ECE103 Winter 2015 Homework #3 Due Tuesday, Feb. 10, 2015 in class at the beginning of the lecture 1. A semiconductor is doped with ND (ND>>ni) and has a resistance R1. The same semiconductor is then doped with an unknown amount of acceptors NA (NA>>ND), yielding a resistance of 0.5R1. Find NA in terms of ND if Dn/Dp=50. 2. The nonuniform doping in the central region of bipolar junction transistors (BJTs) creates a built-in field that assists minority carriers across the region and increases the maximum operating speed of the device. Suppose the BJT is a Si device maintained under equilibrium conditions at room temperature with a central region of length L. Moreover, the non-uniform acceptor doping is such that (a−x )/b for 0 ≤ x ≤ L p x ≅ N A x = ni x e () () () where a = 1.8 µm, b = 0.1 µm, and L = 0.8 µm. (a) Draw the energy band diagram for the 0 ≤ x ≤ L region specifically showing Ec, EF, Ei, and Ev on your diagram. Explain how you arrived at your diagram. (b) Make a sketch of the ξ-field inside the region as a function of position, and compare the value of ξ at x=L/2. (c) Is the built-in electric field such as to aid the motion of minority carrier electrons in going from x=0 to x=L. Explain. 3. Calculate the electron and hole concentration under steady-state illumination in an n-type Si with GL=1016 cm-3s-1, ND=1015 cm-3, and τn=τp=10µs. 4. Assume that an n-type semiconductor is uniformly illuminated, producing a uniform excess generation rate G. Show that in steady state, the change in the semiconductor conductivity is given by Δσ=q(µn+µp) τpG. 5. The total current in a semiconductor is constant and is composed of electron drift current and hole diffusion current. The electron concentration is constant is equal to 1016 cm-3. The hole concentration is given by p ( x ) = 1015 e− x/L cm-3 for x≥0, where L=12 µm. The hole diffusion coefficient is Dp=12 cm2/s and the electron mobility is µn=1000cm2/Vs. The total current density Is J=4.8 A/cm2. Calculate (a) the hole diffusion current density versus x, (b) the electron current density versus x, and (c) the electric field versus x. 6. Excess carriers are injected on one surface of a thin slice of n-type Si with thickness W and extracted at the opposite surface where pn(W)=pno. There is no electric field in the region 0<x<W. Derive the expression for current densities at the two surfaces. If the carrier lifetime is 50 µs and W=0.1 mm, calculate the portion of the injected current that reached the opposite surface by diffusion (D=50cm2/s).