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I've been reading a lot and also been advised to place a series resistor (order of 33 - 45 ohms) on outgoing signal lines. Also called a termination resistor I believe. Example on this topic: Series resistor on digital signal lines

I'm wondering, what is the electrical aspect (theory) behind the resistor making the edges less 'peaky'. And also, it has been mentioned it is important to place the resistor as close as possible to the driver of the signal. What will change, and why, if it will, for instance, be placed more on the receiving side? (in case of a PCB that has no resistor on the driving side, will adding one on the receiving side make any difference?).

I could have responded to those topics but I thought a new topic for this particular discussion would fit better.

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    \$\begingroup\$ In addition to reducing ringing, it's also a bit of short circuit protection -- a 33 ohms resistor will limit 3.3V to 100mA. \$\endgroup\$ Feb 2 '21 at 9:41
  • \$\begingroup\$ What is your signal technically? \$\endgroup\$
    – Andy aka
    Feb 2 '21 at 10:43
  • \$\begingroup\$ Ah clock signal, 12Mhz I believe. And a TX for uart \$\endgroup\$
    – Mart
    Feb 2 '21 at 11:17
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A simple explanation is that wires on PCB and in cables have capacitance. Adding a resistor before capacitance makes an RC filter which is a low-pass filter that removes high frequency content.

It means that the IO pin driving the wire does not have a direct capacitive load, as there is a resistor to limit the current to capacitance, so the current peak at the signal edges is smaller, and voltage changes slower.

And the closer the resistor is to the output pin, the less capacitive load there is directly at the output pin, so the placement of the resistor does matter, and will not help the output pin if it is at the receiving end of the wiring. It can still protect the receiving IO pin from the spiky signals.

A more complex explanation involves transmission line theory, which says that any impedance mismatch and discontinuities in the impedance creates reflections, which is another reason to match the IO pin output impedance with a resistor to the impedance of wires being used to carry the signals.

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As shown in the linked topic, the PCB trace is modeled as a resonant tank circuit (L & C) which will show overshoot (ringing) when not damped with a resistor. So the theory behind the resistor is to dampen the resonant circuit. The damping does not work if the resistor is installed at the receiver side, because then it is after the resonant circuit of the trace.

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