hw5solutions1102

hw5solutions1102 - EECE7201, Homework 5 Solutions 1....

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Unformatted text preview: EECE7201, Homework 5 Solutions 1. n=ND‐NA=2x1016cm‐3 p=ni2/n=small! μn=600 cm2 V‐1 s‐1 σ= qnμn=1.9 Ω‐cm ρ=5.2 Ω‐1cm‐1 Note that the 600 is an estimate using the value for ionized impurity scattering (1.8x1017cm‐3), reduced slightly because of impurity band effects. This is an estimate, but it must the equal to or less than the value for ionized impurity scattering alone, which is about 640 c m2 V‐1 s‐1 . 2. Here the impurity band part is dominant, so we just read μ n=72 cm2 V‐1 s‐1 from the graph. Then n= 2x1019 and ρ=4.34x10‐3 Ω‐1cm‐1. 3. 3.4, Text: The conductivity of intrinsic Ge will be much gr eater than that of Si or GaAs because the r number of carriers will be orders of magnitude greater. T he mobility of the different carriers varies, but not nearly as much as the number of carriers. ni is ab out 1,000 times less for Si and 1,000,000 times less for GaAs. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. ...
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This document was uploaded on 03/22/2012.

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