Calc. # turns to get coil resistence

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

Reply to
Glenn Syborn
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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?

John

Reply to
John Larkin

"John Larkin"

** It's his math homework - d*****ad.

.... Phil

Reply to
Phil Allison

What difference does that make?

John

ps - I didn't answer his question.

pps - why do you think body parts are insults?

Reply to
John Larkin

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

--
Dirk

http://www.neopax.com/technomage/ - My new book - Magick and Technology
Reply to
Dirk Bruere at NeoPax

"John Larkin"

** Being a d*****ad makes a big difference.

.... Phil

Reply to
Phil Allison

I think PA may be jelous that he missed out on taking a length. (o)

Reply to
Dennis

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?

--
Rich Webb     Norfolk, VA
Reply to
Rich Webb

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

You like to use that body part as an insult. And the corresponding female gadget. How charmingly Victiorian of you; or maybe you're just messed up about sex.

Still, it's better than the poop fetish.

John

Reply to
John Larkin

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

John

Reply to
John Larkin

--
Huh???

Isn't that exactly the same procedure you recommended?
Reply to
John Fields

So John L. gave you the length. To get the number of turns you can get close by assuming the wire is square in cross-section. You can then get then number of turns on the first layer, by simple geometery. Then do the second layer. The radius of each circle increases with each layer.

George H.

Reply to
George Herold

According to the Australian electronics newsgroup, he was thrown out of college for having sex with his male teacher.

--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.
Reply to
Michael A. Terrell

The calculated number of turns won't match the actual number of turns.

--
You can't fix stupid. You can't even put a band-aid on it, because it's
Teflon coated.
Reply to
Michael A. Terrell
[snip]

Who? Larkin ?:-) ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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Reply to
Jim Thompson

--
Nobody said anything about numbers of turns.

His: "I went with taking a length and then winding the coil with it."
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Reply to
John Fields

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

Reply to
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.

Reply to
bw

"John Larkin" "Phil Allison"

** Where is a person's " d*****ad " ??
** Better find a good dictionary sometime and read the two definitions.

Fuckwit.

.... Phil

Reply to
Phil Allison

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