Fix 03 — Timer never times out
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Briefing
The motor should start 2 seconds after START is pressed, so people near the conveyor get a moment of warning. STOP cancels a countdown that is still running and stops the motor once it runs. The program has a timer, and it is called every scan, but the motor never starts. Find out why the timer never reaches 2 seconds, and fix it.
Objectives
- After START is pressed and released, the motor stays off for about 2 seconds, then runs
- STOP during the countdown cancels the start
- STOP stops a running motor
- After a stop, the next START waits the full delay again
Hints
- Press and release START and watch the motor. Then ask what the timer's input is doing during the 2 seconds after the button is released.
- A button is on for a fraction of a second, but a timer needs its input on for the whole delay. Something other than the button has to stay on during the countdown.
- Whatever keeps the countdown going must also be cancelled by STOP, otherwise a stop during the delay will be ignored and the motor will still start.
- Check the second start too: after a stop, the elapsed time must be back at zero so the next start waits the whole delay again.
I/O Table
Inputs
STARTStart push-button (momentary NO)
BOOL · %I0.0
STOPStop push-button (TRUE = pressed)
BOOL · %I0.1
Outputs
MOTORMotor contactor
BOOL · %Q0.0
Your program will be tested against:
All test cases run automatically when you submit. Assertions are hidden until you pass.
- #1START → motor off for about 2 s, then on
A momentary START begins a 2 second countdown, then the motor runs
- #2STOP during the countdown → motor never starts
Pressing STOP while the delay is running cancels the start
- #3STOP → running motor stops
STOP switches the motor off once it is running
- #4Start again after a stop → waits the full delay again
The countdown starts from zero on every start
- #5STOP alone → motor stays off
The motor never starts without START
Related scenarios
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