Question

Question: When electrons cross from the n-type to the p-type to equalize the Fermi energy on both sides in an unbiased diode they leave the n-type side with an excess of positive charge and give the p-type side an excess of negative. Charge layers oppose one another on either side of the depletion zone, producing. in essence, a capacitor which harbors the so-called built-in electric

field. The crossing of the electrons to equalize the Fermi energy produces the dogleg in the bands of roughly E_gap , and the corresponding potential differerence

is E_gap /e. The depletion zone in a typical diode is wide, and the band gap is 1.0 eV. How large is the built-in electric field?

Short answer

Answer

The build in electric field is .

Step-by-step solution

Given data

Electric field is uniform.

Concept of electric field

$\mathit{E}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{i}\mathit{c}\mathbf{ }\mathit{f}\mathit{i}\mathit{e}\mathit{l}\mathit{d}\mathbf{=}\frac{\mathbf{\text{Potential difference}}}{\mathbf{\text{width}}}$

Determine the built-in electric field

Determine the built-in electric field.

$$\begin{gathered} \text{Electric field}=\frac{\text{Potential difference}}{\text{width}} \\ =\frac{E_{gap}}{e\times width} \end{gathered}$$

Solve further as shown below.

$$\begin{gathered} \text{Electric field}=\frac{1.0eV}{e\times 1\mu m} \\ =\frac{1.0eV}{e\times {10}^{-6}m} \\ ={10}^{-6}v/m \end{gathered}$$

The build in electric field is ${10}^{-6}v/m$.