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### Topic: Pulldown resistor power requirement (Read 2130 times)previous topic - next topic

#### firehorse

##### May 24, 2013, 12:56 pm
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

#1
##### May 24, 2013, 01:35 pm
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

#2
##### May 24, 2013, 01:42 pm
P(watt)=I*I*R
So you need 50V on the 10k resistor to dissipate 1/4 watt on it..

#### fungus

#3
##### May 24, 2013, 02:14 pm

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 pmLast 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

#5
##### May 24, 2013, 03:02 pm
Pull downs are not normal, are you doing things right?
http://www.thebox.myzen.co.uk/Tutorial/Inputs.html

#### oric_dan

#6
##### May 24, 2013, 07:59 pmLast 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.

#7
##### May 24, 2013, 10:19 pm
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

#8
##### May 24, 2013, 11:19 pm
Duh, half right and half wrong.

#### firehorse

#9
##### May 25, 2013, 05:31 am
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

#10
##### May 25, 2013, 09:12 am
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.

#11
##### May 25, 2013, 02:04 pm
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

#12
##### May 25, 2013, 04:11 pm
OK, makes sense

thanks for the info

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