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