So lets says i have a 16 bit ADC and its going to have an input of 0V -> 20V

I want to know what my LSB resolution is.

So the answer is obviously. (Delta Voltage)/2^(number of bits) or 20/2^16

ok fine....

But now lets say that i have an input range of -10 -> 10V

I dont understand why the conversion is the same. In the first example all 16 bits can be used for data because there is no need for a signed bit.
However, now im trying to represent negative numbers also.

So should the conversion be (Delta Voltage)/2^(number of bits -1)
We need the MSB to be the signed bit so isnt my resolution reduced?

20/2^15??

I dont understand how we can represent the negative number with the same degree of resolution.

Can someone clarify for me??

 

OK. The sign bit on for negative voltages and off for positive, so you have less bit for the magnitude. But the range of the magnitude is now 10V instead of 20. So the resolution remains the same.

 

But the range of the magnitude is now 10V instead of 20. So the resolution remains the same.

I don't understand. They both have a total delta voltage of 20V

 

I don't understand. They both have a total delta voltage of 20V

In other words in one scenario the sign bit has to be used so you only have 15 bits for resolution and in the other example the sign bit can be included in the resolution because its not used for anything

 

Delta voltage is the same. Hence the resolution is the same.

 

In other words in one scenario the sign bit has to be used so you only have 15 bits for resolution and in the other example the sign bit can be included in the resolution because its not used for anything

In one case, there is no sign bit. Hence you have 16 bits to cover a range of 20V.

In the other case, you have 15 bits to cover a range of 10V.

 

In other words in one scenario the sign bit has to be used so you only have 15 bits for resolution and in the other example the sign bit can be included in the resolution because its not used for anything

No, the sign bit chooses between two ranges,
-10-0 and 0-10. Now you have 15 bits for a range 1/2 the size. So the steps are the same.

 

You have 65536 possible states (both 16 bits and 15 bits plus sign give 65536 states)
You have those states divided over a span of 20V in both cases.
Therefore the step size is the same.

 

I suggest carefully reading the actual manufacturer's data pages for the device. NOT the half page summary in some on-line seller's web page. That will provide the correct information.

 

The next question is how much resolution can your circuit achieve.
1 bit is 300uV, and that might be rather smaller than the noise in your circuit.