Calc. # turns to get coil resistence

[Glenn Syborn]

If I have an empty bobbin 20mm ID and 100mm long, using 28 AWG enamel
wire, how can I calculate how many turns to achieve a reistance of 12
Ohms?

Resistance is 0.076 Ohms per foot, but I am not clear about the math
formulae.

Thank you,

Glenn Syborn

 

[Phil Allison]

"John Larkin"

Just measure out 157.9 feet and wind it up. Ignore the number of
turns. That's a lot easier. Better yet, measure out 12 ohms of wire
and then wind that up.

But my wire tables show #28 annealed copper as 0.0649 ohms per foot.
Are you using something else?

Why make a copper resistor?

** It's his math homework - dickhead.


..... Phil

 

[Dirk Bruere at NeoPax]

"John Larkin"


** It's his math homework - dickhead.


.... Phil

Not necessarily.
I had to do a very similar calc recently.
I went with taking a length and then winding the coil with it.

 

[Phil Allison]

"John Larkin"

What difference does that make?

** Being a dickhead makes a big difference.



..... Phil

 

[Phil Allison]

"Dirk Bruere at NeoPax
Phil Allison wrote:
"John Larkin"

Not necessarily.

** Larkin is definitely a dickhead.

But pales into insignificance compared to a bullshitting fuckwit like you.


..... Phil

 

[Dirk Bruere at NeoPax]

"Dirk Bruere at NeoPax
Phil Allison wrote:
"John Larkin"




** Larkin is definitely a dickhead.

But pales into insignificance compared to a bullshitting fuckwit like you.


.... Phil

Keep taking the meds Mr Tourette...

 

[Rich Webb]

If I have an empty bobbin 20mm ID and 100mm long, using 28 AWG enamel
wire, how can I calculate how many turns to achieve a reistance of 12
Ohms?

Not possible.

Unless, of course, you're installing the turns on the inner surface of
the cylinder. The ID is 20 mm but what is the OD? 22 mm? 220 mm?

 

[Glenn Syborn]

Sounds like a trick question since, without knowing the _OD_ of the
bobbin, it's impossible to calculate the number of turns.

Sorry, by "ID" I meant the OD of the unwound bobbin shaft. The OD of
the final coil would depend on the turns number.

Thank you for the general math procedure which is what I was after.

Since another respondant inquired, the reason I need the fixed
resistance is to limit current without having to use an external power
resistor.

And, no, it's not homework. Wish I was still that young.

Glenn Syborn

 

[bw]

If I have an empty bobbin 20mm ID and 100mm long, using 28 AWG enamel
wire, how can I calculate how many turns to achieve a reistance of 12
Ohms?

Resistance is 0.076 Ohms per foot, but I am not clear about the math
formulae.

Thank you,

Glenn Syborn

12 ohms divided by 0.076 ohms per foot equals 157.9 feet

For DC resistance, only the total wire length matters, not the number of
turns.

If you want inductance there are other formulas. Also, "solenoid tables"
already calculated for various sizes.

 

[Phil Allison]

"John Larkin"
"Phil Allison"

You like to use that body part as an insult.

** Where is a person's " dickhead " ??

And the corresponding female gadget.

** Better find a good dictionary sometime and read the two definitions.

Fuckwit.


..... Phil

 

[John Tserkezis]

If I have an empty bobbin 20mm ID and 100mm long, using 28 AWG enamel
wire, how can I calculate how many turns to achieve a reistance of 12
Ohms?

Resistance is 0.076 Ohms per foot, but I am not clear about the math
formulae.

I wrote some software a while back that estimated the number of turns
(or length of wire) providing you knew other critical bits.

The basic requirement was you needed to know what diameter wire it was,
and the physical dimensions of the bobbin, including the diameter of the
core of the bobbin.

The software made the assumption you could neatly roll a layer of wire
to fit the width of the bobbin, calculate that, then do it again on the
second layer, which was now assumed to sit on top of the first layer.
Repeat layers till you get to the outer diameter, or whatever will fit
on the bobbin.

Even with the assumptions, it was close enough to be usable.

