During RS485 transmission, what is the absolute voltage change in the TXD+ and TXD- pins in RS485? Do they swap value (IE if TXD+ is sitting at 4V quiescent, then it goes 4v 0v 4v during transmittion) or do they just move "differentially" (IE TXD+ goes 4v 3.5V 4v and TXD- goes 0v 0.5V 0V)?

Background - but not my actual question:

I'm trying to understand a problem that is being debugged (not by me) on a project that I'm associated with. I only get to hear about it from the side.

The explanation for the problem that I am hearing is that one RS485 device is using voltages (absolute) that are too far apart for the other device. That is to say: that the "master" has it's TXD+ and TXD- sitting at (say) 4v and 0v, wheras the "slave" can only cope with these being at 3V and 0V respectively.

This got me wondering how does the "differential" transmission work in the first place? When this master transmits alternating bits, does it swap TXD+ and TXD- between 0 and 4V, or does it do some small differential change on them, like dropping TDX+ to 3.5V and raising TXD- to 0.5V? Would that be valid?

If it's the latter, then I can understand the explanation. But if its the former, then the explanation makes no sense...


You can find the required voltage difference at the sending and receiving ends in the RS485 standard. As a second best, read the specs of some RS485 transceiver.

What is probably the problem in your project (but it is difficult to reconstruct the original signal after a half-garbling filter) is that the common of your sender and receiver differ too much. That is, there are two requirements for successful RS485 communication:

  • the voltage difference between D+ and D- must be OK. This is probably not your problem.
  • the voltage difference between D+ and ground, and D- and ground must be in the range acceptable by the receiver. If they are not, you have a ground potential difference problem. To solve it, you can either reduce the difference, or find a receiver that can tolerate a higher difference.

The voltage levels for RS485, from The RS485 design guid :

enter image description here

Note that this picture is about the differential voltages, it omits the ground connection.

  • \$\begingroup\$ Thanks for your response. I wonder if I could clarify what I really want to ask. Irrespective of the problem, what would you expect to see as the absolute voltage changes in the TXD+ and TXD- pins in RS485? Do they swap value (IE TXD+ goes 4v 0v 4v) or do they just move "differentially" (IE TXD+ goes 4v 3.5V 4v and TXD- goes 0v 0.5V 0V)? (I've edited my question to ask this more precisely) \$\endgroup\$ – GreenAsJade Mar 16 '14 at 11:23
  • \$\begingroup\$ They more-or-less swap value. \$\endgroup\$ – Wouter van Ooijen Mar 16 '14 at 13:44
  • \$\begingroup\$ Thanks for the picture. As you say, ground is not shown here, so I think that a situation where the green trace starts at 4V and goes to 2.5, and the black trace starts at zero and goes to 1.5 would be valid, eh? \$\endgroup\$ – GreenAsJade Mar 31 '14 at 3:57
  • \$\begingroup\$ As far as this picture is concerned yes, but I think there are also limits to the allowed common mode range, which would be (somewhat) larger in your case. \$\endgroup\$ – Wouter van Ooijen Mar 31 '14 at 5:59
  • \$\begingroup\$ @GreenAsJade I think it is 1(2.5V,1V) and 0(1V, 2.5V). \$\endgroup\$ – richieqianle Aug 18 '15 at 5:18

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