ISRO EC 2023 Q.56: solution and answer

Official answer: CSource: ISRO's revised and final answer key of 16 February 2024

1. The non-inverting input is grounded, so the inverting input is a virtual ground and the op-amp works as an inverting summing amplifier. Each branch current is its DAC voltage divided by its series resistor.

2. Writing $V_{o2}$ for the upper DAC (bits $b_7\dots b_4$) and $V_{o1}$ for the lower DAC (bits $b_3\dots b_0$), with feedback resistor $R_f=16\text{k}\Omega$: $$V_0=-R_f\left(\frac{V_{o2}}{R}+\frac{V_{o1}}{16\text{k}\Omega }\right)=-\left(\frac{16\text{k}\Omega }{R}{V}_{o2}+V_{o1}\right)$$

3. The two 4-bit DACs are identical and share the same reference, so each output is the same constant times its nibble: $V_{o2}=k(b_7{b}_6{b}_5{b}_4{)}_{10}$ and $V_{o1}=k(b_3{b}_2{b}_1{b}_0{)}_{10}$.

4. An 8-bit word has value $16\times (\text{upper nibble})+(\text{lower nibble})$. The output must therefore be proportional to $16N_{\text{upper}}+N_{\text{lower}}$, so the upper DAC must reach the output with 16 times the gain of the lower DAC.

5. The lower branch has gain $\frac{16\text{k}\Omega }{16\text{k}\Omega }=1$. The upper branch must have gain $\frac{16\text{k}\Omega }{R}=16$.

6. Hence $$R=\frac{16\text{k}\Omega }{16}=1\text{k}\Omega$$ The output is then $V_0=-k(16N_{\text{upper}}+N_{\text{lower}})$; the minus sign is only the inversion of the summing amplifier.