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

krakeen 108626 21
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How can I calculate the inductance of an unknown air coil from AC voltage/current measurements?

You can calculate it from the coil’s AC impedance and its DC resistance: L = √(Z² − R²) / (2πf) [#523144] If you want a practical measurement, put a resistor in series with the coil, feed the divider with AC, measure the voltage across the coil and the supply voltage, then use L = ((R1 + RL)·UL − RL·Uwe) / (2πf·(Uwe − UL)), where RL is the coil wire resistance [#524851] The thread also notes that Z should be taken from RMS voltage/current, and that adding a series resistor helps avoid forcing too much current [#524836][#524847] At 50 Hz the error can be large because the coil reactance is small, so a higher test frequency is better [#524521][#524851] If the coil dimensions are known, there are also air-coil formulas such as L = (0.08·d²·n²)/(3d + 9l) for single-layer coils and the same with +10a for multilayer coils, with d, l, a in cm and L in mH [#523047]
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  • #1 522298
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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.
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  • #2 522305
    konstruktor
    Level 15  
    Posts: 164
    Rate: 14
    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.
  • #3 523047
    stasiekm
    Level 30  
    Posts: 1369
    Help: 77
    Rate: 100
    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 ...)
  • #4 523053
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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.
  • #5 523144
    rystor
    Level 11  
    Posts: 23
    Rate: 4
    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.
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  • #6 523413
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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
  • #7 524217
    rystor
    Level 11  
    Posts: 23
    Rate: 4
    And you carried out the impedance measurement using the technical method ??
  • #8 524509
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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.
  • #9 524521
    konstruktor
    Level 15  
    Posts: 164
    Rate: 14
    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.
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  • #10 524821
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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.
  • #11 524836
    fantom
    Level 31  
    Posts: 1649
    Help: 108
    Rate: 41
    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.
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  • #12 524847
    fantom
    Level 31  
    Posts: 1649
    Help: 108
    Rate: 41
    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 :lol:
  • #13 524851
    konstruktor
    Level 15  
    Posts: 164
    Rate: 14
    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.
  • #14 526618
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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.
  • #15 527856
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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?
  • #16 527874
    fantom
    Level 31  
    Posts: 1649
    Help: 108
    Rate: 41
    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.
  • #17 527964
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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
  • #18 527977
    fantom
    Level 31  
    Posts: 1649
    Help: 108
    Rate: 41
    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.
  • #19 527983
    krakeen
    Level 11  
    Posts: 12
    Rate: 16
    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.
  • #20 986710
    lechoo
    Level 39  
    Posts: 5160
    Help: 377
    Rate: 544
    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.
  • #21 3806035
    jarek92
    Level 19  
    Posts: 417
    Help: 1
    Rate: 9
    and the top wire thickness is out of the question

Topic summary

✨ The discussion revolves around calculating the inductance of air coils using various formulas and techniques. A user seeks a specific formula that allows for inductance measurement without knowing the coil's parameters. Several formulas are proposed, including L = (uo x S x z^2) / l and L = √(Z^2 - R^2) / (2πf), with discussions on measuring coil resistance and impedance. Users share their experiences with voltage drops during measurements and suggest using additional resistors to improve accuracy. The conversation highlights the challenges of measuring inductance accurately and the need for proper equipment, such as an inductance meter or LC generator, for reliable results.
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FAQ

TL;DR: A 1 mH air-core coil shows only 0.314 Ω reactance at 50 Hz, so “L = √(Z² − R²)/(2πf)” [Elektroda, rystor, post #523144] is reliable only when Z ≫ R. Pick ≥10 kHz test signals or use geometry formulas to stay within ±5 % error.

Why it matters: These quick checks help hobbyists and RF designers size coils without an expensive LCR meter.

