Uint16 to int(signed) conversion

Amit Aryal said:

Uint holds a 12 bit adc conversion number between 0x0FFF (3.3V adc (4095)) and 0x0000 (0V adc (0)). I want to convert in into in2 so that in2 holds a 12 bit signed number between 2047 and -2048 without overflow.

That is a semantic contradiction.

sorry i mean in not Uint rephrasing,

in is declared as Uint and in2 is declared as int. Variable in holds a 12 bit adc conversion number between 0x0FFF (3.3V adc (4095)) and 0x0000 (0V adc (0)). I want to convert in into in2 so that in2 holds a 12 bit signed integer between 2047 and -2048 without overflow.

Hi!

Conversion, actually math, you re looking for:

in2 = (int)in - 2048;

Note that "blind" typecasting of uint in to int in is safe in your case because max value it holds is <= 4095. It will overflow if value is larger than 32767.

Apart from the simple type cast and the "shift" (subtraction of 2048) - having values from -2048...+2047 representing all-positive ADC values and voltages is a bit dubious. That would fit to truly differential ADCs. However, you need to keep track of your ideas, and be able to read and understand your code later on.

f. m. said:

having values from -2048...+2047 representing all-positive ADC values and voltages is a bit dubious

Depends on ground reference point. What if it is VCC/2? Actually "VCC/2 virtual ground" is how most low-end/low-speed DSP circuits do - use single-rail power supply, single-supply opamps and single-ended ADC/DAC, just lift the ground reference for signals. Note that TMS320 is DSP btw

Do you mean I can get between -2048 and +2047 with manipulating the hardware?I don't quite understand what you mean?

You are right of course. Only the O.P. did not present any hint that the analogue input originates from a differential signal.

My point is - one should keep some consistency with a design, else the next one (or even oneself, a few month later) will be quite confused.