timed conveyer

i'm working on this project like a box is placed in front of a sensor which starts a conveyer that will run for 5 sec then stop under a feedeer . the feeder will feed the box for 2 seconds then stops the conveyer will start up again for two secs then stop after another 5 secs under a second feeder . second feeder will be on for for 3 sec then stop . once the second feeder in complete a cylinder will push the box to another part of the line. after a cylinder will push the box to another part of the line after the cylinder has pushed th box system resets and awaits another box.guys in this i want to know if sensor comes before push botton or after and feeder should eb hard wire close or open

Re read your post and draw out a flow chart for all of these actions.

Assign your inputs/outputs to each action etc...

Sometimes it helps to have something drawn out to code.

Jessica,

You need to break your process down into steps. Each step will be a “cell” in your flow chart. If the chart goes straight down without any branches, (OR conditions) than it may be ok to forgo the flow chart and outline the steps 1, 2 3, and so on. If your process requires that you must branch out from a step in the middle of the process and go back to another step, or skip a step etc.. than you should create a complete flow chart to outline your process for clarity in creating your code. Without flowcharting your process a simple routine may seem more complex than it really is.

Incidentally you only mention the button in your question, and have not indicated where it is in your process. Place the button in the process, than decide how to wire it from there.

From your 1st post I read the steps in your process as follows:

!) a box is placed in front of a sensor which starts a conveyer

2) that will run for 5 sec then stop under a feeder .

3) the feeder will feed the box for 2 seconds then stops

4) the conveyer will start up again for two sec’s then stop

5) after another 5 sec’s under a second feeder .

6) second feeder will be on for 3 sec then stop

7) once the second feeder is complete a cylinder will push the box to another part of the line.

8) the box system resets and awaits another box.

You ask
"I want to know if sensor comes before push button or after and feeder should be hard wire close or open"

Normally to actuate a process you would use a open contact that would close to begin, or run, the process and open to stop the process.

Mike

ok to start my ladder logic i can start with start button do i need to use sensor as open contact to run the conveyer or i can start my conveyer with start botton only and what about stop push botton can i place infront of start as open contact?

If this were a real-world project, assuming that a person is placing the container on the conveyor, it would be safer to have a start button. That would force the person to get his hands out of the way of the conveyor before it started moving. You should always have some method of stopping any system so that when things start to go wrong, an operator can shut down before they get worse. The details of how to go about designging safety into your system depend on both the system and the amount of risk involved in its operation.

Since this is an academic exercise, it doesn't really matter whether you start the conveyor as soon as the container is placed in front of the sensor or whether you wait for a signal from the 'Start' button. However, it's best to develop good habits early rather than have to 'unlearn' bad habits later. Include the Start and Stop buttons in your design.

As to whether a pushbutton should be wired normally open or normally closed, you have to consider what the button does and how it might behave when it fails. You can write your PLC program to handle either choice of contact. Think about your stop button. Its function is to stop the conveyor. Two of the ways it might fail are when the electrical connection between the button and the PLC gets broken or when pushing the button fails to change the state of the contacts. If it's wired normally closed, and the electrical connection between the button and the PLC is cut, from the point of view of the PLC program, that's the same as pushing the button, so the conveyor stops. If it's wired normally open and the electrical connection between the button and the PLC is cut, you can push it as hard and as often as you want, but the PLC will never know, so the conveyor keeps on running. Those facts argue for wiring the stop button normally closed.

For the other mode of failure I described (pushing the button doesn't change the state of the contact), it doesn't matter whether the stop button is wired normally open or normaly closed.

Big honkin' disclaimer
The preceeding discussion is not intended to be interpreted as claiming that all stop buttons wired to PLC inputs should always be wired normally closed. While that is common practice, every situation is unique and should be evaluated on its own merits. The discussion is intended to illustrate some of the factors to be considered.

Jessica

Does your PLC run your conveyor or just the ancillary parts of the process? If the PLC runs your conveyor, it should start when the Start button is pushed because the logic is telling it to. If the PLC is powered, it should be scanning the logic already, waiting on you to tell it to start the conveyors, and perform the other functions you desire. The Stop button should be wired to the PLC input card and programmed to stop the entire process except for scanning. Make sure not to use this Stop button as an Emergency Stop button which must be hardwired.

