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We add a termination resistor at the end of a transmission line in order to prevent its signal from reflecting back. I have deduced from some discussions that the termination resistor is not necessary if the track is short. (for example, this topic).

Have I correctly deduced? Is it a general rule? If so, why does not a signal reflect back in a short track? and how is a short track determined? Is it dependent to the signal frequency?

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If the track is significantly shorter than the rise time (and fall time) of the signal then termination is not neccessary.

This is beacuse the reflection arrives when the signal is still rising (or falling) so it doesn't have much effect on the signal.

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  • \$\begingroup\$ Could you please explain the reason? \$\endgroup\$ – M.H Feb 23 '18 at 21:47
  • \$\begingroup\$ @M.H Imagine what would happen in this animation if instead of a pulse coming, a very VERY VERY wide pulse comes. \$\endgroup\$ – Harry Svensson Feb 23 '18 at 21:50
  • \$\begingroup\$ Thanks @Jasen How is the track length (meters) compared with the rise and fall time (seconds)? For example, suppose I have a 100 MHz signal. What is the maximum length for which the termination resistor can be ignored? \$\endgroup\$ – M.H Feb 23 '18 at 22:40
  • \$\begingroup\$ Thanks @HarrySvensson Could you please explain more. I'm still confused! \$\endgroup\$ – M.H Feb 23 '18 at 22:43
  • \$\begingroup\$ @M.H My last comment: Simulate it. Here a low impedance output and high impedance input with no termination is simulated. \$\endgroup\$ – Harry Svensson Feb 23 '18 at 22:54
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First, imagine a dc voltage driven into the trace after some rising edge. The edge travels down the trace and gets reflected and travels back to the source. in the time before the edge makes it back to the source, the load on the source is the characteristic impedance of the trace (eg: 50R). But after the wavefront reaches the source, the load on the source is equivalent to the load at the other end of the trace (eg: open circuit or a high impedance termination). Before the reflection returns to the source, the driven signal is distorted by the source loading of the characteristic impedance.

This travelling time is obviously dependent on the trace length. If the traveling time is a significant fraction of the total pulse width time, then the signal will be distorted for a significant fraction of the period. If the time of flight of the signal is small compared to the signal period, then the distortion will be only for a short time after the edges.

It gets worse if the signal is reflected so that the reflections are logic high when the source is trying to drive logic low and vice versa. Then the source is essentially driving into an output in antiphase all the time. (This is how damage is done to transmitters when not properly terminated.)

Now, make the slight mental adjustment to sinusoidal waveforms. If the trace length is such that the reflections are 180 degrees out of phase, then the source is driving into the worst possible load (short circuit). if the trace length is doubled, then the reflections are in phase and the source sees no load.(This is ideal, and can be an alternative to termination). if you increase the length a small amount, the reflection will be almost in phase, and the source loading will be minimal, so it's very close to the ideal no loading situation. But of course, a full wave length plus a small section is the same as just the small section, so a very short track is close to the ideal no loading situation. How close is dependent on the ratio between the trace length and the wavelength. At this point, all good engineers just use a rule of thumb rather than analysing each case. Mine is 1/10th of the wavelength., if you have rise time instead of frequency you can use f = 0.35/tr

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If you have a 50 Ohm source and 50 ohm load, the voltage on the load is always going to be 1/2 of the open circuit voltage , for any length up 10% or so of the wavelength.

It's just that if the transmission line (above) is unterminated, the pulse reflects and doubles the amplitude getting back the source where there reflections stop since it was assumed matched.

This is the same as above with twice the voltage for an open circuit or half of open for a matched cct. It's just that with transmission line effects, it is a delayed reaction of about 50 ps/m for typical coax for both directions.

If you dont terminate, then you can expect ringing of fast rising step pulses with a half cycle equal to the prop delay in length but if rise time is slower, then it wont. So it depends on the line being inductive or capacitive when not a long length.

Capacitive means closer to the ground plane.
Inductive means bigger gap or no ground plane.
Example 1m 10:1 probe with a long inductive ground lead causes "apparent" ringing which is a false reading. But if a jumper on a breadboard , could be real ringing from just a small capacitive load.

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