22.13: Questions

Quick questions

You should be able to answer these questions without too much difficulty after studying this TLP. If not, then you should go through it again!

What are the majority carriers of electric current in B-doped Si?

a	Holes
b	Electrons
c	Boron ions
d	Silicon atoms

Answer

A. Boron is an acceptor dopant, and so B-doped Si is a p-type semiconductor. Holes are the majority carriers in p-type semiconductors.

At absolute zero, which of these statements describes the position of the chemical potential in an n-type semiconductor?

a	It is between the top of the valence band and the acceptor levels.
b	It is in the middle of the band gap.
c	It is between the middle of the band gap and the bottom of the conduction band.
d	It is between the middle of the band gap and the top of the valence band.
e	It is between the the donor levels and the bottom of the conduction band.

Answer

E. At absolute zero all the donor states are full and the conduction band is completely empty. Therefore, the chemical potential has to be between these positions at absolute zero. It is often taken to be exactly halfway between these two positions because the Fermi-Dirac function is a step function at absolute zero - see the discussion in the books by Ashcroft and Mermin and Rosenberg in 'Going Further'.

At room temperature, which of these statements describes the position of the chemical potential in an n-type semiconductor?

a	It is between the top of the valence band and the acceptor levels.
b	It is in the middle of the band gap.
c	It is between the middle of the band gap and the bottom of the conduction band.
d	It is somewhere between the middle of the band gap and the top of the valence band.
e	It is at the same energy level as the donor energy levels.

Answer

C. At a finite temperature such as room temperature, some of the electrons in the donor states in the n-type semiconductor are promoted to states in the conduction band. Hence the occupancy of the donor states falls relative to absolute zero, and so the chemical potential must fall in energy relative to its position at absolute zero. If the donor states are half full at a particular temperature, the chemical potential is at exactly the same energy level as the donor states.

E. This is a special case of the correct answer. It is unlikely to be the answer at a particular temperature such as room temperature for a n-type semiconductor selected at random about which no further information is provided.

Which of the following junctions is not rectifying?

a	Metal - p-type semiconductor contact where ΦM < ΦS.
b	p-n junction
c	Metal - p-type semiconductor contact where ΦM > ΦS.
d	Metal - n-type semiconductor contact where ΦM > ΦS.

 

Answer

C. Here the bands will bend upwards at the metal-semiconductor interface, but unlike the case of a metal - n-type semiconductor contact, there is no barrier to hole flow across the junction from the semiconductor to the metal or electron flow from the metal to the semiconductor for suitable biases. Thus, this contact is Ohmic

The work function of gold is 4.8 eV and the electron affinity of silicon is 4.05 eV. Silicon has a band gap of 1.12 eV at 300 K. What is the barrier height at 300 K preventing the majority carriers in n-type silicon from crossing between a piece of n-type silicon and gold where the chemical potential of the n-type silicon is 0.25 eV below the bottom of the conduction band?

a	0.75 eV.
b	0.5 eV.
c	1.12 eV.
d	4.05 eV.
e	0.25 eV.

Answer

B. Before contact, the chemical potential of the n-type silicon is 4.05 + 0.25 eV lower in energy than vacuum. When contact is made, the chemical potential of the metal and the semiconductor have to become equal. The barrier height preventing electrons in the conduction band of n-type silicon (the majority carriers in n-type silicon) from crossing into the metal is therefore 4.8 - (4.05 + 0.25) eV = 0.5 eV.