Thank you Peter -
All good stuff ... let me absorb these concepts, and I will reply after doing some breadboard mockups to get my understanding a little clearer.
One quick question - If I look at the voltage divider diagram on the left, I would think that 5V input going to the digital input pin would only see 68 ohms of resistance (before the branch) and not a total of 168. The 100 ohm resistor is "after the branch" ... what am I not understanding. I can see that anything after the total of the two would see a voltage drop based on the combination of the two resistors.
I will do some mockups and think about it some more.
I=V/R
I=5/68
I=.073
7 mA correct?
Jim
