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I am currently doing a design where I basically have two different types of (digital) signals. The first signal group is fully differential (P/N pair). The other group are single ended signals referenced to ground.

I would like to transfer both these signals via Twisted Pair cabling to another board. The cable producer specifies that the twisted pair cable has a (differential) impedance of 100Ohms, which is what if perfectly fine for the differential signals.

For the single ended signals, lets assume I have an output driver impedance of 50 Ohms and an input impedance of 50 Ohms as well so what I would basically want is a cable impedance of 50 Ohms.

How is this issue addressed in the real world? Is this an issue at all? Is the differential impedance of 100 Ohms "the same" as a single ended impedance around 50 Ohms?

What I understand is that a typical coaxial cable has an impedance of around 50 Ohms, but what about single ended signals on a twisted pair cable?

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  • \$\begingroup\$ How fast are these signals? Do you expect that the track+cable will be more than 1/8 of the wavelength of the highest frequency component? \$\endgroup\$
    – tomnexus
    Feb 14, 2016 at 13:18
  • \$\begingroup\$ You should look abot charactericstics impedance. There is no single ended circuit, if the wires are not equidistant like in a cable then it has no exactly defined characteristics impedance, because it is varying, it is assumed that it will not be used for high data transfer. \$\endgroup\$
    – Marko Buršič
    Feb 14, 2016 at 14:20
  • \$\begingroup\$ Are they digital signals? If so, it might be worth looking at single->differential converter ICs. There are a great many tiny package ICs which will do LVCMOS/LVTTL/etc. to LVDS/PCML/etc. and vice versa. \$\endgroup\$
    – Tom Carpenter
    Feb 14, 2016 at 15:32
  • \$\begingroup\$ @tomnexus: They probably won't but this is also more a general question than one for this specific problem. I've come across this a few times already in the past and usually I didn't have to worry but I think it's important to understand the reasons. Also, I think that it's not the speed of the signal that matters but rise times. \$\endgroup\$
    – Tom L.
    Feb 14, 2016 at 18:29
  • \$\begingroup\$ @TomCarpenter: They are, but this is more a general question than one for this specific problem. We're using differential line receivers and transmitters but there the problem is a different one. \$\endgroup\$
    – Tom L.
    Feb 14, 2016 at 18:32

1 Answer 1

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You can get the effect of a 50\$\Omega\$ impedance line by putting two 100\$\Omega\$ lines in parallel, if you have sufficient number of pairs to do that.

The reason differential lines tend to be 100\$\Omega\$ is they expect the user to use an anti-phase pair of 50\$\Omega\$ drivers and receivers for them. Being anti-phase, the driver impedances are effectively in series, to match the line, and the same for the receivers.

Of course this is all moot if your signal edges are slow enough that the impedance of the line doesn't matter. If the signal rise time is several times the electrical length of the transmission line, then you can ignore proper matching with little problem.

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  • \$\begingroup\$ So if I understand you correctly, you say that simply using the same signal on two physical (100 Ohm) wires will get me my desired 50Ohm cable impedance? What if I don't have that number of wires available? For this very specific problem it doesn't really matter because rise times are not extraordinarily fast and the electrical length rather short. Still, I would like to understand the reasoning. Can you elaborate a bit about the part where you state "being anti-phase, the driver impedances are effectively in series"? Where does that come from? \$\endgroup\$
    – Tom L.
    Feb 14, 2016 at 18:44
  • \$\begingroup\$ A balanced line shouldn't really be connected to an unbalanced feed, even if it has the same impedance. Twisted pairs aren't shielded in any way. By feeding a pair with a single ended signal, and ground, you effectively create an antenna at this point, fed with half the voltage. The average voltage of the pair is half the drive voltage. This voltage excites the pair in common mode against the rest of the chassis, and radiates, and so on. Rather use a balanced line driver. \$\endgroup\$
    – tomnexus
    Feb 14, 2016 at 20:46
  • \$\begingroup\$ @tomnexus You're absolutely correct. I made perhaps an unjustified judgement on the OP's skill, requirements and budget, and answered giving him the least-worst solution using his existing drivers and specified cable. Using balanced mode drivers with balanced cable is of course a better solution. I habitually used just such open cable between FPGAs with S/E drivers going at 100MHz +, and only got into trouble on 500mm extension cables. Mind you, the differential LVDS on the adjacent pairs happily did 800MHz on the same length. \$\endgroup\$
    – Neil_UK
    Feb 14, 2016 at 21:26
  • \$\begingroup\$ @TomL consider a 50 ohm driver, developing +1v into a 50 ohm load, 20mA will flow. Now consider another 50 ohm driver, in antiphase, so developing -1v. This will also drive 20mA into a 50 ohm load. There is 2v between these two outputs. If we put a 100 ohm across these, it will draw 20mA from the 2v. Sketch it out. \$\endgroup\$
    – Neil_UK
    Feb 14, 2016 at 21:29

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