Step 5 Ladder Logic Question

B

Thread Starter

Bill Code

<P>Good day, Siemens experts.</P>

<P>I am supplying a retrofit; replacing Siemens Step 5 PLC with ControlLogix. Generally, the machinery is not difficult to understand with the
exception of a few details I am reverse engineering from the Step 5 PLC program.
I do not have experience with the Step 5, but I thought I was generally able to read and understand the code. Except for the following logic:</P>
<PRE>
Segment 66

-I34.1 -I34.2 -F128.4 -I34.3 -F128.0
+---] [---+---] [---+---(#)---+---]/[---+---------+--( )-+
! !
!-T86 !
+---]/[---+

Segment 67

-T86
-F128.4 +-----+
+---] [---+-!T!-!0!
!KT 050.0 --!TV BI!
! ! DE!
! ! !
! ! !
! +-!R Q!-
! +-----+
</PRE>
<P>My pea brain thinks that F128.4 is an edge trigger that passes power while devices on I34.1 and I34.2 are closed. So, continuing along, F128.0 is true while device I34.3 is open or T86 is not timed out. Seems simple.</P>

<P>But in 67, if I understand the Step 5 manual, I see the timer never timing out. I think that T!-!0 describes an on-delay timer that times for 50 * 0.01 = 0.5 seconds while T!-!0 is true. But it will never see the edge trigger F128.4 long enough
to allow that to happen. If the -] [- instruction in 67 was -]/[- it would make sense to me.</P>

<P>It looks to be a minimum time limit on the period that output F128.0 is true, but I can't see it.</P>

<P>What am I missing here?</P>

<P>Bill Code<BR>
604-513-8049<BR>
KR Consulting Ltd.<BR>
Surrey, B.C., Canada<BR>
 
C

Corne, Chris

Bill,

In segment 66, F128.4 is an intermediate coil. It will be high (or made) when the logic before it is satisfied. It's a method of holding the RLO of that point for use later in the program. So when I34.1 and I34.2 are made, F128.4 will be high so in segment 67 T86 (an ON DELAY timer) will begin timing out. It's "status bit", T86, will be off for .5 seconds, then high for as long as F128.4 is still high. Since segment 66 is scanning for zero on the T86 status bit, .5 seconds after I34.1 and I34.2 are high, F128.0
will be the inverse of I34.3.

Chris
 
W

Wayne Williams

<P>the (#) represents an intermediate coil.</P>

<P>It is the same as putting a coil in the middle of a rung i.e.</P>

<PRE>
--| |------| |-------( )----| |----| |-----( )
</PRE>
 
A

Andrew Hawdon

Hi
F128.4 can be thought of as a coil in the middle of a rung. In your case it will have the status defined by I34.1 & I34.2. In segment 67 the timer will start to time as soon as I34.1 & I34.2 are high. You are correct re the time of T86 and the fact that is an on delay. Re one shots in step 5. There is no one shot contact, which means that one shots have to be built by the programmer. The fact that you are looking at a one shot is therefore not always obviouse. If you need any other info email me at [email protected]

Andrew
 
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