How many number of nodes can be connected on an RS485 Bus

[rajeshkumar123]

Hi,

Can any one tell me how many number of nodes can be connected on an RS485 Bus.

Note: proof needed

 

[blueroomelectronics]

Depends on the IC, 32 is normal but 128 possible. It's in the IC datasheet.

 

[rajeshkumar123]

Re: How many number of nodes can be connected on an RS485 Bu

Hi,


I use a MAX3082 whose data sheet says we can connect upto 256 nodes on the bus. I am not clear about a phrase mentioned in the Datasheet on the receiver input impedance . can u help . Its given here below

"The standard RS-485 receiver input impedance is 12kΩ
(one-unit load), and the standard driver can drive up to
32 unit loads. The MAX3080 family of transceivers have
a 1/8-unit-load receiver input impedance (96kΩ), allowing
up to 256 transceivers to be connected in parallel
on one communication line. Any combination of these
devices and/or other RS-485 transceivers with a total of
32 unit loads or less can be connected to the line."

I have also attached the datasheet for your reference

 

[melc]

Re: How many number of nodes can be connected on an RS485 Bu

Everything is a matter of total impedance on the bus.
The standard maximum load on an RS485 bus is 32UL. That means 12K/32=375 ohms minimal total impedance. This is the minimum impedance which allows a correct common mode for any of the 32 receveirs on the bus with an 120 ohm terminator. Note the current request on such bus, is I.

Increasing the number of UL is possible only if the current is decreased and the impedance of the receiver is increased. For a 96k impedance, 96/375 = 256 UL, with a current much smaller than on the 32UL bus. I1 = I/8.
The common mode voltage for 256UK and 32 UL must remain the same specified by the RS485 spec datasheet.

Now, which is the difference? The 32 UL standard is able to communicate up to 1km. The 256 UL standard will have a smaller noise imunity because the current is 8 time smaller. How long will be the cable ? I don't know, never tested 256UL.

I hope you've understood now.