The photo diode in the figure below has an active sensing area of 10 mm2- a sensitivity of 0-5 A-W and a dark current of 1 - A- The i-to-v converter has a sensitivity of 100 mV-A- For an input light intensity of 4 W-m2- the output V0 in volt is

Output current through photo diode (V_0) =Intensity×Sensitivity×Area V_0 =4Wm^2× 0.5 AW× 10 × 10^ 6 m^2 =20 μA Total current flow through photo diode =I_P + I_D ...

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Standard IX

Physics

Fundamental and Derived Units

The photo dio...

Question

The photo diode in the figure below has an active sensing area of $10 {mm}^{2}$ , a sensitivity of $0.5 A/W$ and a dark current of $1 μ A$ . The $i$ -to- $v$ converter has a sensitivity of $100 mV / μ A$ . For an input light intensity of $4 W / m^{2}$ , the output $V_{0}$ in volt is _______

2.1

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Solution

The correct option is A 2.1
Output current through photo diode $(V_{0})$
$= Intensity \times Sensitivity \times Area$
$V_{0} = \frac{4 W}{m^{2}} \times 0.5 \frac{A}{W} \times 10 \times 10^{- 6} m^{2}$
$= 20 μ A$
Total current flow through photo diode
$= I_{P} + I_{D}$ (Dark current)
Output voltage $i$ to $v$ converter
$V_{0} = (20 + 1) μ A \times \frac{100 mV}{μ A}$
$= 100 mV = 2.1 V$

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The photo diode in the figure below has an active sensing area of 10 mm2, a sensitivity of 0.5 A/W and a dark current of 1 μA. The i-to-v converter has a sensitivity of 100 mV/μA. For an input light intensity of 4 W/m2, the output V0 in volt is _______ 2.1

The correct option is A 2.1Output current through photo diode (V0) =Intensity×Sensitivity×Area V0=4Wm2×0.5AW×10×10−6 m2 =20 μA Total current flow through photo diode =IP+ID (Dark current) Output voltage i to v converter V0=(20+1)μA×100 mVμA =100 mV=2.1 V

The photo diode in the figure below has an active sensing area of $10 {mm}^{2}$ , a sensitivity of $0.5 A/W$ and a dark current of $1 μ A$ . The $i$ -to- $v$ converter has a sensitivity of $100 mV / μ A$ . For an input light intensity of $4 W / m^{2}$ , the output $V_{0}$ in volt is _______

2.1