I need help on how to make coil calculations...

[Cyrano]

Hi People....
I'm inexperienced in this field, so I need your helps to find the right calculations to make a coil of copper wire for a DC electromagnet, that consumes 37,5W (3V at 12,5Amps...).I thank you in advance.
Best Regards,
Cyrano.

[MrAl]

Hi there,


One of the important things about running a coil for use as an electromagnet when you want to get
maximum strength is the resistance of the coil. If you do a calculation that calculates the force
influenced by wire size and battery power and battery internal resistance and then look for the
maximum in the force, you come up with the result that the best possible wire resistance is the
resistance that is equal to the internal resistance of the battery.
This means that you would have to determine the internal resistance of the battery you intend to
use and then knowing the coil form size you would look for the best wire size that would fill the
form completely yet still have that same resistance.

If you dont want to do the extended math required then you can do the math by trial and error.
To do this you simply start with some wire size like maybe #12 and calculate the total resistance
after it is wound around the coil form and fills it completely. If the total resistance comes out
too high then you switch to #11 and try again, but if the total resistance came out too low then
you switch to #13 and try again. In each case you move up or down one wire gauge and try
again. Eventually you find a wire size that works out close to the actual required total resistance
and that's the best wire size.

The reason we do it this way is because the total length to fill the coil form changes depending
on wire size, as well as the wire size itself, so there are two things that influence the total
resistance. It's possible to calculate this outright, but it's just as easy to do it by trial and
error especially if using a computer program to do the calculations. It also makes the process
a little more intuitive as we are always aware that the main point is to search for the best
wire gauge.

Another point to think about is the wire diameter and the malleability of copper.
Some wire sizes will be extremely hard to work with when trying to wind a coil. This precludes
the use of the larger wire diameters like #6 and probably even #8 without special equipment.
#10 would probably be the largest wire size that can be worked by hand around a coil form
without too much trouble, but i might even lean toward #12 as a max. This complicates things
a little as that means that the coil has to be wound bifilar or even trifilar or possibly higher.
This means the calculation has to go a little different if the current is very high, but it's still
the same...it just has to take into account the more wire strands that's all.

Note that if you change batteries later on and they are not the same type, the coil will not be
as perfect as before because the internal battery resistance changed and the coil was designed
for a different resistance.

[Cyrano]

I think that the coil is done better by mathematical calculation...I haven't copper wire to waste.

[Bob Scott]

I'd take a different approach and use the stated requirements and not assume any battery characteristics.

Using R=V/I, resistance of the windings must be 0.24 Ohms.
Using an ampacity table looked up in Google, minimum wire size is about 14 AWG for 12.5 Amps.
Using a copper wire table I find I need 92 feet length of 14 gauge wire in order to have a winding resistance of 0.24 Ohms.

If it were for me, I'd buy a hundred foot spool of 14 AWG magnet wire and leave it on the spool. There! It's wound! Insert an iron core of some type through the hole in the spool and fold it around the winding so that the magnetic poles are close to each other. Maximum magnetism is found between the poles, and when the poles are close together. That's why you should fold the core around the winding, back on itself.

[MrAl]

Hi again,


Bob:
The method i posted is known to give the best coil performance for a given battery type.
That's the best you can get. It's not that hard to calculate so why not?

I think that the coil is done better by mathematical calculation...I haven't copper wire to waste.

Hi Cyrano,

Ha ha, you didnt think i actually meant that you should buy wire and wind it around the form just to see if it fit
and came out right did you?
You do this by CALCULATION, in that you calculate the resistance that you WOULD get IF you wound the wire
around the form. You dont actually get wire and wind it until AFTER you have done the calculations and came up
with the best wire gauge. You then know what to buy too.

[Cyrano]

To Bob Scott:
Ok, the coil is 0,24 ohms.This for a 37,5W coil.
If I calculate the resistance of a coil four times bigger the starting values,ex.12V at 50AMPs (600W...),I got the same resistance:0,24 Ohms!!.This thing confuses me.
I have here a two very different wattage and the same coil is good for both?.Please exlain me what is going on..

[MrAl]

If you use R=V/I and that comes out to 0.24 ohms and the internal resistance
of the battery is 0.24 ohms then 92 feet of 14 gauge wire may not fill the
coil form, and the magnetic strength increases with each turn of wire
so it makes sense to put as many turns as possible.
Going lower in gauge, more turns will be required to get the same resistance
which will result in a much stronger magnet.

This means that if you need to keep the resistance at 0.24 ohms no matter
what then you should at least optimize for the wire diameter to fill the
coil form. This will get the most bang for the current which is the usual
design goal for electromagnets.

[Bob Scott]

To Bob Scott:
Ok, the coil is 0,24 ohms.This for a 37,5W coil.
If I calculate the resistance of a coil four times bigger the starting values,ex.12V at 50AMPs (600W...),I got the same resistance:0,24 Ohms!!.This thing confuses me.
I have here a two very different wattage and the same coil is good for both

No, you cannot use the same wattage coil. You must use wire with a higher current capacity for 50 Amps. You must use this thicker wire and make it longer to make the coil stay at 0.24 Ohms.

Just because the two windings measure the same resistance does not mean that thay have the same wattage.

I have shown you the method of using wire tables for finding a coil's wire gauge and winding length for your specific voltage, current and wattage numbers. If you are going to change the numbers, you must go back to the wire tables and find the appropriate wire gauge to match the current, and wire length for the required resistance of your DC electromagnet.