RS485 Termination Resistance and Reflections

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I am working on a circuit with 2 mega324s that talk to each other via rs485. I choose to use two ISL3179E from Intersil, mainly becuase it comes in a MSOP10 package.

http://www.intersil.com/data/fn/fn6365.pdf

I would like to optimize the termination resistance to reduce power consumption, but I am not sure how to do this. I would like to achieve 250 kbps and my cable is high flex 28 guage twisted pair that is about 18 inch long.

From what I have read, have lower termination resistance minimizes reflections. The datasheet lists 52 ohms and 100ohms as two possible options. I would like to use something more like 1k ohms. I am not sure how to caculate the effect of the higher termination resistance.

Any insight into this is appreciated.

-Daniel

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You could use AC termination by placing a capacitor in series with the termination resistor. Never tried it though.

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You could try the good old-fashioned trial and error :)

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dpaulsen wrote:

I would like to optimize the termination resistance to reduce power consumption, but I am not sure how to do this. I would like to achieve 250 kbps and my cable is high flex 28 guage twisted pair that is about 18 inch long.

From what I have read, have lower termination resistance minimizes reflections. The datasheet lists 52 ohms and 100ohms as two possible options. I would like to use something more like 1k ohms. I am not sure how to caculate the effect of the higher termination resistance.

Any insight into this is appreciated.

-Daniel

Lowest or no reflection happens when termination impedance matches cable impedance. Twisted pair has around 100 or 120 ohm characteristic impedance. When you send a signal pulse to such cable, it appears as if the current flows into this characteristic impedance until signal has traveled to the other end of the cable at about two thirds the speed of light, and then the pulse goes into the termination impedance.

Now, if termination impedance is 1k, it will look like the pulse goes into 100 ohms for the length of the cable and then into 1k, creating sudden positive reflection, and if termination impedance is say 50 ohms, pulse goes first into 100 ohm and then termination creates negative reflection.

AC termination may help you. R and C in series, no DC flows when idle, but AC signals are terminated with characteristic impedance.

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Use the calibration on your scoop.
Connect your cable and probe to it.
Terminate the cable with an 500R, 20t.

Adjust the resistor until the scoop shows an squear pitcure.
Mesure the resistor. That is the cables impedanse. Use half of it in each physkcal end of the cable.

HM

HM

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Sounds like I am not going to predict ahead of time what will work. I am going to have to build something and experiment with my scope.

Thanks for the suggestions.

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At a length of 18", reflections should only barely be an issue. Here is some basic analysis:

Speed of light in space, C, is about 1ns/foot.

Speed of light in a cable is about 0.6C or about 1.6ns/foot

Transit time for 18" = 1.5' is about 2.5ns, round trip about 5ns.

Transmitter rise time is typically 4ns, max 7ns,

The reflection will return just about the time the pulse rise finishes. Thus, it will look like a bit of overshoot or dip, depending on whether the termination is higher or lower than the characteristic impedance of the line.

This will have little or no consequence for data integrity. If the line were much longer, then, definitely an issue.

An AC termination works just fine. It will reduce the DC current flow while keeping the effective signal termination. 120 ohms is the commonly used RS422/485 termination.

Jim

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

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ka7ehk wrote:
At a length of 18", reflections should only barely be an issue. Here is some basic analysis:

Speed of light in space, C, is about 1ns/foot.

Speed of light in a cable is about 0.6C or about 1.6ns/foot

Transit time for 18" = 1.5' is about 2.5ns, round trip about 5ns.

Transmitter rise time is typically 4ns, max 7ns,

The reflection will return just about the time the pulse rise finishes. Thus, it will look like a bit of overshoot or dip, depending on whether the termination is higher or lower than the characteristic impedance of the line.

This will have little or no consequence for data integrity. If the line were much longer, then, definitely an issue.

An AC termination works just fine. It will reduce the DC current flow while keeping the effective signal termination. 120 ohms is the commonly used RS422/485 termination.

