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Understanding Fermi-Distribution Function & Electron Concentration in Silicon, Assignments of Electrical and Electronics Engineering

Georgia Institute of Technology - Main CampusElectrical and Electronics Engineering

A homework assignment for ece 3040, focusing on the fermi-distribution function and electron concentration in silicon. Students are required to calculate the probability of empty states, electron and hole concentrations, and fermi level positions for various doping conditions. They are also asked to understand the concept of partial ionization and why certain impurities are more commonly used as acceptors or donors.

Typology: Assignments

Pre 2010

Uploaded on 08/05/2009

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ECE 3040 Homework 2
1) Purpose: Understanding what the fermi-distribution function is telling us.
The fermi-distribution function depicted in Figure 2.15 describes the probability that a state is
occupied at a given energy. Plot the probability that a state is empty.
This is 1-f(E) . See blue curve under occupancy factors below.
pf3
pf4
pf5

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ECE 3040 Homework 2

  1. Purpose: Understanding what the fermi-distribution function is telling us. The fermi-distribution function depicted in Figure 2.15 describes the probability that a state is occupied at a given energy. Plot the probability that a state is empty. This is 1-f(E). See blue curve under occupancy factors below.
  1. Purpose: Understanding what the electron distribution is within each band - Problem Pierret 2.
  1. Purpose: Understanding the relationships between various energies - Problem Pierret
  1. Purpose: Understanding of simple electron-hole concentration relationships. Find the electron and hole concentrations as well as the fermi level and intrinsic energy positions for a silicon sample with the following conditions: (You may need data from tables 2.1 and 2.3 and assume Total ionization and a 27 C temperature) a) 10 16 cm-3^ Al Al is an acceptor, Si has n (^) i=1e10cm-3^ , since Na >>Nd and Na >>n (^) i : p=Na=1e16 cm-3^ and n=n (^) i^2 /p=1e4 cm-

b) 9x10 17 cm-3^ As and 8.95x10^17 cm-3^ Al Al is an acceptor, As is a donor, Si has n (^) i =1e10cm-3^ , since Na ~Nd but (N (^) d - Na )>>n (^) i : n=(Nd - Na )=5e15 cm-3^ and p=n (^) i^2 /n=2e4 cm- Note: here I used the reduction discussed in class of:

c) What is the majority carrier concentration for parts a, and b? Majority carrier is holes for a) and electrons for b)

d) What is the minority carrier concentration for parts a, and b? Minority carrier is electrons for a) and holes for b)

e) Is the material n-type or p-type in parts a, and b? p-type for a) and n-type for b)

i D A D A i

n ND^ NA ND NA โŽŸ + n โ‰ˆ N โˆ’ N for N โˆ’ N >> n

= โˆ’ + โŽ›^ โˆ’^2

2