RS485FLDPLXDRCEVM: MAX491ESD+T

There are too many termination resistors. On a bus, there should be only two terminations, at the two ends of the bus. (And on a unidirectional bus with exactly one transmitter, the termination at the transmitter can be omitted.)

A bus needs only one set of fail-safe resistors. (But if the devices can be disconnected from the bus, it can make sense to have them at every transceiver.)

But your problem is that there is a collision between the two slaves. With one slave driving high and one slave driving low, the resulting voltage is in the middle. Ensure that a slave's DE is high only when it actually is sending something.

Thanks for the help Clemens!

Mahavir,

Clemens assessment is correct, please let us know if you have any other questions.

Regards,

Eric Hackett 

HI Clemens.. Thanks for your response... I tried to do the same with one Slave's DE is set to high while it is transmitting and other slaves DE set to low. This worked as expected!!!.  However, I fail to understand how to form a RS485 full duplex bus with given transceiver with their DE pins set to high permanently in hardware itself. What actually happens inside transceiver IC which causes the problem when DE pins of both Slaves are kept always ON. There are few full duplex ICs like MAX490ESD which doesn't have DE and RE pins. Since with no DE and therefore not giving user control over driver lines, I assume their Driver lines are always active, will they behave same in bus formation (one master two slaves). just a query...crossed my mind...!!

On a bus with more than one potential sender, you have to ensure that at most one sender is active at the same time. This must be done with the DE signal.

When a sender is always active, then no other sender must exist on the bus. You can connect exactly two full-duplex transceivers to each other (with two buses, one in each direction), but with three devices, you would need three buses.

Thanks Clemens for clarifications...  I appreciate your help.