old s5 converting 95u-cpu

L "MW 180" // Weiterschaltbedingung//register 1
L "MW 140" // alte Schrittnummer uebergeben//register 2
UC "FB 230" // Schrittkette auswerten//pass register 1 and 2

T "MW 140" // neue Schrittnummer uebernehmen//comes out fb 230//goes in fb 229
SET
= "Tag_44" // Keine Aenderung in SK
UC "FB 229"
T "DB 11".D_175 //result from fb 229

i had this code in tia portal 14 you can activate parameters passing via registers and inside the function the first instruction it was a load to a %mw

does any one knows if you pass parameters by registers the fb will recive them?

i altered the funtions to have input and output parameters but i had to change the uc to a call does it chages the way it works??

pls help i'm geting headaches

I would need to see the code in FB230 to answer how the parameters MW180, MW140 are and MW140 are handled.

fb 230 and 229

i print the two of them in a pdf. the project its long but i can send it to you. ty

OK, for FB230, MW140 is being transferred to "Tag_2" (line 0003)
It is being XOW'ed (Exclusive Or) with MW180 (line 0004) etc...

To follow this, every L (Load) instruction places the value in ACCU1 (accumulator 1), and the value that was in ACCU1 goes to ACCU2. Every T instruction (Transfer) takes the contents of ACCU1 to the memory location (or Tag).

Therefore when entering FB230, MW180 is in ACCU2 and MW140 is in ACCU1.

Line 0001 always jumps to line 0003.

In Line 0003 the contents of ACCU1 (MW140) is transferred to Tag_2

XOW is performed on ACCU1 (MW140) and ACCU2 (MW180), and the result is stored in ACCU1 (line 0004)

The L Tag_2 in line 0005 reads it into ACCU1 and the result of the previous XOW is in ACCU2.

XOW is again executed between Tag_2 and the previous XOW result and the new result is in ACCU1 and then transferred to Tag_19.

Same methodology is applied to FB229. MW140 is being transferred to Tag_2.
The T "DB 11".D_175 will depend on the condition of the AN Tag_x conditions.

Example Line 0002, if AN Tag_6 is false, it will not jump, and a value of 1 is loaded (L 1), and BEU exits the block, and this value of 1 will appear in DB 11".D_175

ty very much took into acount your logic and made the changes and it seems to work ty very much

hi again, do you know any form to translate the lir instruction?
in this plc they use

l 6500

lir 0

i have no idea where its supose to go i read the siemens info it dint help much

This is a tricky one...

L 6500 - this is loading a value of 6500 into ACCU1.

LIR 0 - Load Register 0 ACCU1-H (bits 16-31) with the contents pointed to by ACCU1 (in this case 6500)

I need to know your original S5 CPU part number, and a printout of the block in which it is being used in order to explain better.

NAME: SCHRITTK

JU =M000
NOP 0
M000: T FW 250 ;alte Schrittnummer uebernehmen
XOW
L FW 250 ;schneller Akkutausch
XOW
T FW 252 ;Weiterschaltbedingung uebernehm.
L KH 0000
!=F ;wenn Weiterschaltbed. = 0
JC =M001 ;dann zur Ausgabe
A F 252.0 ;wenn Resetmerker gesetzt
BEC ;dann 0 ausgeben und Ende
L FW 252 ;Weiterschaltbedingung als
T FW 250 ;neuer Schritt uebergeben
S F 250.0 ;Merker Verriegelung SK setzen
AN F 250.0 ;VKE auf Null setzen
M001: L FW 250 ;neue Schrittnummer laden
BEU ;und Ende
AW
L FW 250
OW
TAK
XOW
L KH 6500
LIR 0
JU =M002
L KH 6500

part number: 6es5 095-8ma04
e-prom 16kb

On page 573 of the following manual:

https://support.industry.siemens.com/cs/document/1091648/s5-90u-s5-95u-programmable-controller-system-manual?dti=0&lc=en-WW

The area where KH 6500 resides is simply called "Internal Data". This implies to me that the programmer decided to use this memory for his own purposes. You should find all references to this area (LIR/TIR) and change it over to Data Blocks.