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# 40 VAC to energize 24 VAC relay coil?

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When I have 40 VAC present I need a contact closure for a control circuit.  Does anyone know the best way to drop the 40 VAC down to 24 VAC to energize a coil on a cube relay?  I was thinking of putting a resistor in series but do not know how to calculate.  any other ideas?

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Yes, put a resistor in series with it.

Measure the current in the coil and calculate the required resistance using Ohm's law.

Hint: it's the same formula as that used to calculate the series resistor for an LED (this can be found using Google) but replace the forward voltage with 24V.

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Okay, the manufacturers literature states that the 24 vac relay consumes 1.1 VA so that would equal 45.83 mA using ohms law.

Using this online LED resistor calculator... http://ledz.com/?p=zz.led.resistor.calculator i came up with this...

Supply voltage = 40
Voltage drop across LED =24
Desired LED current = 45.83 Ma

Resistor should be:
390 Ohms
1.222 Watt

Did I do that correct?

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That sounds about right to me.

I calculated 349R, the nearest standard E24 value is 360R but any value between 330R and 390R will do.

I = P/V = 1.1/24  = 45.83mA
R = (Vin-Vout)/I = (40-24)/0.04583 = 349R

Your power calculation doesn't look right to me but fortunately your result was conservative.
P = I2R = 0.045832*349 = 0.73W

So use a 1W resistor, minimum. Of course, if the resistor value is slightly higher the current and therefore the power dissipation will be slightly lower.

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Thanks for the help on that.  The online calculator I used came up with .73 watts also but said a 1.22 watt is a safe bet.
I'll give it a try.

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No one makes 1.22W resistors, it's common practise to just use the next power rating up.

Okay, Thanks.

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