atomant: The question regarding \$ 2^N \$ or \$ 2^N-1 \$ in conversion formula, comes back regularly, being a source of confusion when working with ADCs. It's closely related to the classic confusion: Full Scale Voltage (FS) versus Reference Voltage (Vref). See my answer on (the discussion is to 3 bit, so please adapt it to 16 bit):
https://electronics.stackexchange.com/a/287348/22676
EDIT:
Answering your edit about which graph represent the "minor" quantization error, I say, in accordance with my post: There are not two graphs, just one. The use of factors \$2^N\$ or \$2^N-1\$ depends on the interpretation of Full Scale Voltage (FS) and Reference Voltage (Vref). Both lead to same conversion formula. I repeat here a similar treatment I used in my other answer (please, review it with more care), just adapting it to "two references" ADC, instead of "single reference" ADC: For a sake of simplicity, consider a 3 bit "two references" ADC with transfer function (including a \$2^N-1\$ factor): $$ V_d = int(\frac{V_a-V_{FS-}}{V_{FS+} - V_{FS-}}\times7) $$
The \$V_{FS+}\$ and \$V_{FS-}\$ are the positive and negative full scale voltages. Applying the definition of int() function:
$$ V_d \leq (\frac{V_a-V_{FS-}}{V_{FS+} - V_{FS-}}\times7) < V_d+1 $$
or
$$ \frac{(V_{FS+}-V_{FS-})\times V_d}{7}+V_{FS-} \leq V_a < \frac{(V_{FS+}-V_{FS-})\times (V_d+1)}{7}+V_{FS-} $$
How \$V_{REF+}\$ and \$V_{REF-}\$ are related with \$V_{FS+}\$ and \$V_{FS-}\$? Note that lower \$V_a \Rightarrow V_{REF-}\$ and higher \$V_a \Rightarrow V_{REF+}\$. Replacing \$V_d\$ equal to 000 and 111 in these two cases:
$$ V_{REF+}=(V_{FS+}-V_{FS-})\times \frac{8}{7} + V_{FS-} $$ $$ V_{REF-}=V_{FS-} $$
Subtracting these expressions and isolating:
$$ V_{FS+} - V_{FS-}= \frac{7}{8}\times (V_{REF+} - V_{REF-})$$ Returning these values on to original transfer function:
$$ V_d = int(\frac{V_a-V_{REF-}}{V_{REF+} - V_{REF-}}\times8) $$
This form is the expression regularly found in ADC datasheets. There is a question of confusion with \$V_{REF}\$ and \$V_{FS}\$. Procceding in a similar manner:
$$ \frac{(V_{REF+}-V_{REF-})\times V_d}{8}+V_{REF-} \leq V_a < \frac{(V_{REF+}-V_{REF-})\times (V_d+1)}{8}+V_{REF-} $$
If \$V_{REF-}\$= -10 V and \$V_{REF-}\$= +10 V, then:
1 LSB = \$\frac{V_{REF+} - V_{REF-}}{8}\$= 2.5 V
\$V_d\$ = 000, correspond to \$V_a\$ = -10 V
\$V_d\$ = 111, correspond to \$V_a\$= \$\frac{7}{8}\times (V_{REF+} - V_{REF-})+V_{REF-}\$ = +7.5 V
\$ V_{FS-}\$ = -10 V
\$ V_{FS+}\$ = +7.5 V
The resulting graph (transfer function):
