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Topic: Pulldown resistor power requirement (Read 772 times) previous topic - next topic

firehorse

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

fungus

What on earth are you trying to do?

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

pito

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

fungus


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


On a pulldown...?
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

pito

#4
May 24, 2013, 02:29 pm Last Edit: May 24, 2013, 02:31 pm by pito Reason: 1
Quote
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  ?


Quote
On a pulldown...?


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

Grumpy_Mike


oric_dan

#6
May 24, 2013, 07:59 pm Last Edit: May 24, 2013, 11:22 pm by oric_dan Reason: 1
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.

CrossRoads

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.
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

oric_dan


firehorse

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


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 ???











Grumpy_Mike

Ok you never said it was an analogue input.

Quote
They say pullup circuits uses less current traditionally

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

Quote
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.

CrossRoads

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
Designing & building electrical circuits for over 25 years.  Screw Shield for Mega/Due/Uno,  Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at  my website.

firehorse

OK, makes sense

thanks for the info


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