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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?

stevenvh
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D Brown
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2 Answers2

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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.

stevenvh
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  • 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 08:37
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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.

Jay M
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