Annex, Mapping in the bus coupler – BECKHOFF KL5051 User Manual

Page 15

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Annex

KL5051

15


Annex

As already described in the chapter terminal configuration, each bus termi-
nal is mapped in the bus coupler. In the standard case, this mapping is
done with the default setting in the bus coupler / bus terminal. This default
setting can be modified with the Beckhoff KS2000 configuration software or
using master configuration software (e.g. ComProfibus or TwinCAT System
Manager). The following tables provide information on how the KL5051
maps itself in the bus coupler depending on the set parameters.

Mapping in the bus coupler

Standard Format

The KL5051 is mapped in the bus coupler depending on the set parame-
ters. The terminal is always evaluated completely, the terminal occupies
memory space in the process image of the input and outputs.

I/O Offset

High Byte

Low Byte

Complete evaluation

= X

3

MOTOROLA format

= 0

2

D5

D4

Word alignment

= 0

1

CT/ST-3

D2

Default: CANCAL,
CANopen, RS232,
RS485, ControlNet,
DeviceNet

0

D1

CT/ST-0


I/O Offset

High Byte

Low Byte

Complete evaluation

= X

3

MOTOROLA format

= 1

2

D4

D5

Word alignment

= 0

1

CT/ST-3

D1

Default: Interbus,
Profibus

0

D2

CT/ST-0


I/O Offset

High Byte

Low Byte

Complete evaluation

= X

3

D5

D4

MOTOROLA format

= 0

2

-

CT/ST-3

Word alignment

= 1

1

D2

D1

Default: Lightbus,
Bus Terminal Controller
(BCxxxx)

0

-

CT/ST-0


I/O Offset

High Byte

Low Byte

Complete evaluation

= X

3

D4

D5

MOTOROLA format

= 1

2

-

CT/ST-3

Word alignment

= 1

1

D1

D2

0

-

CT/ST-0

Legend

Complete evaluation: The terminal is mapped with control / status byte.
Motorola format: The Motorola or Intel formal can be set.
Word alignment: The terminal is at a word limit in the bus coupler.
CT-0(A0): Control- Byte (appears in the PI of the outputs).
ST-0(E0): Status- Byte (appears in the PI of the inputs).
CT-3(A3): Control- Byte (appears in the PI of the outputs).
ST-3(E3): Status- Byte (appears in the PI of the inputs).
D1, D2, D4, D5 = A1, E1, A2, E2, A4, E4, A5, E5


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