Resistance variation with thickness

N

N_Cook

Guest
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20 mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?



--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
 
"N_Cook" <diverse@tcp.co.uk> wrote in message
news:ghs2vu$ftm$1@news.motzarella.org...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20
mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?
Click to expand...


AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. If this remains unchanged, so does the resistance.



Gareth.
 
"Gareth Magennis" <gareth.magennis@ntlworld.com> wrote in message
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"N_Cook" <diverse@tcp.co.uk> wrote in message
news:ghs2vu$ftm$1@news.motzarella.org...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20 mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. If this remains unchanged, so does the resistance.



Gareth.
Click to expand...

That is correct, but the length also has to remain unchanged The formula for
the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A remain the
same, or change in a manner that produces the same ratio.

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.
 
"DaveM" <masondg4499@comcast99.net> wrote in message
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"Gareth Magennis" <gareth.magennis@ntlworld.com> wrote in message
news:Gsg0l.4221$d11.1850@newsfe25.ams2...

"N_Cook" <diverse@tcp.co.uk> wrote in message
news:ghs2vu$ftm$1@news.motzarella.org...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20
mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. If this remains unchanged, so does the resistance.



Gareth.



That is correct, but the length also has to remain unchanged The formula
for the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A remain
the same, or change in a manner that produces the same ratio.

--
Click to expand...

So that begs the question, how much can a piece of copper wire be
compressed? If you do squash it into a different shape, does or can its
volume change significantly?


Gareth.
 
Gareth Magennis <gareth.magennis@ntlworld.com> wrote in message
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"DaveM" <masondg4499@comcast99.net> wrote in message
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"Gareth Magennis" <gareth.magennis@ntlworld.com> wrote in message
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"N_Cook" <diverse@tcp.co.uk> wrote in message
news:ghs2vu$ftm$1@news.motzarella.org...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil)
=
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20
mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional
Area.
Period. If this remains unchanged, so does the resistance.



Gareth.



That is correct, but the length also has to remain unchanged The
formula
for the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A
remain
the same, or change in a manner that produces the same ratio.

--


So that begs the question, how much can a piece of copper wire be
compressed? If you do squash it into a different shape, does or can its
volume change significantly?


Gareth.
Click to expand...
So it may be an effect of work hardening , relative increase in the effect
of imperfections/micro fractures or some other metallurgical effect.
Mackie speaker voice coil failures due to this flattening/ribboning process
to make the tails to the outside world.
Previous failure at the juncture of round to flat (0.07mm round to about
0.02 x 0.2mm) so at the peak stress point.
This one along the length of the ribbon section, but the whole 50mm or so
run was brittleised and disintegrated on touch, not the slightest sign of
overheating on the remaining 25 turns of round wire.
broken end marked B on this pic
http://home.graffiti.net/diverse:graffiti.net/mackie_horn1.jpg
http://home.graffiti.net/diverse:graffiti.net/mackie_horn2.jpg
Cannot expore the metallurgy as that curve of "wire" as totally
disintegrated to dust.


--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
 
"Gareth Magennis" <gareth.magennis@ntlworld.com> wrote in message
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"DaveM" <masondg4499@comcast99.net> wrote in message
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"Gareth Magennis" <gareth.magennis@ntlworld.com> wrote in message
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"N_Cook" <diverse@tcp.co.uk> wrote in message
news:ghs2vu$ftm$1@news.motzarella.org...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20 mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. If this remains unchanged, so does the resistance.



Gareth.



That is correct, but the length also has to remain unchanged The formula for
the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A remain the
same, or change in a manner that produces the same ratio.

--


So that begs the question, how much can a piece of copper wire be compressed?
If you do squash it into a different shape, does or can its volume change
significantly?


Gareth.
Click to expand...

The shape of the cross section can change to virtually any dimension so long as
the length remains the same. IOW, if you squeeze a bar of 10mmx10mm down to
2mmx50mm, its cross sectional area stayed constant (only the shape of the area
changed). Its length will remain the same, since the volume didn't change;
hence, its resistance will remain the same.
So long as material is not added or removed, the volume will remain the same.
The formula says that the ratio of length to cross-sectional area must remain
the same in order for resistance to remain unchanged. If cross sectional area
is changed, the length must change to maintain the ratio. The volume must
remain constant.

