Help calculating a different electromagnetic coil please...

Do you plan on driving it with the same driver or can you increase your current significantly?

You could simply measure it. I'm not quite sure how you'd measure "magnetic force", but once you get that figured out, get ahold of a current-controlled, variable-voltage bench power supply, and apply a range of currents to the existing coil, and see what "magnetic force" you get - it's usually measured in ampere-turns. Then, hand- wind a coil with your desired wire, and crank up the current until you get the same magnetic force, which should be at the same ampere- turns value, if everything else is the same.

Good Luck! Rich

--- Knowing what you know about the coil, it should be fairly straightforward to do what you want to do.

The most important thing you need to know about the coil is the magnitude of current in the wire and the number of turns in the coil.

Multiply those two numbers together and you wind ;) up with Ampere-Turns, the same number of which you must wind up with in your new coil for it to have the same effect on whatever is in its bore as the old coil did.

For example, let's say that your coil has 10000 turns of #44 AWG wire wound on the bobbin and that when it's operating it's got 1mA of current in the wire. That means that the field in the bore will exhibit a "Force" of:

F = nI = 10000T * 0.001A = 10 Ampere-Turns

Now, if you want to use larger wire and reduce the number of turns, what you'll have to do is increase the current to make up for the fewer number of turns. For example, let's say you wanted to go from 10000 turns to 100 turns. Then you could rearrange:

F = nI

to solve for I:

F 10AT I = --- = ------ = 0.1 ampere n 100T

Not bad, if you've got a supply which can output 100mA at whatever voltage is needed to push that current through the coil, so that's the next step, finding out what the resistance of the 100 turn coil will be. That's the fun part :-)

Just for fun, let's say you have a bobbin which is one inch long, on the inside, from flange to flange, and that the ID of the coil will be 1/4". It shouldn't take a lot of wire to put 100 turns on that bobbin, and it shouldn't take a large diameter wire to be able to handle that 100mA without getting hot.

#24 AWG is pretty easy to work with, and it's got a resistance of about 26 ohms per thousand feet, so that's not going to give us any power dissipation problems, so let's see how much of that we'll need for 100 turns.

From:

we find that #24AWG with heavy formvar insulation has a maximum diameter of 0.0227", so if we have 1" between the flanges we can wind

1" n = --------- = 44.05 ~ 44 turns 0.0227"

on the first layer.

And on the second layer, and that means that since we'll have 88 turns on those two layers we'll only need 12 on the third to get all

100 turns on there.

OK.

Now, since the bobbin has a diameter of 0.25" and the wire has a diameter of ~0.023", the length of each turn of wire on the first layer will be:

C = piD = 3.14 * 0.273" = .857"

and, since we have 44 turns on the first layer, the length of the wire on the first layer will be:

0.857" * 44T L = -------------- = 37.7" ~ 38" T

In the same manner for the second layer, the diameter will be

0.296", the length of each turn ~ 0.930, and the length of the winding ~ 40.92".

For the third layer, the diameter will be 0.319", the length of a turn 1.00", and the length of the winding 3.0", for a total wire length of just about 82".

That's pretty close to seven feet, and with a resistance of 26 ohms per thousand feet (26 milliohms per foot), the resistance of the coil comes out to be 182 milliohms, which means that to push 100mA through it you'll need a voltage of:

E = IR = 0.1A * 0.182R = 0.0182V = 18.2mV

In order to do that properly you'd have to drive the coil with a constant-current supply set for 100mA, ur use a constant-voltage supply and a resistor in series with the coil.

With only an 18mV drop across the coil, you can ignore that and figure the resistance required to pass 100mA at whatever supply voltage you have available.

For example, if you had a 12V supply that you wanted to use, figure:

E 12V R = --- = ------ = 120 ohms I 0.1A

The resistor would have to dissipate:

P = IE = 0.1A * 12V = 1.2W

so a 120 ohm 2 watt resistor would do it.

Finally, the coil would dissipate:

P = IE = 0.1A * 0.018V = 0.0018W ~ 2 milliwatts

so its temperature will be virtually the same when it's operating and when it isn't.

That's about as simple as it gets... ;)

-- John Fields Professional Circuit Designer

...

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You must have missed this part:

"I know the following: Current flow through the coil, number of
wraps"...

