Answered

At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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.

Sagot :

ogorwyne

Answer:

0.31 μm

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

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

= √(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!

Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.