The point being made is that “HIGH” and “LOW” are digital concepts, not analog. If you apply a voltage to it using a pull-up (or down), that voltage will have an impact on the analog reading that will vary as you alter the pull-up resistance or the driving current of the analog circuit.
Given that you are asking about setting it high, I’m going to guess that the analog circuit that you want to measure voltage from is a variable resistance to GROUND. Is that correct?
If you want to measure resistance to ground, you need to first determine the range of possible resistance, and you use that range to calculate the amount of resistance with which to pull the circuit up. This is because the analog input measures voltage, and you need to convert the resistance into a voltage that is scaled evenly across the range of acceptable voltage inputs.
So here is the math;
A voltage divider is made up of 2 resistances. We will call them R1 and R2.
R1 will be your pull-up resistor. It connects from your INPUT PIN to high level voltage (1.25).
R2 will be the resistance-to-ground that you are trying to measure. It will connect from your INPUT PIN to GROUND.
The voltage on the INPUT PIN will be 1.25 * (R2 / (R1 + R2))
Now imagine this; say that your R2 is a variable resistor, and its halfway setting is 100k ohms. You’re going to want your voltage for that point to be about 0.625 volts. Right?
0.625 = 1.25 * (100,000 / (x + 100,000))
Solve for x – that will be the size of the pull-up resistor that you want to use for this circuit.
(Hint: x = 100,000)
Now how about if your halfway point is only 1000 ohms? Well same formula, except this time the pull-up resistor (x) is only going to be 1000 ohms.
What happens if we try to measure 1000 ohms resistance with a pull-up resistor of 100,000?
V = 1.25 * (1000 / (100000+1000))
V = 0.012
And if we try to measure 100,000 ohms with a pull-up resistor of 1000?
V = 1.25 * (100000 / (1000+100000))
V = 1.238
So you see what happens when we use a pull-up resistor of the wrong scale? You end up with your readings being bunched up closely to one of the extremes, which means that don’t have enough sensitivity to make a reasonable measurement of them.
So if you want to set an analog pin as a pull-up, then what size of internal pull-up resistor would you use for it? So you see the problem with an analog input pin having any form of pull-up. You would have to add this resistor externally.
Now if you’re coming from the microcontroller world, you will probably know that they very often do allow you to activate an internal pull-up resistor when set for analog input. I generally don’t recommend using them in that way. The only exception, I would say, is if you are using the pull-up resistor for a “holding” situation (what I mean by that, is if you may or may not have something hooked up to the pin, and need to just make sure that you don’t get ghost readings from it), and you KNOW that the resistance you would have to measure is an order of magnitude lower than the internal. For example, Atmel/microchip MCU’s can have internal resistors anywhere from about 20k to 150k ohm depending on the specific chip. If you’re going to be measuring resistances in the 0-1000 ohm range, then you’re ok to activate the internal resistor, since it’s value will have only a minor impact on the reading that you can compensate for. But if you get into a conflicting range, then again, you need to switch to an external resistor. For instance, 1M.
