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Topic: Arduino as charge controller (Read 602 times) previous topic - next topic

ivars211

Jun 28, 2011, 01:47 pm Last Edit: Jun 28, 2011, 01:50 pm by ivars211 Reason: 1
Hello, I'm very new in electronics and Arduino Duemilanove. I have searched google for many solutions, but nothing helps (or I'm bad searcher). So I have homemade solar panel which makes about 63W (max 3.7A * 19V). I could use analog pin to read voltage (splitting my voltage in half and than one more time in half) that would be 19/4 = around 4.75V. The same I would do with my car battery ( 60Amp*h - 12V).
Then I have to just say to arduino that if battery's reading is under 11V turn the solar panel on(from battery) and if battery is over 13V turn off panel. This sounds easy (but like I'm bad at these things yet).

I have already made Arduino as voltage reader for 0-5V. But when I try to divide voltage with 10k resistors it does not split voltage in half, from 18V battery I get about 7V, not 9V as it should be in internet calculators. If it would be 9V, than it would be easy - 2* input reading and it would show correctly to my LCD screen.

So I am asking for help:
How to read current from my solar panel?
If I put 3.7A current to my arduino input for readings only - it will blow up?
Why 10k resistors does not split the voltage in half? How to fix that? I used this calculator http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html
What kind of resistors I have to use with 3.7A and 19V (just for readings)?
P.S. Sorry for my English
And I hope someone can help me with this.


robtillaart

Welcome on the Arduino forum,

In the upperright there is a search box, I know there have been charge controllers discussed before, so you might search for them.
google "arduino charge monitor site:arduino.cc" might be faster ;)
Rob Tillaart

Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)

MarkT

Splitting a voltage in half with a potential divider (10k + 10k), then in half again means that the second potential divider is loading the first so the result will be different to 1/4.  In fact it will be 1/5.  Potential dividers only work if you draw negligible current from them.

Use four 10k resistors in series and measure the voltage from only the bottom one and you'll get 1/4th of the input voltage.
[ I won't respond to messages, use the forum please ]

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