Pure silicon at room temperature has an electron number density in the conduction band of about $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}{\mathbf{m}}^{\mathbf{-}\mathbf{3}}$and an equal density of holes in the valence band. Suppose that one of every ${\mathbf{10}}^{\mathbf{7}}$silicon atoms is replaced by a phosphorus atom. (a) Which type will the doped semiconductor be, nor p? (b) What charge carrier number density will the phosphorus add? (c) What is the ratio of the charge carrier number density (electrons in the conduction band and holes in the valence band) in the doped silicon to that in pure silicon?

1. The number density of electrons in the conduction band is equal to the number density of holes in the valence band,$n_{si,ec}\left(=n_{si,hv}\right)=5\times {10}^{15}{\mathrm{m}}^{-3}$
2. One of every silicon atoms is replaced by a phosphorus atom.
3. Number density of silicon atoms is,$n_{si}=5\times {10}^{28}/{\mathrm{m}}^3$

If there is an excess of electrons in the band of the semiconductor then the semiconductor is n-type, while, if there is an excess of holes in the band, the semiconductor is p-type. For example- Silica has a valency . So, if silica is doped with a pentavalent impurity, there will be an excess of electrons and it is called n-type. On the other hand if a trivalent impurity is used in the doping, there will be excess of holes and it becomes p-type semiconductor.

Formula:

The ratio of the number of charge carrier density in the doped form to that in pure form,

$Ratio=\frac{\mathrm{Number}\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{atom}}{\mathrm{Number}\mathrm{density}\mathrm{of}\mathrm{the}\mathrm{atom}\mathrm{to}\mathrm{be}\mathrm{replaced}\mathrm{in}\mathrm{both}\mathrm{the}\mathrm{bands}}$ (i)

Each phosphorus atom has one more valence electron; this implies the electron of the semiconductor will have the tendency to move out of the semiconductor.

Hence, the semiconductor is n-type.

The added charge carrier density of the phosphorus atoms is given by:

$$\begin{gathered} n_P={10}^{-7}{n}_{si} \\ ={10}^{-7}\left(5\times {10}^{28}/{\mathrm{m}}^3\right) \\ =5\times {10}^{21}/{\mathrm{m}}^3 \end{gathered}$$

Hence, the value of the charge carrier density is $5\times {10}^{21}/{\mathrm{m}}^3$.

Now, using the given and above values in equation (i), we can get the ratio of the charge carrier density in the doped silicon to that in pure silicon as follows:

$$\begin{gathered} Ratio=\frac{5\times {10}^{21}/{\mathrm{m}}^3}{2\left(5\times {10}^{15}/{\mathrm{m}}^3\right)} \\ =5\times {10}^5 \end{gathered}$$

As there is contribution from both electrons and holes, a factor 2 is multiplied in the denominator.

Hence, the value of the required ratio is $5\times {10}^5$.