I am making a simple Li-Ion battery Tester. Basically its two 5R1 resistor in series and a relay (used to cut off the battery when it reaches cutoff voltage). Arduino is connected with ground to the battery and with A3 to the end of the resistors. But when I try to read the analog value, I receive something like 870, which is about 4.24V. BUT the real voltage is about 3.90V (according to my multimeter). How is that possible? Thanks for any suggestions.
-
I don't understand the description of your schematic. Please draw it instead.– MajenkoCommented Oct 5, 2015 at 18:35
-
Basically I am trying to read the battery voltage, using A3 on Arduino Nano, but it show higher value than it is supposed, which is really weird.– user72028Commented Oct 5, 2015 at 19:02
-
How are you powering the Arduino? How are you calculating the voltage?– MajenkoCommented Oct 5, 2015 at 19:04
-
I am powering it from USB wall socket; I measured it and it outputs 4.92V. Well I am calculation the voltage by multiplying analogRead * 0,00488281– user72028Commented Oct 5, 2015 at 19:06
-
Well, with those figures I calculate it at (870/1023*4.92) 4.184V. How much do you trust your multimeter?– MajenkoCommented Oct 5, 2015 at 19:09
|
Show 5 more comments
1 Answer
I have found the answer. I am not an arduino-newbie, but I was thinking that probably I am making a stupid mistake somewhere. But the problem is in my arduino. I tried setting pin D13 (and then all other pins) to HIGH, and I only measured 4.60V! I measured my USB port and it outputs 5.00V exactly, so there must be a problem in the arduino. So the solution - the ADC takes 4.60V as 1023, and therefore if you calculate it you get that the real voltage is 3.90V, which is exactly the same as my DMM said. Thanks Majenko, you were pointing the right way :) By the way, I was using a cheap Arduino Nano from China, I bought it only for 2.2 USD from Ebay. When I used original Arduino, the code and circuit perfectly works. Do you think that this can limit the functionality of my board?