Calculating Resistivity of Doped Silicon Sample

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SUMMARY

The resistivity of a silicon sample doped with 5.0 X 1019 donor atoms/cm3 and 5.0 X 1019 acceptor atoms/cm3 can be calculated using the relationship between electrical conductivity and resistivity, defined as σ = 1/ρ. This sample is classified as a semiconductor due to the equal concentration of donor and acceptor atoms, indicating it is not intrinsic. The user is advised to consult their textbook for detailed equations and further understanding.

PREREQUISITES
  • Understanding of semiconductor physics
  • Familiarity with resistivity and conductivity concepts
  • Knowledge of doping in silicon
  • Basic mathematical skills for calculations
NEXT STEPS
  • Study the equations for calculating resistivity in semiconductors
  • Learn about the properties of intrinsic vs. extrinsic semiconductors
  • Research the role of donor and acceptor atoms in silicon doping
  • Explore the relationship between conductivity and resistivity in materials
USEFUL FOR

Students studying semiconductor physics, electrical engineers, and anyone involved in materials science or electronics who needs to understand the properties of doped silicon.

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Homework Statement


A silicon sample is doped with 5.0 X 10^19 donor atoms/cm^3 and 5.0 X 10^19 acceptor atmos/cm^3

a)What is its resistivity?
b)Is this an insulator, conductor or semiconductor?
c)Is this intrinsic material?

Homework Equations


No clue.. this is where I need help


The Attempt at a Solution


The question completely threw me off.

I am assuming I can answer b and c once I calculate part a, but I have no clue which equation to use. Can anybody start me off?
 
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Astronuc said:
Please read one's textbook.

Intrinsic semiconductor is undoped.

The electrical conductivity is the inverse of the resistivity: \sigma = 1/\rho.

Read this page - http://www.virginia.edu/bohr/mse209/chapter19.htm

Sorry, Amazon still hasn't sent me the textbook yet, so I had a hard time doing the homework, searching through internet, which led me to this website at end :(

Thanks for your help
 

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