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Aarghhh!!!

12 turns, not three, for a length of 12" instead of 3", and a total
length of 91"

Obviously the thicker the wire just allows you to run more current which allows you to decrease the number of turns.

Ampere's law states that the magnetic field along the axis of a solenoid is

B = mu_0*N/L*I

Therefor if we want another to create another coil with the same B there are

3 independent variables we can change to keep B constant. (as long as the configuration is the same though).

So if we double the current we can cut the Turns per unit length by 1/2.... But on the other hand we have to make sure our wire can handle the new current(which, since it size will increase it will decrease N/L).

So just by wrapping up a new coil with larger wire you should be able to find the right current that will provide the same B as the old coil... assuming your wire can handle the new current.

Theres probably a way to calibrate the coil to the old one using magnetic induction or using a Gauss meter or something similar.

The main issue is that increase the wire size is not linear... i.e., doubling the current does not require you to "double" the wire size since the resistance dependents on its cross sectional area...

R = p*L/(pi*r^2)

hence if we have 2 wires of the same length and same voltage accross them we would have the current's flowing

I1 = V/R1 = V*pi*r1^2/p/L I2 = V/R2 = V*pi*r2^2/p/L

and taking there ratio we have

I1 = (r1/r2)^2*I2

so by doubling the radius of the wire we get 4x the the current flow(simply because the resistance drops by a factor of 4). Ofcourse doubling the area gives 2* the current flow.

Also note that since I = V/R and B = p*N/L*I

B = p*N/L*V/R

so you can also change the voltage across to get the correct magnetic field with the same current...

The conclusion is hopefully that as long as your new coil doesn't drastically differ in configuration as the old one you can vary the current and/or voltage to produce the same magnetic field. (though you might have to add or remove resistance). Ofcourse you have to keep these parameters within constraints(so you don't burn up the wire... if you get, say, a current past the rating of the wire then you need to choose a larger wire(which gives you less resistances but allows you to "fine tune the voltage more)).

Hope that helps some, Jon

I guess my brain-pan skipped over that, because if he knows that already, then what's the question about?

What's he really trying to do? Or what does he really want to know?

Thanks, Rich

Thank you John and everyone that submitted information back. I'm still working all your response information through my head. In the application we are talking roughly 52awg wire with about 6000 wraps.... current CAN be changed in this application, doubled would be just fine or even more possibly. There is no power supply concern for the most part, batteries may eat faster but that's acceptable in this application... we may be able to fit bigger batteries in the battery compartment. All we have to do is actuate a very small lever within a very confined space. Bobbin is 3.8mm tall, about 3 mm wide when empty.... so these are the contraints. I don't have the current draw infront of me i'm away from the bench but I'm going to work over some of these numbers.

NEXT QUESTION: I'm not sure how the 'layers' are layed down, is it a back and forth zig-zag? How are 5000 wraps done for example on a 3.8mm x 3mm bobbin? We haven't torn one appart yet.... and when we do some of our calculations, are there limitations to the number of 'layers' that should be used?

Thanks again for all your time and help it's soooo appreciated!

-B

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You must have missed this part:

"WHAT I WANT TO DO: determine the correct wire size and coil wrap
count to replicate the same amount of magnetic force as the original
coil.... hopefully with less wraps and hopefully thicker wire to
allow me to handle it by hand, the original was probably machine
wound."

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A reed switch?

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I forgot that you mentioned earlier that you knew what the current
draw was.  Could you post that as well, please?

Thanks,

Well, if he knows the number of turns, and the number of amps, that's ampere-turns. All he would need to do is decide how fat of wire he wants to use, pick a number of amps, and divide the ampere-turns from (1) above by amps to get the number of turns.

Or am I missing something?

Thanks, Rich

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Yes, but finding what it is is left as an exercise for the student.
;)

I got to thinking about this a bit too seriously. Just the same, for any available winding cross section, you may change the number of turns by changing the wire size, but you will have little effect on the total magnetomotive force applied (B) (ampere turns is about the same for any gauge wire). Also, inductance varies with the square of the number of turns. This has impacts on the nature of the windings.

--
JosephKK
Gegen dummheit kampfen Die Gotter Selbst, vergebens.  
--Shiller