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'Antwort: Re: basic question'
1999\04\07@103013 by Jochen Feldhaar

Hello Sebastian,

the answer is not as simple as you suggest. Generally a pullup is connected
to a CMOS input to give it a defined state. Because the input impedance of
an unconnected input is VERY high, it may be switching uncontrolled,
causing problems. So the (still rather high) value of the pullup is very
small compared to the CMOS input impedance. Result: the pin is always at a
defined state of "0" or "1" (I always talk of pullups, but pulldowns are
also OK).

So if an external signal (logic output from another IC) is applied to such
an input, again the resistances decide the resulting voltage on the pin:
the CMOS input impedance is seen in parallel withn the pullup, but
generally the source impedance of the applied signal is very much lower
than that. Result: the input will change to that voltage and note the

Now to your problem: if a voltage does not give a defined logic level on an
input, it is very dangerous to have the source impedance in the same region
as the pullup: the resulting voltage is never very much near the far rail,
for any input case. So NEVER use pullups for signal level shifting, signal
forming or similar functions. If this is necessary, then do it in a
separate circuit BEFORE apppying the signal to the input of a PIC!

Hope this adds some clarification to the pullup/pulldown discussion.
Excuse my blabbing about impedances, as I'm a RF guy mainly....

Jochen Feldhaar DH6FAZ

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