I have been tasked with designing what is to me a strange circuit. I have to wire a 12/24 led navigation light with an 'on' indicator and an 'off/extinct' indicator.

This is what i tried - switched 24 V to the nav light with a 24 V, 8-pin relay after it in series. I ran a 24 V red led through the N/C contacts for the 'off' and 'light extinct' indicator and a 24 V green led through the N/O contacts for the 'on' indicator. Everything works on the bench but the nav light is dim and the relay only gets about 18.5v at the coil. Its enough to pull it, but I feel I am missing something in the black magic world of resistors and diodes.

The navigation light is 9 - 33 volts, power is less than 2.5 watts. The relay measures 398 ohms across the coil.

  • \$\begingroup\$ just curious, but do you mean "off/extinguished" ? \$\endgroup\$
    – Daniel
    Jan 5, 2016 at 0:38
  • \$\begingroup\$ The terminology is 'extinct'. To mean its burned out or some other failure, even though the switch is on. \$\endgroup\$
    – Bob H
    Jan 7, 2016 at 22:47

2 Answers 2



simulate this circuit – Schematic created using CircuitLab

It reads as though you have created Circuit 1 above. The relay coil will limit the current through the lamp. I think you intended Circuit 2.

Circuit 2 is not much good as it doesn't prove that the light is connected or working.

Please post some details (in your original post) on the LED navigation lamps (current, voltage, wattage) with links to data sheets and we can suggest solutions.

Update after details added to question

Executive summary

Replace your 24 V relay with a 12 V unit. You'll need to add some parallel resistance to pass enough current to light the lamp brightly.


We know that LEDs brightness is controlled by current. The fact that the light operates from 9 to 33 V suggests that they've got a current regulator in there. We can figure out what current it's drawing by using the 2.5 W power and the 9 V specification.

Since $$P = V·I$$ we can calculate the current required as $$I = \frac{P}{V} = \frac{2.5}{9} = 0.28 A$$

Your relay-in-series with lamp approach is the most reliable as if the wires fall off the lamp or it fails open circuit the relay will drop out so let's see if we can make it work.

If we share the 24 V giving half to the lamp and half to the relay we will have the following.


simulate this circuit

We need 0.28 A going through the LED so we need a relay and maybe R3 to pass 0.28 A as well. From Ohm's law we can calculate that the resistance of the coil and R3 in parallel should be $$R = \frac{V}{I} = \frac{12}{0.28} = 42 Ω$$

Since most suitable 12 V relay coils would have a resistance higher than this we will need R3 in parallel.

$$R3 = \frac{42·Rc}{Rc - 42}$$

Example: 12 V, 400 Ω relay would require $$R3 = \frac{42·400}{400 - 42} = 46 Ω$$

Finally we need to work out the power rating of R3.

$$P = V·I = 12·0.28 = 3.36 W$$

The nearest stock sizes in this example would be 47 Ω, 4 W wirewound. Note that the resistor will get hot.

  • \$\begingroup\$ he needs a current relay in circuit 1 it's basically a few turn of fat enameled wire round a spdt reed switch. \$\endgroup\$ Jan 2, 2016 at 19:47
  • \$\begingroup\$ @Jasen: Agreed. I wanted to check the current. The other option is a low-value shunt in the circuit to generate enough voltage to turn on a transistor ... \$\endgroup\$
    – Transistor
    Jan 2, 2016 at 19:58
  • \$\begingroup\$ Could I run a negative straight from the light through a diode to ground? \$\endgroup\$
    – Bob H
    Jan 4, 2016 at 21:23
  • \$\begingroup\$ No. You'd have no voltage to operate the relay. See updated answer. \$\endgroup\$
    – Transistor
    Jan 4, 2016 at 21:57
  • \$\begingroup\$ @Jasen, I've finished 'the hard way' answer. Do you want to post your reed-switch solution? I haven't got any to experiment with. I reckon you need it to switch at 0.28 A - say 0.2 A for safety. \$\endgroup\$
    – Transistor
    Jan 4, 2016 at 23:12

If this is simply a switch position indicator and isn't being used as a "lamp is drawing current" indicator, this is the simplest solution I can think of:


simulate this circuit – Schematic created using CircuitLab


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