I'll need to sit down with this all later, but MarkT, your advice/explanation to ignore current for this discussion has been extremely helpful. I didn't realize that in this context the current wasn't really important, and thinking about where current was flowing was definitely confusing the hell out of me.
So in general then (at least in situations as simple as the one we're examining), is it true that if part of a circuit is connected to more than 1, what I'll call terminator (ex: +5v, GND), measuring voltage at that part will always reflect the connection to the lowest impedance terminator?
In the ckt being discussed in post #7, the pullup R is in there to bias the "voltage" on the 74HC04 input, not to drive
current into it. The input-impedance of an 'HC input pin is very high, so little current will flow into it.
When the switch is open, pin 1 -> Vcc = logic 1 to the gate. Essentially no current flows through R1 since the only
path for it is into the 74HC04 = high-impedance.
When the switch is closed, pin 1 -> 0V [ground potential] = logic 0 to the gate. Now, current will flow though R1
and to ground, of value I = Vcc/R1.
You're just using the pullup and a simple switch to develop the input voltage to the 74HC04, Vcc in one case and 0V
in the other. You could do the same thing with no pullup and a more complicated, double-throw switch by wiring one
side of the switch to Vcc and the other side to ground.http://www.autolumination.com/images/auto_bulbs/switch_spdt_diagram_wm.jpg