# Determining the coil inductance of a relay when the relay is on

I'm trying to measure the coil inductance of a relay when the relay is on and the normally open pairs of contacts are closed. The relay has a rated AC voltage of 110V and a rated current of 21mA. I know that $V = L\dfrac{di}{dt}$, but I'm not sure how to use the formula to find the inductance for a relay running on sinusoidal voltages and currents. Can somebody shed some light on this please?

• Datasheet of the relay? – stevenvh May 14 '12 at 8:02

You don't specify which relay you're using so I'm taking this one as an example.

Rated voltage = 110V AC
Current = 21mA @ 60Hz
Coil resistance = 932$\Omega$

Now the total impedance

$Z = \dfrac{110V}{21mA} = 5240\Omega$

That's the resistive part (932$\Omega$) with the reactive part at 90°. Then the reactance

$X_L = \sqrt{Z^2 - R^2} = \sqrt{5240^2 - 932^2} = 5154\Omega$

Then, since

$X_L = 2 \pi f L$

$L = \dfrac{X_L}{2 \pi f} = \dfrac{5154\Omega}{ 2 \pi \mbox{ } 60Hz} = 13.7H$

That's a pretty high value, but if you do the same calculation for 50Hz, where the current is 24.2mA, you get a comparable value: 14.1H.

• I think that on AC relays / contactors the inductance increases dramatically when the armature circuit closes. This is an advantage as the coil current will reduce after pull-in and power dissipated will be reduced. This may be the OP's problem. – Transistor Dec 10 '15 at 8:37

To actually measure the inductance (or get some sense that your calculations are correct) you could connect it up such that on opening the back EMF discharges through a resistor placed across the coil terminals. Place an oscilloscope across the resistor and measure the time it takes the voltage to get to, say 20% of it's final value of zero from when the coil power is removed.

Choose a resistor that won't affect your coil driver circuit.

$$L = \frac{-R\centerdot t}{ln(\frac{V(t)}{Vo})}$$

e.g. with 10Vdc and 100R across the coil, it should discharge from 10V to 2V in about 0.22s, if the inductance is 14H.