Help me understand the jump instruction JBI

Hi Everyone

Could anyone of you help me understanding the jump instruction JBI.

It is believed to be working when BR=1

I am going to write my code here---

L 2

L 64

*I

JBI OVER

A I 1.6

A I 1.7

=Q 5.0

OVER:A I 2.1

A I 2.2

=Q 5.1

as I understand that programme should not run which is mentioned between jump instruction if  the BR is one rather it should jump to level over.

I multiplied two values 64 and 2 which gives us 128 decimal and equivlant binary 10000000. it means 8TH BIT br is 1 but it still does not jump to the level OVER and keep scanning all the  program in sequence.

any idea why<?

------------------------------------------------------------------------------------------
Split from PLC Program Help.

Manmohan Sharma

Hi Everyone

Could anyone of you help me understanding the jump instruction JBI.

It is believed to be working when BR=1

I am going to write my code here---

L 2

L 64

*I

JBI OVER

A I 1.6

A I 1.7

=Q 5.0

OVER:A I 2.1

A I 2.2

=Q 5.1

as I understand that programme should not run which is mentioned between jump instruction if  the BR is one rather it should jump to level over.

I multiplied two values 64 and 2 which gives us 128 decimal and equivlant binary 10000000. it means 8TH BIT br is 1 but it still does not jump to the level OVER and keep scanning all the  program in sequence.

any idea why<?

To be precise the BR is the 9th bit. But in this case it doesn't matter. As the BR is placed in the status word of a CPU. And the result of your arthmetics operation is placed in the ACCU1. And these are two different places.

Do some reading in the online help - STL language description on topics: CPU registers, status word and accumulators.

And I'd discourage you to try something like: 

L 2

L 64

*I

T STW

JBI OVER

...

in order to "correct" your code.

Mr. Jacek:

jacek d

And I'd discourage you to try something like:

Perhaps you mean to use the word encourage (positive) instead of discourage (negative)?

Very clever trick for using the Status Word (STW) as you are suggesting.

Mr. Sharma:

Manmohan Sharma

L 2

L 64

*I

JBI OVER

What is the design purpose of your code segment? You are showing us a room through a key-hole so it is a miracle if someone makes a winning suggestion. 

Multiplying two constants with each other makes no sense. The BR (bit result) bit is not effected by multiplying. Could you explain?

Regards.

#dP

Mr. Jacek:

jacek d

And I'd discourage you to try something like:

Perhaps you mean to use the word encourage (positive) instead of discourage (negative)?

Very clever trick for using the Status Word (STW) as you are suggesting.

This was a warning in case he would find this instruction and come up with such a bright idea.

Whatever he is trying to do he is doing wrong IMHO.

jacek d

 Whatever he is trying to do he is doing wrong IMHO.

 I agree with your opinion. At the same time it is an invitation to Mr. Sharma to return to start a brainstorming conversation.

If we talk about humble opinions - I am of the humble opinion that Siemens should promote you to Expert status. Your opinions are valuable and based on expertise.

New question published by EGYMAN is split to a separate thread with the subject STL - When try to convert it to ladder it doesn't happen.

Best regards
Jen_Moderator