Winding a strong electromagnet

[Kator01]

Hi ,

i forgot to attach the physical dimension-table of abailable flat-coils.
This is just an example. As the smallest one is 6 mmm in width I will ask them for smaller width

Regards

kator01

[Xaverius]

Hi Xaverius,

just a short note on your calculation because this also confused me at the beginning of this discussion:

... unless you do not use two cores for the two coils in parallell - the ampere-turns are 1232 x.51 = 628.32 Amp. but at lower resistance as you already stated Two parallel coils on one core act as one coil with reduced resistance, No doubling of the AT.

Is this ( two cores ) what you mean in your calculation ?

Two coils, one core.  Think of the coils as one long wire at a high resistance, it is severed in two, the ends are wired in parallel reducing the resistance.  Now you have the same large number of turns AND an increase in amperage due to the lesser resistance, thereby DOUBLING amperes AND turns.

[TinselKoala]

Two coils, one core.  Think of the coils as one long wire at a high resistance, it is severed in two, the ends are wired in parallel reducing the resistance.  Now you have the same large number of turns AND an increase in amperage due to the lesser resistance, thereby DOUBLING amperes AND turns.

So I have one long wire. I sever it in two lengthwise. Now I have 2 long wires, each with half the cross-sectional area of the original wire. Now I join the ends back together. And I have, magically, created a wire with less resistance than the original wire.

Sorry, there must be something I'm missing.

[capthook]

12' of #30 AWG = 1.4 ohms
cut in half, twist the 2 ends together at each end:
6' x 2 in parallel of #30 AWG = .7 ohms

How does this affect ampturns?  Xaverius has proposed it increases efficiency....

Attach those wires to a power source: 1 'D' battery  1.5v
Measure voltage with DMM
I=V/R

12':  .67V / 1.4ohms = .479A    .67V x .479A = 0.32093 W
6' x 2 parallel: .23V / .7ohms = .329A    .23V x .329A = 0.07567 W

So with the same power source, in the parallel circuit, the RESISTANCE goes DOWN, the VOLTS go DOWN, the AMPS go DOWN and the POWER draw goes DOWN.

Which winding makes for a better EM?  Which will give you the most ampturns per watt?

Let's say 40 turns.
12': 40T x .479A = 19.16 AT  @ .32093W  = 60 AT/w
6'x2: 40T x .329A = 13.16 AT @ .07567W = 174 AT/w

290% increase in AT/w for parallel in this example.

Gonna wind some test coils tommorrow.....

(edit: corrected numbers)

[TinselKoala]

OK, I thought what was meant was to split the wires lengthwise and use the same length (because that's how you double the turns).

But this isn't correct:
"So with the same power source, in the parallel circuit, the RESISTANCE goes DOWN, the VOLTS go DOWN, the AMPS go DOWN and the POWER draw goes DOWN."
Because you really aren't using the same power source. The battery's voltage depends on the load resistance, as you know. So to get correct figures you need to measure the current using a voltage-regulated power supply that won't sag the way a small battery does.

I have often pondered this same question. Taken to the limit, one would expect all electric motors to be wound with extremely fine wire and lots of it. But they aren't, usually. When I was a kid, we used to rewind the armatures of the little Mitsubishi can motors on our slot cars. But we would use fewer turns of thicker wire than original, and it would indeed make the motors more powerful. Of course we didn't care about power consumption, and the motors would generally fail from overheating after a few hours, or "throw a wrap" when a winding came loose from the epoxy at extreme RPM.