ladder help

jtashaffer · May 3, 2012

When writing a program using 3 spst switces to trun on a light, all are interlocked so only one can be on at a time. How would I use a bit to interlock them insted of the inputs?

rootboy · May 4, 2012

jtashaffer said:
When writing a program using 3 spst switces to trun on a light, all are interlocked so only one can be on at a time. How would I use a bit to interlock them insted of the inputs?

If by this you mean that if any two switches are on, then the light is off, then this is how I would do it.



I1 I2 I3 Light

---] [-----]/[-----]/[-----------+------------( ) 

|

I1 I2 I3 |

---]/[-----] [-----]/[-----------+

|

I1 I2 I3 |

---]/[-----]/[-----] [-----------|

jrwb4gbm · May 4, 2012

 

Redrawn diagram that rootboy posted.

 

    I1      I2      I3                        Light

---] [-----]/[-----]/[-----------+------------( ) 

                                 |

    I1      I2      I3           |

---]/[-----] [-----]/[-----------+

                                 |

    I1      I2      I3           |

---]/[-----]/[-----] [-----------|

wimpiesplc · May 4, 2012

this still uses the inputs to interlock.

How would I use a bit to interlock them insted of the inputs?

But what exactly do you mean? Why use a bit instead of the inputs?

daba · May 4, 2012

But the Inputs are "bits" in the PLC data-table.

The logic as posted is sound, and is often used AS WELL AS interlocked switches or pushbuttons wired to inputs to guard against switch contacts shorting out.

TConnolly · May 4, 2012

I would use the method Rootboy has suggested....

However, if this is a homework assignment then the exercise might be intended to help you with developing and simplifying logic.

In which case lets take a minute and develop a truth table for conditions where an interlock bit might be set.



 SW1  SW2  SW3  INTERLOCK

  1    0    0      0

  0    1    0      0

  1    1    0      1

  0    0    1      0

  1    0    1      1

  0    1    1      1

  1    1    1      1

Now lets write the equation:

Interlock = (SW1 + SW2) + (SW1 + SW3) + (SW2 + SW3) + (SW1 + SW2 + S3)

We can simplify that

Interlock = (SW1*(SW2 + S3)) + (SW2 + SW3)

Now you can develop the output equation.

Light = (SW1 + SW2 + SW3) * NOT Interlock.

I'll leave it to you to convert these into ladder code, after all if it is homework I don't want to do it all for you.

jtashaffer · May 4, 2012

I mean use a binary bit. B3:0 as the interlocking bit

daba · May 4, 2012

B3:0 is not a binary bit.

It is a word address - B3:0 contains 16 binary bits

The addressing syntax is....

Filetype Filenumber : Element / Bit

A unique bit address would therefore be B3:0/0

i.e. bit 0 in word 0 of binary file 3.

Alternatively, the programming software allows you to skip the word address... B3/49

is bit 49 in the binary file B3

it is the same address as B3:3/1

the maths is easy - word address is bit address/16, remainder is bit address

B3:0/0 is the same bit as B3/0

TConnolly's post is good thinking, it provides you a means to generate an "interlock" status when more than one input bit is on. I would also program the interlock status should be on if NO inputs are on. You are simply using combinational logic to verify the switches are valid.

jtashaffer · May 4, 2012

Ok the internal relay bit is B3:1/0. It is a class but for work training. I am learning rs5000. I can write the program like you shouwed but I dont understand how to use the internal relay bit B3:1/0 as the interlock.

Lancie1 · May 4, 2012

With switches 1 and 3 ON, Tim's interlock would look like the attached picture.

I think Tim meant this to be the Boolean equation:

(SW1*(SW2+SW3)) + (SW2 * SW3)

Eric Nelson · May 4, 2012

Lancie1 said:
With all 3 switches on, Tim's interlock would look like the attached picture.

Actually, shouldn't the lower rung be I:1/2 AND I:1/3, not I:1/2 OR I:1/3?

With your OR branch, the interlock bit will be on if switch 2 OR switch 3 is on.

-Eric

EDIT: Oh sure, Lancie... Fix your picture on me...

jtashaffer · May 4, 2012

Then use the binary bit on the next rung to turn the light on

iant · May 4, 2012

I am wondering Why you need to do this?
If you are training and learning.
Maybe consider the MVM (move with Mask) command (of the top of my head)
then use the B3:0 bits and interlock them to the output

rootboy · May 4, 2012

jrwb4gbm said:

So after all of these years on PLCS.NET, I *still*can't get the forum to quit eating empty spaces.

What's the secret?

Lancie1 · May 4, 2012

EDIT: Oh sure, Lancie... Fix your picture on me...

Sorry, Eric. I first tried to use Tim's logic, but then realized it had an error, so had to figure out how to fix it.

ladder help

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