Question

Consider a step pn junction made of GaAs at T = 300 K. At zero bias, only 20% of the total depletion region width is in the p-side. The built-in potential ϕi​=1.20V.

acceptor density in the p-side = 5.68x10^16
donor density in the n-side = 1.42x10^16
Determine the depletion region width, xn​, in the n-side in unit of μm.

Answer 1

0.31 μm

Step-by-step explanation:

this question wants us to Determine the depletion region width, xn​, in the n-side in unit of μm. using the information below.

density in the p-side = 5.68x10^16

density in the n-side = 1.42x10^16

`\sqrt{(2*12.7*8.85E-10)/(1.6E-14)((1)/(5.68E16)+(1)/(1.42E16) )(1.2) }`

= √(1.42x10⁵)(1.76056335x10⁻¹⁷ + 7.042253521x10⁻¹⁷)(1.2)

= √150.74x10⁻¹¹

= 3.882x10⁻⁵

approximately 0.39μm

xn = 0.39 x 0.8

= 0.31μm

0.31 um is the depletion region width. thank you!