Homework 3

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).