PROFIBUS Manual
To simplify the estimation of PROFIBUS-DP bus cycle time, findings set out in the preceding chapters can be reduced to the following formula:
TBCycle = (380 + slave x 300 + bytes x 11 ) x tBit + 75 μs
or directly with the bit rate
TBCycle = (380 + slave x 300 + bytes x 11 )/bit rate + 75 μs
where
| TBCycle | = estimated bus cycle time ±10% |
| Slaves | = number of DP slave stations in the network |
| Bytes | = number of bytes for all inputs and all outputs added together |
| tBit | = duration of one bit, corresponds to the inverse of the PROFIBUS bit rate |
This estimation formula results in a magnitude ±10% for bit rates > 500 kBit/s and no minimum or maximum value.
Simple example:
We have a DP master that must control 640 digital inputs and outputs. We consolidate this digital data in the best possible way as 80 bytes and share it among a total of 20 terminal blocks as DP slaves. We want to use a bit rate of 1.5 MBit/s.
Slave = 20, bytes = 80, bit rate = 1500 kBit/s i.e. tBit = 0,667 μs
We obtain an estimated bus cycle time of
TBCycle = (380 + 20 x 300 + 80 x 11) x 0.667 + 75 = 4917 μs ≈ 5 ms
More comprehensive example:
We have a PLC as DP master, which has to control various field devices as DP slaves with a cycle time lower than one ms. What should be the minimum bit rate?
Address |
Number of input bytes |
Number of output bytes |
2 |
3 |
3 |
3 |
5 |
0 |
4 |
10 |
20 |
5 |
0 |
5 |
Total |
18 |
28 |
The minimum bit rate can be determined through
Bit rate > (380 +slaves x 300 + bytes x 11) / (TBCycle - 75 μs )
With slaves = 4, bytes = 46 and a bus cycle time of 1 ms we obtain a minimum bit rate of 2.2 MBit/s. We therefore select the next highest adjustable bit rate of 3 MBit/s. Recalculation with the estimation formula results in a bus cycle time of 0.77 ms, which meets our specification.
