Are standard CMOS logic ICs any more susceptible to damage from external causes than BJT devices via the OUTput terminals? By external causes, I mean things like ESD or a mild leakage current from the mains supply.
As an example, suppose an output from a 4000 series logic gate is intended to drive an external load, but may be left open at times. Assume that it goes to the output terminal through a series resistor (say a few kilohms as a buffer against capacitive loads) but has no resistive path to ground when it's disconnected from the load.
No, the only way you'll damage it is if you end up loading it down too much (like a short to ground, or by trying to supply too much current to a following device), and that's not static related. I suppose maybe a static discharge to the output might jump from the output to the input and kill it that way, but that's a stretch.
The reason CMOS is static sensitive is because it has very high input impedance. When static electricity hits that input, it's a tremendous voltage at miniscule current, and that high voltage blows the input. But if the input was loaded down a bit, it would so easily dissipate the high voltage that the problem would go away.
Thus CMOS is at risk when it's just a stray IC. It may suffer in circuit, if the input is left open to stray input, and no pullup or pulldown resistor (though, that case is less likely to happen since you don't want to leave CMOS inputs open since you then can't rely on the output to be a specific state).
The outputs of CMOS aren't so static sensitive, at the very least they are by nature loaded down by the complementary MOS transistor, so there is finite impedance to any CMOS output.
Yes. That's why I have the series resistor at the output, to serve double duty as a load current limiter *and* for stability with capacitive loads (it's a low-speed circuit that I have not yet constructed or even fully designed).
That /is/ stretching it. I suppose I could make doubly sure by placing bleed resistors between the several output points and ground. After all, resistors cost next to nothing, but I dislike unnecessary clutter.
The inputs are properly terminated.
Thought so, but I wanted confirmation in case there was some aspect I hadn't thought of. Thanks for the reply.
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