PROFIBUS Manual

For single master systems, the minimum bus cycle time TBZyklus can be be calculated more precisely.

 

First we work out the duration of a PROFIBUS telegram:

 

Each byte in a telegram is transferred as 11 bits (see section on UART coding). Cyclic payload is transferred in an SD2 telegram without SAPs. One SD2 telegram without SAPs comprises 9 additional bytes for structure, addresses and error detection, i.e. 99 bits. Status queries are one SD1 telegram (no data) with 6 bytes and therefore 66 bits. A token telegram with SD4 comprises 3 bytes and consequently 33 bits (see section on telegram formats).

 

We therefore have the telegram lengths:

SD1_Telegram = 66 bits

SD2_Telegram = bytes x 11 + 99  bits (without SAPs)

SD2_Telegram = bytes x 11 + 121 bits (with SAPs)

SD4_Telegram = 33 bits (token)

 

The duration of a bit differs depending on bit rate: for 12 MBit/s tBit = 83 ns and for 1.5 MBit/s tBit = 0.67 µs.

 

We now define the cycle time for a DP message cycle:

 

where

TID1 = “Idle Time” until the master is ready for the next message cycle

TID1 ≈ 37 tBit (for masters based on ASPC2 ASICs)

 

TMessage cycle = TSYN + 2 x SD2_telegram + typeTSDR + TID1

TMessage cycle = 33 + 2x(bytes x 11 + 99) + 32 + 37

TMessage cycle ≈ 300 + bytes x 11 tBit

 

The bus cycle therefore becomes

 

TBCycle =  (TToken + TGAP + Slave x TMessage cycle ) tBit

 

where the following applies for individual values:

 

TGAP is the time required to execute a GAP update. A worst-case scenario is assumed here, in which no bus user responds and it is therefore necessary to wait throughout TSL:

TGAP =  TSYN + SD1_telegram + TSL

 

TToken        is the time required to send a token:

TToken =  TSYN + SD4_telegram + TID2

 

TID2 is the idle time until the master is ready to send a token, typically 150 tBit

 

Bytes        = number of input and output data bytes overall

Slaves        = number of slaves

 

With TSL the maximum wait time for a response is approximated. This time depends on the bit rate and is approximated here with an absolute time of  75 µs

 

This formula does not take into account any diagnostic messages, telegram retries, additional acyclic telegrams, nor any additional masters. It therefore only represents an ideal, single master system.