the purpose of
this discussion.
---
Quote:
(As a side note, I'll give you 5E-3F is the same as 5000 microfarads,
but 1e-3A is not the one amp you say your multimeter can supply into a
short.
---
Yes, but that one amp '1A', earlier, was a trypo. It should have
read: "1mA", so the math is correct as it stands. I'm surprised you
didn't catch that, since it's not likely _any_ common ohmmeter is
going to push 1 amp through the resistance it's measuring.
Even my trusty old Simpson 260 on the 'R times 1' range only puts
about 100mA through a 100mA d'Arsonval movement with an internal
resistance of 1.08 ohms.
---
Quote:
If you fix this in the workings, t = 5 milli-seconds, which I
would call "almost zero time" when talking about digital multi-meter
response time.)
---
It doesn't need to be fixed, though. All that needed to happen was
for the typo to be corrected, so the 5 seconds stands at 5 seconds,
which even for a relatively slow digital multimeter with an update
rate of 2 readings per second isn't "almost zero time".
--
John Fields
Professional Circuit Designer
James Thompson
Guest
Tue Mar 14, 2006 1:40 am
Quote:
One minute the Power supply was working fine, then it vibrated for a split
second and then the fuse blew. I replaced the fuse, and the same thing
happened again when I turned it on. I opened the power supply up and I
saw that there was some dried yellow crusty material that appeared to have
oozed out of the capacitor near one of the terminals. It looked like it
had been that way for a while, but I had just recently started using this
power supply after it had been sitting on a shelf for about 12 years.
I checked the resistance between the capacitor terminals (without removing
the capacitor from the circuit), and found that there was no resistance.
I don't know if that is normal.
- D. Boucher
Most likely the crusty yellow stuff is glue and the cap is good.
Caps are sometimes glued in during the assembly. The way you said it went
out ( power supply) , if it were the
cap going bad, you would find paper all over the inside (it would explode).
I would lood for a diode ( bridge or single ) , or a transistor shorted.
Use you ohm meter to check for a near
zero resistance on one of them.
Let us know what you find..
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