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Calculating current draw

Posted: Sat Mar 09, 2013 2:08 am
by u8nc
we are determining a power supply for our R-pi project to provide 5v at 1amp.
Assuming this to be 1 Amp Hour then it will consume 5 Watts over an Hour.

If we use it in a remote location unattended for 1 week it consumes
24 x 7 = 168 hours x 5 watts = 840 watts.
man alive! that's close to 1 kilowatt, can that be right?

assuming it is, if we use a 100 AmpHour GelCell at 12v ( with a step down/regulator ) it provides us with 1200 watts, and some to spare correct?

or do we get more because the the 100 ampHour rating is for a 12v draw?,
does drawing only 5v increase the time use, or alternatively be able to use a lower rated battery?

Re: Calculating current draw

Posted: Sat Mar 09, 2013 7:30 am
by rpdom
You will get a little less than 1200 Watt Hours due to losses in the step-down regulator. A decent one (switched, rather than linear) should be about 85-90% efficient, which gives you about 1000 Watt Hours.

If the RPi is using 1A at 5V, which is 5 Watts as you said, then the regulator will draw about 5.5W to 5.9W at 12V, or 0.46A to 0.49A.

Your battery should last 200 hours if you drain it completely - but that's not good for the battery. It should be ok for the 168 hours you need though :)
u8nc wrote:it will consume 5 Watts over an Hour.
Your terminology is not quite right here.

It will consume 5 Watts for an Hour. Which is 5 Watt Hours.

Watts are a unit of power being used at any one time.
Watt Hours are a unit of power used over time.
The same goes for Amps and Amp Hours, which is current rather than power.

Re: Calculating current draw

Posted: Sat Mar 09, 2013 7:59 am
by u8nc
The 1A value given is conservatively rated. I suspect 800mA will be the actual for the Pi and other bits on the side. Hopefully that will be the margin which leaves some juice in the battery before its gets swapped out. But your calculated 1000 watt hours should be plenty.

I just checked your figures: I need 10 -15% more to overcome the loss due to inefficiency so the mean of that is 12.5% or 1.125 as a multiplier.
5W x 1.125 = 5.625 and thats about right for
(i) the load and the
(ii) regulator's own operating consumption with
(iii) design efficiency issues

my question about 5v being half of 12v sounds like its irrelevant because the regulator is the first point of consumption.

regards the terminology sermon, you are right. Live and Learn - I just lived!


Re: Calculating current draw

Posted: Sat Mar 09, 2013 8:14 am
by rpdom
Could be worse...

If you used a linear regulator (like a 7805 or similar), it would draw 1A at 12V to provide 1A at 5V. That would be taking 12W from your battery and giving off the 7W difference as heat!

Using that type of regulator the 100 AmpHour battery would last 100 hours at most :(