Hello, colleagues.

I need to test roughly a RS485 driver load capacity, since a protective barrier is used on its output. My idea is to test its communication with something (a device generating data) by inserting two resistors equal to expected summarized load of the rest of the channel into each wire (a and b) in series. That would be 375 ohms per wire for full 32 unit loads, 400 ohms for 30 unit loads, etc. Is this a correct way, or I'm missing something here?

ANY response will be appreciated.

Viktor S.
Granch Ltd.

 

Yes, RS485 loads appear in parallel so the current increases with the number so your resistance should be 1/32nd rather than 32 X . But, many newer devices put much less than a standard load on the line. But, since you can't really predict which ones, allowing a standard load would be the way to test.

The current can get really high for a fully loaded terminated bus.

Regards
cww

 

So, basically, I can connect a master device (computer reading data over RS485), a slave device (some data source, say a temperature gauge), and a resistor with the value equal to the total load of the rest of the devices that could be used. Devices get connected - a to a, b to b, resistor - between a and b. That will be all. And the value of the resistor is R = R1term||R2term||Requivalent; and Requivalent = Rload1||Rload2||...||Rloadn for n unit loads between the real devices.

Then, if data gets sent and received without problems, then the load capacity of the tested driver of the master device is sufficient for n+1 unit loads (provided that the slave device is equal to 1 unit load) with termination (120Ohm||120Ohm). For 33 unit loads (yes, I know, 1 extra) R = 120||120||375 (Ohm), which is ~ 51.7Ohms. Then the maximum current would be Imax=Vmax/R=12/51.7=0.232(A). Correct?

>Yes, RS485 loads appear in parallel so the current
>increases with the number so your resistance should be
>1/32nd rather than 32 X . But, many newer devices put much
>less than a standard load on the line. But, since you can't
>really predict which ones, allowing a standard load would be
>the way to test.

 

Termination resistors are intended to stop the reflection of a high frequency signal at the end of the cable.

<p>"Termination resistors should be the same value of the characteristic impedance of the twisted pair and should be placed at the far ends of the cable." - maximintegrated.com tutorial</p>

Please read these:
Guidelines for Proper Wiring of an RS-485 (TIA/EIA-485-A) Network:
http://www.maximintegrated.com/en/app-notes/index.mvp/id/763

Belden EIA-485 Cables:
http://www.belden.com/products/industrialcable/eia-485.cfm

 

Yes, kinda. I can't think through your math at the moment. I take it you understand parallel circuits. But there are other requirements. Each unit load also adds a capacitance which should also go in parallel. And I would also check that the chip doesn't squirt smoke or go into thermal limiting at industrial temps driving that load.

That is worst case, which is fairly pessimistic. And remember, _each_ transceiver on the network has to do that. Modbus is master slave, but the slaves have to drive that load to reply.

You can see why chip makers have designed for less loading.

Regards
cww

 

Thank you,

this is true. But in the context of driver load capacity testing, the termination is just another contribution into the total load of the channel. Therefore, its load equivalent is Rterm1||Rterm2, which is parallel to the rest of the load in the total load formula. Agree?

What I'm wondering about is how to emulate conditions when the increased load matters, say, a common mode voltage within the specified range? In the SLLA166 (http://www.ti.com/lit/an/slla166/slla166.pdf) on page 5 Fig.4 and Fig.5 the principle is described, but how do I implement it? Just hook up the voltage source to the local ground (3-rd wire) and change it throughout the range? But on Fig.5 it shows the useful signal above the ground as it's supposed to be, but the common-mode-noise is bipolar. But on the diagram, still the plus goes to the ground. Can I just hook up two DC power supplies, grounded together exactly on the Fig.5, one unipolar and the second one - bipolar? Then go through the ranges? Then, instead of the unipolar use the real signal from RS485 device, while changing the bipolar one from -7v to +7v?

 

So, Curt,

do you think I can implement it the way described in my response to David Wilson on 17 March, 2015 - 12:54 am?

Regards,
VS

 

You could, If I understand it correctly, or use their corrected model. And you would have some assurance that you can produce the required differential with DC, with 32 "standard" loads. Can that be done driving a real transmission line with 32 stubs and 32 tranceivers of various manufacturers, and less than optimal lead dress? Perhaps, but there is no guarantee. But, unless you really intend to put 32 nodes on the network, which would probably be sub-optimal for other reasons, having the drive capability should give you as good a shot as anyone else. I am a little curious about the putting a barrier downstream part as these current levels would be a little high for safety zones, but that's outside my expertise, I haven't worked in explosive atmospheres, etc.

Regards
cww