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How to Reverse Operation of Photo Eye for Motor using Relay?

How to Reverse the Operation of a Photo Eye Using a Relay, So the Motor Turns on During the Day Instead of at Night?

Suppose we have a setup where a single-phase motor is controlled by a relay for night operation using a photo eye. This scenario is similar to a situation where we need to turn on the motor at night and turn it off during the day. 

Now, consider the reverse case having the same setup. For instance, we want to reverse the motor’s operation, specifically turning it on during the day and switching it off at night. This is what the circuit is about.

The photo-eye circuit operates based on dusk-dawn conditions, turning on the load at night and switching it off during the day. Now, we aim to use the same components to reverse the photo-eye effect (from dusk-dawn to dawn-dusk). For example, the load will turn on during the day and switch off at night using the photo-eye and relay.

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How to Reverse Operation of Photo Eye for Motor using Relay?

Due to the high amperage circuit required for motors, we have used the R04-11A30-120 Series R4 relay. Alternatively, you may use a general-purpose DPDT relay such as LY2, LY3, LY4, rated for 30A at 120V or 240V AC.

The photo eye (photocell light sensor) is directly connected to the 120V AC supply via the Line (Hot) and Neutral.

The L (coil or magnet) terminal of the 30A relay is connected to the red wire (as phase) from the photo eye, and the N terminals of the relay and motor are connected to the Neutral wire from the 120V supply respectively. The COM terminal of the relay (as hot) is connected to the 120V single-phase motor.

The Line wire from 120V AC is also connected to the NC terminal of the relay.

Connect the protected ground wire to all applicable appliances according to the codes.

After completing the wiring circuit, the photo-eye will operate in the reverse direction, i.e., the motor (or any other connected load) will turn on in the morning and switch off automatically at night.

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Rating of 30A Relay (R4) for 120V/240V Circuits

  • A 30-amp relay can be used for a maximum load of 24 amps (30A x 80% = 24A).
  • For 120V circuits, a 30-amp relay can handle a maximum of 2.8 kW (24A x 120V = 2880W).
  • In 240V circuits, the 30-amp relay can be used for a maximum of 5.7 kW (24A x 240V = 5760W).
  • When using a 30-amp relay with a 10-gauge wire, it can support a maximum of 2 horsepower at 100 feet (30 meters) or 1.5 horsepower at 150 feet (45 meters) in 120V circuits.

As the pins configuration of 30A R4 magnetic relay is shown in the fig above. If you are using Omron LY2-3 relay, below are the terminals connection.

  • 1 & 2 = Normally Closed (NC) i.e. power is ON when relay is OFF
  • 3 & 4 = Normally Open (NO) i.e. power is OFF when relay is ON
  • 5 & 6 = Common (COM) terminals
  • 7 & 8 = Magnet or Coil terminals (Connects to the Phase (+) and Neutral (-) supply

Rating of Omron LY2-3 Relay for 120V and 240V Circuits

  • An Omron 8-PIN LY2 relay is suitable for a 7.5-ampere load.
  • LY2 and LY3 relays, rated for 15 amperes, are suitable for 10 amperes in 2-pole or 3-pole configurations, and LY4, a 4-pole relay, is suitable for 5 amperes.
  • For motor loads, it can be used for 1/4 hp at 120V and 1/5 at 240V supply.
  • In 120V circuits, It can be wired for a max of 900W incandescent load (7.5A x 120V = 900W).
  • In 240V circuits, the Omron LY2-3 relay can be wires for max of 1800W incandescent and resistive load (7.5A x 240V = 1800W).

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