Question Number 193065 by MikeH last updated on 03/Jun/23
Commented by MikeH last updated on 03/Jun/23
$$\mathrm{Find}\:{V}_{\mathrm{out}\:} \:\mathrm{from}\:\mathrm{the}\:\mathrm{circuit}\:\mathrm{above}.\: \\ $$
Answered by aleks041103 last updated on 05/Jun/23
$${Golden}\:{rules}\:{for}\:{OpAmps}: \\ $$$$\left.\mathrm{1}\right)\:{I}_{{in}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{V}_{{in}+} ={V}_{{in}−} \\ $$$$\left.\mathrm{3}\right)\:{V}_{+} >{V}_{−} \Rightarrow{V}_{{out}\:} =+{V}_{{c}} \\ $$$$\left.\mathrm{4}\right)\:{V}_{+} <{V}_{−} \Rightarrow{V}_{{out}} =−{V}_{{c}} \\ $$$$ \\ $$$${V}_{+} =\mathrm{0}={V}_{−} ={V}_{{D}} \\ $$$$\frac{{V}_{{out}} }{{R}_{\mathrm{3}} }\:={I}_{{DC}} \\ $$$$\left({R}_{{s}} +{R}_{\mathrm{11}} \right){I}_{{AB}} +{R}_{\mathrm{12}} \left({I}_{{AB}} +{I}_{{DC}} \right)={V}_{{c}} \\ $$$${R}_{\mathrm{12}} \left({I}_{{AB}} +{I}_{{DC}} \right)=−{I}_{{DC}} {R}_{\mathrm{3}} \\ $$$$\Rightarrow{R}_{\mathrm{12}} {I}_{{AB}} =−\left({R}_{\mathrm{12}} +{R}_{\mathrm{3}} \right){I}_{{DC}} \\ $$$$\Rightarrow{I}_{{AB}} =−\frac{{R}_{\mathrm{12}} +{R}_{\mathrm{3}} }{{R}_{\mathrm{12}} {R}_{\mathrm{3}} }{V}_{{out}} \\ $$$$\Rightarrow\left({R}_{{s}} +{R}_{\mathrm{11}} \right)\left(−\frac{{R}_{\mathrm{12}} +{R}_{\mathrm{3}} }{{R}_{\mathrm{12}} {R}_{\mathrm{3}} }{V}_{{out}} \right)+{R}_{\mathrm{12}} \left(−\frac{{R}_{\mathrm{12}} +{R}_{\mathrm{3}} }{{R}_{\mathrm{12}} {R}_{\mathrm{3}} }+\frac{\mathrm{1}}{{R}_{\mathrm{3}} }\right){V}_{{out}} ={V}_{{c}} \\ $$$$−\left(\frac{\left({R}_{{s}} +{R}_{\mathrm{11}} \right)\left({R}_{\mathrm{12}} +{R}_{\mathrm{3}} \right)}{{R}_{\mathrm{12}} {R}_{\mathrm{3}} }+\mathrm{1}\right){V}_{{out}} ={V}_{{c}} \\ $$$$\Rightarrow{V}_{{out}} =−\frac{\mathrm{1}}{\frac{\left({R}_{{s}} +{R}_{\mathrm{11}} \right)\left({R}_{\mathrm{12}} +{R}_{\mathrm{3}} \right)}{{R}_{\mathrm{12}} {R}_{\mathrm{3}} }+\mathrm{1}}{V}_{{c}} \\ $$
Commented by aleks041103 last updated on 05/Jun/23
$${R}_{\mathrm{11}} \:{is}\:{the}\:{resistance}\:{between}\:{B}\:{and}\:{the}\:{center} \\ $$$$\:\:\:\:\:\:\:\:{pin}\:{of}\:{the}\:{varistor}\:{R}_{\mathrm{1}} \\ $$$${R}_{\mathrm{12}} \:{is}\:{the}\:{resistance}\:{between}\:{C}\:{and}\:{the}\:{center} \\ $$$$\:\:\:\:\:\:\:\:{pin}\:{of}\:{the}\:{varistor}\:{R}_{\mathrm{1}} \\ $$$${R}_{\mathrm{1}} ={R}_{\mathrm{11}} +{R}_{\mathrm{12}} \\ $$