# Determining number of turns of a coil?

Discussion in 'Electronic Repair' started by N_Cook, Aug 19, 2012.

1. ### N_CookGuest

Is there an online calculator anywhere ? or failing that what sort of
"packing factor" for 0.08mm (+/- .005mm ) winding on a relay coil.
I have a good idea of the weight , subtracting an estimate of the plastic
former and this would give the length from the density of copper but number
ot turns ?.
Impregnated coil so cannot count-off turns
Rectangular section to coil , on the inside anyway, 14x 16.4mm , 16.8mm
width, outer layer is curved at "corners" and bulging (from the winding not
abuse) so more scatterwound than precise regular lay-up .
Outer dimensions of 21.4mm bulge / 20.3mm , or so, at outer edges one way
and 23.4/22.2mm , the other, a bit of geometry would give a good idea of the
volume of this space but what ratio of that volume would be copper and what
air+varnish, then what sort of weight would be contributed by the varnish?

2. ### mikeGuest

Wind more turns. As many as you can fit in your attention span.
2 is a lot better than 1.
Measure the inductance of each winding.
Or apply volts and measure the ratio.
Accuracy of the result is proportional to the number of
Most anything electrical you do will give more accuracy than weighing stuff.

Curiosity killed the cat...but I gotta ask, "why do you care?"
Sometimes people ask complicated questions when the solution
to their problem is very much simpler.

3. ### N_CookGuest

I forgot to say this coil has a burnt/sputtered patch

4. ### William SommerwerckGuest

Simple approach... Find an ohmeter that can read very low resistances with
high resolution. You can then duplicate the coil by winding a coil with the
same resistance using the same wire on an identical form.

Right?

5. ### mikeGuest

Yet you still refuse to disclose why you care.
Is the number of turns really the defining parameter for this winding?

Cut the winding off the core.
Take a high res picture of the face of the cut.
Calculate the area. Count the turns in a smaller area, multiply.

7. ### N_CookGuest

The copper filling factor would be the ratio of copper as weighed to the
weight if that volume was totally filled with copper. The volume is easy
enough to calculate, but how to get an estimate of the number of turns from
the filling factor. The length is calculatable from the coil weight,
ignoring weight of the impregnation

8. ### William SommerwerckGuest

As others have pointed out (including myself, on many occasions), this is
the sort of question or problem that shows up all-too-often in UseNet groups
(and elsewhere!).

There's a story (probably apochryphal) that Edison asked a new employee to
calculate the volume of a light bulb. The young man sat for some time with
calipers and a slide rule, making little progress. Edison finally
interrupted his work, poured water into the bulb, then emptied the water

On the assumption Mr Cook wants to wind a replacement, why should he not
simply get a form and wind wire of the appropriate thickness, then //test//
the coil * to see if it works the way he wants it to? He could have done
that by now!

I am a //great// believer in theory. You never //really// understand
something until you grasp the principles involved. BUT...! There are some
things you simply go and do, without worrying about theory, because "doing
it" involves a lot less time and trouble.

* Not to be confused with Tesla coil.

9. ### N_CookGuest

Perhaps a bit of integral calculus. In a coil winding manual I have a table
of wire gauge v the minimum excess advance per turn to avoid upsetting of
wound layers. Then sum the layers in terms of the number of turns per width
and this excess , including in a notional paper layer per turn (p), out to
the outer layer set an average value . Including p being negative (layers
settling into the gaps of the previous layer) and determine what value of p
gives the right length of wire to fill that volume, then that gives the
number of turns. At the first approximation ignoring curvature at the
"corners".

10. ### William SommerwerckGuest

You don't need integral calculus to calculate this, any more than you need
calculus to determine the number of "layers" in a recording-tape "pancake".
The approximation assuming the "discreteness" of each layer is good enough.

11. ### N_CookGuest

typical Usenet , most of the thread ,so far , is off-beam.

I doubt the increase in "layer length" of wire per layer goes up lineally
with each layer but again first approximation could assume so and so simple
arithmetic series summation should be all that is required. I just measured
the outer layer "circumference" ie rounded rectangle and not as much
difference as I thought, 77mm around the centre bulge and 74mm at the edges
..

Hopefully I can now do the maths and then if I mess-up grinding across a
section of the original coil ,
so unable to count wire endings, then will at least I will have something to
go with . I don't know how the impregnation will react to a cutting disk ie
smearing.

12. ### N_CookGuest

I found that tables from wire suppliers are a great guideline. But,
nothing beats having to fight 'stacking factor' on your own.

I use 0.5 for wrapping 36 Awg enameled wire. I know I should be able
toget better than 0.7.

So take the winding area multiply by SF and divide by wire cross
section area and you'll get very close..

And, from experience, if there are a few layers, don't count on nice
neat, high density layering and stacking, doesn't work that way.

Oh, also watch out for stretching the wire, making it slightly thinner
and 'appear' to be able to put on more turns, you'vve actually reduced

++++

By stacking factor do you mean the ratio of copper volume to the volume
occupied by the copper ? ie the last layers always need squahsing into the
available (calculated) space between former and iron core etc.

I will try that , also calculation via derivation of a formula for such
situations , and also try cutting through the coil mass and counting
directly , hopefully . For future reference as to accuracy of each method
(in one instance anyway)

13. ### William SommerwerckGuest

There's a story (probably apochryphal) that Edison asked
I might have misquoted it. Edison might have said "find".

Regardless, one of the points of this anecdote is that you should look for a
good solution -- not necessarily the solution you were asked for.

14. ### N_CookGuest

I found that tables from wire suppliers are a great guideline. But,
nothing beats having to fight 'stacking factor' on your own.

I use 0.5 for wrapping 36 Awg enameled wire. I know I should be able
toget better than 0.7.

So take the winding area multiply by SF and divide by wire cross
section area and you'll get very close..

And, from experience, if there are a few layers, don't count on nice
neat, high density layering and stacking, doesn't work that way.

Oh, also watch out for stretching the wire, making it slightly thinner
and 'appear' to be able to put on more turns, you'vve actually reduced

++++

By stacking factor do you mean the ratio of copper volume to the volume
occupied by the copper ? ie the last layers always need squahsing into the
available (calculated) space between former and iron core etc.

Or do you mean 1 is the mathematical theoretical ideal packing of first
layer n, next n-1, next n etc
all precisely registered over the previous layer , within the hollows, that
in itself a fraction of the total volume occupied by the coil , again some
preciese mathematical value for a given wire gauge.
Then 0.7 is the real world ratio of copper for a machine wound coil plus
skilled worker ratio of actual to this ideal.
Then .5 is the likely ratio value for anyone else doing layup of fine wire
with hand cranked coil winder.

15. ### N_CookGuest

My formula uses the assumption that the length per turn increases, layer to
layer, linearly by 2 wire diameters per "corner" so 8 x .08mm per turn. For
44SWG .08mm wire I assumed the manual coil winder excess turn advance of 8
percent. From the weight of wire then the length. Then using my formula ,
came to 30 layers and a p value of -.013 mm
That gave 5790 turns

I disc ground through the coil and because I can weigh to 0.01 gm counted
off a sample of 952 turns and then ratioed by weight to 6420 turns .
Perhaps for machine wound coil can drop the 8 percent to 7 percent for 44
swg to give a better result.

So now know the inductance as well as the DC impedance (plus a useful
formula for the future , for 44 swg anyway)  