I'm reverse-engineering an electromagnetic coil that's very very small,
made
from very small wire. I'm not an electrical engineer so this is tough,
but I have
some background enough to be dangerous. Some help would be greatly
appreciated.
I know the following: Current flow through the coil, number of wraps,
size of wire,
area of the bobbin it's wound on, length of finished coil.
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.
Is there any way to estimate this and avoid all the math, or do I need
to
go the calculation route? If I must calculate it, do I calculate to
Gauss for
the orignal one and then solve for number of wraps given specific wire
sizes?
Any suggestion on the formulas to keep it as simple as possible would
be greatly appreciated, I hate to chase my tail when I don't have much
experience in this area.
---
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:
http://www.mwswire.com/awgsearch2.asp
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...
