Omron CQM1H - PROGRAM Programming Manual page 221

Basic Ladder Diagrams
TR 0
00000
Diagram B: Corrected Using a TR bit
TR 0
00000
196
00001
00002
In terms of actual instructions the above diagram would be as follows: The
status of IR 00000 is loaded (a LOAD instruction) to establish the initial execu-
tion condition. This execution condition is then output using an OUTPUT
instruction to TR 0 to store the execution condition at the branching point. The
execution condition is then ANDed with the status of IR 00001 and instruction
1 is executed accordingly. The execution condition that was stored at the
branching point is then re-loaded (a LOAD instruction with TR 0 as the oper-
and), this is ANDed with the status of IR 00002, and instruction 2 is executed
accordingly.
The following example shows an application using two TR bits.
TR 1
00001 00002
00003
00004
00005
In this example, TR 0 and TR 1 are used to store the execution conditions at
the branching points. After executing instruction 1, the execution condition
stored in TR 1 is loaded for an AND with the status IR 00003. The execution
condition stored in TR 0 is loaded twice, the first time for an AND with the sta-
tus of IR 00004 and the second time for an AND with the inverse of the status
of IR 00005.
TR bits can be used as many times as required as long as the same TR bit is
not used more than once in the same instruction block. Here, a new instruc-
tion block is begun each time execution returns to the bus bar. If, in a single
instruction block, it is necessary to have more than eight branching points that
require the execution condition to be saved, interlocks (which are described
next) must be used.
When drawing a ladder diagram, be careful not to use TR bits unless neces-
sary. Often the number of instructions required for a program can be reduced
and ease of understanding a program increased by redrawing a diagram that
would otherwise required TR bits. In both of the following pairs of diagrams,
the bottom versions require fewer instructions and do not require TR bits. In
the first example, this is achieved by reorganizing the parts of the instruction
Address Instruction
00000
Instruction 1
00001
Instruction 2
00002
00003
00004
00005
00006
Address Instruction
00000
Instruction 1
00001
00002
Instruction 2
00003
00004
Instruction 3
00005
00006
Instruction 4
00007
00008
00009
00010
00011
00012
00013
00014
Section 4-3
Operands
LD
OUT TR
AND
Instruction 1
LD TR
AND
Instruction 2
Operands
LD 00000
OUT TR
AND 00001
OUT TR
AND 00002
Instruction 1
LD TR
AND 00003
Instruction 2
LD TR
AND 00004
Instruction 3
LD TR
AND NOT 00005
Instruction 4
00000
0
00001
0
00002
0
1
1
0
0