I have an 12V electromagnet that draws 350mA at full hold strength and I want to put a relay in series with it so that when the magnet is energized the coil of the relay is too, which will turn on a lamp that indicates this. Is it possible to do this or will the relay interfere too much with that? I was thinking of using a T9AP1D52-12. If it's not possible is there some way to hook up an indicator light that will stay on as long as the electromagnet is on? This is the circuit I would like to build.This is the circuit

This is a modification of a pre-existing circuit that is identical except it doesn't have the relay or the light.

I have tried putting the LED in series but it is too low of a wattage and it doesn't let the electromagnet turn on

  • 1
    \$\begingroup\$ Use the relay to switch power to the electromagnet, that's what everyone else does. Put the indicator across the electromagnet with a suitable series resistor. \$\endgroup\$
    – Finbarr
    Jul 1, 2021 at 16:28
  • \$\begingroup\$ ok so then how would I know if the electromagnet is on? I want to know if there is a break in the line anywhere and the electromagnet is not getting power \$\endgroup\$
    – joshua0823
    Jul 1, 2021 at 16:29
  • \$\begingroup\$ Updated my comment. Put the LED across the coil, it's complete overkill to drive the LED with a relay. \$\endgroup\$
    – Finbarr
    Jul 1, 2021 at 16:32
  • \$\begingroup\$ Sorry I updated my question after your comment. I tried that but the LEDs I have are too low of a wattage to allow the electromagnet to turn on. I tried it and it didn't work \$\endgroup\$
    – joshua0823
    Jul 1, 2021 at 16:34
  • 2
    \$\begingroup\$ ACROSS the coil, not in series with it. \$\endgroup\$
    – Finbarr
    Jul 1, 2021 at 16:35

2 Answers 2


You're trying to use the relay as a current sensor. That's of course possible, but it requires a low voltage relay that has a low coil resistance.

For example, suppose that your sensing relay would be Axicom/TE IM00 model, with 1.5VDC 16Ω coil. The coil expects about 100mA when operating. Since the current you're sensing is 350mA, you'd need to bypass 250mA around the coil. For that, you can use a resistor with 2.5x the conductance of 16Ω, that is 6.4Ω. Even a lower value, like 5Ω, would work, since the relay cuts-in 75% of the operating voltage. This would further reduce the voltage drop across our current sensor.

Thus, your circuit would be:


simulate this circuit – Schematic created using CircuitLab

D1 and D2 should be close to the coil L1 - ideally mounted right across its terminals. D3 can be any signal/switching diode, e.g. 1N4001 would work just fine. D4 can be most any modern LED. C1-R3 protect the contacts of the pushbutton switch from arcing, prolonging its life.

Note how there's no need for a diode parallel to RL1's coil. Also, the electromagnet's L1 should have two anti-paralleled 12V Zener diodes to equalize the turn-on and turn-off time constants. A single diode has the unfortunate effect of making turn-off up to an order of magnitude slower than turn-on. Sometimes you don't care about it, but sometimes you do.

Now, this circuit isn't just a voltage indicator: it is current detector, and e.g. if the electromagnet coil L1 fails open, you'll get immediate feedback: RL1 will stay open in spite of the button SW1 being closed, and D4 will stay off.

If all you need is a simple indicator without current detection - you don't need any extra relays! Just connect D3-R2-D4 in parallel with the electromagnet coil L1.

  • \$\begingroup\$ thanks. that really clears it up \$\endgroup\$
    – joshua0823
    Jul 1, 2021 at 18:32
  • \$\begingroup\$ If I don't care about how long it takes for the magnet to turn off/on do I need the zener diodes? This magnet holds a pretty heavy load and as soon as power gets cut off it'll end up dropping it which is all I care about. Also, since i'm dropping ~1.22V across the resistor/relay won't I only end up with 10.78V at the magnet? Is this ok? \$\endgroup\$
    – joshua0823
    Jul 12, 2021 at 18:01

A) Put a reed switch near the electromagnet, wire it to the LED.

Your circuit can work, but you need a different relay:

B) use a current relay instead of a potential relay.

C) pick a relay that operates at 350mA - eg: Panasonic DS1E-M-DC1.5V (267 mA on paper but most relays are OK at up-to 50% coil overload)

D) build you own relay by rewinding an existing relay or wrapping magnet wire round a reed switch capsule, oe on a spool,


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