cablescharacteristic-impedancedifferential
I am hunting around for a cable with twisted-shielded pair conductors with the correct differential characteristic impedance for RS-485 (120 ohm)
Most of the cable spec sheets for cables with twisted shielded pair just say "Characteristic Impedance = … " (here is an example: http://www.alphawire.com/Home/Products/Cable/Alpha-Essentials/Communication-and-Control-Cable/6365?device=pdf)
Is this referring to the differential impedance of the two conductors or the impedance of a single conductor?
Scheme #1 is terminating only the differential mode signal, not the common mode.
Scheme #2 is terminating both differential and common mode.
Even with a perfectly symmetrical differential output signal you will have what we call "differential to common mode conversion" in the cable. So at the receiver you will have both common mode and differential mode.
One source of this is the different propagation delay for the two signals of the pair (length mismatch and other effects). You measure this to 2-3ns, so you know it's there.
At the receiver, the common mode signal sees no termination and is reflected 100% (voltage doubling) with scheme #1. With scheme #2 some of that energy is absorbed by the termination resistors (note that the common mode impedance match may not be perfect, but it's definitely better than in scheme #1).
I did a quick simulation to show the effect of the two termination schemes with a 2ns skew in an otherwise perfect setup. See for yourself how much of a difference it makes.
Scheme #1 with only differential mode termination.
Scheme #2 with both differential and common mode termination.
Update:
There is a bit more details in this blog post I wrote while I was at it:
http://www.ee-training.dk/tip/terminating-a-twisted-pair-cable.htm
Update 2:
I swapped the plot for scheme #1 for the correct one. Guess you won't notice the difference, but the simulation was not done correctly.
From your comment:
How do you calculate 28*0.10mm into 38?
You don't. A single 0.10mm strand is AWG38.
The formula is based on AWG36 wire being exactly 0.005" (5 mils) in diameter. All other values are based on an exponential curve:
$$diam = 5\text{ mils} \cdot 92^{\frac{36 - AWG}{39}}$$
Solving for AWG as a function of diameter gives:
$$AWG = 36 - 39\cdot log_{92} \frac{diam}{5\text{ mils}}$$
28 strands of AWG38 has the same cross-sectional area as AWG24.
Best Answer
You don't want the 6365 Alphawire for the connection: while there are 5 twisted pairs, the pairs are not shielded individually. You might have unwanted interference. Better use Belden 9842, 120-Ohms. But it has common shield. If 1-pair cable, it has 120 Ohms in one shield..
For RS-422 (100 Ohm) there is Belden 8162, with individually-shielded pairs.
I am sure if you search deeper, you will find 120-Ohm individually-shielded multi-conductor cable.
CORRECTION: since the RS-485 is supposed to be completely balanced, the cross-talk is very diminished, which allows to use non-individually shielded twisted pairs. The classic cable for two full-duplex links is Belden 3109A, pictured in TI guide.