Inductance Equations using SI Units

When I googled "inductance calculator + si units", I eventually navigated to this:

Thanks, Ian. What I was looking for, though, were some equations or formulas. I don't learn much from a calculator, and I don't really have a good way to even know whether it's accurate. But I appreciate your response.

googling solenoid equation meters seems pretty good.

This is also not the DIY equations that you're looking for but there's a handy Windows calculator over at

For the DIY question, the ARRL Handbook is a great resource but it's very ... inchy. I can't reproduce all of the goodies, but for your basic single layer air-core inductor, try L = (d^2 * n^2) / (18 * d + 40 * l) where L is in micro-henries, d is diameter of the winding circle (center of wire to center of wire), l is inductor length, and n is the number of turns, where length is >= 0.4 * d. And of course, it's all in inches. Unit conversion left as an exercise, or something. ;-)

..

If it's a long solenoid then you can get a guesstimate from physics.

L*i =3D N*flux. (N=3D number of turns) For a long solenoid the B field is roughly constant. B~ mu(sub zero)*i*N/L (where L is the coil length) (MKS units) So the flux is B*A (A is the area of the coil). Putting it all together,

L =3D mu(sub zero) * N^2 * A/L

For a single loop there should be another (simple) solution, but the integral will be a lot harder. (There's an equation in Terman that I could quote.... It happens to be sitting on my desk.) A page down on here looks like the same equation.

George H.

Rich, George, and John, thanks for the ideas. It looks like I'm going to have to suck it up and do the conversions. I hesitate to do that, because I often multiply when I should divide (or vice versa), and wind up with something like furlongs per fortnight as units.

I don't think many of us understand exactly what you're looking for? At least I am a bit confused, if not for others here ;)

Doing induction calculations seems to be a black art it seems. For years i've seen a variation of formula's to represent the value of a coil once all the data is known.

For example..

In a long single coil, a formula of this type is used and there are others, too.

u n^ A L =--------- l

L = uH

u = permeability of air, some where around 1.26-05

A = cross section area of the coil in "m"^

l = Length of the coil in "m"

N = number of turns^

And now for the big HOWEVER>

If you were doing magnetic cores.. the math changes just a little.

In this case, the "u" permeability for air is 1.0

Note the constant 0.012? This was from a formula I got some where, it was a note slide in one of my books that is so old it's turning yellow.

and "A" cross section area is now cm^ not "m"

and Lc is your magnetic size of the field, the physical length of it, which can extend a bit depending on the form you're on.

I also have some math for inches.

Like I said, it's a black art. for the last few months I've been playing around with a concept that involves using reluctance alterations to monitor surface changes. This has forced me to dig out some older references in my library.

It seems the internet is becoming a junk yard and is hard to find a agreed method of doing certain things, like this for example.

Jamie

Lessee. 18 / 25.4 = oh, this is too complicated.

OK, it's pretty simple. I can find all manner of equations, formulas, calculators, etc., that calculate inductance but that use non-SI units such as inches, feet, pounds (or whatever).

Of course I could just convert the various values into SI units, but... it's a pain, and I tend to mix things up a bit and reverse the calculations I should be doing (multiply when I should divide, etc.).

Same here. To my understanding, the actual calculations (accurate ones) require diffeq's. In order to simplify, accuracy is sacrificed and they come up with various approximations that work under a certain set of conditions (long solenoid; coil with thickness For example..

I don't know what you mean by magnetic cores. Do you mean using cores that contain iron, that would affect the inductance?

You're right about that. Back in the olden days (when I was learning this stuff) the problem was a lack of information. You had to either have it in books at home, or go to the library or school for it.

Now there is an endless amount of information, but so much of it is crap that you've got to sort through lots of chaff to find the wheat. Way too much information, often unreliable, and too much to process in a reasonable time.

Ah, well. Thanks for your ideas. Maybe I'll eventually figure this out somehow. Or I *could* just wind the stupid coils and measure the inductance, and try to figure out a relationship myself.

At least SI units have got rid of a lot of pi ;-)

When I started it was CGS, then we had to change to MKS, then to "Rationalized" MKS, then SI.

It has already become a junkyard, not helped by search engines that seem to make up their own mind about what it is you're looking for, with "advanced" searches that just plain don't work. A good proportion of the "information" out there is seriously flawed, anyway. Give me a good old fashioned textbook.

Look for the Ferrite Magnetic Design Tool from Epcos. It really works, for their cores, obviously.

For coils with no magnetic material, it's mostly freshman Physics. (Didn't you keep your text book?) The N*flux(B) =3D L*i is like the Q=3DC*V formula. It's some geometrical constant. It's sometimes useful to also think about inducatnce in terms of energy. The energy in a coil is 1/2L*i^2 and this is equal to the magnetic energy density ( B^2 /(2* mu(sub zero)) integrated over all space. Most of the time a messy integral, but it helps sometimes when thinking about how the inductance will change.

George H.

Thanks, Fred. Actually, I was only going to use air-core coils, but I'll file this away for future reference. If I can *find* it when I need it, of course...

or

I kept all my books; but had a fire that took everything I had. Kind of like an enforced spring cleaning. By that time I had about 2000 books, which made moving a *HUGE* pain. Suddenly, it became quite easy, so I'm not complaining.

Thanks. That should be helpful... I'll need to review the math, though, since I don't remember much of it. I've got more books now, so this shouldn't be too difficult.

Thanks.

--=20 Kindness is a language which the deaf can hear and the blind can read. -- Mark Twain

This works if you put the "18" and "40" into units of inches/henry, then convert to millimeters:

L = (d^2 * n^2) / ((475 mm/H)*d + (1016 mm/H)*l)

Or with meters:

L = (d^2 * n^2) / ((0.475 m/H)*d + (1.016 m/H)*l)

Or if you want to be demented about units:

L = (d^2 * n^2) / ((0.475 A^2/N)*d + (1.016 A^2/N)*l)

(Where, as any school child can figure out in a matter of seconds, one amp squared per Newton is, of course, one meter per Henry)