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.
 
On Dec 12, 11:55 am, "DaveM" <masondg4...@comcast99.net> wrote:
"Gareth Magennis" <gareth.magen...@ntlworld.com> wrote in message

news:Uto0l.27356$ja3.12004@newsfe11.ams2...







"DaveM" <masondg4...@comcast99.net> wrote in message
news:0r-dnfQdzaS7JdzUnZ2dnUVZ_qadnZ2d@giganews.com...
"Gareth Magennis" <gareth.magen...@ntlworld.com> wrote in message
news:Gsg0l.4221$d11.1850@newsfe25.ams2...

"N_Cook" <dive...@tcp.co.uk> wrote in message
news:ghs2vu$ftm$1@news.motzarella.org...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) > >>>> 0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20 mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?

AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period.  If this remains unchanged, so does the resistance.

Gareth.

That is correct, but the length also has to remain unchanged  The formula for
the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A remain the
same, or change in a manner that produces the same ratio.

--

So that begs the question, how much can a piece of copper wire be compressed?
If you do squash it into a different shape, does or can its volume change
significantly?

Gareth.

The shape of the cross section can change to virtually any dimension so long as
the length remains the same.  IOW, if you squeeze a bar of 10mmx10mm down to
2mmx50mm, its cross sectional area stayed constant (only the shape of the area
changed).  Its length will remain the same, since the volume didn't change;
hence, its resistance will remain the same.
So long as material is not added or removed, the volume will remain the same.
The formula says that the ratio of length to cross-sectional area must remain
the same in order for resistance to remain unchanged.  If cross sectional area
is changed, the length must change to maintain the ratio.  The volume must
remain constant.

--
Dave M
MasonDG44 at comcast dot net  (Just substitute the appropriate characters in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.- Hide quoted text -

- Show quoted text -
Click to expand...
What I believe Norm is questioning/proposing is that the wire may have
been made more "dense" by being compressed without lengthening, and
that would probably decrease its resistance.

Bob Hofmann
 
On Fri, 12 Dec 2008 14:50:40 -0800 (PST), "hr(bob) hofmann@att.net"
<hrhofmann@att.net> wrote:

What I believe Norm is questioning/proposing is that the wire may have
been made more "dense" by being compressed without lengthening, and
that would probably decrease its resistance.
Click to expand...
IMHO, if the wire is squashed it will defin8itely get work hardened
and this will mean an increase in resistivity

--
Thanks for your time

Archer
 
Ardent <iam@here.com.invalid> wrote in message
news:b95ak4lvglio8ufvoivc2t99ug56kldmla@4ax.com...
On Fri, 12 Dec 2008 14:50:40 -0800 (PST), "hr(bob) hofmann@att.net"
hrhofmann@att.net> wrote:

What I believe Norm is questioning/proposing is that the wire may have
been made more "dense" by being compressed without lengthening, and
that would probably decrease its resistance.

IMHO, if the wire is squashed it will defin8itely get work hardened
and this will mean an increase in resistivity

--
Thanks for your time

Archer
Click to expand...

I think a little experiment is in order. A length of fine wire, measure R
and diameter , then squash between shim spaced flats and re-measure R and
dimensions. Whether work-hardening (xtal structure change ?) or
micro-fractures or whatever does not really matter , just some data on
resistance change.


--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
 
Needs repeating with a proper milli-ohm set up rather than DVM 0.1 ohm
resolution or preferably actively with some higher current through it.
Probably really only becomes manifest close to the current carying limit of
the original round wire, replicating the problem in speaker voice coil
production and use.

Tried about 1m of 0.09mm (including varnish) wire around a 15mm a side flat
and squashed between 2 flats in a protected vice and made no difference to
4.0 ohm, despite 32 theoretical, not obvious, flats.
Repeated with just a single 15mm length of that 1m long wire, reducing 0.09
mm to about 0.03 mm , and repeated further along, so 2 squashed bits of 15mm
.. May have increased to about 4.05 ohm overall but not as obvious a change
as I was expecting.


--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
 
Again 1m length , ps set for constant voltage and unlimited current.
Set V for 0.5 amps , 2 flats and no change, set for 1A and 4 flats added, no
change ie less than 0.01 amp change, if any.
Set to give 1.7 amp and varnish burnt off. So no further forward.


