test2 2010 - Page 1 Materials 1M03, Test #1 2010 March 2...

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Unformatted text preview: Page 1 Materials 1M03, Test #1 2010 March 2 2010 Version (1) Page 1 of 7 Last Name:_______ Model Solutions First Name: __________________ Student ID#: __________________ Section: _____________________ MATERIALS 1M03 TERM TEST #2, 2010 Time: 90 Minutes This test contains 13 questions. You are responsible for ensuring that your copy of the paper is complete. Bring any discrepancy to the attention of your invigilator. SPECIAL INSTRUCTIONS: Aids allowed: 1 double-sided page crib sheet, nominally 8 by 11 in. McMaster standard Casio FX-991 calculator Answer all multiple choice questions on the special answer sheet only. There are 12 multiple choice questions each is worth 2 marks. There is one written question, worth 6 marks. Answer this question on the examination copy. Be sure to show your work so that part marks can be awarded for incomplete answers. Be sure to include units with all answers that require them. Useful constants: Gas constant, R = 8.31 J/mol-K Charge per electron, e = 1.6 x 10-19 C Avogadros number, N A = 6.02 x 10 23 Boltzmanns constant, k = 1.38 x 10-23 J/atom-K = 8.62 x 10-5 eV/atom-K 1cal =4.184J MC ___/24 Q13 ___/6 Total ___/30 Page 2 Materials 1M03, Test #1 2010 March 2 2010 Version (1) Page 2 of 7 Part A: This part consists of TWELVE multiple choice questions, worth 2 marks each. Put your answers on the scan sheet (marked McMaster University Examination Answer Sheet). Read the Marking Directions on this sheet carefully before starting. 1. Consider a boron doped (p-type) semiconductor. It is well-known that the conductivity vs. T curve of such a semiconductor will have three regions; freeze-out region (low T), the saturation region (intermediate T) and intrinsic region (high T). Estimate the temperature at which the saturation zone ends and the intrinsic zone begins given that the concentration of boron is 10 20 /m 3 , E acceptor is 0.01 eV/atom, E gap is 1.6 eV/atom, the number of electrons in the valance band at 0 K is 10 30 /m 3 , the mobility of electrons is 0.14 m 2 / V s and that of holes is 0.05 m 2 / V s . [Hint: Set the conductivity due to extrinsic carriers equal to that due to intrinsic carriers and solve for T ] (a) 273 K (b) 328 K (c) 380 K (d) 401 K To solve this question equate the conductivity in the saturation region to the conductivity in the intrinsic region. This leads to: ) ( | | | | h e h e n e p But, kT E p a exp 10 20 , where Ea = 0.01 eV/atom. and, kT E n g 2 exp 10 30 , where Eg = 1.6 eV/atom Therefore ) ( | | 2 exp 10 | | exp 10 30 20 h e g h a e kT E e kT E Solve for T to get: 380 K. Page 3 Materials 1M03, Test #1 2010...
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test2 2010 - Page 1 Materials 1M03, Test #1 2010 March 2...

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