0
\$\begingroup\$

Good morning. I am hoping to get some insight on a setup I am working on.

Current setup: 24V system. An area sensor when flagged by hand does two things. First, it sends a signal to the network that lights up a bin location and second, it takes the empty tray away that the product was in via small divert conveyor. The area sensor is connected to a relay that has a 2ms timer delay. After 2ms, the tray takes off.

Problem: Some items in the tray are heavy and causes the tray to take off while item is being removed. I can't add more delay as this will cause added time to the whole process.

Solution idea: I want to add a foot pedal which temporarily disables the area sensor when pressed in which will allow item from tray to be removed and when foot pedal is released bin location will light up and empty tray will take off. Also the area sensor should be back in normal use.

Solution problem: I am having a hard time figuring out how to light up bin location after foot pedal is released and putting the area sensor back in normal use.

Attempts made: I connected foot pedal in series/ normally closed with area sensor which cuts off signal when foot pedal is pressed in and when item is removed from tray, foot pedal is released and area sensor has to be flagged again in order to get bin location and tray to divert to return line.

Question: How can I set this up where after foot pedal is released I get a bin location and put area sensor back in normal use?

Thank you in advance.

Timing diagram 1. Existing system. (OP to edit.)

                        ____
Area sensor:    _______|    |_________________
                         ___
Light request:  ________|   |_________________
                            ____
Conveyor move:  ___________|    |_____________

Timing diagram 2. Desired operation. (OP to edit.)

Step1.                        ____
Footswitch pressed:   _______|    |__________
Hold conveyor no light bin:  ________________________                                   
Area sensor input deactivate:    ________________________                                       
Step2.                                 ___
Footswitch released: _________________|   |_________________
                         ___
Light request:  ________|   |_________________
                                     ____
Conveyor move:  ____________________|    |____
                                        ___     
Area sensor back to normal :    _______|   |_________________
                       
Footswitch:     ________________
Step3:                  ___
Area sensor:    _______|   |_________________
                         ___
Light request:  ________|   |_________________
                                     ____
Conveyor move:  ____________________|    |____
```
\$\endgroup\$
15
  • \$\begingroup\$ I'm having great difficulty understanding both the existing operation and the desired operation. 2 ms delay on a conveyor system seems pointless. I have added two sample timing diagrams for you to edit to help clarify the required timing. Can you edit those? \$\endgroup\$
    – Transistor
    Jul 9 at 16:13
  • 1
    \$\begingroup\$ Is the foot pedal being used as an on/off switch that supplies power to the area sensor? Is the power not being restored to the area sensor when the foot pedal is released? What about using the foot pedal as an enable logic input and AND it with the area sensor signal? \$\endgroup\$
    – tim
    Jul 9 at 16:43
  • \$\begingroup\$ Hi, so the foot pedal is an addition to the system. I have no issue with power delivery. Area sensor is made up of photo eye sensors. When light aray is broken by flaging it by hand, it sends a input to the relay which has small delay before sending input to the controller. Once controller gets the input, it turns on a bin location where the product needs to go and moves empty tray to next lane. I want to use pedal to temporarily cancel area sensor operation when pressed and when unpressed I want it to tell controller to send bin location and then resume area sensor back to normal operation. \$\endgroup\$ Jul 9 at 17:57
  • \$\begingroup\$ Eddie, that's basically repeating what you wrote in your question. Can you edit the timing diagrams I provided? \$\endgroup\$
    – Transistor
    Jul 9 at 18:03
  • \$\begingroup\$ It would help all of help you if you posted a schematic, not a frizzy thing with links to technical information on the hardware devices. \$\endgroup\$
    – Gil
    Jul 9 at 18:33
0
\$\begingroup\$

This is not an answer. It will be deleted when the question has been fixed.

Here's a timing diagram.

                        ____
Area sensor:    _______|    |_________________
                         ___
Light request:  ________|   |_________________
                             ___________
Conveyor move:  ____________|           |______

Here's how to read it.

enter image description here

  1. Area sensor signal is low.
  2. Area sensor signal switches high (on).
  3. Area sensor signal switches low (off).
  4. Light request signal turns on slightly after the area sensor turns on.
  5. Light sensor turns off at the same time as 3.
  6. The conveyor move turns on when 5 turns off.
  7. Some time later the conveyor move command turns off.

If you do this properly the relationship between your signals will be clear to us and will probably clarify your own thinking too.

\$\endgroup\$
0
\$\begingroup\$

A ladder diagram would look something like this (untested because I don't have a PLC to hand).

  • The first two rungs create a latch for the area sensor.
  • The third rung creates a 2 ms on timer.
  • Rung 4 uses the timeout to indicate that the area sensor latch should be reset.
  • Rungs 5 and 6 assumes a 10 s delay for the light and conveyor to remain on before going off.
  • Rungs 7 and 8 create a 2 ms pulse to reset the area sensor latch. The purpose of the pulse is to allow it to be reactivated if another tray arrives before the current tray has finished travelling along the conveyor to the bin location. If that is not the case, these rungs can be excluded and Reset area sensor used directly.

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.