Spiral Strip Inductor

[Gary Pace]

Hi :

I take a strip of copper, width w and thickness t, wind a spiral coil of N
turns with an inner diameter d1 and an outer diameter d2 (no iron, just
plastic spacers and air).

Does anyone know a method (numerical or closed-form) for determining the
inductance ?

There are a few Java calculators for spiral coils online, but it's not clear
if they refer to wire or strip in a spiral.

Thanks
Gary

 

[John Popelish]

Hi :

I take a strip of copper, width w and thickness t, wind a spiral coil of N
turns with an inner diameter d1 and an outer diameter d2 (no iron, just
plastic spacers and air).

Does anyone know a method (numerical or closed-form) for determining the
inductance ?

There are a few Java calculators for spiral coils online, but it's not clear
if they refer to wire or strip in a spiral.

Thanks
Gary

As long as the thickness of the coil (the width of the strip) is less
than the inner diameter of the coil, it doesn't make much difference
if its wire or strip.

 

[Gary Pace]

As long as the thickness of the coil (the width of the strip) is less than
the inner diameter of the coil, it doesn't make much difference if its
wire or strip.

Thanks John.
That's not what I expected.
My thoughts went like this :

- I'm using 3" x 0.125" thick strip
- If I wound a spiral with 0.125" diameter wire I'd get a figure for L
- If I wound 24 of these (i.e. 3" wide strip) and connected them in
parallel, but no magnetic coupling I'd get a really low L
- Obviously, when I stack these side by side, I get some coupling
- So the geometry of the strip really matters

I tried to work it out from first principles (flux per amp), but university
is just too long ago.

I had naively hoped that a quick Google would yield L = MagicFunction
(N,w,t,d1,d2)

 

[John Popelish]

Thanks John.
That's not what I expected.
My thoughts went like this :

- I'm using 3" x 0.125" thick strip
- If I wound a spiral with 0.125" diameter wire I'd get a figure for L
- If I wound 24 of these (i.e. 3" wide strip) and connected them in
parallel, but no magnetic coupling I'd get a really low L
- Obviously, when I stack these side by side, I get some coupling
- So the geometry of the strip really matters

It certainly does. Paralleling lots of wires that produce essentially
the same field pattern produces essentially the same inductance, but
with lower series resistance. The strip and the wire would produce
essentially the same field if the inner diameter were (much) larger
than the width of the strip (3 inches, in this case). This
requirements means that the thickness of the pancake is small compared
to the diameter of the coil, so the diameter dominates the field shape.

I tried to work it out from first principles (flux per amp), but university
is just too long ago.

I had naively hoped that a quick Google would yield L = MagicFunction
(N,w,t,d1,d2)

Good key words help.
http://home.earthlink.net/~jimlux/hv/wheeler.htm
http://f3wm.free.fr/tesla/equations.html

 

[The Phantom]

Hi :

I take a strip of copper, width w and thickness t, wind a spiral coil of N
turns with an inner diameter d1 and an outer diameter d2 (no iron, just
plastic spacers and air).

Does anyone know a method (numerical or closed-form) for determining the
inductance ?

There are a few Java calculators for spiral coils online, but it's not clear
if they refer to wire or strip in a spiral.

Thanks
Gary

Does your news provider give you alt.binaries.schematics.electronic?

If so, I'll post a formula (a complicated one) over there.

And, give me the details for a typical coil that you will wind; number of
turns, thickness of strip, width of strip, spacing (insulation thickness)
between turns, inner radius of coil, and outer radius of coil so I can work one
example for you.

 

[Chris Jones]

Hi :

I take a strip of copper, width w and thickness t, wind a spiral coil of N
turns with an inner diameter d1 and an outer diameter d2 (no iron, just
plastic spacers and air).

Does anyone know a method (numerical or closed-form) for determining the
inductance ?

There are a few Java calculators for spiral coils online, but it's not
clear if they refer to wire or strip in a spiral.

Thanks
Gary

Spiralmodel from MIT (free, open source) works well and is accurate in my
experience. Creating the model file to use as input can be a little
tedious. There was (maybe still is) a viewer for this file format at
fastfieldsolvers.com.

Chris

 

[Gary Pace]

Does your news provider give you alt.binaries.schematics.electronic?

If so, I'll post a formula (a complicated one) over there.

And, give me the details for a typical coil that you will wind; number of
turns, thickness of strip, width of strip, spacing (insulation thickness)
between turns, inner radius of coil, and outer radius of coil so I can
work one
example for you.

Yes it does - that would be helpful.

Here is my coil :

N = 11
t = 0.125" (3.175mm)
w = 3" (76.2mm)
spacers = 0.125" (3.175mm)
IR = 1.75" (44.5mm)
OR = 3.87" (98.4mm)

Thanks y'all

 

[The Phantom]

Yes it does - that would be helpful.

Here is my coil :

N = 11
t = 0.125" (3.175mm)
w = 3" (76.2mm)
spacers = 0.125" (3.175mm)
IR = 1.75" (44.5mm)
OR = 3.87" (98.4mm)

Thanks y'all

Ok, I've posted the formula(s) over on ABSE.

I notice an inconsistency with your measurements. Imagine starting at the
inner radius and moving outward. You should pass through 11 strips of .125
thickness and 9 insulating spaces of .125 thickness. This adds up to 2.625
inches, which when added to the inner radius of 1.75 inches should give an outer
radius of 4.375 inches, not 3.87 inches.

Or, working backwards, if you have an outer radius of 3.87 and an inner radius
of 1.75, this gives a winding depth of 2.12. Subtracting the total thickness of
11 strips of .125 gives a remainder of .745 to be divided into 10 insulating
spaces of .0745, not .125. I used .0745 in the example calculation, so the
winding pitch (distance from center-to-center of two adjacent turns) is .0745 +
..125 = .1995" = .50673 cm.

A subtlety in the use of these formulas is that when there is insulating space
between the turns (denote each space by d), you must imagine an insulating space
d/2 on the inside of the inner radius, and d/2 on the outside of the outer
radius. So if you measure the inner radius (call it r1') right up against the
copper and the outer radius (call it r2') similarly, the "true" inner radius to
be used in an inductance formula would be r1 = r1'- d/2 and the "true" outer
radius would be r2 = r2'+ d/2. But notice that the average radius, a, is the
same whether calculated as (r1 + r2)/2 or (r1' + r2')/2. Since the formulas I
posted don't explicitly use r1 and r2, you can use r1' and r2' to calculate the
variable a.

But the winding depth c must be calculated as r2 - r1, not r2' - r1'. Another
way to do it which always avoids errors is to measure the pitch (p) of the
winding and calculate winding depth c as n * p.