Simple Lader Logic Question! PLEASE HELP!!

[ankitj]

RUNG 1 : [A]-----(B)
RUNG 2 : [B\]----[A\]-----(C)
RUNG 3: [C]-----(DLatch)
RUNG 4: [D]------(A)
RUNG 5: [C]-------(DUnlatch)
RUNG 6: [E]--------(E)

Scan
​

A
​

B
​

C
​

D
​

E
​

0
​

False
​

False
​

False
​

False
​

False
​

1
​

2
​

3
​

4
​

5
​

I need the values for A,B,C,D,E after 5 scans. I don't understand how if works with multiple scans. PLEASE HELP!!

 

[mvandyke]

This has been asked before:
http://www.plctalk.net/qanda/showthread.php?t=25722&highlight=scans

 

[goatom]

Scan

I followed the provided link and my questions is; are we to assume that no one changes the state of any of the inputs? I am a bit new to this stuff too and although the ladder is simple the question leaves me puzzled on what I am to assume.

 

[ankitj]

Hi goatom,

Thanks for replying. That's exactly why I am confused. I have not been given any more information about this either. I was wondering the same thing as to what actually causes the states to change at each scan. Does the latching and unlatching change the values?

 

[ankitj]

I followed the provided link and my questions is; are we to assume that no one changes the state of any of the inputs? I am a bit new to this stuff too and although the ladder is simple the question leaves me puzzled on what I am to assume.

Hi goatom,

Thanks for replying. That's exactly why I am confused. I have not been given any more information about this either. I was wondering the same thing as to what actually causes the states to change at each scan. Does the latching and unlatching change the values?

 

[ankitj]

thanks mvandyke. that's a great start.

 

[rsdoran]

I followed the provided link and my questions is; are we to assume that no one changes the state of any of the inputs? I am a bit new to this stuff too and although the ladder is simple the question leaves me puzzled on what I am to assume.

Does the assignment state something changes externally? Organize your thoughts, hopefully you have the ladder displayed better then it was above.

The idea may be to evaluate how much you have learned (at least retained) of the information provided.

LETS ASSUME, the plc has this code in it, noone has done anything i.e. no inputs or outputs are attached nor did the instructor or anyone else state anything was done. What will happen each scan? What happens after 5 scans? Here is a picture to show what it would look like when plc is started i.e. at beginning of first scan and each sucessive scan til the 5th scan?

 

[goatom]

I am mostly trying to learn PLC programming on my own and with the help of this site ( thanks to folks who take the time to pound things into my head) soooo. It is my understanding that nothing changes in the rung unless there is a I/O change of state. So if Input A is closed then output B fires. Just using the first rung I cannot see what would change after 5 scans or 50 scans if input A does not change state. Please explain if I am overthinking this problem or not thinking at all.

 

[Martin T.]

Look at rung 0001 closely. What will happen on the first scan?

 

[Steve Bailey]

Think of each rung as an equation. Starting with all five bits off at the beginning of the first scan, solve each equation. Some of the bits may change state more than once within a single scan. Don't worry, that's OK. The condition of the bits at the end of the first scan is the condition at the beginning of the second scan.

Can you see how the condition of the bits at the end of the first scan will be A and C TRUE, the rest FALSE? If you can see that, keep going until you either reach a stable state or you can demonstrate that the bits will keep changing ad infinitum.

 

[Ken Moore]

First Scan:
Read input/output image table, then scan logic.
A is False, so B is False
A & B are both False, so C is True
C is True, so D is Latched
D is True, so A is True
C is still True, so D is unlatched
E is False, so E remains False
Update image table

Second Scan:
A is True, so B is True
A & B are both true, so C is False
C is False, D is not latched
D is False, so A is False
C is False, so D remains false
E is False, so E remains false
update image table

Third Scan:
A is false, so B is False.....
Same as first scan

See the pattern?

 

[Ron Beaufort]

Greetings goatom ...

for educational purposes, let me answer your question with a question ... suppose that in the rung below, "D" has a status of "1" ... here's the question: what will the status of "A" be immediately AFTER the processor executes the rung? ...

[D]------(A)

...are we to assume that no one changes the state of any of the inputs?

hint: just because no ONE changes the state doesn't mean that noTHING does ...

 

[goatom]

Don't know where the orginal poster went and I did not mean to take over his question, but I hope you are following this. To answer Martin's question: If A and B are open then C will fire ( activate) or whatever the correct terminology is.

Steve: Thanks for writing " view each rung as an equation." I got a D in high school algebra but straight A's in college Alebra and Trig. Must have been because I had to pay for the class?????

Ken and Ron I will study your input next.

Thanks gentlemen and/or any not so gentle men.

 

[Ron Beaufort]

... from the boot camp ...

Greetings goatom ...



sometimes the terminology fights us ... I’d STRONGLY suggest that you avoid using the terms “True” and “False” and/or “Open” and “Close” to refer to the status of the bit/boxes on your processor’s data tables ... let’s just say that each of those bits/boxes can contain either a “1” or a “0” ...



with that in mind, here’s a better way to phrase exactly what’s going on as a rung gets executed ... consider the following sample rung ...

step-by-step ...



the XIC instruction tells the processor to: “go look for a 1 in the bit/box with the address B3/15” ... does the processor find a “1” in that bit/box? ... yes ... therefore the processor evaluates the XIC as “True” ... (note: if the processor does NOT find a “1” in the bit/box, then the processor will evaluate the XIC as “False”) ...



the XIO instruction tells the processor to: “go look for a 0 in the bit/box with the address B3/11” ... does the processor find a “0” in that bit/box? ... yes ... therefore the processor evaluates the XIO as “True” ... (note: if the processor does NOT find a “0” in the bit/box, then the processor will evaluate the XIO as “False”) ...



since “True” comes into the OTE instruction, the OTE instruction tells the processor to: “go write a 1 in the bit/box with the address B3/3” ... (note: if “False” came into the OTE instruction, then the OTE instruction would tell the processor to: “go write a 0 in the bit/box with the address B3/3” ...



(note: to any of my more-experienced colleagues who are completely comfortable with some other terminology (such as “true” and “false” for the status of bit/boxes in the data tables), I say: “if it works for you, then party on” ... but personally I’ve found over the years that MANY beginning students have been needlessly confused by anything other than the easy-to-explain status of “the bit/box contains a 1” or “the bit/box contains a 0” ... it’s hard to get any simpler than that - and then the “True/False” analysis of the XIC/XIO instructions works out correctly every time ...

 

[Ken Moore]

You're right on Ron.
I never considered that using "True" and "False" might confuse a beginner. I've been using those terms for sooooo long, that they are just second nature to me.
That's probably why you are an excellent instructor and I just muddy the waters.