I have a hand-wound coil I'm using in an LC Tank circuit. Using my multi-meter, I found out that the coil is drawing .6mA at 9v. Using Ohm's Law...

$$R = V/I$$
$$R = 9/.0006$$
$$R = 15kΩ$$

Is there a way to calculate the inductance of my coil from the DC resistance?

  • \$\begingroup\$ Are you sure of your measurement anyway? Below you say your coil is 8 turns * 9.5mm diameter that is some 25cm. Did you actually use a 60kohm/meter wire? \$\endgroup\$ – Vladimir Cravero May 13 '15 at 7:28
  • \$\begingroup\$ I was shorting the inductor and measuring resistance that way. I've since discovered my mistake. \$\endgroup\$ – Allenph May 13 '15 at 9:53
  • \$\begingroup\$ That makes even less sense to me. \$\endgroup\$ – Vladimir Cravero May 13 '15 at 9:56
  • \$\begingroup\$ In a word; NO. \$\endgroup\$ – EM Fields May 13 '15 at 10:18

An ideal coil has an inductance \$L = mew \times turn\text_density^2 \times Area \$.

The resistance of your coil is \$R = \rho \times \frac{length\text_wire}{cross\text_section\text_area\text_of\text_wire}\$. This can be simplified to \$R = constant\times length\text_wire\$

Naturally the if the coil has a greater turn_density or area, then the length of wire would be longer, so they do go up together, but the inductance is highly dependent on the geometry of the coil, so you are probably better off trying to calculate the inductance with the equation.

For instance if you increased the length of the coil while keeping the same density and area, the resistance would go up, but the inductance would not. So they cannot be related

  • \$\begingroup\$ I just re-wound my coil as to measure the specifications. The coild is 12.5" in length. There are two "leads" which are 1" in length before the coil starts bending. The spacing between each loop is (As consistently as possible) 3/16". How would I go about calculating the inductance? There are no units in your equations. \$\endgroup\$ – Allenph May 13 '15 at 5:56
  • \$\begingroup\$ Inductance units are Henrys mew = 4pi*10^-7 Henry/meter length of coil = 12.5" = 0.3175 meters spacing = 3/16" = 0.0047625 meters number of turns = N = length of coil/length used in each turn = 0.3175/0.0047625 = 66.66666... turns. I will say there are 67 turns, would you say that is correct? Anyway the turn density will be the reciprocal of 3/16" 1/0.0047625meters = 210 turns/meter \$\endgroup\$ – steve May 13 '15 at 7:38
  • \$\begingroup\$ I would need one more piece of information to tell you the inductance, what is the area of the circle the loops make? Lets just call this Area for now L = mew*density^2*Area L = 4pi*10^-7 * 210 * 210 * Area L = 0.0554 Henrys/m^2 * Area(in square meters) \$\endgroup\$ – steve May 13 '15 at 7:38

No. If you have an LC tank circuit, I would recommend using an oscilloscope to measure the voltage across it and then use a function generator to inject a sinewave. Find the resonance in the form of a minimum or maximum in the response and calculate what the L should be, given the C that you are using.

It may also be possible to get your tank to ring if you give it a whack electrically. Try connecting your scope probe across the tank circuit, put it in single shot mode with auto triggering turned off, and then connect and remove a power source across the tank. You will likely get a few cycles of oscillation due to the voltage step. It may also take a few tries to capture it, try screwing around with the trigger level. Measure the period of the oscillations, and that should be the resonant frequency. Work backwards to find L given the C that you used.


I found an online calculator.

L =Inductance in uH
D = Coil Diameter in Inches
L = Coil Length in Inches
N = Number of Turns

$$L= (D^2 * N^2)/(18D+40I)$$

My Coil

D = .375"
L = .75"
N = 8

L ≈ 0.24490 uH ≈ .245 uH

  • \$\begingroup\$ This assumes that the coil is air filled (which it might be). If you've got a ferrite or iron core in the middle, then it's different. \$\endgroup\$ – Dean May 15 '15 at 17:42
  • \$\begingroup\$ Should have added that to my answet, but yes I'm using an air core indcutor. \$\endgroup\$ – Allenph May 16 '15 at 1:51

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