### Ensc 324 | Electronic Engineering homework help

ENSC 324 HOMEWORK #2 Fall 2015

DUE: **Monday **October 19, 2015 at **2 PM **(note new time!)

Please note that unless you show work in the derivations and solutions you will get no credit for the

answers. Obviously copied answers from study partners or other sources, etc., will also receive no credit.

Please do all parts of all eight problems. *It is suggested that you make a copy of your homework *before

turning it in in case it cannot be returned before Exam 2 (solution key will be provided).

Problem #1

In a particular sample of n-type silicon, the Fermi energy level varies linearly with distance over a short

range. At x=0, the difference between the Fermi level and the intrinsic Fermi level is 0.4 eV. At x=10-3

cm, the same difference is only 0.15eV. The electron diffusion coefficient is 25 cm2/s.

a) Write an expression for the Fermi level with respect to the intrinsic fermi level versus distance.

b) What is the electron concentration versus distance?

c) What is the electron diffusion current density at x=0, and at x=5 x 10−4 cm?

Problem #2

A rectangular bar shaped silicon semiconductor resistor with a cross-sectional area of 100 μm2,

and a length of 0.1 cm, is doped with a concentration of 5 x 1016 cm−3 arsenic atoms (and no

acceptors). Let *T *= 300 K. A bias of 5V is applied across the length of the silicon resistor.

a) Calculate the current in the resistor.

b) Repeat part *a*) if the length is reduced by 0.01 cm.

c) What is the drift current in parts *a*) and *b*)?

Problem #3

Assume we have silicon with typical mid-range doping levels of donors (no acceptors) at 300K. Assume

that the mobility for electrons is limited by lattice scattering and has a temperature dependence of

approximately T-3/2. Determine the electron mobility at: a) T=200K; b) 400K.

Problem #4

A silicon bar has a length of *L *= 0.1 cm and a cross-sectional area of *A *= 10−4 cm2. The

semiconductor is uniformly doped with phosphorous at a level 5 x 1016 cm−3 (no other doping).

A voltage of 5 V is applied across the length of the material which is at T=300K. The minority

carrier lifetime is 3 x 10−7 s. For *t *< 0, the semiconductor has been uniformly illuminated with

light, producing a uniform excess carrier generation rate of *g*‘ = 5 x 1021 cm−3 s−1. At *t *= 0, the

light source is turned off. Determine the current in the semiconductor as a function of time for -∞

≥*t *≥ ∞.

Problem #5

An n-type GaAs semiconductor at 300 K is uniformly doped with donors (no acceptors) at a

level of 5 x 1015 cm−3 (mid level doping). The minority carrier lifetime is 5 x 10−8 s. A light

source is turned on at *t *= 0 generating excess carriers uniformly at a rate of *g*‘ = 4 x 1021 cm−3 s−1.

There is no external electric field.

a) Determine the excess carrier concentrations versus time over the range 0 ≤ *t *≤ ∞.

b) Calculate the conductivity of the semiconductor versus time over the same time period as

part (*a*).

Problem #6

Consider a bar of p-type silicon material at T=300K that is homogeneously doped to a value of 3

x 1015 cm−3 (ND=0), which may be considered mid level doping. The applied electric field is

zero. A light source is incident on the end of the semiconductor on one end (where x = 0, with x

increasing into the silicon bar). The excess carrier concentration generated at x = 0 is 1013 cm−3.

The lifetimes for electrons and holes are 0.5 μs and 0.1 μs, respectively.

a) Calculate the steady-state excess electron and hole concentrations as a function of

distance into the semiconductor.

b) Calculate the electron diffusion current density as a function of x.

Problem #7

Calculate the position of the quasi-Fermi level with respect to the intrinsic level for the following silicon

crystal that is steadily illuminated with an excess carrier generation rate of 1021 /cm3S:

NA = 1016 /cm3 , ND = 0

τn0 = τp0 = 1 μS

Problem #8

A silicon sample at 300K has the following impurity concentrations: ND = 1015 /cm3 and NA = 0 . The

equilibrium recombination rate is Rp0 = 1011 cm−3 s−1. A uniform generation rate produces an excess

carrier concentration of 1014 /cm3 .

a) What is the excess carrier lifetime?

b) By what factor does the total recombination rate increase?