--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
 
"N_Cook" <diverse@tcp.co.uk> wrote in message
news:gi8gog$7ee$1@news.motzarella.org...
Again 1m length , ps set for constant voltage and unlimited current.
Set V for 0.5 amps , 2 flats and no change, set for 1A and 4 flats added, no
change ie less than 0.01 amp change, if any.
Set to give 1.7 amp and varnish burnt off. So no further forward.


--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
Click to expand...
All your testing seems to have verified that the formula for resistance vs.
cross-sectional area does work. No matter how many 'flats' you make on the
wire, you haven't changed its resistance. The small amount that it changed can
easily be contributed to variations in meter connections and/or minute changes
in length due to the squeezing of the wire to make the 'flats'. Barring any
crystalline structure changes in the metal itself, so long as the
cross-sectional area and length doesn't change, the resistance doesn't change.

I suggest that the speaker winding that you're trying to diagnose failed because
of metal fatigue, possibly due to loose mounting, broken adhesive or just an
imperfection in the coil at manufacture.

Time to move on?

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.
 
DaveM <masondg4499@comcast99.net> wrote in message
news:Tf2dnc225sXSTdrUnZ2dnUVZ_s7inZ2d@giganews.com...
"N_Cook" <diverse@tcp.co.uk> wrote in message
news:gi8gog$7ee$1@news.motzarella.org...
Again 1m length , ps set for constant voltage and unlimited current.
Set V for 0.5 amps , 2 flats and no change, set for 1A and 4 flats
added, no
change ie less than 0.01 amp change, if any.
Set to give 1.7 amp and varnish burnt off. So no further forward.


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




All your testing seems to have verified that the formula for resistance
vs.
cross-sectional area does work. No matter how many 'flats' you make on
the
wire, you haven't changed its resistance. The small amount that it
changed can
easily be contributed to variations in meter connections and/or minute
changes
in length due to the squeezing of the wire to make the 'flats'. Barring
any
crystalline structure changes in the metal itself, so long as the
cross-sectional area and length doesn't change, the resistance doesn't
change.

I suggest that the speaker winding that you're trying to diagnose failed
because
of metal fatigue, possibly due to loose mounting, broken adhesive or just
an
imperfection in the coil at manufacture.

Time to move on?

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters
in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the
faster
it goes.
Click to expand...

If it was just a one off , then fair enough. But 2 separate Mackie amps with
the same problem in the same 2 squashed percent of the wire of the speaker
voice-coils ?

--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
 
N_Cook wrote:
DaveM <masondg4499@comcast99.net> wrote in message
news:Tf2dnc225sXSTdrUnZ2dnUVZ_s7inZ2d@giganews.com...
"N_Cook" <diverse@tcp.co.uk> wrote in message
news:gi8gog$7ee$1@news.motzarella.org...
Again 1m length , ps set for constant voltage and unlimited current.
Set V for 0.5 amps , 2 flats and no change, set for 1A and 4 flats
added, no
change ie less than 0.01 amp change, if any.
Set to give 1.7 amp and varnish burnt off. So no further forward.


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




All your testing seems to have verified that the formula for resistance
vs.
cross-sectional area does work. No matter how many 'flats' you make on
the
wire, you haven't changed its resistance. The small amount that it
changed can
easily be contributed to variations in meter connections and/or minute
changes
in length due to the squeezing of the wire to make the 'flats'. Barring
any
crystalline structure changes in the metal itself, so long as the
cross-sectional area and length doesn't change, the resistance doesn't
change.

I suggest that the speaker winding that you're trying to diagnose failed
because
of metal fatigue, possibly due to loose mounting, broken adhesive or just
an
imperfection in the coil at manufacture.

Time to move on?

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters
in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the
faster
it goes.



If it was just a one off , then fair enough. But 2 separate Mackie amps with
the same problem in the same 2 squashed percent of the wire of the speaker
voice-coils ?
Click to expand...

Poor quality control, or a defective manufacturing process.


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There are two kinds of people on this earth:
The crazy, and the insane.
The first sign of insanity is denying that you're crazy.
 
