Overshoot and ringing

Short lines ring just as well, just at higher frequencies (than your scope may be able to see), and you CAN kill both ends of the connection by driving excessive energy into them.
The solution is a little transmission line theory. If you have a transmission line (trace on pcb) longer than 1/10 the risetime (for 1ns risetime about 2 cm, quite short), then you need to think.

OP's scenario where you get the double voltage on the output is correct, and shows a problem: Four times the required ENERGY to change the state of the wire was fed into it, and now the excess energy has to dissipate somewhere, and it does so in the receiver, transmitter, and as radiated emissions. That leads us to the art of termination.

First the problem circuit described. Suppose you have a 3.3V driver with 20 ohm output impedance, and a 100 ohm transmission line connected to it. If connected directly to the driver, the driver, when switching, sees a 120 ohm load (the impedance of the driver plus the impedance of the transmission line), and drives a 2.75V change into the wire at a current of 27.5 mA. After some time this step has propagated to the other end of the wire, and is reflected since there's nothing there to absorb the energy. It is reflected back at the transmitter, and will arrive there in a little while, but right now we have the 'outbound' current, plus the reflected current, giving us 5.5V at the output. If there's some poor device there, its protection diodes will take some of the energy. If not, then when the pulse returns to the transmitter, the transmitter sees 5.5V, and may not like that. The pulse will bounce back and forth untill SOMETHING has absorbed the energy.

There are two solutions. First, if the line was infinitely long, the pulse would never be reflected. This is what you do in end termination: Place a 100 ohm resistor (aka. infinitely long transmission line simulator) to ground, or some other voltage. Now the receiver sees the 2.75V step, and there is no reflection or ringing.

The second option is to insert 80 ohms in series with our driver. When it switches, it sees a (20+80+100) 200 ohm impedance, and drives 16.5 mA into the 200 ohm load. This means that the voltage across the input of the transmission line is half of the 3.3V, and the voltage step that propagates down the line is 1.6V. When it reaches the end of the transmission line (where there's only a high impedance receiver), it is reflected back at the transmitter, but first look at the receiver's end: The sum of the incident and reflected steps is 1.6+1.6=3.3V, so it sees a sharp 0V-3.3V transition. Back on the driver end, for two times the line propagation delay, the driver has been sourcing 1.6V across the input of the transmission line. When the reflected pulse arrives, it's also 1.6V, giving a total of 3.3V, and current stops flowing. It's now at steady state with no additional reflections or ringing.

The latter is called source termination, and is best when you have point to point connections, or a few receivers at the end of the line, since a receiver at the midpoint will see a two-step transition. The upside is that it dissipates less than half the energy of the first solution.

End termination is what is used when you send 1Gbps across a hundred meters of cheap twisted pair, or in old style 10Mbps RG58 ethernet.

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