RS485 terminating resistor 2

[techzone12]

Recently I installed a "RS485" connection between our controller and a third party controller. It was specified that one must connect terminating resistor on each end of the RS485 loop (because it's a long run). It says that resistor is designed to match the electrical impedance characteristic of the twisted pair loop, and will prevent signal echoes from corrupting the data on the line. I also hear that some times you need to install a capacitor in series with a resistor?

I am trying to understand the theory behind this. Can some one explain this in some level of detail? Or maybe point me to a link on the web where it's explained?
I need to understand how does the signal get echoed? And what does the resistor exactly do?

Thanks

 

[JLSeagull]

Review TIA STANDARD TIA-485-A; Electrical Characteristics of Generators and Receivers for Use in Balanced Digital Multipoint Systems. Also review the TIA TSB-89, TELECOMMUNICATIONS SYSTEMS BULLETIN with Application Guidelines for TIA-485-A.

"Each signal pair will normally have a resistor, equal to the characteristic impedance between the pair, at the extreme ends of the backbone" ... "maximize ac signal power
transfer from the interconnect media" ... "conditions are compensated for by providing a steady-state bias
via an external resistor network" ...

visit tiaonline.org; or Global Engineering Documents, +1 303 397 7956

 

[Skogsgurra]

The first secret lies in the impedance of the cable.

It is not something you can measure with an ordinary bridge and it is a constant that is characteristic for the cable construction. You will have the same impedance in a 10 m cable as you have in a 100 m cable. The definition is sqrt(L/C), where L is inductivity/length unit and C is capacitance/ length unit. Normal values are between 50 ohms (coax cables) to 300 ohms (old-fashioned TV cables).

The next secret is that waves travel at limited speed. The ultimate speed is the speed of light (c), but waves in a cable travel at lower speed - typically 60 - 70 percent of c. When a wave front travels down a cable and hits a discountinuity (end of cable, for instance) it is reflected back and that is where your trouble starts. Instead of having one clean wave front, you have one coming down the cable and one travelling back. The result is bit error (in digital systems), ghost pictures (in TV systems) and problems of all sorts in any system.

The trick is to avoid the discontinuity. By terminating the cable with its own characteristic impedance (resitance), you trick the signal into believing the cable never ended. All signal energy is absorbed by the termination resistor and nothing is reflected back. Result: no wave front reflected - no problems. Simple as that!

Gunnar Englund

http://www.gke.org

 

[ScottyUK]

Type "transmission line impedance", including the "" into Google. There are numerous hits, many of which explain the principles with varying levels of mathematics. A university-level text book on electrical principles will almost certainly also explain this, but the maths will be heavier.


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I don't suffer from insanity. I enjoy it...

 

[techzone12]

Thank you for the valuable feedback. It is starting to make sense to me. I have a question however:
If the purpose of using the resistor is to absorb the signal energy on the transmission line, what happens if someone install a resistor (at the end of the line) with higher value that the characteristic impedance?

Thanks

 

[Skogsgurra]

Answer: You do not get complete absorbtion and some of the energy is reflected back. That is quite common and is not a problem if the mismatch isn't too big. A 20 or 30 percent deviation usually isn't too bad.

Gunnar Englund

http://www.gke.org

 

[techzone12]

I know the answer might be very involved, but I cann't help but asking why ?
Why is a bigger resistor is unable to absorb (dissipate) all the induced enegry by the signal?

Thanks

 

[ScottyUK]

There are two extremes of mismatch: one is an open circuit termination, and the other is a short circuit termination. Both cause 100% reflection of the input signal: the difference is in the relative position of the nodes and antinodes where the signal amplitude drops to zero and reaches double amplitude respectively. A short circuit causes a node to exist at the termination; an open circuit causes a node to exist at the termination. If your signal was a sinusoid you see a response similar to that in the link below. When it is a digital pulse stream containing many frequencies, the result is a chaotic mess which can't be analysed by simple means.

The following link gives a reasonable explanation without too much maths:

http://www.allaboutcircuits.com/vol_2/chpt_13/7.html

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I don't suffer from insanity. I enjoy it...