Question 5 of 6: Weighted-Resistor (Binary-Ladder) Digital-to-Analog Converter

Question 5: Weighted-Resistor (Binary-Ladder) Digital-to-Analog Converter (5+10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Four weighted branches $b_3V_{REF}$ through $R$, $b_2V_{REF}$ through $2R$, $b_1V_{REF}$ through $4R$, $b_0V_{REF}$ through $8R$, all summing at the virtual-ground inverting input of an ideal op-amp with feedback $R_f=R/2$.

Find. (a) $V_o$ in terms of $b_3b_2b_1b_0$. (b) $V_o$ for $R=10\text{k}\Omega$, $V_{REF}=10\text{V}$, word $=1101$.

Approach. Sum the four branch currents at the virtual-ground summing junction and multiply by $-R_f$ (standard inverting summing amplifier).

1. Branch currents at the virtual ground. With the $(-)$ input held at $0\text{V}$, each branch contributes $i_k=b_kV_{REF}/R_k$, and all of it flows through $R_f$ to the output (no current into the ideal op-amp input): $$V_o=-R_f\left(\frac{b_3V_{REF}}{R}+\frac{b_2V_{REF}}{2R}+\frac{b_1V_{REF}}{4R}+\frac{b_0V_{REF}}{8R}\right).$$
2. Substitute $R_f=R/2$. $$\boxed{V_o=-\frac{V_{REF}}{2}\left(b_3+\frac{b_2}{2}+\frac{b_1}{4}+\frac{b_0}{8}\right)}$$ — a standard 4-bit binary-weighted DAC transfer function, MSB $b_3$ down to LSB $b_0$.
3. (b) Numeric evaluation. Word $1101\Rightarrow b_3=1,\ b_2=1,\ b_1=0,\ b_0=1$: $$b_3+\frac{b_2}{2}+\frac{b_1}{4}+\frac{b_0}{8}=1+0.5+0+0.125=1.625.$$ $$\boxed{V_o=-\frac{10\text{ V}}{2}\times1.625=-8.125\text{ V}}.$$