Question 29.1: Design a 3-bit resistor-string ladder using a binary switch ......
Design a 3-bit resistor-string ladder using a binary switch array. Assume that V_{{{R E F}}} = 5 V and that the maximum power dissipation of the converter is to be 5 mW (not including the power required by the digital logic). Determine the value of the analog voltage for each of the possible digital input codes.
The power dissipation will determine the current flowing through the resistor string by
{I}_{M A X}=\frac{5 \times10^{-3}\,\mathrm{W}}{5\,\mathrm{V}}=1\,\mathrm{mA}Since a 3-bit converter will have eight resistors, the value of R is
R=\frac{1}{8}\cdot\frac{5~\mathrm{V}}{1~\mathrm{mA}}=625~\OmegaThe converter can be seen in Fig. 29.3. Examine the switch array if the input code is D_{2}D_{1}D_{0}=100 or 4_{10}. Since D_{2} is high, the top switch will be closed and the lower switch, \overline{{D_{2}}}, will be open. In the row corresponding to D_{1}, since D_{1}=0, both of the switches marked \overline{{D_{1}}} will be closed and the other two will be open. The LSB controls the largest number of switches; therefore, since D_{0} is low, all of the \overline{D_{0}} switches will be closed and all of the D_{0} switches will be open. There should be only one path connecting a single tap on the resistor string to the output. This is bolded, with the resistor string tapped in the middle of the string. Therefore, \mathcal{v}_{_{O U T}} = ½ V_{R E F}=2.5\ {{V}}. The remaining outputs can be seen in Fig. 29.4.
