Estimating/determining a likely value of a small low voltage polystyrene cap
that has no value printed on it or as in this case has melted due to being
near a hot R and foils shorted.
Ground down the ends of a few such caps , stripped the covering off and
unwrapping the foils.
2 foils per strip length L, width W and separating film thickness d.
120pF, 22x4.5 x .04mm
2220pF, 60x4 x .05
680pF, 28x4.5 x .06
1800pF, 200x5.5 x 0.03 mm
Averaging out gives an approximate fomula of
Capacity in pF approximately = (L x W )/ (420 d^2)

Anyone aware of a more general formula,refinement or other method?

--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/

N Cook:
Unless you know the "working voltage" rating of the cap, value
estimates based on physical are just about meaningless.
As you undoubtably know, with the same value cap, the higher the
voltage, the bigger the physical size and visa-vera.
Hopefully, somewhere else on the board you can find a similar cap with
intact markings to get some kind of idea of what you are dealing with.
electricitym

wrote in message
oups.com...
N Cook:
Unless you know the "working voltage" rating of the cap, value
estimates based on physical are just about meaningless.
As you undoubtably know, with the same value cap, the higher the
voltage, the bigger the physical size and visa-vera.
Hopefully, somewhere else on the board you can find a similar cap with
intact markings to get some kind of idea of what you are dealing with.
electricitym


Implicitly allowed for in the value of d

--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/

On Wed, 12 Oct 2005 11:36:54 +0100, "N Cook" put
finger to keyboard and composed:

Estimating/determining a likely value of a small low voltage polystyrene cap
that has no value printed on it or as in this case has melted due to being
near a hot R and foils shorted.
Ground down the ends of a few such caps , stripped the covering off and
unwrapping the foils.
2 foils per strip length L, width W and separating film thickness d.
120pF, 22x4.5 x .04mm
2220pF, 60x4 x .05
680pF, 28x4.5 x .06
1800pF, 200x5.5 x 0.03 mm
Averaging out gives an approximate fomula of
Capacity in pF approximately = (L x W )/ (420 d^2)

Anyone aware of a more general formula,refinement or other method?

The units are wrong. I would expect something like ...

C = k * (L * W / d)

But then that doesn't fit nicely with your empirical data. shrug

The formula for a parallel plate capacitor is ...

C = k eo A / d

k = dielectric constant
eo = permittivity of free space = 8.85 x 10 -12 F/m
A = plate area
d = distance between plates

Plugging your data into the above formula gives:

120pF k=5.5
680pF k=37
1800pF k=5.5
2220pF k=52

See this URL:

"On Capacitor Dielectric Materials - A Chemist's View":
http://www.audience-av.com/on_capaci...c_material.htm

According to the author, polystyrene has a k of ~2.55 which is in
disagreement with your results, although not by much. However,
something looks very wrong with the 680pF and 2220pF data. I like the
way you've approached this problem, though.

-- Franc Zabkar

Please remove one 'i' from my address when replying by email.

"N Cook" wrote in message
...
Estimating/determining a likely value of a small low voltage polystyrene
cap
that has no value printed on it or as in this case has melted due to being
near a hot R and foils shorted.
Ground down the ends of a few such caps , stripped the covering off and
unwrapping the foils.
2 foils per strip length L, width W and separating film thickness d.
120pF, 22x4.5 x .04mm
2220pF, 60x4 x .05
680pF, 28x4.5 x .06
1800pF, 200x5.5 x 0.03 mm
Averaging out gives an approximate fomula of
Capacity in pF approximately = (L x W )/ (420 d^2)

Anyone aware of a more general formula,refinement or other method?


Is this just a supply decoupling cap, or something mre critical like a
timing circuit or tuned circuit? If it's a supply decoupling cap, it's not
hard to make an educated guess on a suitable replacement. If it's the
latter, a capacitance box or substition of likely values until it works
satisfactorily might be good enough.

Try 1nf or 10nf and work up/down from there!

Dave

"Franc Zabkar" wrote in message
...
On Wed, 12 Oct 2005 11:36:54 +0100, "N Cook" put
finger to keyboard and composed:

Estimating/determining a likely value of a small low voltage polystyrene
cap
that has no value printed on it or as in this case has melted due to
being
near a hot R and foils shorted.
Ground down the ends of a few such caps , stripped the covering off and
unwrapping the foils.
2 foils per strip length L, width W and separating film thickness d.
120pF, 22x4.5 x .04mm
2220pF, 60x4 x .05
680pF, 28x4.5 x .06
1800pF, 200x5.5 x 0.03 mm
Averaging out gives an approximate fomula of
Capacity in pF approximately = (L x W )/ (420 d^2)

Anyone aware of a more general formula,refinement or other method?

