A photodiode is connected in an ideal op-amp signal conditioning cirucit as shown in the figure. When ligth falls on photodiode the reverse saturation current changes from $100 μ A$ to $200 μ A$ . Then the magnitude of change in output voltage $V_{o}$ of circuit is ____ $V$

11

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Solution

The correct option is B $1$
Redraw the circuit

Since ideal op-amp takes zero input current, Hence no voltage drop across $2 k Ω$ resistnace
$∴ V_{A} = 9 V$
using virtual ground concept
$V_{A} = V_{B} = 9 V$
Case - I: Photodiode current $I_{p_{1}} = 100 μ A$
Apply KCL at node $B I_{P} - I_{P}$
$\Rightarrow 100 μ A = \frac{V_{o} - 9}{10 k}$
so $V_{o} = 10 V$

Case II: Phototodiode current $I_{p_{2}} = 200 μ A$
Apply KCL at node $B$ , $I_{P} = I_{i}$
$I_{p_{2}} = 200 μ A = \frac{V_{0} - 9}{10 k}$
so $V_{o} = 11 V$
Change in output voltage $V_{0} = 11 - 10 = 1 V$

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