Notes on Coils from the PSN Archive

Subject: Re: Coil winding formula ?
From: ChrisAtUpw@.......
Date: Mon, 24 Jul 2000 11:53:23 EDT

Dear Arie,

<< Hi, A long time ago, far, far away, I used to remember a simple formula
that gave the length of wire for a coil, given the number of turns, internal
diameter, length of coil and gauge of the wire. But alas, I can't remember
it, so if anyone knows such a "winding" formula, I would be most 
appreciative. 

The fields on the coil axis are easily calculated. Off-axis fields 
involve solving elliptic integrals. I will give the field H in Oersteds, all 
dimensions in cm, the current i in amps and Pi= 3.142. To change from 
Oersteds to Tesla, multiply by 10 ^ -4

For infinite solenoids H = 0.4 x Pi x N x i N is in turns per cm

For shorter solenoids H = 0.2 x Pi x N x i x (cos(theta1) - cos(theta2)) 
N is in turns / cm. theta1 and theta2 are the angles between the axis and 
the ends of the coil at the point being considered. You may need to be a bit 
careful if using a computer programme to check how theta = arctan(a/d) is 
evaluated as d goes through zero and becomes negative at the end of the coil, 
where a is the coil radius and d is the disance from the point considered to 
one end of the coil. 

For short large diameter coils you have a choice. H = 0.2 x Pi x Nt x i x 
sin(theta) / a where Nt is the TOTAL number of turns and theta is the 
angle the coil makes with the axis at the point d from the coil centre. The 
alternative formula involves substituting a and d to give sin(theta). H = 0.2 
x Pi x Nt x a^ 2 x i / (a^2 + d^2)^1.5 The '^' is used for 'raised to the 
power of' and 'x' for 'multiplied by'.

For Helmholtz coils, if EACH has Nt turns separated by distance a between 
the coils, the formula is H = 3.2 x Pi x Nt x i / (5^3)^0.5 x a where 
(5^3)^0.5 = sqrt(125) 

For the length of wire used, this is 2 x Pi x a x N per cm for a solenoid 
or 2 x Pi x Nt x a for each Helmholtz coil. 

For a short fat close wound coil of length l, internal radius a1 and 
external radius a2, the wire length is 2 x Pi x N x l x (a2 - a1) x N x (a1+ 
a2)/2 = Pi x N^2 x l x (a2^2 - a1^2)

If the dia of the wire is d cm and the coil is close wound, N = 1/d turns 
/ cm, but 'achieved values' may be a few % less.

Always allow some excess for connections at the ends. The dia of 
enamelled wire is usually quoted as the dia of the copper, but the thickness 
of the enamel or the actual turns / cm, in, etc may be quoted. For fine 
cotton coated enamel wire, I suggest that you see just how many turns you can 
actually wind per cm. I found that it varies a bit. If you 'wave wind' the 
coil for low interwire capacitance, you need to be able to estimate the 
'filling factor', which is determined by the coil winding machine. Check in 
the instruction manual.

Hope that this helps,

Chris Chapman

Subject: coil formula
From: sean@...........
Date: Mon, 24 Jul 2000 13:19:28 -0500 (CDT)

Arie,

If you are considering a coil/magnet as for a seismometer, as 
compared to an air-inductor as in a radio transmitter, perhaps
the handiest formula, which I have posted once-upon-a-time, is

G = B * L * 10^6 Newtons/Ampere (or volts/meter/second)

Where B is the magnetic field in gauss, and L is the total length of
the winding in centimeters. If one plays with the math (from below),
all the other physical dimensions cancel out. You can get L by knowing
the coil resistance and the wire size.

If you know the coil parameters and can measure the generator constant,
this can be used to determine the stength of your magnet(s).

The more elaborate formulas for the generator constant of a multi-layer
coil in a magnetic field are in the appendix of Riedesel: 

"Limits of Sensitivity of Inertial Seismometers with
Velocity Transducers and Electronic Amplifiers";
by Mark A Riedesel, R.D.Moore, and J.A.Orcutt; Bulletin of
the Seismological Society of America, Vol. 80, No. 6, December 1990.

Or would you like me to post the formulas.?

Regards,
Sean-Thomas

Subject: coil winding
From: sean@...........
Date: Mon, 4 Sep 2000 13:39:49 -0500 (CDT)

Barry,

I haven't made a formless coil for a speaker magnet because I haven't
found a temporary form of the exact dimensions to wind it on. (see my 
write up on making the formless coil.... or do you need a repeat of it).
But I haven't tried either, since I can make a reliable, reproducible 
rare earth magnet assembly with McMaster parts for $30, rather than 
dismanteling speakers. (also written and posted).

But when you play with the formulas, the output is a function of
the total winding length. This means that the output increases directly 
as the number of turns increases. However, to get more turns into the 
same winding volume (the cylinder of wire in the magnet gap), requires
smaller wire, obviously. Wire of 1/2 the cross section area, like
# 38 is half the area of #32 (from wire tables), will result in twice
the number of turns so twice the wire length and output. But half the
area is twice the resistance per length, so the winding resistance
increases by a factor of 4.

For example, increasing the turns conveniently by using wire of 1/2 the 
diameter with the same coil dimensions, if I use #42 enameled wire, which
is about half the diameter of #36, I will get twice the turns per layer, 
and twice the number of layers, for 4 times the turns and output. With 
the same mean length per turn, this will be four times the wire length, 
but it will measure 16 times the resistance. So I get only 4 times the 
output at 16 times the resistance, BUT I also have 16 times the Johnson 
noise due to the resistance, and it may be difficult to damp (as a moving
coil sensor) with a resistor if the magnet is not strong enough.

size diameter resistance NEWARK roll length, ft cost
#32 0.0088 164.1 
#34 0.0069 260.9
#36 0.0055 414.8 ohms/1000 ft. #36E1321 1/2lb 6400 $27.33
#38 0.0044 659.6 #36F779 1 lb 19300 $73.49
#40 0.0034 1049.0
#42 0.0028 1659
#44 0.0023 2593

And another consideration for a coil used for a broadband feedback
sensor, high coil resistance leads to instability. Anything much over
100 ohms is a problem. This became quickly evident when manufacturers
tried to add feedback to existing seismometers with constants of 50
to 200 Newtons/Ampere (Volts/meter/second) but thousands of ohms.

Increasing the coil magnet output is an advantage to a point
for a simple moving coil sensor, but for feedback it does NOT increase
the output: in fact it decreases it, where k=M/(G*C). However, it
does improve the high frequency response of the feedback loop, so
there are tradeoffs.

Regards,
Sean-Thomas

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