The units are wrong. I would expect something like ...

C = k * (L * W / d)

But then that doesn't fit nicely with your empirical data. shrug

The formula for a parallel plate capacitor is ...

C = k eo A / d

k = dielectric constant
eo = permittivity of free space = 8.85 x 10 -12 F/m
A = plate area
d = distance between plates

Plugging your data into the above formula gives:

120pF k=5.5
680pF k=37
1800pF k=5.5
2220pF k=52

See this URL:

"On Capacitor Dielectric Materials - A Chemist's View":
http://www.audience-av.com/on_capaci...c_material.htm

According to the author, polystyrene has a k of ~2.55 which is in
disagreement with your results, although not by much. However,
something looks very wrong with the 680pF and 2220pF data. I like the
way you've approached this problem, though.

-- Franc Zabkar

Please remove one 'i' from my address when replying by email.

Going by memory I thought the relation was inverse square law with "d".

The other factor is they are spiral wrapped with the same dielectric
material on the
other 2 surfaces, not nessarily same thickness and when wrapped these become
double
thickness - 3 dielectrics with 2 "plates" all spirally wrapped -
Ive no idea what the design formula for such construction is

On Thu, 13 Oct 2005 06:37:45 +1000 Franc Zabkar
wrote:

On Wed, 12 Oct 2005 11:36:54 +0100, "N Cook" put
finger to keyboard and composed:

Estimating/determining a likely value of a small low voltage polystyrene cap
that has no value printed on it or as in this case has melted due to being
near a hot R and foils shorted.
Ground down the ends of a few such caps , stripped the covering off and
unwrapping the foils.
2 foils per strip length L, width W and separating film thickness d.
120pF, 22x4.5 x .04mm
2220pF, 60x4 x .05
680pF, 28x4.5 x .06
1800pF, 200x5.5 x 0.03 mm
Averaging out gives an approximate fomula of
Capacity in pF approximately = (L x W )/ (420 d^2)

Anyone aware of a more general formula,refinement or other method?

The units are wrong. I would expect something like ...

C = k * (L * W / d)

But then that doesn't fit nicely with your empirical data. shrug

The formula for a parallel plate capacitor is ...

C = k eo A / d

k = dielectric constant
eo = permittivity of free space = 8.85 x 10 -12 F/m
A = plate area
d = distance between plates

Plugging your data into the above formula gives:

120pF k=5.5
680pF k=37
1800pF k=5.5
2220pF k=52

See this URL:

"On Capacitor Dielectric Materials - A Chemist's View":
http://www.audience-av.com/on_capaci...c_material.htm

According to the author, polystyrene has a k of ~2.55 which is in
disagreement with your results, although not by much. However,
something looks very wrong with the 680pF and 2220pF data. I like the
way you've approached this problem, though.

I like his approach, too.

If you consider that the wrapped construction uses both sides of the
foil strips, this doubles the effective area, which effectively halves
the calculated k values above. This starts to look much more
reasonable now, except for the 37 and 52, for which there must be some
other explanation.

-
-----------------------------------------------
Jim Adney
Madison, WI 53711 USA
-----------------------------------------------

On Wed, 12 Oct 2005 22:57:47 +0100, "N Cook" put
finger to keyboard and composed:

"Franc Zabkar" wrote in message
.. .
On Wed, 12 Oct 2005 11:36:54 +0100, "N Cook" put
finger to keyboard and composed:

Estimating/determining a likely value of a small low voltage polystyrene
cap
that has no value printed on it or as in this case has melted due to
being
near a hot R and foils shorted.
Ground down the ends of a few such caps , stripped the covering off and
unwrapping the foils.
2 foils per strip length L, width W and separating film thickness d.
120pF, 22x4.5 x .04mm
2220pF, 60x4 x .05
680pF, 28x4.5 x .06
1800pF, 200x5.5 x 0.03 mm
Averaging out gives an approximate fomula of
Capacity in pF approximately = (L x W )/ (420 d^2)

Anyone aware of a more general formula,refinement or other method?

The units are wrong. I would expect something like ...

C = k * (L * W / d)

But then that doesn't fit nicely with your empirical data. shrug

The formula for a parallel plate capacitor is ...

