# ADC output and voltage resolution question

I've made an attempt at this question, but the answer was wrong. Could someone please tell me where I was wrong in my calculation?

• The error might be rounding 14.8 to 15, so the answer should be 14. Jan 27, 2022 at 7:21
• should i not round up to cover every step? Jan 27, 2022 at 7:27
• Your values are all correct. For an ideal ADC the transfer function, the value is rounded down (if your LSB is 0.5V, 0V=0, 0.1V=0, 0.2V=0, 0.3V=0, 0.4V=0, 0.5V=1), and input of or is offset by half a bit (0V=0, 0.1V=0, 0.2V=0, 0.3V=1, 0.4V=1, 0.5=1). The question must be assuming that the transfer function is not shifted over. Bad question IMO. embedded.com/… Jan 27, 2022 at 7:30
• NB: if voltage reference is 2.56 V and ADC (not shifted) is 4 bits -> voltage resolution is (typo error) 2.56/16=0.16 V. So, the voltage for 1111 is 15*0.16= 2.4 V. Jan 27, 2022 at 11:04

## 2 Answers

You wrote:

but the answer was wrong

So please provide the correct answer.

In my opinion, when you got 14.82 you cannot change it to 15 which is the nearest; you are to decide the lowest natural number which is 14.

• It depends on which ADC transfer function is used. embedded.com/… Jan 27, 2022 at 7:31
• I think you're using 'Answer' on this Q&A site like 'Reply' is used on discussion sites. Although well-intended, this is a comment as it just requests more information then presents an opinion, not a definite Answer. I appreciate that you don't yet have sufficient Reputation points to have posted this as a comment. Jan 27, 2022 at 8:39

The voltage resolution is wrong — 4 bits gives $$\2^k\$$ steps = 16, not $$\2^k-1\$$ (that would be the highest code, 0b1111 = 15).

The binary value sounds correct though, regardless of whether the transfer function is offset by 1/2 LSB or not.