So when talking to one board, the other 9 are high impedance;
This isn't quite right. The input impedance of the receiver doesn't change when it is listening or not listening. So all 10 loads will be high impedance (or capacitive).
If this is the case, should I series terminate each line according to the transmission line impedance (ribbon cable, so ~100 Ohms)?
This won't do any good. 100 ohms in series with a high impedance is still a high impedance. If you are going to terminate these lines at the receiving end, you would need the termination to be in parallel with the load. But be careful before you do that and make sure your driver can actually drive a 100 ohm load.
Series terminations are more often seen at the source, since the driver tends to be low impedance, and, say, 95 ohms in series with the driver might match a 100-ohm line reasonably well.
If you have no series termination and high impedance, you get up to 2*Vin at the source on the back reflection. That was just one wire; I just scaled that up for 9 more reflections. Is my logic flawed?
Yes, your logic is flawed. Because if you split up the signal to lead to 9 (or 10) loads, only a fraction of the energy would travel down each line. If all the lines were the same length, you'd end up with a total reflection of 2*Vin (or probably a little less because some of the signal would have been reflected back at the fanout point, and returned out of phase with the other reflections).
So what should you actually do?
Depending on your design constraints, you could try
- Connecting the loads in daisy-chain configuration and provide a parallel termination only at the end.
or
- Use a fan-out buffer to drive the signal to each load separately
When probing high frequency signals, the standard way to allow an arbitrary length of cable between the device under test (DUT) and the scope, is to make the scope 50\$\Omega\$ input impedance, and use 50\$\Omega\$ cable.
In the ideal world, that will be good enough, Because the cable is terminated by the scope correctly, no reflections will occur at the scope, so no reflections will make it back to the driven end of the cable. The input to the cable will present a 50\$\Omega\$ load to the device being measured. We can choose to drive that load how we like.
However, in the real world, both scope and cable have a tolerance, and there will be some reflection. At very high frequencies, that could be quite large. Making the drive to the cable approximately 50\$\Omega\$ absorbs whatever does come back, improving the frequency response dramatically.
The 'tidiest' way to make this happen is to arrange for your DUT to have a 50\$\Omega\$ output impedance, to a connector. If the source of signals is low impedance, like the output of a power supply for instance, then a series 50\$\Omega\$ resistor will do nicely. If it's not convenient to use a connectered jig, then solder a 50\$\Omega\$ in line at the end of the cable.
Knowing what I did about matching, I was then surprised on my first day in a microwave lab to be shown how they probed circuits. A 50\$\Omega\$ cable, with a 470\$\Omega\$ carbon resistor soldered to the end. This was the -20dB probe.
Remember I said the input to a cable properly terminated by the scope looks like 50\$\Omega\$. The 470\$\Omega\$ resistor in series with this gives a roughly 10:1, or -20dB pot-down. It doesn't need to be matched at the sending end. It would have a flatter frequency response if it were, but another 50\$\Omega\$ resistor at the probe end would complicate the probe (obviously the cable ground is grounded to the circuit at the 'same' point, size matters!), and decrease signal or increase circuit loading for the same pickoff. For most measurements it was flat enough, and was the right price!
Best Answer
The datasheet is quite clear.
First I need to correct your assumptions.
Note that this chip has a 1ns (typ) rise time and max skew of 500 ps and 1" was recommended max for path length for best performance.
What does this mean?
It means signal integrity of clocks degrades as path length increases beyond this without impedance matching but is preserved with matching.
Transmission Line Rules for impedance matching become important when the risetime , Tr is < 15% of the propagation delay. i.e. Wavelength is \$\lambda =~0.35/Tr\$ thus if Tr is greater than 5% of a \$\lambda\$. Others use 10% as a rule of thumb with less margin on signal integrity.
Otherwise inductive tracks with distributed capacitance can cause overshoot and ringing. Parallel Loading can reduce the Q and decay time.
A true matched impedance with stripline and termination resistor is the ideal scenario at the expense of 50% reduced Vpp swing.
Obviously compromises can be made between ringing and load impedance if you know the level of ringing, but usually this is not needed as terminations can be split from Vcc to gnd to maintain proper bias around input thresholds.
Essentially these rules of Transmission Line design apply to all CMOS for rise time vs path length when these rules need to be applied. This is because is the rise time from higher output impedance drivers with higher load capacitance results in lower slew rates, longer path lengths can be used without terminations using careful controlled impedance tracks.