Jim

Thanks Jim. That is exactly the explanation I was looking for. When you put numbers to it, it makes it a lot easy to understand what the effect is.

I think I will try a 120 ohm resistor with a 22 pf capacitor and see how that works.

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22pF. How did you work that out? In a square wave you do have some fundamental there. 9600 baud is about 1ms or 1KHz. I'd look at something like:

C(uF) = 1/(2*PI*frequency in MHz * 120 Ohms) = 1.3uF, say 1uF

Then check it with an oscilloscope.

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Transmission line effects start becoming problematic at cable length of 1/4 wavelength of the fundamental freq? Usually kilometers. I think this was all worked out in the early days of telegraphy? Some old Prof tried to teach me how to use a Smith Chart. It was Real Wierd.

Imagecraft compiler user

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I see the power EE students fawning over those things all the time... also Real Weird.

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On 18" distance you don't need RS485 at all. Even if you put there RS485 interface, you don't need any termination, at least for 250kbps. At the worst case, you can try to use so called "diode termination", or slowing the edges a little.
Here in Bulgaria we have saying: "He want to tie his pants." for someone that wants to play safe - so, don't tie your pants too tight. ;)

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I should have realized the length. 18 inches does not require RS-485 so the system may be a bit over-engineered. Even RS-232 should do it, but not all tranceivers go up to 250kbps. Perhaps you can do it without any tranceivers, just logic level signal?

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While we are on the subject, what length would be a problem for rs485 at 250kbits, even with good termination? 500 ft? 5000 ft?

Imagecraft compiler user

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davef wrote:
22pF. How did you work that out? In a square wave you do have some fundamental there. 9600 baud is about 1ms or 1KHz. I'd look at something like:

C(uF) = 1/(2*PI*frequency in MHz * 120 Ohms) = 1.3uF, say 1uF

Then check it with an oscilloscope.

Do you want the fundamental to pass through the cap? I was assuming that you want only the reflection to attenuate, not the fundamental. Am I think of this wrong?

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With good termination, the useable distance depends on cable loss. At low baud rates, it is primarily wire resistance. So, wire gauge is important. The amplitude required for reliable detection is also important and that could vary depending on the chip; I seem to recall seeing specs for minimum reliable amplitude.

As baud rate goes up, cable has additional losses. National Semi has some app notes on RS422/485 that shows distance vs baud rate. I seem to recall that the break is in the 100s of kbaud range but will check.

The reference I am thinking of is here:
http://www.national.com/an/AN/AN...

Figure 3 shows baud rate vs cable length for RS232 & RS422/485. For 24AWG copper wire, 16pf/foot, terminated in 100 ohms, it shows 4000' at low baud rates, breaking at (about) 150Kbaud, dropping to about 60' at 10Mbaud which is the upper limit.

These numbers MIGHT be exceeded with newer chips that have better low-amplitude detection, but if you want to maintain RS422/485 compatibility with any chip that happens to come along, those are the numbers you should be able to rely on.

Jim

 

Until Black Lives Matter, we do not have "All Lives Matter"!

 

 

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60' = 18m
4000' = 1.2km = 1200m

;)

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jayjay1974 wrote:
You could use AC termination by placing a capacitor in series with the termination resistor. Never tried it though.

This is an interesting idea. Would this work for CAN? It seems like you'd want the impedance of the RC network (at the frequency of your communication) to be equivalent to the twisted pair impedance. But you'd still get reflections from the harmonics - so it's not a perfect solution I think.

So you'd want twisted pair impedance = (R^2+(1/(w*C)^2))^0.5.

edit: apparently I was way off: http://www.national.com/an/AN/AN...

Last Edited: Fri. Jul 23, 2010 - 08:00 PM
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If you had a RC combination set up properly at the fundamental the harmonics would be taken care of. The reactance of the capacitor becomes less at higher frequencies, which effectively leaves you with just the (DC blocked) resistor.