Hi Bill,

The electrolyte is kind of like a salt water soaked paper, if you will,
it is very low resistance. Its purpose is twofold: first, it creates
an intimate contact with the aluminum oxide, which is the only
dielectric in the cap, and second, it provides a method of repairing
flaws in the aluminum oxide layer. The electrolyte IS the second plate
of the capacitor.

If the electrolyte wasn't there, (eg. it was paper instead), the
dielectric constant of the paper would dominate and the capacitance
would be similar in magnitude to that of a paper capacitor. It is very
important that there not be any nonconductive material (other than the
aluminum oxide layer, that is) between the plates.

Another way an aluminum electrolytic capacitor could be made would be to
build up the oxide layer on one of the plates, and then coat the oxide
with a liquid metallic layer, such as silver epoxy paint, or "nickel
print" to form the other plate. That should give you the high
capacitance of a wet aluminum electrolytic cap, but the oxide wouldn't
ever degrade.

-Chuck



--exray-- wrote:
Chuck Harris wrote:

Hi Alan,

Because of the physical construction of an electrolytic cap, it MUST
change capacitance if the oxide grows thinner in storage, or thickens
thru reforming...

But, I too notice that sometimes the change is large, and othertimes it
is not.

I suspect that what is happening is the oxide layer thins out only in
spots (probably around impurities) in some caps. These spots are large
enough to readily affect the leakage current, but are small with respect
to the total surface area of the plates. Because they are a small
percentage of the total surface area, they only minimally affect
the total capacitance.

-Chuck Harris


Let me pose a question...not knowing inimately how electrolytics were or
are made.
Seems to me that the 'extra' oxide, ie thicker plates, are taking up
some of the physical space that was formerly the electrolyte (part of
the dielectric, so to speak) thereby leaving the plates closer together.
That would indicate more oxide=more capacitance.
In the case of thin oxide (not holes) I don't see how thick or thin
would relate to leakage as long as there was something there. Running
with the same thought, if the oxide is totaaly absent what makes it
redeposit on the wax paper/mylar?
Maybe the leakage is a result of the metallic compounds being absorbed
by the electrolyte and the reforming process sends them back to the
original metal, albeit somewhat randomly.
Does this make any sense?

-Bill

"Chuck Harris" wrote in message
...
Hi Alan,

Because of the physical construction of an electrolytic cap, it MUST
change capacitance if the oxide grows thinner in storage, or thickens
thru reforming...

But, I too notice that sometimes the change is large, and othertimes
it
is not.

I suspect that what is happening is the oxide layer thins out only in
spots (probably around impurities) in some caps. These spots are
large
enough to readily affect the leakage current, but are small with
respect
to the total surface area of the plates. Because they are a small
percentage of the total surface area, they only minimally affect
the total capacitance.

-Chuck Harris


That makes sense. It would be interesting to know how many amp-seconds
per square foot is needed to form the initial oxide layer vs. how many
are needed to reform a cap. Generally, the reforms run a few mils for
less than an hour. A few minutes is typical.

Frank Dresser

"Chuck Harris" wrote in message
...
Hi Alan,

Because of the physical construction of an electrolytic cap, it MUST
change capacitance if the oxide grows thinner in storage, or thickens
thru reforming...

But, I too notice that sometimes the change is large, and othertimes
it
is not.

I suspect that what is happening is the oxide layer thins out only in
spots (probably around impurities) in some caps. These spots are
large
enough to readily affect the leakage current, but are small with
respect
to the total surface area of the plates. Because they are a small
percentage of the total surface area, they only minimally affect
the total capacitance.

-Chuck Harris


That makes sense. It would be interesting to know how many amp-seconds
per square foot is needed to form the initial oxide layer vs. how many
are needed to reform a cap. Generally, the reforms run a few mils for
less than an hour. A few minutes is typical.

Frank Dresser

FWIW, 5 pages from the 1962 edition of the "Electronic Experimenter's
Handbook", starting at page 115, showed the design and construction of
"The Restorer ... gives your electrolytics a new lease on life", by
H. E. Sanders, W4CWK. It used a string of 8 NE-2's across a 720-volt
DC power supply (with dropping resistors, of course!-) to produce
selectable voltages in approximate 70-volt steps. Toward the end was
the sentence "Relatively new capacitors will form in a few minutes;
very old ones may take several hours."

