Electronic – Why does a Pull down resistor not affect AC

From a TI datasheet, this is the equivalent circuit for a 7407 buffer:

Note that the input is relatively low impedance (having a 6k pullup to Vcc through the emitter-base junction (which effectively looks like a diode) of the transistor shown). The datasheet says it requires 1.6mA to pull the input low (parameter IIL). This is a lot of current compared with the few microamps, nanoamps, or picoamps typical of CMOS or other high-impedance inputs.

Q: Since the pull-down resistor is supposed to create a low-level at the buffer's input, why doesn't it show something closer to 0V at A?

A 5.1k pulldown resistor is too large to pull the low impedance input down. With 1.6mA flowing through 5.1k, the voltage drop (calculated via Ohm's law) would be 8.16 volts. To pull the input down below 0.8 volts, you would need a 500 Ohm or smaller resistor. I would suggest trying 330 Ohms.

Q: If a buffer's output is supposed to mirror its input, why does it not read the same 1.9V as A?

The output is floating. The voltage you are measuring is just stray charge. Unless the output is actively pulled down by the 7407 or pulled up by an external circuit, it will float.

Also, it should only "mirror" its input in terms of logic state, not voltage. When the input is pulled to a logic low level (below 0.8V), the output transistor is fully switched on. When the input is at a high logic level (above 2V), the output transistor is fully switched off. At input voltages between 0.8V and 2V, the output transistor may be fully on or fully off or partially switched on.

Yes, in your circuit, the resistors form a voltage divider when the switch is closed, assuming that the input impedance (how much current flows into it) of the Trigger pin is much higher than the value of the resistors.

When the switch is open, the 10K resistor R1 acts to keep the voltage on the Trigger Pin from floating. If, say, the Trigger Pin was to high impedance input (that is, it lets very little current flow though it, like the gate of a MOSFET), then without R1 it is possible that just stray electrical charges and fields could be enough to make the voltage on the Trigger Pin go high enough to trigger even though it is not connected to anything.

While you probably do not need the R2 resistor, it could in theory function to reduce the amount of current used when the switch is closed, or specifically to make sure the voltage reaching the Trigger Pin is always a little lower than (rather than equal to) V1 when the switch is closed.

Moving the resistor R2 before or after the switch could in theory also have effects. For example, if V1 is sensitive to voltage transients, then putting the resistor between the switch and V1 could help mitigate any static discharge that happens when someone physically touches the switch, at least compared to having the switch connected directly V1 and using resistor in position B2. Similarly, if you were worried about, say, a screwdriver falling into the switch and shorting the contacts to ground, then adding resistor R2 between the switch and V1 could limit the current in flowing in that case and prevent the short from blowing out the power supply or the connecting wires.

None of these theoretical effects probably matter in your circuit, but in practice they can be a real consideration. Check out this question (of mine!) where the location of the resistor is important for similar reasons...

What is the purpose of adding a 300 ohm to 500 ohm resistor on the Neopixel data line?