Chapter 10: Q65E (page 470)

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?

Expert verified

Answer

The build in electric field is .

Step by step solution

Electric field is uniform.

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

Determine the built-in electric field.

$Electric field = \frac{Potential difference}{width} = \frac{E_{g a p}}{e \times w i d t h}$

Solve further as shown below.

$Electric field = \frac{1.0 e V}{e \times 1 μ m} = \frac{1.0 e V}{e \times 10^{- 6} m} = 10^{- 6} v / m$

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

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