Help with voltage divider and measuring a 9V battery

Hi All,

A project I'm working on I need to measure the charge of a 9V battery. Essentially it's a voltmeter.

Using the analogue inputs on a Uno (the end project will just be a stand alone ATmega328), I need to scale back the 9V to under 5V. I was told to use a voltage divider as per this site:

I used the values 6.8 and 4.7 ohm for the resistor. According to the calculator on the website this would scale back 12V to 4.9V hence allowing for a way overcharged battery to be connected without damaging the input pins.

However, upon connecting the battery to the resistors (arduino not connected), the 6.8ohm resistor started to smoke and burn. I'm at work right now and don't want to set off the fire alarms (I'm right next to our major data centre!).

Can anyone advise what I have done wrong? Or what I should be doing?

Thanks, Ross.

your going to need higher resistance than that

try like 6.8k and 4.7k or 680k and 470k, you just dumping a ton of current and power to ground, which creates heat, which creates fire (like 1 amp and 12 watts)

analog readings dont need much current to function, I wouldn't be surprised if you didnt insta-drain a battery doing that

and when you get some down time look up ohms law, and a online calulator, it will help you avoid those types of issues (like dropping 12 watts into a 0.25 watt resistor hehe)

and dont worry, everyone smokes something here and there... just part of it

Thank you! Didn't even think about power/current.... that's my brain freeze moment of the week.

Used a 47k and 67k - works great. No heat whatsoever.


[quote author=Magic Ross link=topic=125570.msg943993#msg943993 date=1349325470] Used a 47k and 67k - works great. No heat whatsoever. [/quote]

Actually 10000 times less heat, not no heat ;)