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I have a relay which is rated for 5v. My power supply is 12v.

I have measured the resistance across the coil as 400 ohm.

If I place the relay in series with a resistor that is 560 ohm, would it correctly lower the voltage to 5v?

Or is there something more complicated going on, given that the relay is electromagnetic and not a passive component?

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  • \$\begingroup\$ You can think of the relay's coil as a resistor. So yes, a resistor can be used. But check coil power first then coil resistance from datasheet. This will tell you required coil current (P=I^2 × R). From Ohm's Law, you can calculate required series resistance for voltage drop. The power dissipation of that series resistor is important. Don't forget to take it into account. \$\endgroup\$ – Rohat Kılıç Aug 8 '17 at 21:16
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Your relay is rated for 5 V and you know its DC resistance is 400 Ω. That means it draws (5 V)/(400 Ω) = 12.5 mA.

Yes, a resistor would work to drive this relay from 12 V. The resistor needs to drop 7 V with 12.5 mA going thru it. From Ohms law it needs to be (7 V)/(12.5 mA) = 560 Ω. Your calculation is therefore correct.

You should also consider the power dissipation of the resistor. It will be (7 V)(12.5 mA) = 85 mW. That's within what a "⅛ W" resistor can handle, like something in a 0805 case.

Yes, that's about all there is to it.

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    \$\begingroup\$ Thank you very much, I appreciate your detailed response. \$\endgroup\$ – CaptainCodeman Aug 8 '17 at 21:37

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