Michael A. Terrell <mike.terrell@earthlink.net> wrote in message
news:UZednYr_MvLoZ9rUnZ2dnUVZ_g2dnZ2d@earthlink.com...
N_Cook wrote:

DaveM <masondg4499@comcast99.net> wrote in message
news:Tf2dnc225sXSTdrUnZ2dnUVZ_s7inZ2d@giganews.com...
"N_Cook" <diverse@tcp.co.uk> wrote in message
news:gi8gog$7ee$1@news.motzarella.org...
Again 1m length , ps set for constant voltage and unlimited current.
Set V for 0.5 amps , 2 flats and no change, set for 1A and 4 flats
added, no
change ie less than 0.01 amp change, if any.
Set to give 1.7 amp and varnish burnt off. So no further forward.


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




All your testing seems to have verified that the formula for
resistance
vs.
cross-sectional area does work. No matter how many 'flats' you make
on
the
wire, you haven't changed its resistance. The small amount that it
changed can
easily be contributed to variations in meter connections and/or minute
changes
in length due to the squeezing of the wire to make the 'flats'.
Barring
any
crystalline structure changes in the metal itself, so long as the
cross-sectional area and length doesn't change, the resistance doesn't
change.

I suggest that the speaker winding that you're trying to diagnose
failed
because
of metal fatigue, possibly due to loose mounting, broken adhesive or
just
an
imperfection in the coil at manufacture.

Time to move on?

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate
characters
in the
address)

Life is like a roll of toilet paper; the closer it gets to the end,
the
faster
it goes.



If it was just a one off , then fair enough. But 2 separate Mackie amps
with
the same problem in the same 2 squashed percent of the wire of the
speaker
voice-coils ?


Poor quality control, or a defective manufacturing process.


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There are two kinds of people on this earth:
The crazy, and the insane.
The first sign of insanity is denying that you're crazy.
Click to expand...

Its more than that.
In both Mackie cases the main active coils showed absolutely no overheating
let alone burning. In all other odd o/c speakers I've poked my nose in, have
had large sections of charring or completely burnt and broken coil formers.
Not failure in the tail sections because they are usually different / larger
conductors soldered or welded to the main coil. These Mackie tails are
acting as fuses, protecting the voice coil, which is ridiculous.

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

Poor quality control, or a defective manufacturing process.

Its more than that.
In both Mackie cases the main active coils showed absolutely no overheating
let alone burning. In all other odd o/c speakers I've poked my nose in, have
had large sections of charring or completely burnt and broken coil formers.
Not failure in the tail sections because they are usually different / larger
conductors soldered or welded to the main coil. These Mackie tails are
acting as fuses, protecting the voice coil, which is ridiculous.
Click to expand...

No, it isn't. Who would manufacture speakers that way for Mackie? If
they did, all of them would fail. They would have to pay extra, and get
no warranty. It sounds like the tinsel wire was the wrong type for the
power level and was likely got through due to poor QC. Poor quality
tinsel wire also suffers from work hardening, and broken strands. When
enough break, the rest burn. Also, if they are a few percent too short,
they get more mechanical abuse, which destroys them. Since you didn't
see them when they were brand new, you have absolutely no idea what
really happened.

I have worked in failure analysis in electronics manufacturing, and I
can tell you that production people can be some of the biggest idiots in
the world. Monday mornings they have hangovers, and Fridays they don't
give a damn what they do, as long as they can leave on time to drink
their paychecks.


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There are two kinds of people on this earth:
The crazy, and the insane.
The first sign of insanity is denying that you're crazy.
 
Michael A. Terrell <mike.terrell@earthlink.net> wrote in message
news:6aqdnTK3c9d8l9XUnZ2dnUVZ_sTinZ2d@earthlink.com...
N_Cook wrote:

Michael A. Terrell wrote:

Poor quality control, or a defective manufacturing process.

Its more than that.
In both Mackie cases the main active coils showed absolutely no
overheating
let alone burning. In all other odd o/c speakers I've poked my nose in,
have
had large sections of charring or completely burnt and broken coil
formers.
Not failure in the tail sections because they are usually different /
larger
conductors soldered or welded to the main coil. These Mackie tails are
acting as fuses, protecting the voice coil, which is ridiculous.


