Is the pullup resistor on the reset pin of the TCA9548A multiplexer necessary if I do not intend to ever use the reset pin? The datasheet says to "Connect [RESET] to VCC... through a pull-up resistor, if not used." If the pin is not used do I really still need a resistor? Will the IC ever try to assert that pin low for any reason causing a short? I can't think of any other reason why they would recommend a resistor.
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\$\begingroup\$ Possibly to protect the reset input during power cycling. If the datasheet advises you to do it, not doing it you're on your own. \$\endgroup\$– AlmostDoneMay 12, 2018 at 2:08
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2\$\begingroup\$ You may need to debug with a reset in future. It is called good design practice (DFT) which of course you are free to ignore. \$\endgroup\$– Tony Stewart EE75May 12, 2018 at 2:20
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\$\begingroup\$ Fair enough -- I guess it's no harm just to add it in there \$\endgroup\$– Murey TasrocMay 12, 2018 at 2:20
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1\$\begingroup\$ Are you asking if you should use a pull-up resistor, or if you can just short it out to Vcc? \$\endgroup\$– Ale..chenskiMay 12, 2018 at 2:50
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\$\begingroup\$ @Tony Stewart do you mind making your comment an answer so I can accept it and people won't think this is an unanswered question \$\endgroup\$– Murey TasrocMay 12, 2018 at 3:12
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The datasheet actually says, in section 8.4.2:
The RESET input must be connected to VCC through a pull-up resistor
Please note the level of assertion: MUST. It is therefore the condition when manufacturer guaranteed the IC to function correctly.
According to the datasheet, Table 6.5 DC characteristics, the RESETn pin is listed a having leakage anywhere from +1 uA to -1 uA. If you leave this pin unconnected, it might assume some middle level in vicinity of switching threshold, and can begin to oscillate like hell, screwing all things. The effect might be spurious, depend on ambient humidity and temperature, affecting maybe 20-30% of your boards.
If you decide to connect this pin to VCC directly, you might run into another trouble. Again, according to the datasheet, it takes only 6 ns for the RESET to go active and reset all internal circuits. Only 6 ns !!!. You might be even unable to see this glitch even with a good oscilloscope. It is well known that power rails can have glitches due to logic switching around, and bypass caps can't be placed at every centimeter of board. So glitches on power rails do happen, and they might reset your mux configuration in the middle of operations. But if you use say, a 10k pullup, together with 5 pF input pin capacitance you will have a 50-ns filter at least.
But instead of listening to all this musing about glitches, leakages etc. you could simply follow manufacturer's requirement and get a working device without wasting your time. Always assume that there is a reason behind every manufacturer's requirement.
answered May 12, 2018 at 3:21
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\$\begingroup\$ 5ns glitch to Vcc/2 demands very high current on 0.1uf of about> 50A which is unlikely Ic=CdV/dt \$\endgroup\$ May 12, 2018 at 6:17
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\$\begingroup\$ @TonyStewartolderthandirt, 50ns glitch would need only 4A, but will reset the IC in no time, better than a 6-ns glitch. Also, rails are inductive, and can be perceptive to external EMI. Switchers can easily ring at 10-20MHz as well. Equipment with an IC with sensitive RESET connected directly to power rail is a recipe for trouble, it won't pass immunity to EFT (61000-4-4) test. Whatever the reasons are, RESETs should not be connected to power rails, period. \$\endgroup\$ May 12, 2018 at 18:13
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\$\begingroup\$ Of course most newbies ignore magnetic radiated and conducted impulse noise vs ESR of shunt caps if they exist here or not \$\endgroup\$ May 12, 2018 at 19:48