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I have a relay of coil resistance 200 ohm whose input voltage is 12V DC.

Now, under the same 12V DC supply, will the relay be activated when:

  1. The relay coil is decreased, say by 100 ohm i.e. new coil resistance is 100 ohm.
  2. The relay coil resistance is INCREASED to 200 i.e. new coil resistance is 400 ohm.

How does the relay behave in both cases under 12V DC? The same question for 5V DC relay.

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    \$\begingroup\$ Explain how you're changing the coil resistance. \$\endgroup\$
    – Bimpelrekkie
    Commented Nov 29, 2021 at 12:18
  • \$\begingroup\$ Explaining what you are trying to achieve is vital to get proper answers. Spehro's answere is excellent but as we don't know what your intention is there may be a way of helping you that has not been made clear. \$\endgroup\$
    – Russell McMahon ♦
    Commented Nov 30, 2021 at 9:56

1 Answer 1

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The force that makes the armature in a relay move is created by ampere-turns.

Looking only at the coil, you have one degree of freedom when you fill the winding area fully (which, as it turns out, is the most efficient). You can have more turns of thin wire or fewer turns of thick wire. It also turns out that the power drawn by the relay will be approximately constant as you vary that degree of freedom. Your example is 720mW, so if you rewind the coil so that it is 100\$\Omega\$ then the nominal operating voltage will be more like 8.5V, and if you rewind the coil with thinner wire so that it's 400\$\Omega\$ then the operating voltage will be more like 17V.

Of course that's based on filling the winding space more-or-less completely. If you choose not to do that, then the operating power of the relay will be higher than optimal and you may need more voltage than calculated to reliably operate the relay, and the coil will get hotter (further increasing the resistance of the coil).

For example, if you simply removed turns of wire to halve the resistance, then you would need twice as much current to operate the relay. So the relay might operate on the same voltage (ignoring temperature coefficient of copper, and ignoring the fact that removing outer turns would reduce the resistance a bit faster than reducing the number of turns) but draw double the power.

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  • \$\begingroup\$ ok. So there is no way to make the relay work under 12V by increasing the coil resistance by 200 Ohms??? \$\endgroup\$
    – Nar T
    Commented Nov 29, 2021 at 12:43
  • \$\begingroup\$ @NarT Re-wind the coil with thicker wire (not easy). There is no easy way to do it from outside the relay. You can, of course, build a boost circuit to generate a higher voltage. But it's probably easier to just buy a relay that is the proper voltage. If you have too much voltage, on the other hand, you can just add a resistor in series and throw away the power. To operate a 720mW 5V relay on 12V you would be using 720mW for the relay and 1W for the dropping resistor. \$\endgroup\$
    – Spehro Pefhany
    Commented Nov 29, 2021 at 12:48

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