GATE 2026 IN Q.65: solution and answer

GATE 2026 · INQ.65 · 2 marks · NAT

A photodiode has a responsivity 0.8 A/W. If the input optical power to the photodiode is 2 mW, the power delivered to a 50 Ω load is _______ µW. (rounded off to the nearest integer)

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Answer and solution

ANSWER

The power delivered to the 50 Ω load is 128 µW.

Official answer: 128Source: Official answer key, GATE 2026 IN (IIT Guwahati)

Responsivity is photocurrent per unit incident optical power: $R = \frac{I _{p h}}{P _{o pt}}$ .
$I_{p h} = R P_{o pt} = 0.8 A/W \times 2 \times 1 0^{- 3} W = 1.6 mA$ .
The photodiode acts as a current source of value $I_{p h}$ , so all of this current flows through the $50 Ω$ load.
Power in the load: $P_{L} = I_{p h}^{2} R_{L} = (1.6 \times 1 0^{- 3} A)^{2} \times 50 Ω$ .
$P_{L} = 2.56 \times 1 0^{- 6} \times 50 = 128 \times 1 0^{- 6} W = 128 μ W$ .

KEY IDEA

A photodiode behaves as a current source whose current is responsivity times optical power. Find that current first, then the load power from $I^{2} R$ .

WATCH OUT FOR

Confusing optical power with electrical power, for example reading 0.8 × 2 mW = 1.6 mW as the load power. The optical power must first be converted to a photocurrent, and the electrical power then comes from $I^{2} R_{L}$ .

From the GATE 2026 IN paper. Every question of this paper · Download the full paper · All 11 Optical questions, with notes

GATE 2026 IN Q.65: solution and answer

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