Spiral Strip Inductor

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.

Good key words help.

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

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)

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.

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

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.