Photo Diode as current source

Fig. 1: Inverting Op-Amp: Photo Diode as current source

$U_{\rm DD}{\rm~=10~V},~U_{\rm SS}{\rm~=-10~V},~R_{\rm 1}{\rm~=10~k\Omega}$

Use the values from figure ## for $U_{\rm IN},~U_{\rm OUT},~R_{\rm 2}$.


Complete the arrows in the scematic of the circuit.
Take the values for $U_{\rm 1},~U_{\rm 2},~U_{\rm OUT}$ from figure ##.
Use these values to calculate the sum of the voltages at node ${\rm N_{12}}$.
Compare your result by measuerement.


$U_{\rm 1}{\rm~=}$


$U_{\rm 2}{\rm~=}$


$U_{\rm OUT}{\rm~=}$


Calculated $U_{\rm 12}{\rm~=}$


Measured $U_{\rm 12}{\rm~=}$


What are your results?

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What will happen if you short-circuit $R_{\rm 2}$?
Try it and explain your results.


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