In a p-n junction diode,the thickness of the | vedclass

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(C) Given: Barrier potential,$V = 0.3 ~V$. Thickness of depletion layer,$d = 2 	imes 10^{-6} ~m$. The electric field $E$ is given by the relation $E

Vedclass · Accepted Answer

$1.5 	imes 10^5 ~Vm^{-1}$ from $n$ to $p$ side

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(C) Given: Barrier potential,$V = 0.3 ~V$. Thickness of depletion layer,$d = 2 	imes 10^{-6} ~m$. The electric field $E$ is given by the relation $E = \frac{V}{d}$. Substituting the values: $E = \frac{0.3}{2 	imes 10^{-6}} = 0.15 	imes 10^6 = 1.5 	imes 10^5 ~V/m$. In a $p-n$ junction,the depletion layer is formed by the diffusion of electrons from the $n$-side to the $p$-side and holes from the $p$-side to the $n$-side. This creates a positive charge on the $n$-side and a negative charge on the $p$-side. Therefore,the electric field is directed from the $n$-side to the $p$-side.

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

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