C = k eo A / d

k = dielectric constant
eo = permittivity of free space = 8.85 x 10 -12 F/m
A = plate area
d = distance between plates

Plugging your data into the above formula gives:

120pF k=5.5
680pF k=37
1800pF k=5.5
2220pF k=52

See this URL:

"On Capacitor Dielectric Materials - A Chemist's View":
http://www.audience-av.com/on_capaci...c_material.htm

According to the author, polystyrene has a k of ~2.55 which is in
disagreement with your results, although not by much. However,
something looks very wrong with the 680pF and 2220pF data. I like the
way you've approached this problem, though.

-- Franc Zabkar

Please remove one 'i' from my address when replying by email.

Going by memory I thought the relation was inverse square law with "d".

The other factor is they are spiral wrapped with the same dielectric
material on the
other 2 surfaces, not nessarily same thickness and when wrapped these become
double
thickness - 3 dielectrics with 2 "plates" all spirally wrapped -
Ive no idea what the design formula for such construction is

I would think the capacitance would roughly double.

If the diagram on the left represents a cross section of the spiral,
where T & B are the top & bottom plates, and D is the dielectric
material, then I would think that the diagram on the right would be
its electrical equivalent.

T B T B T B B T T B
T D B D T D B T D B B D T T D B
T D B D T D B T D B---B D T---T D B
T D B D T D B T D B B D T T D B
T B T B T B B T T B

V+ V- V+ V- V+ V- V- V+ V+ V-

Let's say C is the capacitance of a single wrap of the spiral. It
follows then that the capacitance of 2 adjacent wraps is 3C. Therefore
the capacitance of n wraps would be C x (2n-1), ie roughly double,
given a sufficiently large number of wraps.

So the figure of k in two of the calculations above now approaches 2.8
which is in the ball park for polystyrene.

BTW, the value of 2220pF is strange. Shouldn't that be 220pF? If so
then k = 5.2 /2 = 2.6 which is once again looking good.

That only leaves 680pF. Was the cap marked 680? If so, how can you be
certain that this means 680 rather than 68 x 10^0? If the latter, then
k=1.9 which is again pretty close.

-- Franc Zabkar

Please remove one 'i' from my address when replying by email.

"Franc Zabkar" wrote in message
...
On Wed, 12 Oct 2005 22:57:47 +0100, "N Cook" put
finger to keyboard and composed:

"Franc Zabkar" wrote in message
.. .
On Wed, 12 Oct 2005 11:36:54 +0100, "N Cook" put
finger to keyboard and composed:

Estimating/determining a likely value of a small low voltage
polystyrene
cap
that has no value printed on it or as in this case has melted due to
being
near a hot R and foils shorted.
Ground down the ends of a few such caps , stripped the covering off
and
unwrapping the foils.
2 foils per strip length L, width W and separating film thickness d.
120pF, 22x4.5 x .04mm
2220pF, 60x4 x .05
680pF, 28x4.5 x .06
1800pF, 200x5.5 x 0.03 mm
Averaging out gives an approximate fomula of
Capacity in pF approximately = (L x W )/ (420 d^2)

Anyone aware of a more general formula,refinement or other method?

The units are wrong. I would expect something like ...

C = k * (L * W / d)

But then that doesn't fit nicely with your empirical data. shrug

The formula for a parallel plate capacitor is ...

C = k eo A / d

k = dielectric constant
eo = permittivity of free space = 8.85 x 10 -12 F/m
A = plate area
d = distance between plates

Plugging your data into the above formula gives:

120pF k=5.5
680pF k=37
1800pF k=5.5
2220pF k=52

See this URL:

"On Capacitor Dielectric Materials - A Chemist's View":
http://www.audience-av.com/on_capaci...c_material.htm

According to the author, polystyrene has a k of ~2.55 which is in
disagreement with your results, although not by much. However,
something looks very wrong with the 680pF and 2220pF data. I like the
way you've approached this problem, though.

-- Franc Zabkar

Please remove one 'i' from my address when replying by email.

Going by memory I thought the relation was inverse square law with "d".

The other factor is they are spiral wrapped with the same dielectric
material on the
other 2 surfaces, not nessarily same thickness and when wrapped these
become
double
thickness - 3 dielectrics with 2 "plates" all spirally wrapped -
Ive no idea what the design formula for such construction is

I would think the capacitance would roughly double.

If the diagram on the left represents a cross section of the spiral,
where T & B are the top & bottom plates, and D is the dielectric
material, then I would think that the diagram on the right would be
its electrical equivalent.

T B T B T B B T T B
T D B D T D B T D B B D T T D B
T D B D T D B T D B---B D T---T D B
T D B D T D B T D B B D T T D B
T B T B T B B T T B

V+ V- V+ V- V+ V- V- V+ V+ V-

Let's say C is the capacitance of a single wrap of the spiral. It
follows then that the capacitance of 2 adjacent wraps is 3C. Therefore
the capacitance of n wraps would be C x (2n-1), ie roughly double,
given a sufficiently large number of wraps.

