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4 Pages

### Homework%206_Solutions

Course: ECE EE339, Fall 2009
School: University of Texas
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Word Count: 350

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#6, Homework Due 11/5/09 I. (20 points) Problem 5.13 of text. In (a) Charge storage capacitance is another term for the diffusion capacitance. Part (b) refers to the minority carrier injection/diffusion currents. Assume identical minority carrier lifetimes. Also dominate might be too strong a word. Just figure out which would be greater. II. (25 points) Problem 5.14 of text. Fig 5-22 may be helpful for part (b)....

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#6, Homework Due 11/5/09 I. (20 points) Problem 5.13 of text. In (a) Charge storage capacitance is another term for the diffusion capacitance. Part (b) refers to the minority carrier injection/diffusion currents. Assume identical minority carrier lifetimes. Also dominate might be too strong a word. Just figure out which would be greater. II. (25 points) Problem 5.14 of text. Fig 5-22 may be helpful for part (b). Assume that the Fermi level on the p+ side is approximately at the valence band edge. d III. (25 points) Problem 5.40 of text. (Note that the potential drops through the depletion region here and below arent really linear.) IV. (30 points) Consider putting down a metal electrode with a 4.50 eV work function on a piece of Ge which has an electron affinity of 4.00 eV, an intrinsic carrier concentration at 300K of 2.5 x 1013/cm3, and that is doped with 1.0 x 1016 donors/cm3. Assume an ideal contact. (a) What is the work function in the Ge? (b) Is this a Schottky barrier or an ohmic contact? (c) & (d): repeat parts (a) and (b) assuming that the is Ge doped with 1.0 x 1016 acceptors/cm3. Take the intrinsic Fermi level as being precisely mid gap in the Ge. Neglect tunneling through any barrier and assume that any barrier over a few kBT in height is a significant barrier. (a) q s q E c E F q ( E c E i ) ( E F E i ) q Eg k BT ln( n / ni ) 2 16 3 1 10 / cm 0.67 eV 4.00 eV 0.0259 eV ln 2.5 1013 / cm 3 2 4.34 eV 0.16 eV 4.18 eV (b). This is a Schottky barrier diode: The barrier height to majority electrons injected from the semiconductor toward the metal is qm qs = 4.50 eV 4.18 eV = 0.32 eV = positive and significant. (c) q s q Ec EF q ( Ec Ei ) ( Ei E F ) q Eg k BT ln( p / ni ) 2 1 1016 / cm 3 0.67 eV 4.00 eV 0.0259 eV ln 2.5 1013 / cm 3 2 4.34 eV 0.16 eV 4.50 eV (actually closer to 4.49 if I kept more didgits) (d) This is an ohmic contact: The barrier height to majority holes injected from the semiconductor toward the metal is qs qm = 4.50 eV 4.50 eV = 0.00 eV. No barrier, and anything negative or small leads to the conclusion of ohmic.
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