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I have a 24 VDC sensor that connects to a switching amplifier. When the sensor detects a fault it should send a signal to a relay which is connected to a green LED and a red LED. The Green LED should be on all the time unless there is a fault sensed by the sensor. If there is a fault the red LED should come on and the green one should turn off. Is it possible to do this with just one relay? And how would the wiring schematic go? The input voltage is 120 VAC. The switching amp connects to the sensor on one side through an intrinsically safe barrier.

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  • \$\begingroup\$ Are your LEDs powered from 120Vac? If you can power them from 24Vdc it makes things much easier and safer. Basically all you need is a relay with a change-over contact, also known as Single Pole Double Throw (SPDT). What sort of output do you have from the "switching amplifier" that can be used to drive the relay? \$\endgroup\$ – Steve G Oct 25 '17 at 22:47
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Indicator relay schematic.

How it works:

  • The relay is chosen by coil voltage and contact arrangement. You need a 24 V coil and a single changeover contact.
  • The green (OK) LED is powered from the normally closed contact.
  • The red (FAULT) LED is powered from the normally open contact.

I've shown resistors on the LEDs as a reminder that current limiting is required. It's not clear why you might want to use 120 V AC for the LEDs. If you do then you need 120 V AC LED indicators with built in diodes and resistors.

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You just described a relay with a Normally Closed and a Normally Open connection. A SPDT or DPDT relay will do this.

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