So the figure of k in two of the calculations above now approaches 2.8
which is in the ball park for polystyrene.

BTW, the value of 2220pF is strange. Shouldn't that be 220pF? If so
then k = 5.2 /2 = 2.6 which is once again looking good.

That only leaves 680pF. Was the cap marked 680? If so, how can you be
certain that this means 680 rather than 68 x 10^0? If the latter, then
k=1.9 which is again pretty close.

-- Franc Zabkar

Please remove one 'i' from my address when replying by email.

yes bouncing key , should read 220pF
i will try repeating another unwrapped 680pF measurements today and check
on ac cap meter also

--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/

"Franc Zabkar" wrote in message
...
On Wed, 12 Oct 2005 22:57:47 +0100, "N Cook" put
finger to keyboard and composed:

"Franc Zabkar" wrote in message
.. .
On Wed, 12 Oct 2005 11:36:54 +0100, "N Cook" put
finger to keyboard and composed:

Estimating/determining a likely value of a small low voltage
polystyrene
cap
that has no value printed on it or as in this case has melted due to
being
near a hot R and foils shorted.
Ground down the ends of a few such caps , stripped the covering off
and
unwrapping the foils.
2 foils per strip length L, width W and separating film thickness d.
120pF, 22x4.5 x .04mm
2220pF, 60x4 x .05
680pF, 28x4.5 x .06
1800pF, 200x5.5 x 0.03 mm
Averaging out gives an approximate fomula of
Capacity in pF approximately = (L x W )/ (420 d^2)

Anyone aware of a more general formula,refinement or other method?

The units are wrong. I would expect something like ...

C = k * (L * W / d)

But then that doesn't fit nicely with your empirical data. shrug

The formula for a parallel plate capacitor is ...

C = k eo A / d

k = dielectric constant
eo = permittivity of free space = 8.85 x 10 -12 F/m
A = plate area
d = distance between plates

Plugging your data into the above formula gives:

120pF k=5.5
680pF k=37
1800pF k=5.5
2220pF k=52

See this URL:

"On Capacitor Dielectric Materials - A Chemist's View":
http://www.audience-av.com/on_capaci...c_material.htm

According to the author, polystyrene has a k of ~2.55 which is in
disagreement with your results, although not by much. However,
something looks very wrong with the 680pF and 2220pF data. I like the
way you've approached this problem, though.

-- Franc Zabkar

Please remove one 'i' from my address when replying by email.

Going by memory I thought the relation was inverse square law with "d".

The other factor is they are spiral wrapped with the same dielectric
material on the
other 2 surfaces, not nessarily same thickness and when wrapped these
become
double
thickness - 3 dielectrics with 2 "plates" all spirally wrapped -
Ive no idea what the design formula for such construction is

I would think the capacitance would roughly double.

If the diagram on the left represents a cross section of the spiral,
where T & B are the top & bottom plates, and D is the dielectric
material, then I would think that the diagram on the right would be
its electrical equivalent.

T B T B T B B T T B
T D B D T D B T D B B D T T D B
T D B D T D B T D B---B D T---T D B
T D B D T D B T D B B D T T D B
T B T B T B B T T B

V+ V- V+ V- V+ V- V- V+ V+ V-

Let's say C is the capacitance of a single wrap of the spiral. It
follows then that the capacitance of 2 adjacent wraps is 3C. Therefore
the capacitance of n wraps would be C x (2n-1), ie roughly double,
given a sufficiently large number of wraps.

So the figure of k in two of the calculations above now approaches 2.8
which is in the ball park for polystyrene.

BTW, the value of 2220pF is strange. Shouldn't that be 220pF? If so
then k = 5.2 /2 = 2.6 which is once again looking good.

That only leaves 680pF. Was the cap marked 680? If so, how can you be
certain that this means 680 rather than 68 x 10^0? If the latter, then
k=1.9 which is again pretty close.

-- Franc Zabkar

Please remove one 'i' from my address when replying by email.

680 pF measured 705pF without leads C compensation so 680pF

Must have mangled the wrong "680pF" cap
The construction is different being 2 foils 48mm * 4.5mm with 0.015mm
dielectric film to one side only.
Plugging those figures in gives the 5.5 value
Spiral structure factor 2 seems to not come into play unless the plate area
A is defined as the combined area of both plates.

So general formula is C in pF = 5.5 * 8.85 * L * W / (1000 * d )

C = 0.0487 * L * W / d, C in pF and dimensions mm