Question

Stop I:0/1 HE Start I:0/0 JE Counter Done C5:0/DN JE Motor 0:0/3 M1 0:0/3 HE M1 0:0/3 НЕ CTU Count Up Counter Preset Accum Hcu) C5:0 10 Hon) Reset I:0/4 НЕ C5:0 HRESH

In the first rung, what is the purpose of the branch, which has “M1 O:0/3” as the input?

Explain this program. That is, explain what’s happening. Include all rungs in your answer. Be brief!

How does the motor “come on”?

In the first rung, why is the “Counter Done, C5:0/DN” input an XIO, and how does this work here?

What makes the CTU increment?

If input I:0/4 is activated, what happens?

When will the counter DN but be “on”?

Answer 1

The purpose of the branch M1 O:0/3 parallel to I:0/0 is to keep the circuit closed after the Start button I:0/0 is rung and released, usually the start button switch will be push button which is pressed once and then released. That one press energizes M1 O:0/3 and keeps the current flow.

Thus, the "Start" button I:0/0 turns 'ON' the Motor O:0/3.

The M1 O:0/3 also energizes the CTU.

CTU also known as Count Up, is an instruction attached to C5:0. The counter has a "Preset" of 10. The M1 O:0/3 energizes the CTU and increments the "Accum" to 1. This increment keeps the CU bit high and it continues till "Accum" reaches "Preset" 10. When "Accum" == "Preset" DN bit goes high.

The DN bit turns the C5:0/DN "Counter Done" (Normaly Closed) switch OFF to cut the supply and turn 'OFF' the motor. It also activates I:0/4 to Reset (RES) the "Accum" of C5:0 back to 0.

The "Stop" I:0/1 (Normaly Closed) functions as a fail safe button to stop the motor immediately. The parallel branch M1 O:0/3 also ensures that the Motor will not start until the "Start" button I:0/0 is pressed. This eliminates the accidental starting of motor as and when the power is restored after an accidental failure.

To sum up, the Motor will run till the CTU counts up to 10.