If a pull up resistor is to make the voltage on an input pin high, how does that work since there is a voltage drop across the resistor? Doesn't that mean the voltage coming into the pin would be near 0v since it dropped when going across the resistor? I'm sure I have something backwards i my head.
Thanks for the help
Brian

 

Depends on the input. For many devices such as op-amps, the impedance is very high, say 10MΩ. A pull-up resistor of say 10K would indeed drop a bit of voltage, but a tiny amount compared to the 10MΩ inside the IC. If your "input" is one pole of a lightbulb, then the same 10K resistor would drop almost the full voltage and very little would be seen across the lightbulb.

 

In what context TTL, CMOS, OPTO?
Generally pull-up implies a high impedance input, so there would be next to no drain through the pull up, and register a HIGH.
Max.

 

You have to consider the input impedance of the circuit the pull-up resistor is connected to. It is usually much higher than the pullup resistor value.

Think about it as a resistor divider:

VCC - pullup resistor - high impedance input - GND

Almost all voltage will be on the high impedance input, almost no voltage on the pullup.

 

Ok I was using an Arduino Uno and just trying one of the beginning circuits. I'm not sure what the impedence on the input pin on the Arduino is. I realize the Arduino also already has a pull up resistor built into the controller, I was just confused when I started reading around on pull up resistors and could not get things straight in my head. It sounds like then that the impedence on the Arduino pull up resistor must be greater than the resistor that is being used in the circuit I am trying.

 

I'm not sure what the impedence on the input pin on the Arduino is.

Well... it depends. As long as you leave it as an input, and keep the voltage on the pin within the range of Vss (or ground) to Vdd (Vcc?) (the supply voltage pin) then it will look like essentially an open circuit.

If you go above Vdd (Vcc?) the input will look like a diode to Vdd.

If you go below Vss (ground) the input will look like a diode to Vss.

And if you make the pin an output it has a very low impedance.

 

Just to clarify one more way, an input pin should have a default value, you cannot just leave it floating. That is, you can't have a switch that connects/disconnects the input to ground and assume the pin is high if the switch is open (pin not connected to anything). You can not assume the reverse either, that the pin is connected to ground if the switch open and connected th +5 volts when the switch is closed. Therefore, you must connect the pin to ground (pull down resistor) or +5v (pull up resistor). Since no current is flowing (very little current is flowing) into the high impedence input, you can assume the pin voltage is the same as the rail it is pulled to. Them if you connect a copper wire between pin and the opposite rail, you insure (a) you don't have a short between common and power rail and (b) that the new state of the input pin is what ever rail the bater copper wire (or switch) is connecting to.