Balluff BTL7-T500 Series, Configuration Manual

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4
Configuration (continued)
4.5 Baud rate and transmission time
PROFIBUS systems can be operated at baud rates of
between 9600 bps and 12 Mbps. The higher the baud rate
the shorter the permissible cable lengths and the more
sensitive the system is to reflections and interference
(connector and cable quality). The lower the baud rate the
less data can be sent within a particular time. Estimating
the amount of data helps in setting the optimal baud rate.
Baud rate (Kbps) Permissible cable
length / m
(cable = BKS-S103…)
Sum of all stub
cable lengths / m
Length of a word
T
Word
/ µs / Word
Basic overhead for
cycle T
MC
/ µs
9.6 1200 500 2290 41700
19.2 1200 500 1140 20800
45.45 1200 100 484 8800
93.75 1200 100 235 4270
187.5 1000 33 117 2130
500 400 20 44 800
1500 200 6.6 14.7 267
3000 100 not allowed 7.33 133
6000 100 not allowed 3.66 66.7
12000 50 not allowed 1.83 33.3
Tab. 4-4: Cable lengths and transfer times as a function of the baud
rate.
An exact calculation of the message/bus cycle time can be
found in IEC 61158. Using the above table the transfer
time for operation with just one master can be estimated.
Acyclic communication, multi-master operation, internal
processing times for the master as well as telegram
repetition when there are interference and reflections are
not considered. If there are multiple slaves on the bus, a
similar calculation must be performed and all times added
together.
To request a data telegram from the BTL7-T500… at least
228 bits are required for Header, Request and Response.
In addition there are 8 bytes of output data (STW2 &
G1_STW1) and around 108 bit times for the pauses
between telegrams. Also the length of the input telegram
which needs to be separately determined – a total of 400
bits. A byte is sent on the bus with 11 bits (8 data bits +
start bit + stop bit + parity bit), so that 22 bits are needed
for a word.
Thus the time for cyclic data exchange can be calculated
using the following formula:
T
cycle
= T
MC
+ (number of InputWords × T
bytes
)
1st example:
Query standard telegram 81 cyclically at a baud rate of
12 Mbps. 2 bytes of output are then sent to the slave and
6 bytes of input data to the PLC. The time for the net data
is then 33.3 µs + (1.83 µs / word × 6 words) = 44.3 µs.
2nd example:
When transferring 16 positions the data quantity is
38 words. At 9600 bps you require
41.7 ms + 38 words × 2.29 ms / word = 128.7 ms.
BTL7-T500-…
Configuration Guide