It is a well-known fact that ISDN S0-Bus must be terminated with 100ohm resistors to match the transmission line. This is the case even for everyday household installation with cable runs shorter than 50m. Why is this so?

Some back of the hand calculations:

According to this, ISDN has a bandwidth of 80 kHz. Assuming propagation in vacuum, it follows from c=f*lambda that the typical wavelength is about 3.8 km. Reducing this by a factor of two to account for the shorter wavelength in copper wires gives 1.9 km.

Assuming that interference due to cable reflections is non-deleterious for path differences below lambda/10, this would mean that for cable runs of less than 150m, no terminating resistors should be necessary.

This is clearly not the case however. Where is the mistake in above reasoning?


Simply put: The old rule of thumb of "you don't need termination if the wire length is less than xx% of the wavelength" is not at all a good rule. There are many cases where it doesn't apply and ISDN is certainly one of them.

The rule is wrong, not ISDN.

Another possibility (which does not rule out what I said above) is that some signaling standards operate on what we call a "current loop". Basically it's the change in current, not the change in voltage, which encodes the data. In these signaling standards the termination resistor is what provides a path for the current. With no current, no data is transmitted.


Let me elaborate on why I think that old rule is wrong...

  1. People misinterpret the rule. They think that "frequency" means the clock or data rate while it really means the highest frequency in the spectrum of the signal. Other people will often say things like "it's not the data/clock rate that matters but the edge rate", which is technically correct but doesn't really help either. The pedantic among us will point out that this is a problem with the person, not the rule-- so sue me.

  2. The rule assumes that you know the frequency content of your signal. For digital signals with reasonable edge rates you don't really know this. Further, since good square waves contain an infinite number of harmonics you could interpret this to mean that you always need termination. Of course putting termination on every signal is silly. The point is, for digital signals the rule doesn't give you any useful guidance on if termination is needed or not.

  3. The rule assumes that this one signal is the only important signal. It is possible that crosstalk on this signal will couple noise into another signal, and that adding termination (regardless of trace length) will minimize that crosstalk.

There might be other reasons why this rule of thumb doesn't work, but that's all I have right now.

  • \$\begingroup\$ "The old rule of thumb is wrong". Yes, but why exactly. In fact in many situations the rule of thumb is right. Interesting point about current loop signalling. I was not aware of this. Do you know whether it applies to ISDN? \$\endgroup\$
    – ARF
    Mar 14 '12 at 9:56
  • \$\begingroup\$ @ArikRaffaelFunke I'm 90% sure that ISDN is a current loop, but it has been a while since I designed an ISDN interface and I don't have my books handy right now. As for the rule of thumb, I added more to my answer about that. \$\endgroup\$
    – user3624
    Mar 14 '12 at 14:47
  • \$\begingroup\$ Thanks for elaborating why the rule of thumb for transmission line reflections is not useful for digital signals. I was considering the effect of modulation on the frequency content but the importance of the "edge rate" completely escaped me... Thanks again! \$\endgroup\$
    – ARF
    Mar 14 '12 at 21:37

Would it be due to the exactly defined Voltage levels that comes with the 2B1Q coding?

From your provided web page: Each dibit has it's specific voltage amplitude. The combinations of dibit is illustrated in the next table.

Dibit   Voltage     Symbol
00           -2.5V          -3
01           -0.833     -1
10           +2.5       +3
11           +0.833     +1

The terminator may quite probably influence the voltages.

  • \$\begingroup\$ I was considering that the modulation scheme modifies the calculation. I was hoping somebody could explain exactly why and how. \$\endgroup\$
    – ARF
    Mar 14 '12 at 9:53

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