Hi
Is there specification regarding bit rate performance for SN65HVD72 and SN65HVD75 RS485 transcivers at 800m cable length ?
yoni W
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Hi Yoni,
Below is a typical graph and explanation for showing the relationship between bus length and data rate for a transceiver with maximum 30ns rise and fall times. Fortunately the SN65HVD72 and SN65HVD75 also have a 30ns maximum rise and fall times so this graph applies to them as well. Please see the explanation below and let me know if you have any questions.
If we consider the flat section of the graph from 100bps to 10kbps, this section of performance is dictated by the physical bus wires. For example, the RS485 standard states that a transceiver should output 1.5V across a 54Ω load. If typical twisted pair cable will have resistance on the order of 50mΩ per foot, therefore at 1000 feet the bus will be contributing 50Ω of impedance. If you take this series resistance in conjunction with the termination resistance the receiver will only be receiving half of the originally 1.5 volt differential signal. The cable and termination resistors have essentially created a resistor divider. As the bus gets longer, the voltage drop across the bus will get larger, and the receiver will receive a smaller and smaller signal. Eventually the signal level at the receiver will not have a high enough amplitude to be seen. In summary, this section of the typical graph is dependent on differential output voltage and series resistance of the bus.
Now if we consider the far right vertical portion of the typical curve at 10MBPS, this performance is dictated by the driver's rise and fall times. As the bit times get shorter and shorter the driver needs to be able to drive the bus faster so that the signal gets to its settled value in time for the receiver. A rule of thumb is take the max rise / fall time of the driver and divide by 0.3. You then invert this to get the data rate. For example for the SN65HVD1786 RS485 transceiver has a maximum rise and fall time of 30ns. The calculation would be 1/(30ns/0.3s). This results in 1/100ns, which corresponds to 10MBPS.
Thanks,
John