Question 2.6: Show a layout that might be used to match two capacitors of ......
Show a layout that might be used to match two capacitors of size 4 and 2.314 units, where a unit-sized capacitor is 10 μm × 10 μm.
Four units are simply laid out as four unit-sized capacitors in parallel. We break the 2.314-unit capacitor up into one unit-sized capacitor in parallel with another rectangular capacitor of size 1.314 units. The lengths of the sides for this rectangular capacitor are found from (2.33), resulting in
\mathrm{y}_2=\mathrm{x}_1\left(\mathrm{~K} \pm \sqrt{\mathrm{K}^2-\mathrm{K}}\right) (2.33)
\mathrm{y}_2=10 \mu \mathrm{m}\left(1.314 \pm \sqrt{1.314^2-1.314}\right)=19.56 \mu \mathrm{m} \text { or } 6.717 \mu \mathrm{m}
Either of these results can be chosen for y_2 , and the other result becomes \mathrm{x}_2 ; in other words, the choice of sign affects only the rectangle orientation. Thus, we have the capacitor layout as shown in Fig. 2.31 Note that the ratio of the area of the rectangular capacitor to its perimeter equals 2.5, which is the same as the ratio for the unit-sized capacitor.
