Resistivity Ratios: Germanium Homework Statement

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SUMMARY

The discussion focuses on calculating the resistivity ratio of n-type Germanium with 1023 ionized donors per cubic meter at room temperature (300K) compared to high purity intrinsic Germanium. The electron-hole product for Germanium is given as 1038 m-6, leading to the conclusion that the relative resistivity can be expressed as PExtrinsic/PIntrinsic. Additionally, the discussion addresses the condition for vacancy density in solids, stating that the relative density of vacancies will remain below 10-8 percent unless the melting point exceeds 1000K.

PREREQUISITES
  • Understanding of semiconductor physics, specifically n-type and intrinsic materials
  • Familiarity with resistivity and its dependence on doping concentration
  • Knowledge of the electron-hole product in semiconductors
  • Basic principles of thermodynamics related to vacancy formation in solids
NEXT STEPS
  • Study the calculation of resistivity in semiconductors, focusing on n-type materials
  • Learn about the impact of temperature on semiconductor properties, particularly at 300K
  • Explore the relationship between vacancy density and melting point in solid-state physics
  • Investigate the role of mobility in determining electrical conductivity in semiconductors
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Students and professionals in materials science, semiconductor physics, and electrical engineering, particularly those working with Germanium and its applications in electronics.

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Homework Statement



(a) A sample of n-type Germanium contains 1023 ionized donors per cubic meter. Estimate the ratio at room temperature (300K) of the resistivity of this material to that of high purity intrinsic Germanium. The Electron hole product for Germanium is 1038m-6 and it can be assumed that the mobility of the electrons and the mobility of the electrons in the hole are similar at 300K.

(b) Suppose it takes an energy of 2eV to create a vacany in a certain solid. Show that the relative density of the vacanies of the atoms will always be less than 10-8 per cent unless the melting point is higher than 1000K.

The Attempt at a Solution



Well, I thought since the electron hole product is 1038 then that gives:

Intrinsic - Ne=Nn = 1019
Extrinsic - Ne=1023 therefor Nn=1015

relitive resistivity can be given as PExtrinsic/PIntrinsic.

I'm not really sure how I continue from there. Do I disregard the mobilities because the are similar?

(b) I'm really unsure how to complete (b), I think it might be something to do with manipulation of n=Ne-E/kT.

Any help would be much appreciated
 
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