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hw6_1102

# hw6_1102 - EECE7201 H.W#6 Due 1 5.1 text 2 5.2 text 3 A...

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Unformatted text preview: EECE7201 H.W. #6, Due March 15, 2011 1. 5.1, text. 2. 5.2, text. 3. A silicon step junction NA = 4X1018 cm‐3 (assume non‐degenerate), ND = 1016 cm‐3, is at 300K. Calculate the following in equilibrium, assuming the depletion approximation. a. Vbi b. xn, xp, and W c. npo and pno d. Percentage of the depletion region on the p‐side 4. A silicon step junction NA = 4X1018 cm‐3 (assume non‐degenerate), ND = 1016 cm‐3, is at 300K. (Same as problem 3) a. Determine the junction capacitance for an applied voltage of 0 V. b. Determine the saturation current, assuming tn = tp = 10‐6 sec. c. Recompute a. and b. for a change in NA to 4X1017. d. Recompute a. and b. for a change in ND to 1015, keeping NA at 4X1018. e. Which change has a larger effect on the characteristics? 5. An attempt to measure the internal contact potential of a p‐n junction is made by connecting a voltmeter with copper leads to contacts on opposite sides of the junction. The voltmeter reads zero. Explain the failure of this method. 6. Derive the I‐V characteristics of the short‐base diode shown, where the boundary condition at the contact on the n‐side is given by Sp very large (equivalent to saying that the excess minority carrier concentration is zero). Note that this is not in either limit discussed in the text! You need to do the diffusion problem including both positive and negative exponentials. xn' ‐ xn is on the order of Lp, xp' ‐ xp >> Ln. Lp and Ln are the minority carrier diffusion lengths. ...
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