PLC Ladder Logic for simple ON-OFF of Motor and Lamp

Write ladder logic for the simple ON/OFF of motor and lamp.

Consider the following cases:


  • Switch 1 ON —— conveyor motor ON
  • Switch 2 and switch3 ON ——- lamp 1 ON & lamp2 ON
  • Switch 4 or switch5 ON —– conveyor motor OFF
  • Switch 6 ON —– lamp 1 OFF, lamp2 OFF

When you turn ON a switch 1 for the first time, conveyer motor should ON. When Switch 2 ON and switch 3 ON (Both are in ON Condition) should make Lamp 1 & 2 should go ON.

When Switch 4 or Switch 5 (Anyone ON) will make conveyor motor (Which was already turned ON by switch 1) should go OFF. When Switch 6 ON, will make lamps 1 & 2 (Which was already turned ON by switch 2 and switch 3) should go OFF.

Ladder Logic

Ladder Logic for simple ON OFF of Motor and Lamp

Logic Explanation

Network 1:

When switch 1 is pressed, the motor will turn on. Switch 4 and 5 are connected as normally closed contact in series with the output which will make output to go off when anyone goes ON.

Network 2:

Switch 2 and 3 are connected as normally open contact in series with the output which will make output lamp 1 and 2 to go ON only if both are ON. Switch 6 is connected as normally closed contact to break the power line when it goes ON.

Author: Hema

Training Course: Siemens PLC Training

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  1. To my knowledge the circuit isn’t correct, for the fact that the Switch 4, Switch 5 and Switch 6 are already ON but the statement doesn’t say that.
    The Conveyor Motor should be turned off when either the Switch 4 or 5 are ON, not when they are OFF, same thing for the Switch 6.

    So I would rather do the following logic.

    I connect I0.0 in series with M0. 2 to the output Q0.0 this is the first network and I0.0 turns ON the output Q0.0 the Conveyor Motor runs.

    In the network 2

    I connect I0.1 and I0.2 in parallel then both in series with M0.3 to the output Q0.1

    Then the third rang I0.3 and I0.4 in parallel then these two switches supply the output Q0.2, then M0.2 will deactivate the Conveyor Motor in the first rung.
    And finally I0.5 supplies Q0.3 in its turn will deactivate M0.3 in the rung 2.