adcconversionmicrocontroller
While working on my project, I got to a point where I don't know the answer. I have a 10-bit ADC (the one incorporated in the ATMEGA328P) and it returns values up to 1023 (210 – 1).
The official documentation states that the equation to convert the values from the ADC to the corresponding voltage is:
$$
V = \frac{V_{in} \cdot ADC}{2^{10}}
$$
But, if the values go up to 1023, then wouldn't it be better to replace the 210 by 210 – 1?
Thanks.
The solution provided by IgnacioVazquez-Abrams was correct. Apparently, I was incorrectly clearing the lower 4 bits in ADCSRA when I should have been doing it to ADMUX. I've updated the code above to reflect this and the errors I've made in the code have been commented out (they are still there for reference).
In your ADC initialisation, change this line
ADCON1bits.ADFM = 0;
to this
ADCON1bits.ADFM = 1;
As the datasheet explains on page 242
Your configuration effectively resulted in a number that was shifted 6 bits to the right, so was 64 times bigger than expected.
This is illustrated in the following figure from the same section of the datasheet:
Best Answer
Ignoring the harsh reality of the performance of ADCs which can be cheaply built into MCUs, in the ideal case each output count represents a possible well-defined range of input voltages.
Here is an ideal (according to Atmel's definition) 3-bit ADC.
There are \$2^N\$ steps. In the above example, that is 8. With an ideal unipolar ADC, an input of lower than \$\frac{V_{REF}}{2^{N+1}}\$ will give an output of 0- as you can see above, below 1/16 of the 2V reference voltage the output code will be 0.
Between \$\frac{V_{REF}}{16}\$ and \$\frac{V_{REF}}{16}+\frac{V_{REF}}{8}\$ the output count will be 1 and so on, with each step the same.
When you get up to the top of the range, a count of \$2^N-1\$ (maximum count) will represent a voltage greater than \$V_{REF}\cdot \frac {13}{16}\$.
So- the best guess (minimizing error) for a given count x (where \$ 0 \lt x \lt 2^{N-1}\$) is \$ x \frac{ V_{REF}}{2^{N}}\$, as given in the original question.
Usually MCU ADCs will saturate so ideally in the special case of x=0 you can say the input voltage is less than \$0.5 V_{REF}\cdot \frac{1}{2^N}\$ = \$V_{REF}\cdot \frac{1}{2^{N+1}}\$ and if x = \$2^{N-1}\$ then the input voltage is greater than \$V_{REF}\cdot\frac{ 2^{N+1}-3}{2^{N+1}}\$
Of course with a typical MCU of 10 bits or greater, the ADC error and noise will usually exceed 1 LSB, and Vref will usually not be accurate to 1 LSB either so it doesn't matter that much.
TL;DR
Use the original equation.