In a p-n junction diode, the thickness of depletion layer is 2×10−6 m and barrier potential is 0.3 V. The intensity of the electric field at the junction is

Here, the barrier voltage, V = 0.3 volt and the width of depletion layer, d=2×10−6mElectric field at the junction, E=Vd=0.3 V2×10−6m=1.5×105V/mThe direction of electric field is from n-type to p-type.

In a p-n junction diode, the thickness of depletion layer is 2×10−6 m and barrier potential is 0.3 V. The intensity of the electric field at the junction is

Here, the barrier voltage, V = 0.3 volt and the width of depletion layer, d=2×10−6mElectric field at the junction, E=Vd=0.3 V2×10−6m=1.5×105V/mThe direction of electric field is from n-type to p-type.

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In a p-n junction diode, the thickness of depletion layer is $2 \times 10^{- 6} m$ and barrier potential is 0.3 V. The intensity of the electric field at the junction is

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$0.6 \times 10^{- 6} {Vm}^{- 1} from n to p side$

$0.6 \times 10^{- 6} {Vm}^{- 1} from p to n side$

$1.5 \times 10^{5} {Vm}^{- 1} from n to p side$

$1.5 \times 10^{5} {Vm}^{- 1} from p to n side$

answer is C.

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Detailed Solution

Here, the barrier voltage, V = 0.3 volt and the width of depletion layer, $d = 2 \times 10^{- 6} m$
Electric field at the junction,
$E = \frac{V}{d} = \frac{0.3 V}{2 \times 10^{- 6} m} = 1.5 \times 10^{5} V / m$
The direction of electric field is from n-type to p-type.