# Pull up resistor calculation

http://i.imgur.com/TbYIHnb.png?1

So, if I have a pull up and apply a current source. How would I need to read/calculate resistance?

V=IR
I = 5V/8k = .0006A would result in a voltage drop of close to 5V

Using a current source/Voltage reader I supply .0006A which should equate to ~5V drop

But when I read my voltage reader I get 9.83V. Is this correct?

Then my resistance would be 9.83/.0006 ~ 16.4k ?

These calculations do not seem correct.

## Answers and Replies

berkeman
Mentor
http://i.imgur.com/TbYIHnb.png?1

So, if I have a pull up and apply a current source. How would I need to read/calculate resistance?

V=IR
I = 5V/8k = .0006A would result in a voltage drop of close to 5V

Using a current source/Voltage reader I supply .0006A which should equate to ~5V drop

But when I read my voltage reader I get 9.83V. Is this correct?

Then my resistance would be 9.83/.0006 ~ 16.4k ?

These calculations do not seem correct.

Sounds like your current source is not very precise. Keep in mind that real-world current sources need to have some voltage drop across them in order to operate correctly. Typically you need a couple of volts across a current source in order for it to say "in compliance".