PCB Design – Pull-Up and Pull-Down Resistor on One Trace

Pull-up and Pull-downs are normally used to ensure a line has a defined state while not actively driven. They are used on inputs to prevent floating lines, rapidly switching between high and low and a middle "undefined" region. Outputs normally do not need them.

But most mcu pins are GPIO, and sometimes on startup are defined as inputs instead of outputs. As you said, sometimes you don't want an IC pin input floating on startup, especially like a reset pin that you would normally drive with your microcontroller's GPIO.

This is when you use a Weak Pull-up or Pull-down on the line. Because they are weak, and you choose the default state, they provide no interference with your circuit (If the input is supposed to be low at all times, then pulled high, you choose a weak pull-down, and vis versa), but they do consume a bit of current. This is why you choose a resistor weak (Higher the value, the weaker) enough for the job.

Another normal output setup that uses pull-ups (or pull-downs, rarer) is Open Drain or Open Collector connections. These only drive a connection low, or release the line, leaving it floating. The pull-ups are used to bring the line to a high logic state.

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.