Is doing this as simple as adding a resistor in series with the coil? I know that would work with DC, but I have never tried doing it with AC. The relay that I want to use is rated 6.3mA, 8100 ohms coil resistance. My guess would be a 4W resistor (.0063 x 240 = ~1.5) sized as close to 8100 ohms as possible. Does that sound correct?

 

Forget about the dc coil resistance. The inductance of the coil will determine the ultimate coil current, in this case 6.3mA.

Dropping 120V at 6.3mA: R=120/6.3mA

You would need a 18K~20K 2W or 3W resistor in this case.

You can try an experiment using a normal 0.5W 20K resistor in series with the 240V supply and the relay, with a voltmeter connected across the relay coil. Make all the connections before power ON. If you then read about 120V on the coil the resistor value is about right. Don't keep the supply ON for too long before using a proper 2W resistor.

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L.Chung

 

An other point is to check that the relay contacts can safely switch off 240 Volts, unless you switching a lower voltage.
Your relay contacts may be rated for 120 Volts and may burn in after a couple of switching events.

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I am only talking about the coil, not the contacts. It doesn't need to be rated for 240VAC because I am changing the voltage with the resistor.

 

hi,
If my calculations are close, I would say the inductance of the relay coil is around 25mH, you say the 'dc' resistance is 8100hm: .

As you require only 6.3mA, you could use a low value series capacitor, rated at 250Vrms to limit the current to about 6.3mA
when connected to 240Vac

Use the Xc = 1/( 2*pi*50Hz*50mH) formula

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Be careful to avoid a capacitor value which would resonate with the inductance. This would maximize the current in the coil.

25 mH doesn't sound right for an 18k reactance at 50 Hz.... more like 50 Henries? That's just a guess, better get the calculator out. ;-)

 

Don't forget that as the relay is inductive you'll need to take it into account otherwise you could end up with a resonant circuit that will meltdown.

Also, a 1W resistor should be sufficient, 120^2/20000 = 0.72W

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Just a quick thought (probably of no consequence):

If the relay is 120V then the current rating may be at 60Hz.

 

Or, if you have the space and another identical relay, put two relay coils in series. This would be especially effective if you had two SPDT relays and needed a DPDT relay.

Dean

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Dean Huster, Electronics Curmudgeon
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