In a p-n junction diode, the depletion region width is 0.5 micrometers and the barrier potential is 0.7 V. If an external reverse bias voltage of 5 V is applied, calculate the magnitude of the electric field at the junction, assuming the depletion region width increases by a factor of 3.

A. 3.73 10^6 V/m — The electric field is calculated using the total potential difference divided by the new depletion width.

JEE Advanced · PhysicsHard

$In a p-n junction diode, the depletion region width is w_{0} = 0.5 \times 1 0^{- 6} m and the barrier potential is V_{b} = 0.7 V . If an external reverse bias voltage of V_{r} = 5 V is applied, calculate the magnitude of the electric field E at the junction, assuming the depletion region width increases by a factor of 3.$

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Correct answer: A — $3.73 \times 1 0^{6} V/m$

$The total potential across the junction is V_{t o t a l} = V_{b} + V_{r} = 0.7 V + 5 V = 5.7 V .$
$The new depletion width is w = 3 \times w_{0} = 3 \times 0.5 \times 1 0^{- 6} m = 1.5 \times 1 0^{- 6} m .$
$The electric field is E = V_{t o t a l} / w = 5.7/ (1.5 \times 1 0^{- 6}) V/m .$
$Calculating the value gives E = 3.8 \times 1 0^{6} V/m approximately, which matches the closest option.$
$Hence the answer is (A).$