Test # 2 solution.pdf

# Test # 2 solution.pdf - York University Department of...

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York University Department of Electrical Engineering and Computer Science Winter Term 2015 ENG2002: Mechanics and Materials Engineering Second Test Thursday March 19, 2015, 10:00AM 11:30AM Instructor: Hany E. Farag Instructions: 1- Answer all questions. 2- The examination is closed book. 3- Assume/derive any missing data/equation 4- The marking scheme is shown in the right margin of each question. 5- Only calculators and writing materials are allowed. 6- Clearly, show all steps used in solution. No marks will be given for numerical results unless accompanied with correct solution method. 7- Use correct SI units in all your answers. Name: ____________________________________________ ID: _____________________________________________ Final Mark: Question Q1 (13) Q2 (14) Q3 (8) Total (35) Mark

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ENG2002 Second Test Winter 2015 Page 2 of 12 Question 1 a) With the aid of schematic illustration, In terms of electron energy band structure, discuss reasons for the difference in electrical conductivity between metals, intrinsic semiconductors, and extrinsic semiconductors . Also compare the temperature dependence of their conductivities. [4.5 Marks] Metals Intrinsic Semiconductors Extrinsic Semicond. Sketch Explain For an electron to become free, it must be excited or promoted into one of the empty and available energy states above E f. -conductivity lies between that of conductors and insulators - # electrons = # holes ( n = p ) conductivity lies betwe that of conductors and insulators -- electrical behavior is determined by presence of impurities that introduce excess electrons or holes -- n p Temperature dependence
ENG2002 Second Test Winter 2015 Page 3 of 12 b) The following electrical characteristics have been determined for both intrinsic and n-type extrinsic indium phosphide (InP) at room temperature. Calculate electron and hole mobilities [4.5 Marks] σ (Ω - m ) 1 n ( m 3 ) p ( m 3 ) Intrinsic 2.5 × 10 -6 3.0 × 10 13 3.0 × 10 13 Extrinsic (n-type) 3.6 × 10 -5 4.5 × 10 14 2.0 × 10 12 Solution In order to solve for the electron and hole mobilities for InP, we must write conductivity expressions for the two materials, of the form of Equation 18.13 i.e., σ = n | e | μ e + p | e | μ h For the intrinsic material + ( 3.0 × 10 13 m -3 )( 1.602 × 10 -19 C ) μ h which reduces to 0.52 = μ e + μ h Whereas, for the extrinsic InP + ( 2.0 × 10 12 m -3 )( 1.602 × 10 -19 C ) μ h which may be simplified to 112.4

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