Example schematic showing MAX14890
(source: maximintegrated.com)

In the above diagram of a differential line receiver for an encoder, resistors are place between each line and its inverse. I don't see how this would lower the current of each line, so I assume it has some other purpose.

What is the resistor's role in this scenario? Is it at all related to a pull up/down resistor?

  • 5
    \$\begingroup\$ Search term: "Termination Resistor" \$\endgroup\$ Commented Jun 4, 2018 at 19:28
  • 3
    \$\begingroup\$ So electricity is more complex than just connecting wires... \$\endgroup\$
    – user76844
    Commented Jun 4, 2018 at 19:40
  • 6
    \$\begingroup\$ @Gregory Kornblum, I didn't realise I had suggested it wasn't... \$\endgroup\$
    – 19172281
    Commented Jun 4, 2018 at 19:48
  • 1
    \$\begingroup\$ ^^ @GregoryKomblum, It's like a thermos... keeps cold things cold, hot things hot, but, "how does it know?" :) \$\endgroup\$
    – CapnJJ
    Commented Jun 4, 2018 at 20:06
  • 1
    \$\begingroup\$ @CapnJJ yes, like many magic items we have \$\endgroup\$
    – user76844
    Commented Jun 4, 2018 at 20:11

3 Answers 3


The purpose is to match the characteristic impedance of the twisted pairs that carry the differential signals.

By matching the impedance, reflections from the termination can be minimized, which helps with signal integrity. This is elementary transmission line theory.

120 ohms is about the characteristic differential impedance of a typical twisted pair. For example, nominally two AWG 38 wires with PTFE insulation twisted together will be around 115 ohms.

If you use wires with a much different characteristic impedance, you can modify the terminating resistors to get better results. For example, if you chose to use CAT6 cable, 100 ohms might be more appropriate.

In many cases we're happy to be within 10 or 15% of the nominal.

If you want get intuitive, you can think of a lossless transmission line as sort of the sum of a large number of series inductances with parallel capacitances .


simulate this circuit – Schematic created using CircuitLab

if you had a very long twisted pair (say a few light-seconds long) and put a ohmmeter at one end you would measure about 120 ohms, briefly, as the wave traveled down the pair at perhaps 0.6 or 0.7 of the speed of light. Notice there is not a single resistor. There is initially no voltage across the (say it is open) far end, because of the speed of light. When the wave hits the end the voltage I * 120 appears across the end, but there is nowhere for the current to go so a reflected wave begins back toward the ohmmeter. Eventually it will settle down to show open-circuit (if the far end is open), or short if the far end is shorted. If you terminate the far end with 120 ohms, the reflection does not occur and you see a constant 120 ohms, intially and forever.

In practice the wave nature becomes important as the dimensions start to approach a wavelength of the highest frequency involved. If you have say a 1MHz square wave, if you consider components up to 11MHz the wavelength on a twisted pair would be around 17m, so for lengths above about 1/20 (1/10 for round trip) of that, or about 1m, it starts to be become significant. Those are just rough rules of thumb, others may have slightly different rules. Even a few inches of traces on a PCB can lead to significant ringing for a relatively low frequency (say 25MHz) square wave.

It's not unusual for precision position encoders to work up to MHz and beyond.

Anyway, if you want to learn about it, there is plenty of information available on the internet, Wiki and course notes. Or buy this paperback from a used dealer, and work the 165 solved problems and you'll be up to a solid 2nd or 3rd year undergraduate Engineer's level in understanding.

enter image description here

  • \$\begingroup\$ When you refer to impedance, is that interchangeable with resistance, considering it is a DC circuit? Are reflections caused by the presence of the other line in the twisted pair? Where is the termination? I think I now understand what it does, but I'm still not to sure on how, or why it works. Thanks :) \$\endgroup\$
    – 19172281
    Commented Jun 5, 2018 at 12:13
  • 1
    \$\begingroup\$ No, it's not resistance. The small resistance of the copper contributes a loss but in an ideal transmission line that's not a factor. It is not a DC circuit, the signal levels switch as the encoder position changes. \$\endgroup\$ Commented Jun 5, 2018 at 12:24
  • \$\begingroup\$ The signal levels switch, but current is always flowing in one direction, so doesn't that make it DC? \$\endgroup\$
    – 19172281
    Commented Jun 5, 2018 at 14:38
  • \$\begingroup\$ Current changes direction when they switch. \$\endgroup\$ Commented Jun 5, 2018 at 14:43
  • \$\begingroup\$ Hmmmm, I'm not sure I understand. If the encoder is being powered by a DC supply, how can an AC output be delivered? By switch, do you mean the transition from high to low and vice versa? \$\endgroup\$
    – 19172281
    Commented Jun 5, 2018 at 14:48

Wired pairs should have an impedance equal to the terminating resitance to transfer the maximum power and minimize reflections. The impedance of wires must be balanced from source to termination to also reject common mode current noise and voltage noise (B&H field) from interference as this Signal to Noise ratio affects error rate.

Bit rate is high enough for this simple inexpensive interface for short 10m lengths and exceeds 10Mb/s or long haul 1,000 m at 100Kbps. The limit is due to the product of bit rate and distance due to line capacitance.

Additional intermediate bus tranceivers are permitted without terminator R as long the stub length is short and meets spec a the high impedance intermediate node (properly designed) adds no interference.

There are many industries which use this; music control, automation, etc.


All signal lines are transmission lines.

However, some we have to treat as transmission lines, the electrically 'long' lines. We have to explicitly terminate them with resistors or we get reflections that mess up data transfer.

Electrically 'short' lines can be driven and sensed without bothering about terminations or reflections.

What's the difference between 'long' and 'short'?

The difference is the relationship between the signal rise-time produced by the driver into the line, and the signal propagation time along the line.

If the rise-time is so long, that is the rate of rise of the signal is so leisurely, or the end-to-end line propagation time so short, that the whole line stays more or less the same voltage, then the line is 'short'. Any reflections are small, and only serve to modify slightly the slow change of voltage on the line.

If the rise time is fast, so that the driven end of the line has risen in voltage while the remote end is still low, then the line is 'long'. When that wave hits the far end, it could reflect full height, and travel back down the line again being mistaken for another logic edge.

In the case of your LVDS interface ICs, LVDS drivers are very fast, which is the whole point of the system, being able to transmit a high data rate. For any practical inter-cabinet distance, this means the lines are 'long', and so must be terminated.


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