Discussion in 'General Electronics Discussion' started by Leek1001, Feb 29, 2012.

1. ### Leek1001

2
0
Feb 29, 2012
Good day all,

I've got 2 times 12V lead accid battery in serie (24V) and I want to measure the voltage of it.
I need to use an voltage divider (of 0.1 (24V / 10 = 2.4V)) before the ADC.
the resistors used in the voltage divider are: R1 / R2 = (3k3 + 39k) / 4k7 with an accuracy of 0.01%.

Reference voltage of the ADC is: 4.096V

How can I calculate the resolution needed to obtain accuracy of 25mV?

I think I need to do it like this:

Reference voltage / accuracy = resolution (M).
4.096V / 25mV = 163.84 --> 2^M = 163.84
M = log(163.84) / log(2) = 7.56 = 8bit.

Could I manage it with this resolution [bits] or do I need to get a higher resolution?

Could I have some distortion?

Thank you all for your help!

Greets,
Leek.

2. ### Harald KappModeratorModerator

11,522
2,654
Nov 17, 2011
Leek,
your calculation is right on the spot. I recommend that you average a few samples to minimize noise in the digital output signal. The batteries's voltage will not change that rapidly under normal conditions (unless you want to observe spikes and dips due to switched loads).

Harald

3. ### Leek1001

2
0
Feb 29, 2012
Hehe, I use a 16bit ADC to do this now... woops ^^

How to calculate the effect of the voltage divider on the ADC?

4. ### BobK

7,682
1,688
Jan 5, 2010
The voltage divider needs to have a low enough output impedance to meet the specs of the ADC. The impedance of your divider is roughly 3.3K. It this is not good enough, you can use a capacitor parallel to the 3.3K resistor, which lowers repsonse time, but is not a problem for this application.

Bob