Analog Inputs as Switches

a. Input	You have pushed the PBNO plunger, connecting the input pins 12 & 13 to what? Hold the plunger down. Look back to TP1.
b. Processor	i. With the plunger held down, what state (high or low) are pins 12 & 13, the inputs to the first gate. And the Output (pin 11) is the opposite because it is a NAND Gate. ii. The capacitor fills almost instantly (TP3) iii. RC1. Review the answer to #6 in Lesson 19. iv. Both inputs to the second gate (pins 1 & 2) are now High, and the Output (pin 3) of the second gate is Low. That means that pin 3 is connected by the switches inside the chip to ground (pin 7).
c. Output	(TP4) Now because there is a voltage difference between V+ and Pin 3 (output) the power flows through the LED through the IC, to ground (pin 7) and the LED turns on.

a. AT Rest	At Rest, the LED turns ON. Return to the explanation of TP4 to explain why it is on now.
b. Press the PBNO Plunger	When you push & release the plunger, the LED actually turns off for a timed period, then turns on again automatically.
c. Thinking	Again, Review TP4. The LED was off when there was no voltage difference ( 9v = 9v) In this situation, when the circuit is active, what voltage is pushing against ground? That's not as hard as you imagine, because pin 3, the output of the NAND Gate, is Low when the inputs are High. So, nothing is pushing against nothing. What's the result?