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I Have A 28Vdc Small Relay And Need To Replace It, But Can Only Find A 24Vdc One. .

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lucky131 | 13:46 Wed 07th May 2014 | Technology
6 Answers
Asthis is the coil voltage do you think I will damage the 24V coil by subjecting it to 28V. If so what would the resistor need to be to convert from 28 to 24Vdc. Bear in mind that we are only talking of powering up a small coil with little amp draw. Thanks
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The resistor will have to provide a drop of 4 volts at the current which the relay actually takes. All you say is that it's a 'little amp draw' but we need actual numbers. If it's 100 milliamps, you need a 40 ohm resistor. If it's only 20 milliamps, you'd need a 200 ohm resistor. Whatever the figure, you'll need Ohms = 4000 divided by milliamps.
Almost a 17% increase. I'm unsure you'd get away with it but you could try. Personally I'd go buy the correct size.

If you must try it with a resistor in series bear in mind the coil is inductive and the resister will be, well resistive. Maybe it won't make a lot of odds though. Been too long to work it out at my age. Memory fails.

Bear in mind the wattage of the resistor too. 4² / 40Ω = 400mW.
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When I was a lad aircraft used to use 28V - perhaps somewhere selling aircraft spares may help.
The power dissipated by the relay is proportional to the square of the voltage – by increasing the voltage by 4V (17%) you will increase the power dissipation by 36%, probably nothing to worry about unless the ambient temperature of the relay is over 50C.

Rather than use a resistor to drop the 4V, the value of which will depend on the current – why not use a 3.9V zener diode, that way you will be supplying 24V to the relay regardless of current flow.
Adding a resistor to a relay coil isn't always as simple as it sounds.

Relays draw an inrush current which is three to five times the holding current when they are first energised because the magnetic circuit is not yet energised. It is possible that a resistor could cause such a voltage drop during inrush that it does not pull the armature hard enough.

I would be more inclined to used Hymie's zener diode solution.

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