I've got a circuit which is 5v with a 1k pull-up resistor. I've been trying to get my head around the math for a while but I'm failing.
I'm trying to use the variable resistor to give a linear voltage output from 0-5, please could someone advise on the best size to use?
What does the statement have to do with the question? There is nothing confusing about I = E / R.
Your question has all the trappings of an x-y problem as any value will work so long as the element resistivity is linear and there is no output load current.
Telling us what you’re actually atempting to do will probably get you a far more useful answer then the above response.
If you want a linear voltage adjustable from 0-5V, just connect the pot across the 5V supply and take the voltage from the wiper.
Optimum resistance depends on what you plan to attach to the wiper and how much current you want to draw from the 5V supply.
More details = better suggestions.
You cannot get a linear output using a pull-up resistor. You would need a constant-current source.
That is not impossible, but far easier just to use it wired as a potentiometer.
I've got a circuit which is 5v with a 1k pull-up resistor.
Is that the only thing in the "circuit"? If not, give us a schematic or tell us what you know. That fixed resistor may make things harder. Can you get rid of it?
Does the [u]Read Analog Voltage Example[/u] help you?
That example uses a [u]potentiometer[/u] (AKA "pot") which is a standard 3-terminal variable resistor used as a variable [u]voltage divider[/u]. A standard linear pot will give you a linear voltage (2.5V at the center position, within tolerances).
I'm trying to use the variable resistor to give a linear voltage output from 0-5,
You won't find a "plain" variable resistor. You CAN use just 2-terminals of a pot to make a variable resistor, and you CAN put that in series with your 1K resistor to make a variable voltage divider. But, it's NOT LINEAR and you can't get the full 0-5V range because there is always some voltage drop across the fixed 1K resistor.
A pot + pull-up will also never give 0-5V output unless the supply voltage is >5V.
Thanks for the help, I've been doing it wrong, I've been using only 2 legs of the pot, and using a pull-up resistor. I'll do it without a pull-up, and connect:
first leg - 5v
2nd leg - output /wiper
3rd leg - ground.
Sorry for being a numpty
Jimster:
I'll do it without a pull-up, and connect:first leg - 5v
2nd leg - output /wiper
3rd leg - ground.
What are you attaching to the wiper? You don't want it to draw more than 10% of the current in the pot (stiff voltage divider rule of thumb); less is better.
I know it's a pain to attach images on this forum, but when discussing circuits, it's the best way to convey information. That avoids having to ask so many clarifying questions.