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I am designing a circuit to replace a problematic computer controlled raise/lower operation for our mobile home bed. The bed uses a 30 amp DC motor (similar to a wiper motor) and simply reverses the polarity to raise/lower the bed. Quite simple really. There is a limit switch that cuts power when the bed hits an upper limit (thereby opening the normally closed switch) and a master on/off switch controlling the main power supply from a vehicle battery.

I am confident the circuit will work fine however I would like to place an LED in the circuit somewhere to indicate when the limit switch has been activated i.e to light when the limit switch is open.

I have looked at How to create a circuit to turn off LED when switch is "ON" and turn on when "OFF" but this and other answers I can find do not have a relay or other similar load attached to the circuit so I'm unsure how to proceed.

I am considering simply bridging the level limit switch with an LED and appropriate value resistor figuring that with the switch open current will flow through the LED/resistor lighting it. With the switch closed currect will flow through the closed switch instead (path of least resistance) turning the LED off. However I am unsure of the effect the relay will have on the LED when the switch is open. Will it (in effect) try to use power through the LED/resistor thereby burning them out?

Would appreciate some tips/comments/suggestions please.

power bed schematic

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    \$\begingroup\$ Is the limit switch really a SPDT as per your drawing? If it is you could use the unused terminal to connect the LED and a appropriate series resistor to ground. \$\endgroup\$
    – Unimportant
    Commented Dec 31, 2018 at 10:39
  • \$\begingroup\$ Bridging the limit switch contact has a risk- if the LED current exceeds a few percent of coil current the relay may not drop out and result in mechanical damage or a burned out motor. This also applies to the unused contact suggestion since LED current has to flow through the relay coil. What is the relay coil current rating? \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Dec 31, 2018 at 10:41
  • \$\begingroup\$ True, but a 30A DC motor suggests a beefy relay. A modern indicator LED can do with a single mA. That's unlikely to be anywhere close to the holding current for such a relay. \$\endgroup\$
    – Unimportant
    Commented Dec 31, 2018 at 10:46
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    \$\begingroup\$ @Unimportant Sure, 1mA would likely be fine. Must-release current for the common DC12V T-90 30A relay is less than 8mA. If it's a typical automotive relay there probably isn't an issue, even at 15mA. Without guidance I fear OP could aim for 20mA or something. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Dec 31, 2018 at 10:51
  • \$\begingroup\$ @Unimportant Thanks for the suggestion but the limit switch is SPST. Sorry, I should have mentioned this. I just couldn't find an appropriate SPST graphic. The relays are automotive relays rated 30amps. \$\endgroup\$
    – Drunkencelt
    Commented Dec 31, 2018 at 13:28

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I have carried out some breadboard tests and find that bridging the switch with an LED and 780ohm resistor does not work properly enough to be reliant. The LED lights when the switch is open but the coil is energised too as suggested by @Spehropefhany meaning the relay switches intermittently. With the switch closed the relay is energised and the LED goes out.

Not quite the effect I was looking for. I consider the only way to do this properly is to use a separate bi-pole low current relay for the LED only - connected to GND after the limit switch - so the LED is on when the switch is open and off when the switch is closed.

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  • \$\begingroup\$ 780 ohm results in roughly 15mA. That is way too much and enough to keep the relay energized. Try 10K for approximately 1mA and use a modern high brightness LED that only needs 1mA to light up brightly. \$\endgroup\$
    – Unimportant
    Commented Jan 1, 2019 at 12:19

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