# Pulldown resistor power requirement

Hey guys

Yipeeee, I got it working (after 2 bloody hours dammit :))

I am using a 10k resistor to "pull_down" ground as the sensor needs stability

I have a 10k connected now but its 1/4 watt

is 1/2w better in case it gets hot ?

Yes No ?

I don't know how to calculate power on that 10k resistor

I know I=V/R, but the rest is rocket science to me..

:(

Any thoughts ?

cheeers

What on earth are you trying to do?

P(watt)=I*I*R So you need 50V on the 10k resistor to dissipate 1/4 watt on it..

pito: P(watt)=I*I*R So you need 50V on the 10k resistor to dissipate 1/4 watt on it..

On a pulldown...?

I am using a 10k resistor to "pull_down" ground as the sensor needs stability

I have a 10k connected now but its 1/4 watt

is 1/2w better in case it gets hot ?

On a pulldown...?

I do not know what firehorse does, but you can have a pulldown 10k resistor with more than 50Volts on it.. :)

Pull downs are not normal, are you doing things right? http://www.thebox.myzen.co.uk/Tutorial/Inputs.html

Edited per Bob:
P = V^2 * R = V * V / R also works.
P = V^2 / R = V * V / R also works.
P = 5V * 5V / 10K = 0.0025 watts

To dissipate 1/4W at 5V, R = V^2 / P = 5V*5V/0.25W = 100 ohms.
So, your 1/4W 10K R is cool [literally], for voltages in the range 0…5V.

little typo there:

P = I^2*R = V^2/R

Comes from P = I*V V=IR or I=V/R, subbing in for I or V yields the above.

Duh, half right and half wrong.

Wonderful, now rocket science isn't so bad, Ohms law works really well here thanks guys

Grumpy_Mike: Pull downs are not normal, are you doing things right?

I dunno, it is working for the situation I need

Here's the story:

I have the 10k resistor connected to ANALOG INPUT and ground because of a floating input, when the sensor open circuits,

ie: INPUT goes LOW

Thanks for the link, I tried both pullup & pulldown, both work fine

They say pullup circuits uses less current traditionally but does my hardware need pullup or will pulldown work OK too ?

Just not sure why I would need, to use pullup if that's the case ???

Ok you never said it was an analogue input.

They say pullup circuits uses less current traditionally

No I don't think there is anything in that.

just not sure why I would need, to use pullup if that's the case ?

You need one or the other but as I said in that page it is a touch safer having a long lead with a ground rather a long lead with power.

http://www.atmel.com/Images/doc8349.pdf Atmel uses pullups here as part of power reduction technique. I think Nick Gammon also did some testing on this http://www.gammon.com.au/forum/?id=11497

OK, makes sense

thanks for the info