A much-simpler circuit just for 450-volt caps was given in "Nuts & Volts
Magazine" in the last few years, but the page I tore and filed doesn't
have a date so I can't give a citation. But if you can cobble-up an
appropriate voltage source, it only used four additional components
(plus the capacitor to be formed); I'll try an ASCII schematic:

+------+--220K--+---NE2---+
| | | |
|+ | +--0.22C--+
450- | |
volt +---68K------------+
source 2 |
|- Cx (to be formed)
| |
+-------------------------+

Operation: "At initial power-on, the voltage across Cx is at zero and the
voltage across R2 is 450 volts, which lights NE2, the neon bulb. If Cx
is anywhere near healthy, it will slowly start to reform and charge up.
As the process continues (which can take hours), the voltage across R2
falls to the point where it is insufficient to keep the neon continuously
lit, at which time it begins to flash at a rate proportional to the amount
of current flowing through Cx. Once the neon lamp stops flashing, the
voltage across R2 is too low to light the lamp, and it can be assumed Cx
is fully charged and successfully reformed. For lower-voltage electro-
lytics, adjust the [source] voltage [to match the working voltage of Cx]."

--Myron, W0PBV.

--
Five boxes preserve our freedoms: soap, ballot, witness, jury, and cartridge
PhD EE (retired). "Barbershop" tenor. CDL(PTX). W0PBV. (785) 539-4448
NRA Life Member and Certified Instructor (Home Firearm Safety, Rifle, Pistol)

FWIW, 5 pages from the 1962 edition of the "Electronic Experimenter's
Handbook", starting at page 115, showed the design and construction of
"The Restorer ... gives your electrolytics a new lease on life", by
H. E. Sanders, W4CWK. It used a string of 8 NE-2's across a 720-volt
DC power supply (with dropping resistors, of course!-) to produce
selectable voltages in approximate 70-volt steps. Toward the end was
the sentence "Relatively new capacitors will form in a few minutes;
very old ones may take several hours."

A much-simpler circuit just for 450-volt caps was given in "Nuts & Volts
Magazine" in the last few years, but the page I tore and filed doesn't
have a date so I can't give a citation. But if you can cobble-up an
appropriate voltage source, it only used four additional components
(plus the capacitor to be formed); I'll try an ASCII schematic:

+------+--220K--+---NE2---+
| | | |
|+ | +--0.22C--+
450- | |
volt +---68K------------+
source 2 |
|- Cx (to be formed)
| |
+-------------------------+

Operation: "At initial power-on, the voltage across Cx is at zero and the
voltage across R2 is 450 volts, which lights NE2, the neon bulb. If Cx
is anywhere near healthy, it will slowly start to reform and charge up.
As the process continues (which can take hours), the voltage across R2
falls to the point where it is insufficient to keep the neon continuously
lit, at which time it begins to flash at a rate proportional to the amount
of current flowing through Cx. Once the neon lamp stops flashing, the
voltage across R2 is too low to light the lamp, and it can be assumed Cx
is fully charged and successfully reformed. For lower-voltage electro-
lytics, adjust the [source] voltage [to match the working voltage of Cx]."

--Myron, W0PBV.

--
Five boxes preserve our freedoms: soap, ballot, witness, jury, and cartridge
PhD EE (retired). "Barbershop" tenor. CDL(PTX). W0PBV. (785) 539-4448
NRA Life Member and Certified Instructor (Home Firearm Safety, Rifle, Pistol)

put some DC thru it. If you get DC current thru a capacitor it may be
defective. Measure with a suitable ampmeter and resistor.(ohm meter
check) Capacitors look leaky when checked using AC. :^

gil wrote:
I always heard about leaky caps being a problem with boatanchors, other than
checking for the capacitor value is there a way to check if its a "leaky
cap" using just a multitester or voltmeter?

Thanks.....gil

put some DC thru it. If you get DC current thru a capacitor it may be
defective. Measure with a suitable ampmeter and resistor.(ohm meter
check) Capacitors look leaky when checked using AC. :^

gil wrote:
I always heard about leaky caps being a problem with boatanchors, other than
checking for the capacitor value is there a way to check if its a "leaky
cap" using just a multitester or voltmeter?

Thanks.....gil