Thomas said:
Thanks for the input. Would the circuit vary if 12V instead of 5V would be
used? Also, is that a 22 Ohm resistor in series with the pot? And again, I
really appreciate the help!
If everything was in proportion, then no, the circuit could be the
same. IOW, if the pull-in voltage of the 12V relay was (12/5)*4 =
9.6V, and the its DC resistance was (12/5)*400 = 960 ohms, then the
timing would follow a similar curve.
Yes, it is a 22 ohm resistor. With the pot at its maximum, the total
resistance in series with the big cap is therefore 90 ohms.
Dealing with your other post's query, a formula (which would be
complex, as we're dealing with exponential voltage over time) would be
of little practical value to you. As I said, your relay will not have
the specs I've assumed.
There are also other factors I didn't raise earlier, which will also
have a bearing on the practical results, and reinforce my suggestion
to use trial and error (assuming you want to use this ultra-simple
circuit). In practice I suspect the lid won't be as co-operative as in
my simplistic assumptions. IOW, it won't falsely open for 0.1 - 1
second and then stay closed again for a second or so before opening
again (giving the 4700uF capacitor plenty of time to discharge through
the relay's coil). It will probably be opening and closing to some
extent constantly - almost vibrating - with some jumps higher and
therefore lasting longer than others. That sort of behaviour presents
quite a complex signal at the input to the series resistor.