HW1 - Solution

HW1 - Solution - Page 1 of 9 UNIVERSITY OF CALIFORNIA...

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Unformatted text preview: Page 1 of 9 UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences EE 105 Prof. Salahuddin Fall 2009 Homework Assignment #1 Solution Due at the beginning of class on Friday, 9/8/09 Problem 1 [10 points]: Intrinsic Semiconductor Material a) The first transistors were fabricated using germanium (Ge) as the semiconductor material. Ge has a much smaller bandgap energy ( E g,Ge = 0.66 eV) than silicon ( E g,Si = 1.12 eV). Would you expect the intrinsic carrier concentration in a Ge sample to be larger or smaller than that in a Si sample maintained at the same temperature? Explain your answer qualitatively, i.e. without resorting to any equations. b) A semiconductor sample is said to be “intrinsic” if its electron and hole concentrations ( n and p ) under thermal equilibrium conditions ( i.e. with no electric field, magnetic field, or radiation applied) are each equal to -- or approximately equal to -- the intrinsic carrier concentration ( n i ). This is because its electrical conductivity is (approximately) equal to that of an undoped semiconductor sample. A doped semiconductor sample can become “intrinsic” if it is heated to a sufficiently high temperature, such that the concentrations of carriers due to thermal generation is much (>10×) greater than that due to dopant ionization. Estimate the temperature at which this occurs for a Si sample doped with phosphorus at a concentration of 10 15 cm-3 , by finding the temperature at which n i is 10 × greater than the dopant concentration. (Use the formula for n i given in Lecture 1, Slide 7. You may assume that the band-gap energy has no temperature dependence.) Solution : a) The lower melting point of Ge indicates that the covalent bonds are weaker in Ge. Therefore, it is easier to free an electron from a covalent bond in Ge, to generate an electron-hole pair. Thus, at a given temperature, the intrinsic carrier concentration in a Ge sample is larger than that in a Si sample. b) Referring to Lecture 1, the intrinsic carrier concentration n i is given by − ⋅ = kT E T n g i 2 exp 10 2 . 5 5 . 1 15 [electrons/cm 3 ], where T is the temperature in K , E g is the band-gap energy in eV, and k is Boltzmann’s constant in eV/K. Given E g,Si = 1.12 eV at T = 300 K, k = 8.62 ∙ 10-5 eV/K, & D = 10 15 cm-3 , assuming that the band-gap energy has no temperature dependence, we want to find the temperature at which n i = 10 ∙ & D = 10 16 cm-3 . Substituting the parameter values into the equation for n i , ⋅ ⋅ ⋅ − ⋅ = − T T 5 5 . 1 15 16 10 62 . 8 2 12 . 1 exp 10 2 . 5 10 , from which T ≈ 708 K = 435 o C Page 2 of 9 Problem 2 [10 points]: Doping and Carrier Concentrations Consider a Si sample maintained under thermal equilibrium conditions at T = 300K, doped with boron at a concentration of 2×10 16 cm-3 ....
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HW1 - Solution - Page 1 of 9 UNIVERSITY OF CALIFORNIA...

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