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Unformatted text preview: Final Exam ECE103 3/19/2009 ALLOWED: Calculators and two 8.5”x11 page of notes. Note you are required to turn in your cheat sheet of notes with our exam. NOT ALLOWED: Textbooks, class notes, problem sets, etc. ATTEMPT ALL PROBLEMS AND SHOW ALL WORK! This exam has a maximum score of 100 points. You have 170 minutes to complete this exam. q = 1.60x10-19 C; k = 8.617x10-5 eV/K; m 0 = 9.11x10-31 kg; ε 0 = 8.85x10-14 F/cm; 1 μ m =1x10-4 cm =1x10-6 m; n i (Si, T=300K) = 1.0x10 10 cm-3 ; E g (Si at T=300K) = 1.12 eV; Ks (Si) = 11.8, K O (SiO 2 ) = 3.9. 1. (10 points) Answer True or False for each question. (a) [2 points] Introducing impurities – dopants - can change the conduction of a semiconductor due to generation of free carriers in conduction and/or valence band ( ). However, the dopants need thermal energy to be activated or so called ionized ( ). Typically at low temperature or intrinsic region, the dopants are not activated ( ), and at sufficiently high temperature or extrinsic region, all the dopants are activated ( ). (b) [2 points] For a p-n junction under thermal equilibrium, the Fermi energy is constant throughout the entire diode. ( ) When a p-n diode is under applied bias, quasi-Fermi levels are introduced, but however the quasi- Ferm Levels remain to be constant across the diode. ( ) The depletion width of a p-n junction changes with dopant concentration proportionally, ( ) but does not change with applied voltages. ( change with applied voltages....
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- Spring '11
- Bipolar junction transistor, Base transport factor, P-n junction, light emitting diode, Emitter Base Collector