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.
Michael
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).
> 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.
>
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 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 inputs are properly terminated.
> 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.
>
Thought so, but I wanted confirmation in case there was some
aspect I hadn't thought of. Thanks for the reply.
If you're paranoid, you could put a couple of 1N4148's or 1N914's to
Vdd and ground, "downstream" of your load limiting resistor.
Cheers!
Rich