Came across this question recently and saw some answers too, but need a simple answer which is easy to understand.
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2\$\begingroup\$ Can you edit the question to include a link to the existing question and answers, to provide more context. \$\endgroup\$– Chester GillonCommented Feb 24 at 10:18
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1\$\begingroup\$ Share some more information on what is the exact thing that you want to know about interrupts being active low. Searching for 'Why are interrupts active low' provides a lot of links. \$\endgroup\$– Im GrootCommented Feb 24 at 10:32
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2\$\begingroup\$ What part of the existing answers were not easy to understand? What kind of answer you would find easy to understand? \$\endgroup\$– JustmeCommented Feb 24 at 10:35
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\$\begingroup\$ Open drain or open collector interrupt sources don't need to know the supply voltage of the interrupt receiver and are compatible between mixed supply systems. \$\endgroup\$– JensCommented Feb 25 at 15:27
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2 Answers
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That's not always the case. Or it's configurable.
To be like RESET, or to signal an error? Then why is RESET often active-low? To be active during power-up.
In old times, having active-low signals allowed to reduce a little bit the amount of transistors and have faster logic. Transistors (bipolar, NMOS, CMOS...) gates are inverters, it could be better to use an NAND or a NOR gate connected to your input.
Logic levels in TTL circuits allowed open collector, wired NOR connections with NPN easier than the inverse (PNP transistors).
Why not?
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\$\begingroup\$ There is no reason why any signal could not be active high. Not even reset. 8051 MCUs had active high reset. The same resistor-capacitor circuit with cap to supply will guarantee that the reset voltage rises with supply voltage, so it will be active during powerup, and resistor brings it low. So the active state of any reset signal does not really have anything to do with interrupts being active low. \$\endgroup\$– JustmeCommented Feb 24 at 11:06
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Interrupts are active low because this way an unlimited and not fixed number of interrupt sources can be connected to a single input by the so-called "wired OR".
answered Feb 24 at 10:20
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\$\begingroup\$ I will ignore the unrealistic part about unlimited number of interrupt sources, and would call it a wired AND because one pulling low would pull low, and even if so, what prevents you from putting an unlimited amount of high active interrupt sources in theory for wired OR? So there is really no explanation why active low is used. \$\endgroup\$– JustmeCommented Feb 24 at 10:38
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\$\begingroup\$ @Justme, I mean we are connecting many (okay, several) open collector outputs to one INT input. If they were conventional outputs, we would have to have a separate INT input for each of them, and we would have to know in advance how many sources there are. In other words, it gives us flexibility. \$\endgroup\$ Commented Feb 24 at 10:47
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\$\begingroup\$ Yes but what prevents from having a system, where an interrupt output is an open collector or open drain which can only pull an output high, and you have a single pull-down resistor? It is basically identical but mirrored for being active high. \$\endgroup\$– JustmeCommented Feb 24 at 10:49
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\$\begingroup\$ @Justme, To be honest, that crossed my mind, but then I decided that "pulling to the ground" was preferable. \$\endgroup\$ Commented Feb 24 at 10:54
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1\$\begingroup\$ Pulling to ground actually is more preferable, at least in my opinion. \$\endgroup\$– JustmeCommented Feb 24 at 11:00