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GATE 2016 EC – Question 51

Analog Circuits · Op-amp Circuits · 2 marks · Numerical answer

A $p$-$i$-$n$ photodiode of responsivity 0.8 A/W is connected to the inverting input of an ideal opamp as shown in the figure, $V_{CC} = 15$ V, $-V_{CC} = -15$ V, load resistor $R_L = 10\ \text{k}\Omega$. If 10 μW of power is incident on the photodiode, then the value of the photocurrent (in μA) through the load is ________

The photodiode, reverse-biased through a 1 MΩ resistor to $+V_{CC}$, feeds the inverting input of an ideal op-amp. A 1 MΩ feedback resistor connects the output $V_0$ to the inverting input, the non-inverting input is grounded, and the load $R_L = 10\ \text{k}\Omega$ connects the output to ground.

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Correct answer: 790 to 810

Explanation

The photocurrent is $0.8\ \text{A/W} \times 10\ \mu\text{W} = 8\ \mu$A. This current flows through the 1 MΩ feedback resistor, so the output is $V_0 = -8\ \mu\text{A} \times 1\ \text{M}\Omega = -8$ V, which is within the $\pm 15$ V supply limits. The current through the $10\ \text{k}\Omega$ load has magnitude $\frac{8\ \text{V}}{10\ \text{k}\Omega} = 0.8$ mA, which is 800 μA.