Question

A semiconductor has equal electron and hole concentration of $3 \times 10^8$ m$^{-3}$. On doping with a certain impurity, the hole concentration increases to $6 \times 10^{10}$ m$^{-3}$.

(a) What type of semiconductor is obtained on doping?
(b) Calculate the new electron concentration of the semiconductor.
(c) How does the energy gap of semiconductor change with doping? Draw the energy band diagram for it.

Hint:

Holes became the majority carriers, so it is p-type. Use n_e n_h = n_i squared. The gap stays almost the same and an acceptor level appears near the valence band.

Answer 1

Step 1: Understanding the Concept:
In a pure semiconductor the electron and hole concentrations are equal, \(n_e = n_h = n_i\).
Doping changes one of them a lot. Even so, the product \(n_e n_h\) stays equal to \(n_i^2\) at the same temperature. This is the law of mass action.

Step 2: Part (a): Type of semiconductor:
The hole concentration rises from \(3 \times 10^8\) to \(6 \times 10^{10}\ \text{m}^{-3}\).
Holes are now the majority carriers. So the semiconductor is p-type. It was doped with a trivalent impurity such as boron or aluminium.

Step 3: Part (b): New electron concentration:
Use the law of mass action: \(n_e n_h = n_i^2\).
\[ n_e = \frac{n_i^2}{n_h} = \frac{(3 \times 10^8)^2}{6 \times 10^{10}} \]
\[ n_e = \frac{9 \times 10^{16}}{6 \times 10^{10}} = 1.5 \times 10^{6}\ \text{m}^{-3} \]
The electron concentration falls a lot, because the extra holes recombine with electrons.

Step 4: Part (c): Energy gap and band diagram:
The energy gap \(E_g\) of the semiconductor stays almost the same on doping. The gap is a property of the host crystal.
Doping only adds a new energy level for the impurity.
Energy band diagram of a p-type semiconductor (described in words):
1. Draw the conduction band at the top, with its lowest edge marked \(E_c\).
2. Draw the valence band at the bottom, with its top edge marked \(E_v\). The gap between them is \(E_g\).
3. Draw a thin dashed line \(E_A\), the acceptor level, just above \(E_v\). The gap between \(E_A\) and \(E_v\) is very small (about 0.01 to 0.05 eV).
4. Show electrons from the valence band jumping up to the acceptor level. They leave holes in the valence band.

Final Answer:
(a) The semiconductor is p-type. (b) The new electron concentration is \(1.5 \times 10^{6}\) m\(^{-3}\). (c) The energy gap stays almost unchanged, and an acceptor level appears just above the valence band. \[ \boxed{n_e = 1.5 \times 10^{6}\ \text{m}^{-3}} \]