Termination Resistor - RS-485 line.

Sounds bad to me. If the data rate is low, just don't terminate it, or use a series R-C. At high rates (relative to line length) any non-resistive termination can cause pattern-dependent distortion, "deterministic jitter" and possible errors.

John

John, the idea of the poster was AC termination. Meaning a 100 Ohms in series with perhaps a 100nF.

Rene

I mentioned series R-C termination. Again, that can cause pattern-dependent timing errors and may snarf the data. If the data has no DC or low-frequency components (like NRZI or Manchester), the series R-C will work, but saves no power!

John

Nice answers.

I think both are right, and I meant an RC termination implicitly. Now I am realizing is the most convenient termination for my network.

One more question to John, why you said "saves no power"?

Cheers...

Franco.

I guess I really don't see why you would want to use such a termination in an RS485 system. Would you not want to use a terminating resistor that closely matches the characteristic impedance of the cable?

The cable and terminating resistor would form a divider circuit. By matching the terminating resistor to the impedance of the cable, you would eliminate or at least greatly dampen any reflections on the lines.

You mention not wanting such a big load as a 120 ohm resistor. I am curious as to what your line of thinking is here.

If the data is balanced (NRZI, Manchester) the DC voltage across the cap is zero, so it may as well be a short; so all the differential signal swing appears across the resistor.

If the data is unbalanced, like async serial ASCII or anything that pauses, the cap can charge up to almost the full driver swing in either direction, which can severely skew the receive timing when transitions start again.

Why not use a 120 ohm resistor?

John

Ok. The point is than I am going to try to join (by using analog switches) two or more "little" RS-485 buses with a transceiver of 54ohms max drivers load. If I use two terminator resistors of 120ohm on each bus, I think I cannot drive more than two of these "little" buses. All of this, not considering the load of the receivers (1/8).

Maybe is not possible, or maybe the way to do it is with bidirectional buffer kind of circuitry, but, so far, I am just trying to get just an overall view.

Comments, please...

Best Regards...

Franco.

What sort of data? What rates? Are drivers gated?

John

As you switch in different bus configurations you want to make sure that the "little" busses do not join into non-linear structures. You want two distinct ends (and those two places are where your two 120 ohm terminators go) and have a straight path to the other end going past all intermediate nodes without side branches and stubs. If the switching topology cannot be done with fixed end terminated nodes then you may have to also put switches to turn the termiminators on and off at some nodes that swap back and forth between end and intermediate type nodes.

- mkaras

Double check the data sheets. When you say that the drivers can handle a max load of 54 ohms are you certain that this means that it can't handle a larger impedance rather than a smaller one? The smaller (less ohms) the load on the drivers, the more current they will have to pump out to cause a signal to reach full height as it propagates on the transmission line.

The Texas Instruments application note SLLA070C discusses the various terminations and list the 100 Ohms plus 1nF betwwen the two lines as feasible alternative termination for baudrates below 100kbit. It saves power as soon as the bus is idle in either state.

Rene

That's a 100 ns tau for 10 microsecond bits. In that case, you may as well not terminate at all. There might be some corner cases where such a termination will reduce edge ringing.

But why not terminate with a resistor?

John

The system is battery powered. I think pure termination resistor (parallel termination) is not an option. Moreover, I want to join little buses, to expand a kind of global bus. As far as I understand, if one driver is going to send the signal towards two little buses, both terminated at 120 ohm (multipoint buses), I am over-clocking the transceivers (let said, min 54 load - not less). The transcievers are: SP3078E.

I want high rates, as high as I can. The micro-controllers driving each transceiver attached to the bus are fast enough to all the span of rates (up to 10Mbs).

It is a distributed system.

I know I am going to break the guidelines of linear buses, when joining two little buses. I cannot do anything about that, more than just analyze the effects of that. In this moment I am trying to figure out what happen when placing AC termination at both end of the buses.

Cheers...

Franco.

more stuff here

martin

here

Very good information indeed ;).

Gracias...

Franco.

This is true if the length of the cable is greater than (roughly) 1 fifth of the wavelength of the signal. It might be easier to use slew-rate controlled (read: bandwidth limited) RS485 drivers because they allow longer unterminated transmission lines.

Well, the swing is the differential voltage across the resistor of 3V with either sign. With 100 Ohms, this is a loss of 90mW. For a battery operated device, these 30mA are a lot. The reflection of the signal is a reflection of the slope, and it is the slope that dissipates in the AC termination. I'd possibly go for a tau in the order of a tenth of a bit.

Rene

not really.

if the 120 ohms is hurting go to a lower voltage or use a different cable. with a higher characteristic impedance.

or go to a lower bitrate - low enough for one bit to flood the whole bus.

you ran data wires but forgot to run power wires?

Bye. Jasen