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# 25 x 104 cm 3 ee 332 spring 2013 example 2 problem

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Unformatted text preview: 104 cm-3. EE 332 Spring 2013 Example 2 (Problem 3.11) A semiconductor device has Nc = 1019 cm-3, and Eg = 2eV. If it is doped with 1017 cm-3 donors.11 (fully ionized), calculate the electron, hole, and Prob. 3 intrinsic carrierCalculate electron, hole, and intrinsic carrier concentrations. Sketch band diagram. concentration at 627 0C. Sketch the simplified band diagram. 19 18 17 Nc-10 ^ cm' Er-EF n = Nc-e ^=5-10 kT ^ E =2e V r T=627°C=900K n =10 - V \ ^ 10 17 ^ = 0.36eV -+ EC-EF = - k-T-l n n = -0.078eV-ln > 1019 , vNcy E F -E V = [ (E C -E V )-(E C -E F ) ] = [ E g -(E C -E F ) ] = [2eV-0.36eV] = 1.64eV EF-EV p = N v -e " kT 5-10 18 1.64eV ^ ^ °-078eV = 3.7-109-½ cm cm f n ; =yjn^p = 1.9-10 ,13 1_ _ Eg note: n ; = ^ N c • N v • e 2kT A may also be used v 0.36eV -C "> >2eV 1.64eV •E V ^ EE 332 Spring 2013 Temperature Dependence of Carrier Concentration Temperature Dependence of Carrier Concentration EE 332 Spring 2013 ntrinsic Carrier Concentration Temperature Intrinsic Carrier Concentration vs.vs. Temperature EE 332 Spring 2013 Carrier Concentration vs. Temperature vs....
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## This document was uploaded on 02/18/2014.

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