No, it isn't. Who would manufacture speakers that way for Mackie? If
they did, all of them would fail. They would have to pay extra, and get
no warranty. It sounds like the tinsel wire was the wrong type for the
power level and was likely got through due to poor QC. Poor quality
tinsel wire also suffers from work hardening, and broken strands. When
enough break, the rest burn. Also, if they are a few percent too short,
they get more mechanical abuse, which destroys them. Since you didn't
see them when they were brand new, you have absolutely no idea what
really happened.

I have worked in failure analysis in electronics manufacturing, and I
can tell you that production people can be some of the biggest idiots in
the world. Monday mornings they have hangovers, and Fridays they don't
give a damn what they do, as long as they can leave on time to drink
their paychecks.


--
http://improve-usenet.org/index.html

aioe.org, Goggle Groups, and Web TV users must request to be white
listed, or I will not see your messages.

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There are two kinds of people on this earth:
The crazy, and the insane.
The first sign of insanity is denying that you're crazy.
Click to expand...
The problem is not at the tinsel wire, ie the flexible bridge. But before
that, a flattening of the voice coil wire into ribbon. Production seems to
be a precise length of say 2.5m with the ends squashed giving some precise
run of round voice coil wire. One end has the ribbon fixed to the tinsel in
the ideal spot but the other end arrives at maybe half a turn from ideal and
has to make a half turn to join the tinsel for the other termination.
Tinsel ribbon is presumably higher current carying than the crresponding
voice coil wire, so no problems there.
This Mackie problem is something to do with a run of same gauge but
squashed wire.
Remember the whole half-turn of ribbon disintegrated so not due to a nick or
rubbing on metalwork.


--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
 
Perhaps it is chemistry . Bear in mind that the wire is silver and not
copper and the surface area of a squashed wire is more than when it was
round and any corrosion on that surface will have proportionally more effect
on the thin flats than the bulky round.

The voice coil wire looked like copper because it was under a coppery brown
lacquer. But, before it disintegrated, the ribbon section was darker brown
than the round section.
Now if air could get under the lacquer and tarnish the silver to black
copper sulphide, or whatever that blackening is, then that would explain
it.
Unfortunately none of that section remains as it literally turned to dust
after photographing it, you could not pick it up even with fingers, it was
little more than a wraithe.


--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/
 
N_Cook wrote:
Perhaps it is chemistry . Bear in mind that the wire is silver and not
copper and the surface area of a squashed wire is more than when it was
round and any corrosion on that surface will have proportionally more effect
on the thin flats than the bulky round.

The voice coil wire looked like copper because it was under a coppery brown
lacquer. But, before it disintegrated, the ribbon section was darker brown
than the round section.
Now if air could get under the lacquer and tarnish the silver to black
copper sulphide, or whatever that blackening is, then that would explain
it.
Unfortunately none of that section remains as it literally turned to dust
after photographing it, you could not pick it up even with fingers, it was
little more than a wraithe.
Click to expand...
Are you sure that voice coil isn't aluminium? Some aluminuium v/cs are
copper coated btw.

Ron(UK)
 
Ron Johnson <ron@lunevalleyaudio.com> wrote in message
news:Oaqdnb81Z7ugSdXUnZ2dnUVZ8oydnZ2d@bt.com...
N_Cook wrote:
Perhaps it is chemistry . Bear in mind that the wire is silver and not
copper and the surface area of a squashed wire is more than when it was
round and any corrosion on that surface will have proportionally more
effect
on the thin flats than the bulky round.

The voice coil wire looked like copper because it was under a coppery
brown
lacquer. But, before it disintegrated, the ribbon section was darker
brown
than the round section.
Now if air could get under the lacquer and tarnish the silver to black
copper sulphide, or whatever that blackening is, then that would
explain
it.
Unfortunately none of that section remains as it literally turned to
dust
after photographing it, you could not pick it up even with fingers, it
was
little more than a wraithe.

Are you sure that voice coil isn't aluminium? Some aluminuium v/cs are
copper coated btw.

Ron(UK)
Click to expand...

The voice coil remains unaffected, do you know of a simple test for silver v
aluminium ? I would say it looked brighter, more silvery indeed, than
aluminium but I am not familiar with seeing .07 to 0.09 mm diameter Al wire
or silver wire for that matter.


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