Inductance Calculation for Air Coils: Expert Tips, Formulas & Techniques | Electronics Forum

HELLO
I used to calculate the inductance of the coil using such a formula and it indented me somewhere, and in my literature I can't find it. It was a formula with such a large root, it could only be used to calculate the air coils, because it had magnetic permeability. I remember having to measure the coil resistance (with constant voltage), and i also know i needed an ac ammeter.

If anyone knows what pattern I mean, I would like to ask for it, because I have several coils that I could use, and I do not know their L.

The only thing that comes to mind is:

L = (uo x S xz ^ 2) / l

L - coil inductance
uo - magnetic permeability of vacuum;
S - coil cross section;
z - number of turns;
l - thread length.

Measuring the inductance of the coil by measuring the resistance of the wire ??
Hmm, a very interesting concept.

the formula provided by the constructor is a book version (not very precise)
in the elf there are the following patterns:
air single layer:
L = (0.08 * d ^ 2 * n ^ 2) (3d + 9l)
air multilayer:
L = (0.08 * d ^ 2 * n ^ 2) (3d + 9l + 10a)
L-inductance in mH
l-length
d-diameter (average)
a - winding thickness
everything in cm
n-number of turns
using an AC voltmeter and an ohmmeter, you can measure the inductance using the technical method (taking into account the resistance of the coil, you need to count a little ...)

The thing is, there was no number of turns or wind length. It was such a formula thanks to which it was possible to measure the inductance of an air coil with completely unknown parameters.

The formula is as follows
L = the root of Z to 2 - R to 2/2 (pi) f
First, determine the coil resistance, then the impedance and substitute it in the formula.

Yes, that was the formula, but something wrong, because the number under the square root is negative for me. I measured the resistances, it came out 1.6 Ohm
J = 2.4A
U = 3.7V
Is it supposed to be the supply voltage or the voltage with the coil connected?
It may be that my luck is weak and the tension is about to drop, which is quite possible.
something is wrong with me, maybe I am doing something wrong.
Help me

And you carried out the impedance measurement using the technical method ??

Yes, the impedance measurement was carried out using a technical method, I think I'm lucky, because in idle I have 5.1V and after connecting the coil it is 3.5V. There shouldn't be such a drop in tension, and that's not the end of the story.

Inductive reactance (inductive passive resistance):

X [ohms] = 2 x pi xfx L.

f [Hz] - current frequency (in this case 50Hz - mains);
L [mH] - coil inductance.

Assuming that the coil has e.g. inductance L = 1mH, we will get
X = 0.314Ohm.

Well, I know this pattern well, I determined Z with U / J, but actually there is some phase shift there and we would have to count it on complex hyb. It turns out I am under-educated.
How am I supposed to measure the impedances without knowing L so as to substitute it for the above formula, because now I do not know it myself.

No complex because it is not an impedance but an impedance module, because you have the effective values of the current and voltage. Maybe connect a resistor in series with this coil so that you do not have to force such a current and then derive the formula.

Oh, if you don't know how to derive it, just add this "new" R to the measured coil. By the way: this formula is nothing magical, just transforming the formula into an impedance module. wakes up at 12 o'clock

Exactly as the phantom said.
Put the resistor in series with the coil and connect the divider in this way to the transformer (although 50Hz is a bit too little for my taste). Measure the voltage at the coil terminals and the inductance can then be calculated from the formula:

L = ((R1 + RL) x UL - RL x Uwe) / (2 x pi xfx (Uwe - UL))

where:
R1- additional series resistor;
RL - coil wire resistance;
UL - voltage measured at the coil terminals;
Uwe - voltage applied to the divider (alternating current !!!);
f - current frequency;

In the case of measurements at f = 50Hz, the measurement result will be burdened with a large error due to the small value of the reactance of the XL coil.

Good luck.

I will try to do it tomorrow, because I haven't had time lately.
As for my knowledge of electronics, I'm still learning, because so far I'm titled as a tech. electrician.

Hmm, I get 2.7mH, only negative, I don't know what's going on, I was counting exactly according to the formula. Gives the measurement results:
R1 = 5.6 Ohm
RL = 1.6 Ohm
UL = 1.4V
Uwe = 12V
J = 1.35 A.


2.7 that would even match what I expected, but why a minus?

How powerful do you have this luck? Reduce this current even more. After all, you can see immediately that if you have 1.4V on the coil, about 6V on the resistor, it still lacks up to 12.

I took another trafo with a voltage of 5.4V The drop on the coil is 0.55V and on R1(11.2Ohm)- 4.8v. Now it would agree, but after substituting into the formula it's some miracles come out
Rl=1.6 Ohm

Well, unfortunately, this may be the effect that the constructor talked about. If you think that the coil has 2.7m, then after substituting it, the reactance will be about 750 million ohms. More frequency would be needed, but this can be a problem. The kind of accuracy your calculator gives you might then come out of it. Cheers.

I haven't rounded up anything.
I guess without the right equipment there's nothing to play with.
I just need to get someone with an inductance meter.
Thanks for your help, though.

Sorry for reheating the topic - it may be out of date, but it may be useful to others ...

With a frequency counter, the inductance of the coil can be measured indirectly using an LC generator, preferably built in a Colpitts circuit. Knowing the capacitance and frequency, it is easy to calculate the inductance.

and the top wire thickness is out of the question

And you saw the date, I'm closing!