I used it mainly to determine how many *turns* I could get onto a
bobbin using a particular wire type. Great when I needed a certain
turns ratio, but had no idea determining the amount of wire the bobbin
can hold.


Unfortunately, this was back in the DOS days, I've lost it since then,
and I've moved into IT so I don't need to re-write it. So can't help.

It's certainly not hard though, nothing that early basic high school
maths can't sort out.
Even if you have to factor in resistance.

 

[Dirk Bruere at NeoPax]

Unless you carefully layer wind it, the solution won't be very exact.

It did not need to be, but then again, the coil diameter was about a
metre. I wanted a coil that would provide an 8 Ohm resistive load to an
audio amp in worst case of very low frequency driving [although I
planned to drive it with 1mS spikes ]

 

[Dirk Bruere at NeoPax]

Unless you carefully layer wind it, the solution won't be very exact.

It did not need to be, but then again, the coil diameter was about a
metre. I wanted a coil that would provide an 8 Ohm resistive load to an
audio amp in worst case of very low frequency driving [although I
planned to drive it with 1mS spikes ]

I should add that I knew what approximate inductance I needed and worked
out the number of turns, estimated the length of wire needed and then
chose its diameter based on resistance per metre. So maybe not quite
Larkins method, more a kind of inverse. But because resistance was more
important than the inductance I used that as the primary measure.

 

[Rich Grise]

If I have an empty bobbin 20mm ID and 100mm long, using 28 AWG enamel
wire, how can I calculate how many turns to achieve a reistance of 12
Ohms?

Resistance is 0.076 Ohms per foot, but I am not clear about the math
formulae.

Why? #28 AWG is 64.9 ohms/1000 ft. So do the arithmetic. Wind that many
feet on the coilform and let it decide how many turns get wound. Make it
enough longer than your calculation, to provide for a couple of flyleads.

Have Fun!
Rich

 

[Shaun]

"John Larkin"
"Phil Allison"


** Where is a person's " dickhead " ??

Maybe Phils dickhead is so small, he can't find it?

 

[ehsjr]

Read the subject line, John.

I think JF has in mind the "Larkin method", in which JL
specifies that you ignore the number of turns.

As to # of turns, I suppose that might be useful information
if you have a winding machine with a turns counter, and were
winding onto a big form like the OP is using, wanted to
wind 12 ohms worth of wire, and didn't want to measure the
length up front. That's a helluva lot of suposin'.

A client of mine produces wire. All of that information, and
more, is important to them, and measured. Turns count, form
dimensions, length, weight, resistance/ft, material, tensile
strength, wire diameter and on and on. But AFAIK, their final
product is always so many feet of specified wire, not so
many ohms of it.

Ed

 

[Spehro Pefhany]

Close, but you need the wire's EXTERIOR diameter (which will vary
according to the insulation applied, is NOT the same for all '28 AWG'
copper wire), and you need the packing fraction (the ratio of the
wire to wire-plus-gap cross section areas). Then you know how
much of the winding throat is filled with wire (this gets you the
necessary info on how high the radius gets as you wind the bobbin
fatter and fatter...).

Packing factor Ku of 0.6 (ratio of _copper_ cross-sectional area to
bobbin fill cross-sectional area) is one rule of thumb for
transformer/inductor design (Billings).

 

[Dirk Bruere at NeoPax]

I was just trying to show Glen a way to attack the problem. He said
he was not sure of the math formula. The general idea was to get wire
length, convert this to volume , convert that to turns on the bobbin.

To me it is pretty obvious that wire diameter meant actual wire
diameter, but maybe I should have emphasized that point. I did say to
regard this as a starting point assuming perfect winding. He did not
say how accurate an answer he needed , how he was planning on doing
the winding , or even why he wanted to know the number of turns.

Dan

Well, if he wants an equation he can start by calculating the overall
height of one circle sitting between two of the same diameter in a
triangular formation.

 

[Jamie]

---

Lets not get all melodramatic and teary-eyed!


Jamie

 

[Dirk Bruere at NeoPax]

Nah, It works just fine. (though it's a PITA to get it all to lie
straigth) the only problem is at the edges where the wire turns over.
You always get a bump there. I've been wondering what coils made from
copper tape are like? Is the resistance too low?

We need square wire.