# Bidirectional termination from one end only

## Problem

I have been trying to figure out if it is possible to terminate a bidirectional bus from one end only. I have found techniques for terminating a bidirectional bus using both sides of the line, but haven't found one that just uses one side.

This is a circuit I came up with that attempts to terminate the bidirectional bus:

simulate this circuit – Schematic created using CircuitLab

The design is pretty simple, Rseries1 is a series termination for when XCVR2 is driving, and Rparallel1,2 are a parallel termination for when XCVR1 is driving.

Will this work? Is there is a better way that I'm missing? It doesn't need to perfectly eliminate reflections, it just needs to be better than an unterminated line.

## Background

This is a debug bus for a SiLabs microcontroller. The board on the left is the programmer, and the board on the right is my board. The cable is ~6" ribbon with a characteristic impedance of ~80 ohms.

The vendor recommends just populating Rparallel1, but some people have had issues with that configuration. Running at a lower frequency is possible, but I'm more interested in this problem now :)

• Cable length: 6"
• Prop delay: 1.25 ns/ft
• Maximum bus freq: 12 MHz (~16 ns rise time max)
• Maximum part freq: 50 MHz (~4 ns rise time max)

Sorry, but no.

Rseries1 is a series termination for when XCVR2 is driving, and Rparallel1,2 are a parallel termination for when XCVR1 is driving.

When XCVR1 is driving, Rseries1 does not magically disappear. So, assuming XCVR2 is a good voltage source when driving, the effective series resistance that it sees is indeed Rseries1. However, when XCVR1 is active, the effective series resistance is Rseries1 + Rth, where Rth is the Thevenin equivalent of the parallel resistors. So, for instance, if Rseries1 is 100Ω, and each parallel resistor is 200Ω, XCVR2 will see 100Ω (which is what you want), but XCVR1 will see 200Ω.

If the prop delay is less than the rise time then return loss won't be a problem.

But when it is longer in prop delay than the rise time, ringing may cause ringing with inductive lines.

With a controlled impedance transmission you get better noise immunity but still get end reflection undershoot.

So it depends on your bandwidth requirements, prop delay, signal integrity requirements and tolerance to EMI.

• I added some more information about the cable characteristics. I wasn't able to find a rise time in the datasheet anywhere, but I'm handwaving about 2-4ns. If that's the case, then prop delay < rise time, and I should be alright. That said, I wouldn't be surprised if the drivers were faster than that. Are there any pads I can put in during layout in case something doesn't work quite right? – user35740 Sep 11 '16 at 21:46

Consider the standard termination options: -

All the above work but focus on (a), the series termination at the driver end. It relies on the receiver input impedance being relatively high compared to Zo so that if the driver put out 5Vp-p, it would become 2.5Vp-p as it enters the cable and, due to the open end termination at the receiver, the voltage would double back to 5Vp-p.

Now consider the same when there is a series termination at both ends; the receiving end would still function and get a decent signal but when the roles reversed everything would still be OK due to symmetry.

Having said all of that, a 4 ns rise time implies a "top" frequency of about 100 MHz and the wavelength of 100 MHz is 3 metres. The rule of thumb is that you can avoid a termination if the cable /trace length is less than one tenth the wavelength of the highest useful frequency. So one tenth of 3 metres is 30 cm or about 12 inches. It's borderline because the speed of signals is closer to 50% that of light so effectively, this makes the cable appear longer.

I'd put a termination of 40 to 60 ohms at both ends.