Wiring Diagram

I'm working on setting up an 120V indicator light on a fill auger motor that's only on when the motor is actually running. The main power goes through a paddle switch that stops the motor once the hopper is filled with feed, then starts it again when the feed is carried away by a separate line. This happens multiple times in an minute. I put a wire on the load side of the limit switch (the side that turns off when hopper is full), to power the indicator light, but the light is always getting voltage. What am I doing wrong?

  • \$\begingroup\$ Be mindful of the risk of a dust explosion if you're using an incandescent bulb, which is a potential ignition source. Sealed, explosion-proof light fixtures are available for use where organic dust flies. \$\endgroup\$
    – MTA
    Commented Feb 6, 2022 at 14:08

2 Answers 2


Compared to the current demands of the indicator light, the motor looks like a dead short. When the switch is open, the motor side of the switch will have a voltage very close to that at L2, which will power the light. In fact, I would expect that the light would see 208V in that condition, and brighten up.

I would connect the indicator across the motor leads. From there, I would do one of four things -- you'll have to determine on your own which is the best in terms of cost and safety:

  • Use a 208V (or 240V -- it should work) bulb. If these are common and cheap, it is by far and away the best approach. Even if it's a bit uncommon, if you use an LED bulb then it should last nearly forever -- if you put the indicator behind a panel and put a spare bulb inside next to it, then whoever changes your bulb will have a spare right where it's needed, and will hopefully be smart enough to order a replacement and put it where you did.
  • Use two 120V indicator bulbs in series. When the motor is energized, each one will get 104V and should light up just fine.
  • Use one 120V indicator bulb and a resistor in series. Size the resistor to drop (208V - 120V) = 88V at the bulb's rated current. If you're using an incandescent bulb, the resistor will get hot. If you're using an LED indicator, this may be problematical. If you design for an incandescent bulb and someone puts in an LED bulb, it could get really problematical.
  • Use a 208 to 120V dropping transformer, and a 120V bulb.
  • \$\begingroup\$ Thanks so much! Will probably just go with a 208V lamp. \$\endgroup\$
    – Geo
    Commented Feb 5, 2022 at 17:37


simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) What you've done. (b) What you need.

  • In Figure 1a the lamp is lit by L2 feeding back through the motor to neutral.
  • In Figure 1b this is prevented by avoiding the neutral connection.

If the lamp won't take 208 V then power it via a transformer with its primary in parallel with the motor.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.