<P>I'm in a controls class in college, and I'm trying to figure out the logic for our first homework. Heres the problem statement.

<P>1. After pressing start button (R000.2) the program will begin running motor #1 (R015.0)

<P>2. Motor #1 runs until limit switch 1 (R 000.0) is pressed. When LS1 is detected, motor #1 turns off and motor #2 turns on.

<P>3. Motor #2 runs until LS2 (R000.1) is detected. When LS2 is detected, motor #2 turns off and motor #1 turns back on.

<P>4. The cycle must be self sustaining until stop button(R000.3) is pushed. All switches are Normally Open, momentary switches.

<P>Here's the logic that I have come up with... but we only have one shot at getting it right due to time constraints in class. If we get it wrong we have to get this one working, plus have another one done for the next week also.

<P>Here's what I've come up with, can anybody tell me if my logic is correct??
<PRE>
| Start Stop
|--| |------|\|----(run)--------|
| | |
| run | |
|---| |----| |
| |
| |
| |
| run LS1 |
|--| |------|\|----(Motor 1)----|
| | |
| LS2 | |
|---| |----| |
| |
| |
| run LS2 |
|--| |------|\|----(Motor 2)----|
| |
| LS1 |
|---| |----|
|
|
</PRE>

 

That is a good start, but as it is BOTH motors will start right away.
After LS1 is made Motor 1 will shut off, but only until the limit is no longer made, then it will continue running along with Motor 2 again.

It looks like you need to think this through just a bit more. You are very near the solution so that is all the advice I will give for now.
If you need more after trying something else let us know.

-mjv

 

Hello Mark:

Being a homework assignment, I should give you the opportunity to learn, rather than just giving you the answer. Here's what will happen with your
current circuit:

Rung 1:
When Start is pressed, the run bit is set and latched. Pressing the Stop button will break the circuit holding the run bit set, and therefore turn it off. So far so good.

Rungs 2 and 3:
If the run bit is set or LS2 is true, and LS1 is not true, motor 1 runs.
If the run bit is set or LS1 is true, and LS2 is not true, motor 1 runs.
Ignoring the or segments for the moment, this means that any time both limit switches are open (which is probably most of the time), both motors
run. With the or segments, a motor will run if a limit switch is true, regardless of the status of the run bit. That means the motors could
possibly run even if the run logis says that they shouldn't, which is an unsafe situation.

Look closely at rung 1 for a better understanding of how rungs 2 and 3 should work.

Best of luck,

David Wooden
Senior Software Engineer, Systems Integration
Automation and Enterprise Solutions Group
TAS Division of Omron Electronics LLC
Office: (847) 884-7034 Extension 432
Fax: (847) 884-9383
E-mail: [email protected]

 

The first problem I see is that if you push start and neither limit switch is made, both motors will run.

You might be better off with a run bit and a motor select bit.

Make sure the motor select bit is correct when you stop and restart running, but I'll leave that to you as to how it's done.

Rufus

 

Moin Mark
moin all,

> I'm in a controls class in college, and I'm trying to figure out the logic
for our first homework. ...

Why don't you try it yourself ?

>From Modicon/Schneider you can download the Concept 2.5 Trial Software.
http://www.modicon.com/

For a PC-based IEC 61131 software look at
http://groups.yahoo.com/group/oats1131

There are more.

Friedrich

 

I think that the first rung of your diagram is correct, since it allows you to activate the whole cycle pressing START, and to interrupt it using STOP; However, in the following lines, your active ones both motors in simultaneous form, for what your control philosophy won't be carried out in the requested form.
I suggest you that you build other more rungs, where you consider that the operation of each motor requires of the activated cycle.
It is well that you consider the contact run in both rungs, but this it should not be in parallel with another contact. This doesn't allow to this to be a required condition, since the logic can be completed although this it is not energized.
So that the coil of the motor 1 are energized, whenever LS1, and the motor 2 are disabled (that these they inhibit the operation of M1). This will allow you the logic to begin activating this motor like first option. Do you remember that the motor 2 and the LS1 are inactive during this first sub cycle.
This way you will be able to energize the motor 2 with the appearance of the same condition that stopped the motor 1, the activation of the LS1. The operation of this motor should be conditioned to that the motor 1 off are and that run is active.
My answer would be more clear in Spanish

 

I think you should build a SFC first,then you can see the control process clearly. It's easier for you to finish your work.

 

The problem statement says the limit switches are all normally open contacts.....I see some normally closed contacts in the logic????

Gerald Beaudoin

 

You don't have the right inputs. The problem states that all inputs are normally open, momentary contact. You used normally closed switches for Stop, LS2 and LS1.

 

The problem with your description is, you haven't defined the behavior of the motors under all of the possible states of the limit switches.

There are four possible states of the two limit switches; both on, both off, and either one on with the other off.

Furthermore, you haven't defined the cause/effect relationship between the motors and the limit switches. For example, does motor #1 cause limit switch #2 to open before limit switch #1 closes? Or are the actuations of the limit switches independent of the state of the motors? Can both motors be on simultaneously?

The easy part of designing ladder logic is when everything happens in its normal sequence. The real challenge lies in designing logic that gracefully handles all of the ways things can go wrong.

 

I understand now that the way I have it, once a limit switch is no longer being pushed it turns right back on... but I've been racking my brain trying to figure out how to keep the power off and I can't seem to figure it out....

 

You should assume the field contact is normally open, not the PLC internal contacts.

 

How's it going? If you notice point 4, "......All Switches are normally open, momentary switches" This will change your ladder some what. Your ladder also would allow M#1 and M#2 to run at the same time, assuming SL1 and/or SL2 were not high.

Gus

 

Thanks guys!!

Figured it out!! It took a while, but I finally got it. Thanks for your help, I may be back on future assignments if I need more help.

Thanks again!
Mark

 

Mark,

I'd be interested in your final approach, and I'm probably not the only one. Care to post it?
(There's more than one way to rung a ladder...)

Rufus

P.S. Be sure to say:

This is just to show what I did; not to spawn new threads about how it could have been done.

 

Glad to see that Mark got his program figured out ..."with a little help from my friends". It was refreshing to see that he put effort into solving the problem first, and THEN asked for help. Quite often there are posts that want the gang to do their assignments for them.
Good luck, Mark, .....as my mother told me....."the world always has place for a worker"

Gerald Beaudoin

 

<P>I don't have a copy of it with me right now.. but I'm pretty sure it went something like this:
<PRE>
| Start Stop
|--| |------|\|----(run)--------|
| | |
| run | |
|---| |----| |
| |
| |
| |
| motor1 Run LS1 |
|--| |-----| |-|\|---(Motor 1)--|
| | |
| LS2 | |
|---| |----| |
| | |
| Start | |
|---| |---- |
| |
| motor2 Run LS2 |
|--| |-----| |-|\|--(Motor 2)---|
| |
| LS1 |
|---| |----|
|
|
</PRE>
<P>P.S. I have another problem now. I have posted it as a fresh topic.</P>