Question 5 of 6

Question 5 (15 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.

Approach. An $n$-bit DAC produces $V_{out}=V_{ref}\cdot D/2^n$, where $D$ is the binary word's decimal value; $V_{LSB}=V_{ref}/2^n$ is the step size (one LSB's worth of output).

1. (a) Convert $B_{in}$ to decimal. $$B_{in}=10110100_2=1(128)+0(64)+1(32)+1(16)+0(8)+1(4)+0(2)+0(1)=180$$
2. Compute $V_{out}$. $$\boxed{V_{out}=V_{ref}\cdot\frac{D}{2^8}=5\,\text{V}\times\frac{180}{256}=3.5156\ \text{V}}$$
3. (b) Compute $V_{LSB}$. $$\boxed{V_{LSB}=\frac{V_{ref}}{2^8}=\frac{5\,\text{V}}{256}=19.53\ \text{mV}}$$

Check: $180\times19.53\,\text{mV}=3.5156\,\text{V}$, i.e. the output is simply the code times one LSB. Full scale (code 255) would be $5\times255/256=4.980\,\text{V}$, one LSB short of $V_{ref}$, and this input sits at about 70 % of that range, as expected for a code whose MSB is set.