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Kilcoy Australia
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Hey guys

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

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..  

smiley-sad

Any thoughts ?

cheeers
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Valencia, Spain
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What on earth are you trying to do?

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

On a pulldown...?
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Rapa Nui
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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..  smiley
« Last Edit: May 24, 2013, 07:31:31 am by pito » Logged

Manchester (England England)
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Pull downs are not normal, are you doing things right?
http://www.thebox.myzen.co.uk/Tutorial/Inputs.html
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the land of sun+snow
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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.
« Last Edit: May 24, 2013, 04:22:04 pm by oric_dan » Logged

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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.
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the land of sun+snow
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Duh, half right and half wrong.
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Kilcoy Australia
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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 ???










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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.

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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.
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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
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Kilcoy Australia
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OK, makes sense

thanks for the info

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