Boolean Convert to Integer value

Hi, Vince,

What PLC are you using?

I am using siemens S7-300. Sorry for late reply :)

000000000000 = 0

000000000001 = 1

000000000010 = 2

000000000011 = 3

000000000100 = 4

000000000101 = 5

000000000110 = 6

000000000111 = 7

000000001000 = 8

000000001001 = 9

000000001010 = 10

000000001011 = 11

000000001100 = 12

?????

Regards JD.

I understand exactly what you meant, it was also what our Blogger intended us to understand I believe too, but some of our other CR4 brethren got confused with normal binary!!!!!!!

***Forgive them as they know not what they do!!!***

Or you could call them Chumps!!! But the Yanks among them may not understand that either!!!

Start with your output device. What indication do you want for your integers? If you want something like a 7 segment display then you will need to generate the correct code for each integer. Then set up a table like this:

000000000001 = code for 1

000000000010 = code for 2, etc.

Then you can do a Karnaugh map or Quine-McCluskey logic reduction for each bit of the code, and that will tell you what goes in your PLD. Assuming you don't have PLD sw that will do that reduction for you.

http://en.wikipedia.org/wiki/Karnaugh_map

http://en.wikipedia.org/wiki/Quine%E2%80%93McCluskey_algorithm

Do you only use only the PLC instruction set or are you different software to set the PLC?

the easiest way i to add it with a masked bit

Use a BCD to integer converter function in the PLC. This will solve your problem.

Move all the 12 bits into a temporary register. Then use the BCD to integer converter function to convert the Temporary register value to Integer value.

Here's an algorithm using SHIFT RIGHT (SRW). No error checking is done; it is assumed that only one bit is set in INPUT.

[Note that this will loop forever (or until RESULT overflows & causes a crash) if INPUT is negative (i.e. high order bit is set) on entry. Edit - zeros are shifted in from the left.]

1. RESULT=0
2. INPUT =0 ? Y, DONE
3. RESULT = RESULT + 1
4. SRW INPUT
5. GO TO 2

There are lots of ways of doing it in a loop - e.g. you could use CC1, which holds the bit shifted out after the SRW.