Question 7. *5.88 Figure 5.106 shows an instrumentation amplifier driven by a bridge. Obtain the gain vo/ Vi of the amplifier. + 25 ΚΩ w 500 ΚΩ ΑΛΛ 20 ΚΩ 30 ΚΩ Vi o Ο- WW 10 ΚΩ νο 40 ΚΩ 80 ΚΩ 2 ΚΩ 10 ΚΩ w 25 ΚΩ W + 500 kΩ WW

by virtaal grand. concept inverting ferminal \left(\underline{V}=V_{I}\right)(1) cor apamp is equal to nominverting termimal \left(V_{t}=V_{N I}\right) & in opamp-1 \left.V_{a}\right) oxemp,2 \left(v_{b}\right) & oparma 3\left(v_{c}\right) are V_{I} \& V_{N I}also intermal carrent eb all opamp. i=09pply \mathrm{KCL} ab V_{C} rode (opamp-3) of V_{-}=V_{I}=V_{C}\begin{aligned}& \frac{V_{c}-V_{0}}{500 k^{c}}+\frac{V_{c}-V_{O_{1}}}{25 k}+i=0 \\\Rightarrow & V_{c-V_{0}}+20\left(V_{c}-V_{O_{1}}\right)+0=0 \\\Rightarrow & 21 V_{c}=V_{0}=20 V_{01} \quad, \quad(1) \Rightarrow V_{0}=21 V_{c-2}-2 V_{O_{1}}\end{aligned}Now \mathrm{ccl} of V_{C} at V_{t}=V_{I}\frac{v_{c}}{500 k}+\frac{v_{C}-v_{02}}{251 c} \neq i=0\begin{array}{l}v_{c}+20\left(v_{c}-v_{O_{2}}\right)=0 \\21 v_{c}-20 V_{2}=0 \\v_{O_{2}}=\frac{21 v_{c}}{20}-2\end{array}(2)\mathrm{KCl} of v_{a}\left(\right. opamp-1) v_{t}=v_{N_{I}}\begin{array}{l}\frac{v_{a}-v_{m}}{20 k^{c}}+\frac{v_{a}-0}{30 k}+i=0 \\3 v_{a}-3 v_{i n}+2 v_{a}=0 \\5 v_{a}=3 v_{i n} \\v_{a}=\frac{3}{3} v_{i n}\end{array}\mathrm{KCL} at V_{q} at V_{-}=V_{I} (oparom-1)\begin{array}{l}\frac{v_{a}-v_{0_{1}}}{10 k}+\frac{v_{a}-v_{b}}{2 k}+i=0 \\v_{a}-v_{0_{1}}+5 v_{a}-5 v_{b}=0 \\6 v_{a}-5 v_{b}=v_{01}\end{array}\mathrm{KCl} af V_{b} im (opamp-2) at V_{+}=V_{N_{I}}(3)\frac{V_{b}-0}{80 k}+\frac{V_{b}-V_{i n}}{40 k}+i=0\begin{array}{l}\Rightarrow \quad V_{b}+2 V_{b}-2 v_{i n} t_{0}=0 \\\Rightarrow \quad 3 v_{b}=2 v_{i n} \\\Rightarrow \quad v_{b}=\frac{2}{3} V_{i n}-3\end{array}\mathrm{kCl} at V_{b} in \left(\right. oparmp-2) at V_{-}=V_{I}\begin{array}{l}\frac{v_{b}-v_{a}}{2 k}+\frac{v_{b}-v_{02}}{10_{k}}+i=0 \\5 v_{b}-5 v_{a}+v_{b}-v_{02}=0 \\6 v_{b}-5 v_{a}=v_{02} \\6\left(\frac{2}{3} v_{i n}-5\left(\frac{3}{5} v_{i n}\right)=v_{02}\right. \\4 v_{i n}-3 v_{m}=v_{0_{2}} \\v_{i n}=v_{02}\end{array}Now from eqn & put value ab v_{a}=\frac{3}{5} \mathrm{Vm}_{\mathrm{m}} \& v_{b}=\frac{2}{3} v_{\mathrm{m}}\begin{array}{l}G v_{a}-5 v_{b}=v_{0 j} \\6\left(\frac{3}{5} v_{i n}\right)-5\left(\frac{2}{3} v_{i n}\right)=v_{01}\end{array}(4)from equn (2) v_{O_{2}}=\frac{21}{20} v_{C} but value ab v_{O_{2}}\begin{array}{ll}\Rightarrow \quad V_{i n}=\frac{21}{20} V_{c} \quad & \text { from ean (6) } \\\Rightarrow \quad & \left(V_{02}=V_{m}\right)\end{array}Now from ean 7 , (8), put value ab v vo, v_{1}, \$ v_{c} in term ab v_{i n} in eqn (\begin{array}{c}\Rightarrow 21 V_{c}-V_{0}=20 V_{0_{1}} \\\Rightarrow 21\left(\frac{20}{21} V_{i m}\right)-V_{0}=20\left(\frac{4}{5} V_{i m}\right) \\\Rightarrow 20 V_{i n}-V_{0}=16 V_{i n} \\\Rightarrow 20 V_{i n}-16 V_{i n}=V_{0} \\\Rightarrow 4 V_{i n}=v_{0} \\\Rightarrow 4 \frac{V_{0}}{V_{i n}}=4\end{array}thank you ...