7 segment display on step 7 using ladders

[MIDD]

I have a basic understanding of truth tables and the addresses. My issue is that for every letter that I have to make, I need a truth table that represents A B C therefore the combinations of the word "CONEJO" have to be parallel to the times I find which is true and false. For example, letter C in boolean has to be 1001110 corresponding to abcdefg but as I go down to the other letters, O N J... they have to be accounted for and then simplified. In the symbol table, I need to include ABC from the truth table for how long my word is and the outputs have to (I believe) be equal to the 7 segments. The simplified version of C or as I'm calling it F(a)=A'B'C'+A'BC+AB'C. From here, I have to consider the display format, checking which segments will turn on and other will keep off. In C, A F E D will turn on while B G and C will keep off. My other issue is that for the amount of F(x) are simplified and added to the Networks, I don't know the next procedure.

 

[drbitboy]

Welcome to the forum!


What is your question?


What brand of PLC are you using?

 

[MIDD]

It's more towards simulation, how would you create a word from ladders and convert it to a one seven segment display, simplifying the letters without causing too much chaos.


I don't recall, I think it was festo but Im not sure which model

 

[JesperMP]

I think you are using S7-200 and Microwin.
Anyway.....
This seems to be a homework assignment.

I need a truth table that represents A B C therefore the combinations of the word "CONEJO" have to be parallel to the times I find which is true and false

What has this to do with a 7-segment display ?

In the symbol table, I need to include ABC from the truth table for how long my word is and the outputs have to (I believe) be equal to the 7 segments.

????

I suggest you post the entire assignment. It is OK if it is in spanish. There are people here that understand spanish very well.
Do not expect to get a finished solution. You will learn nothing if someone just gifts you the solution.
But hopefully someone can give you a push in the right direction.

One thing I can say already, the way to find a solution is to have the task clearly and logically defined to begin with.

 

[MIDD]

Display step 7

True, it's an assignment that I will be making physically. It's been difficult getting help from the teacher and my classmates. Anyways, I hope this clarifies what I've been trying to understand.

 

[drbitboy]

what are A (or AA), B, and C?

what is "CONEJO?" Is it spanish for rabbit?

What is the relationship between [A(or AA),B,C] and "CONEJO?"


What do you want to display on the 7-segment display? Do you want to display either "A" or "B" or "C" only, or are O, N, E, and J other possibilities?

How many 7-segment displays are involved?

 

[drbitboy]

Ah, the penny drops: CONEJO is your name!


And you want to put C in the 7-segment display when the input pattern is 1010b (0xA), O when the pattern is 1011b (0xB), etc.?

 

[MIDD]

1.It's one segment display that has to produce every individual letter. so to simplify the process ABC of every boolean combination. just your typical truth table. ABC=000 and final number 0; 001 ->1; 010 ->0; 011-> 1 and there even more for abcdefg.

2. It's rabbit, I had to make a word longer than 4 letters. some did seven and even more. so im behind on work.

3. I just want to understand the organization of the simplified versions of F(a), F(b), F(c) and so on. where should I place them and why.

 

[drbitboy]

The assignment lists four input bits (ABCD => 0-16 decimal => 0x0-0xF => 0000b-1111b), and displays the decimal digits 0 through 9 for the first ten combinations.

You

• are only using three input bits (ABC => 0-7 decimal => 000b-111b), and
• are not displaying the decimal digits 0 through 9?

 

[JesperMP]

I also get it now.
There are some assumed limitations of the assignment. Only a small subset of the alphabet can be displayed.
For example "M" is impossible, and "N" only possible if you assume that П is an "N".
I.e. "COПEJO"
To distinguish between D and O, you may have to display as "d" and o".
K is also tricky. As is W.

It is fairly simple. You create a truth table with the letters on one side and the 7 segments A-G on the other side. Fill in with zeroes and ones that gives the best representation of the letters (with its limitations mentioned above). You don't necessarily have to reduce the truth table.
You can create a function with a ladder of the truth table; the function receives the single letter and spits out the 7 segment bit pattern.
Then call that function as many times as there are letters in the word. Let there be a delay time after each call so you have time to see each individual letter.

 

[drbitboy]

It is fairly simple. You create a truth table with the letters on one side

I wonder if this is the confusing part: do you mean the displayed letters (e.g. C in CONEJO) or the discrete inputs' tag names' letters (e.g. C in ABC)?

The discrete input bits' tag name letters A, B, and C have nothing to do with, or stated another way are in a separate domain than, the letters displayed on the 7-segment display. So the "C" displayed has nothing to do with to the third discrete input bit, which has the tag name of "C" assigned to it.



Maybe that is not the confusing part, but it may be simpler to name the discrete input bits I2, I1, and I0.

 

[JesperMP]

Maybe the task is to only generate the 7-segment bitpattern for the fixed word "CONEJO" when a button is pressed.
But I am imagining that it could be any 'word', i.e. STRING that must be able to be displayed.
If that is the case, then you need to take all the letters, from A to Z, then try to find a 7-segment combination that fits the 'best' for each letter. Granted some of these will be difficult.

One thing that is confusing that the source of the word (i.e. STRING) is not described at all.
I tried to read the 1st post, but it doesnt make much sense. What are AA, B and C ?
Instead I imagine what it could be. And I imagine that in some way a STRING is input in some way on an HMI, or one or several existing STRINGs are selected by a button or switch.

With the active STRING, you can loop through each character and feed it to the generic converter to get the 7-segment combination for each letter of the STRING.
i.e.
"mystring"[0] --> converter --> bitpattern
"mystring"[1] --> converter --> bitpattern
"mystring"[2] --> converter --> bitpattern
etc.

 

[widelto]

Ascii to seven segment table exists just look for it.
In Allen Bradley PLCs , there is an instruction called SQO
https://control.com/technical-articles/common-sequencer-functions-sqo-sqi-sql-sqc/

You can use it from CONEJO letter by letter, you can even use just the part you need, C,O,N,E,J. ( O is repeated)
The logic is very easy just take letter by letter ( use compare the use SQO) and the seven segment display will show letters as needed.

In step 7 I do not know if that instruction exist, if negative . You have to create your own logic. Use JESPER advise, that is the way to go.