Today we do some practice on timer and counters.
Problem 1 :
Develop a program that will turn on light B 20 second after input A is pressed. After A has been pressed 3 times B will be turned off and D will be turned on.
Solution:
Here let us suppose A is X0 (i0.0), B is Y0 (Q0.0) and D is Y1 (Q0.1).
We gave supply to timer T39, on delay timer which is set as 200 means it turns on after 20 sec. As shown below…
Timer instruction: TON (On delay timer), TOFF (OFF delay timer)
Timer: - Time Base Number
1ms T32, T96
10ms T33-T36, T97-T100
100ms T37-T63, T101-T255
When we press X0 for 20 sec then Y0 will be turned on. Now We give same supply X0 for Counter up C0 which is set at 3 means When X0 pressed thrice C0 will be turned on as shown below…
Counter instruction: CTU (For counter up), CTD (For counter down)
Counter number: C0 to C255
The NC of Counter preset is connected before Y0 so when C0 turned on, NC of C0 will be turned off.
Problem 2 :
A motor will be controlled by two switches. The GO switch will start the motor and the STOP switch will stop it. If the stop switch was used too stop the motor, the GO switch must be pressed twice to start the motor. When motor is active, A light should be turned on. The STOP switch will be wired as normally closed.
Solution:
Let us suppose GO is a switch X0 (i0.0) and Stop is another switch X1 ( Q0.1). There is two loads as shown above one is Motor (Q0.0) and another is Light (Q0.1). Go switch gives the supply to counter up C5 which is set at 2. When Go is pressed counter preset C5 turns on and then Motor turns on so light also turns on. Now if we press Stop switch Motor and light both turns off. We can see at reset of counter, there is NC of stop switch so Counter reset at that stage. If we want to turn on Motor Go button is pressed twice cause counter has set value 2.
There are some problems for you guys. If you have any trouble in solving then inform me. I will post solution of these problems.
Practice 1:
A conveyor is run by switching on or off a motor. We are positioning parts on the conveyor with an optical detector. When the optical sensor goes on, we want to wait 1.5 seconds, and then stop the conveyor. After a delay of 2 second the conveyor will start again. We need to use start and stop button, A light should be on when the system is active.
Practice 2:
A conveyor is run by switching on or off a motor. We are positioning parts on the conveyor with an optical detector. When the optical sensor goes on, we want to wait 1.5 seconds, and then stop the conveyor. After a delay of 2 second the conveyor will start again. We need to use start and stop button, A light should be on when the system is active.
Gages are attached in above question too indicate good or bad. If the part is good, it continues n. If part is bad, We do not want delay of 2 second, but instead actuate a pneumatic cylinder.
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