Analog to digital scale

Question

Say I have a 2-bit ADC operating on a range [-2 V, +2 V] to make it easy. After the quantization I have a sequence of values [01, 11, 10]. Do these values represent voltage intervals or specific voltage values?

As far as I understood they represent intervals so [[-1 V, 0 V], [1 V, 2 V], [0 V, 1 V]] but I was watching this explanation (link to example minute) and it confused me a little (since real values are mapped to 5 values rather than 4 intervals). After the ADC inside a microprocessor, how would the corresponding digital signal of my example look?

Answer 1

n bits = 2

If using a bi-polar converter

 Voltage  Code
 3 1/3         Vref
 2        11   Vmax
 2/3      10 
-2/3      01
-2        00

$$V_{step} = 4/3$$ $$V_{step}= \frac {V_{ref}-V_{min}}{2^n} = \frac {V_{max}-V_{min}}{2^{n-1}}$$

Notice that 0V cannot be measured.

If there is noise greater than 1 LSB for most ADCs, a simple moving average can be applied to extract a zero value.

Many ADCs transition to the next code at the 1/2 LSB mark to improve quantization noise. This also places the desired code at the midpoint of the step-shelf.

See the link in @Transistor's comment.

Answer 2

If you have a 2-bit unipolar ADC operating on range -2 V, + 2 V then you will have a sequence of values:

00 = -2 V
01 = -1/3 V
10 = +1/3 V
11 = +2 V

You could expect the digital reading to change midway between each point.

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. A 2-bit ADC with resistor chain, comparators and binary encoding.

• When AIN < -1.333 V all comparators' outputs will be low.
• When -1.333 V < AIN < 0 V CMP3 will switch high.
• When 0 V < AIN < 1.333 V CMP2 will switch high (as well).
• When 1.333 V < AIN < V1 CMP 1 will switch high (as well).

U1 encodes these as follows:

CMP1 CMP2 CMP3   b1 b0
 0    0    0      0  0
 0    0    1      0  1
 0    1    1      1  0
 1    1    1      1  1