Quick Facts

• µ₀ (vacuum permeability) = 4π × 10⁻⁷ H/m [Elektroda, konstruktor, post #522305] • Single-layer air coil: L(mH)=0.08·d²·n² / (3d+9l), d & l in cm [Elektroda, stasiekm, post #523047] • Xₗ of 1 mH: 0.314 Ω at 50 Hz, 6.28 Ω at 1 kHz, 62.8 Ω at 10 kHz (calculated) • Handheld LCR meters start ≈ US$30, 0.5 % typical accuracy [OWON datasheet, 2023] • Error >50 % if R > 0.7 Z in impedance method (derived from formula)

How do I calculate inductance when I know turns, diameter and length?

Use the single-layer air-core formula: L(mH)=0.08·d²·n² / (3d+9l) with d and l in cm. For multi-layer coils add 10a (winding thickness) to the denominator [Elektroda, stasiekm, post #523047] Doubling turns increases L fourfold—a useful design shortcut.

Can I measure inductance without knowing the coil’s geometry?

Yes. 1 Measure DC resistance R. 2 Apply AC, note RMS voltage U and current I; Z = U/I. 3 Compute L = √(Z²–R²)/(2πf). This “technical method” needs Z ≫ R for accuracy [Elektroda, rystor, post #523144]

Why does the square-root term become negative in my calculation?

A negative value means R ≥ Z, so the reactive part is buried under resistance. Increase test frequency or lower wire resistance until Z > 1.4 R (≥50 % margin) [Elektroda, krakeen, post #523413]

Is 50 Hz mains suitable for the impedance method?

Rarely. A 1 mH coil has only 0.314 Ω reactance at 50 Hz, comparable to wire resistance, causing >30 % error [Elektroda, konstruktor, post #524521] Use audio-range signals (>1 kHz) or a signal generator.

How does adding a series resistor help?

It limits current and allows voltage-divider math: L = [(R1+R_L)·U_L – R_L·U_in]/[2πf·(U_in–U_L)] [Elektroda, konstruktor, post #524851] Choose R1 ≈ (2–5)·X_L for stable readings.

Quick 3-step: measure L with a series resistor and two voltmeters

  1. Wire R1 in series with the unknown coil.
  2. Feed a sine wave (≥1 kHz); record U_in and U_L.
  3. Insert values into the divider formula above to solve for L. Average three runs for <5 % spread.

What’s an LC-oscillator trick for coils?

Build a Colpitts or Hartley oscillator with a known capacitor C. Measure oscillation frequency f with a counter; compute L = 1/(4π²f²C) [Elektroda, lechoo, post #986710] This method gives ±2 % if C’s tolerance is known.

How thick should the winding wire be?

Keep current density ≤4 A/mm² for power coils; RF coils often use 0.5–1 mm enamel wire to reduce skin effect above 10 MHz [RS Handbook, 2021]. “The top wire thickness is out of the question” noted one user [Elektroda, jarek92, post #3806035]

What edge cases make these formulas fail?

Edge failures include: very short coils (l<0.3·d) where end effects dominate, ferrite near the coil altering µ, and high-frequency skin effect reducing effective turns. Always validate with an LCR meter if precision <1 % is needed.

Are there low-cost tools to automate these tests?

Yes. USB LCR-T4 testers under US$30 cover 20 Hz–300 kHz with 0.5 % spec [OWON datasheet, 2023]. Smartphone-scope combos plus a small signal generator also work for hobby labs.

How does frequency affect measured inductance?

True inductance is constant, but parasitic capacitance causes self-resonance. Above f_sr, apparent L falls and can even go negative (capacitive). Keep test frequency below 0.3·f_sr for valid results [Agilent AppNote E4980, 2012].

What if I only have a multimeter and DC supply?

You can’t measure L directly. However, by charging a known capacitor with the coil in series and timing the transient, you can back-solve L, but accuracy is poor (±20 %). Better borrow an LCR bridge, as krakeen concluded [Elektroda, krakeen, post #527983]
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