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I have been looking at which TVS diode to select for a protecting a set of RS-422 driver lines, however I am struggling to understand how to choose them based on the diode capacitance. The signal will be 2MHz and be driven from a AM26C31IDBR quad RS-422 driver.

The first option I picked was this SM712, which is designed for RS-485. However, I noticed that the diode capacitance is 75 pF, which seems high however I am not sure what scale to use.

How do I calculate the maximum diode capacitance for a certain frequency signal? The lines will be 120 ohm twisted pairs

If I am correct, higher capacitance increase ESD protection but reduces frequency.

Thanks for any help

enter image description here

Edit: As requested here is how I will use the diodes

enter image description here

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  • \$\begingroup\$ Please show a schematic extract of how you propose to use the TVS diodes. \$\endgroup\$
    – Andy aka
    Commented Oct 27, 2023 at 10:12
  • \$\begingroup\$ @Andyaka I have edited the post to show the schematic extract \$\endgroup\$
    – Friizzle
    Commented Oct 27, 2023 at 10:15
  • \$\begingroup\$ That could depend on what level of protection you want. Higher ESD levels require larger diodes that have larger capacitance. Also, your output 4 is a short circuit. And if you plan to use common off-the-shelf CAT5 cables, you might want to consider the pinout to actually use the differential pairs. \$\endgroup\$
    – Justme
    Commented Oct 27, 2023 at 10:29
  • \$\begingroup\$ @Justme I want the highest protection available while being able to pass 2Mhz. Thank you I will take a look at the pinout I haven't considered that. I am not using the 4th channel because I am going to pass the ground reference in one of the twisted pairs \$\endgroup\$
    – Friizzle
    Commented Oct 27, 2023 at 11:51
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    \$\begingroup\$ Here is an approximative table from Toshiba, But it seems that they put the bar very low. Between 2 and 5 pF. toshiba.semicon-storage.com/ap-en/semiconductor/knowledge/faq/… \$\endgroup\$
    – Fredled
    Commented Oct 27, 2023 at 18:08

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I'm not that familiar with the voltage levels of RS422 but, I would suggest that the diodes will always be reverse biased and, this graph from the data sheet is worth considering: -

enter image description here

So, digging into what the voltage levels might be reveals this image from here: -

enter image description here

So, you could make a decent argument that the average reverse bias is 2.5 volts and, this would make the TVS reverse biased capacitance more like 50 pF. Then, knowing that the most likely signal disruptor is an impedance mismatch, you could argue that it is the series capacitance of two 50 pF that should be considered.

This means something like 25 pF and, at 2 MHz, that's an impedance of 3183 Ω. However, you should consider that the 7th harmonic is relevant (it could be a square wave) and now, the capacitive impedance drops to 454 Ω. Compare this with a typical termination resistor of 120 Ω.

Is that significant or can it be ignored. I'd favour accepting that it won't be a problem. If in fact you mean that the data rate is 2 Mbits/sec then, the effective frequency drops to 1 MHz and this becomes even more likely not to be a problem.

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  • \$\begingroup\$ Thank you for the help. Could you explain to me abit more about how you considered the 7th harmonic here? To be more specific, it needs to be able to propagate 1.25 us pulses but differentiate between a 33% and 66% duty cycle pulse (I am targeting WS2812B LEDs) \$\endgroup\$
    – Friizzle
    Commented Oct 27, 2023 at 14:44
  • \$\begingroup\$ The 7th harmonic is a good rule of thumb for ensuring you cover the bases when dealing with signals that have a decent rise and fall time (i.e. are square-ish). If in doubt, try simulating the effect @FergusRooney \$\endgroup\$
    – Andy aka
    Commented Oct 27, 2023 at 14:59
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    \$\begingroup\$ @FergusRooney if we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. \$\endgroup\$
    – Andy aka
    Commented Nov 3, 2023 at 13:31

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