Two for one--Steve and I posted at the same time.

My 2 cents,

Jessica,

The answier to your question depends on if your START button is to be used to set up the unit for normal run, or to be used as an emergency "push to stop" button.

We all think about safety alot and dont like blood & guts on our personel files. So in you excersize the Start button just might be the Start button and not an emergency stop device.

An emergency stop button can be a momentary button that will close only while you are holding it, (like a door bell button). Or it can be a big fat button that will be a "push to stop" type button. Either type of buttons will cut most or all power to the driving, or moving, units of the conveyer so that nothing happens even though the PLC may be outputting to a motor or coil. This can be done buy wiring the E-Stop button to feed the PLC output commons. That way even though the outputs may be on, there's no power to apply to the drive.

In the case that your "start button" is actually a start button. It would be wired so that when it is open the process wont run via your PLC logic alone. And when you close it, the process can start. So the start button when closed will place power on the desired PLC input.

Within your PLC code, the start button contact will close a circuit that will also require the photo sensor to look for the object start your timers etc...

Did you do your flowchart yet??? Lets see it.

Mike

The way I've often seen it done.

jessica said:

...i want to know if sensor comes before push botton or after and feeder should eb hard wire close or open

Click to expand...

Conceptually, the sensor comes after the button. On a bare bones system, the start button, in concert with a few other items, drives a 'system on' bit (coil), though it goes by many names. This SYSTEM ON bit then enables all the other logic as needed.

What do we need to do this?

1. stop circuit normally closed input - It sounds goofy but it may help to think of it like this: "If it is true that the stop signal is not asserted"
2. start signal normally open input
3. system on output

How is it constructed? Convert this to ladder logic:

IF {start input TRUE .OR. IF system on output TRUE} .AND. stop input TRUE (remember, it's normally closed) THEN system on output TRUE

BTW, this circuit is one which looks identical to its hardwired relay counterpart. It even has a name.

in this project i dont care about safty all i want to think is its perfect world so no ifs and buts so in this

1. i start the system when i step in my office
2. placed the box infront of sensor
3. conveyer runs for 5 sec
4. fedder #1
5. 2 sec conveyer
6. 5 sec delay
7. 3 sec fedder #2
8. 2 sec cylinder
9.reset and wait for box (back to step 2.
my question is how do i set step number 3 and 4 and #6 (5 sec delay

i like to keep my stop anf start buttons normally open contacts is that ok ? give me suggestions plz

jessica said:

i like to keep my stop anf start buttons normally open contacts is that ok ?

Click to expand...

Based solely upon your previous "in this project i dont care about safty" statement, the answer is yes.

beerchug

-Eric

Do you have a plc and software to attempt this?

You have alot of ideas already. Can you post your code?

If this is a class assignment you should say so. You have to understand that there are 10 million ways to do this.

You will not get code here for this if you do not try.

P.S. Please do not attempt code on real world applications.

This is not that much different than your previous traffic light project. The traffic light had to continuously cycle. This one runs through its cycle once and then waits for the external sensor before starting through it again.

i like to keep my stop anf start buttons normally open contacts is that ok ?

Click to expand...

As I mentioned in my earlier answer, you can write ladder logic to control the 'Run' state no matter which way the start and stop buttons are wired.

Steve Bailey said:

This is not that much different than your previous traffic light project. The traffic light had to continuously cycle. This one runs through its cycle once and then waits for the external sensor before starting through it again.

Click to expand...

Steve, This is the traffic light program. Did you ever notice the stop buttons they are installing at crosswalks these days? :nodi:

Jessica, Can you post your code so we can comment on it? We'd all love to see it. If your unsure how to draw code check here: http://www.plcs.net/ladder.txt

Jessica,


In your program the input will effect the output as below:

Stop button open = PLC output off = Run On



  Stop Btn

-----|/|-----------------(RUN)







Stop Button Closed = PLC input on = Run on



  Stop Btn

-----| |------------------(RUN)

Does that clear things up a little?