HW5_Fa11_solutions_final

HW5_Fa11_solutions_final - MatE153:HW5Solutions Reading...

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1 MatE 153: HW 5 Solutions Reading Kasap: 5.2 and 5.3 Be sure to look over the reading review notes on D2L. This will help you prepare for the final. Total Points: 70 Points 1. How would the saturation temperature change if you increased the dopant concentration? if you decreased the band gap? (5 points) If you increased the dopant concentration then the saturation temperature would increase (it would take a greater amount of energy to ionize them all). The second part of the question was supposed to ask what would happen to TI, the intrinsic temperature, if you decreased the band gap. If you decreased the band gap, the intrinsic temperature would decrease. It would require a lower temperature (with a smaller badngap) for the intrinsic carriers to equal the carriers from the ionized dopants. 2. What is a degenerate semiconductor? What would be a use of such a semiconductor? (5 points) As more dopants are added to a semiconductor, the Fermi energy rises towards the conduction band for n-type or lowers towards the valence band for p-type. A degenerate semiconductor is one that is so heavily doped the Fermi level is in the conduction band for n- type or valence band for p-type. A degenerate semiconductor behaves almost like a “metal”. Heavily doped degenerate semiconductors are used in various MOS (metal- oxide- semiconductor) devices where they serve as the gate electrode (substituting for a metal) or interconnect lines. They have an advantage over a different metal in that they form better interfaces with silicon semiconductor. 3. Si is doped with 10 18 atoms/cm 3 of B. a. Using a spreadsheet program, plot the ln p versus T for 0 to 2000K. For full credit, clearly indicate the equation you used in each region. (10 points) In region 1 (freeze-out, ionization regime) Δ = kT 2 E exp N N 4 1 p 2 / 1 A V In region 2 (extrinsic, saturation region), p=N A = 10 18 atoms/cm 3 () = = = kT 2 E exp N N n p n G 2 / 1 V C i 2 / 3 2 * h V h kT m 2 2 N ⎡ π =
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2 The plot below uses the equations above in the appropriate temperature ranges. The data plotted is highlighted in red.
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This note was uploaded on 03/13/2012 for the course MATE 153 taught by Professor Staff during the Fall '08 term at San Jose State University .

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HW5_Fa11_solutions_final - MatE153:HW5Solutions Reading...

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