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On a development board, I found this schematic for an up to 100 MHz clock input/output pin hooked up to a 2.54 mm pin header block:

series termination

The 22 ohm series termination makes sense if the pin is configured as an output, but how should this be driven as an input? I tested driving it from a ~20 ohm gate with no further termination through the header block (~3 cm of trace including header) and while everything works, there is a fair amount of over/undershoot visible on the scope due to reflections as compared to parallel terminated pins. What is the best practice in this situation?

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  • \$\begingroup\$ Add ~33 ohm back (source) termination resistor at the output of the driver. This should improve the overshoot & undershoot you said is occurring. Note, I'm assuming that the traces are designed and built for ~50 ohm impedance. \$\endgroup\$
    – SteveSh
    Jun 2, 2021 at 23:39
  • \$\begingroup\$ We did such series "termination" to reduce transmitted noise for EMI compliance. \$\endgroup\$ Jun 4, 2021 at 7:27

2 Answers 2

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I suspect they are using 3.3V logic which can vary around 22 Ohms +/-25 to 50% depending on the chip. The R makes no difference going into a 10K ESD protection resistor or a 1pF ESD diode or the 2pF gate capacitance.

You only need to match the driver to the transmission line, but if there is a mismatch then load capacitance will make a difference.

You can play with my simulation with transmission line length and impedance mismatch, mouse click switches etc. I added ESD protection which also clamps ringing.

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  • \$\begingroup\$ Thanks, that simulation is really helpful. After simulating a few different digital traces on my board and calculating the overshoot, I'm wondering when it becomes a problem? The design I have works, but some of my digital buses apparently have over a volt of overshoot (at least for a few hundred picoseconds). Should I trust that the clamp diodes handle that or do I really want (for example) 32 series resistors on my bus driver IC? \$\endgroup\$ Jun 5, 2021 at 1:21
  • \$\begingroup\$ add the 2nd stage 10k + clamp diodes to simulation if you want to see how well clamps work. It's the margin near threshold that determines SNR BER \$\endgroup\$ Jun 5, 2021 at 8:34
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You say that's a clock input/output -- presumably, the 22\$\Omega\$ resistor is there to terminate it when it's used as an output.

If I were driving it, I'd put the proper series termination on my output. The extra series termination probably slows things down, so I'd verify that things still work. If I were certain that I'd never use that pin as an output I might either replace it with a zero-ohm jumper, or see if there's an opportunity to replace the series termination with a parallel one (official or not). That kind of depends on having a handy ground plane available at the right place though.

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