HardPhysicsHigh School (11–12)ElectronicsNumericalConceptual

A silicon wafer is doped with both kinds of impurity…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A silicon wafer is doped with both kinds of impurity. It has $3.0 \times 10^{21}\ \text{m}^{-3}$ donor atoms and $1.0 \times 10^{21}\ \text{m}^{-3}$ acceptor atoms. At room temperature, all of them are ionised. The acceptors take electrons from the donors, so only the difference counts for the carriers. Use $n_i = 1.0 \times 10^{16}\ \text{m}^{-3}$ , $\mu_e = 0.135\ \text{m}^2/\text{V s}$ and $\mu_h = 0.048\ \text{m}^2/\text{V s}$ . Find the type of the wafer, the number of free electrons and of holes, its conductivity and its resistivity in $\Omega$ cm. Then an engineer adds $4.0 \times 10^{21}\ \text{m}^{-3}$ more acceptors. Find the new type and the new conductivity. The terms for the minority carriers can be left out.

Need a reminder? The key ideaShow

A donor gives one electron, so $n_e \approx N_D$ . An acceptor gives one hole, so $n_h \approx N_A$ . The law of mass action is $n_e\, n_h = n_i^2$ , with $n_i = 1.0 \times 10^{16}\ \text{m}^{-3}$ for silicon at 300 K. Conductivity is $\sigma = e\,(n_e \mu_e + n_h \mu_h)$ , with $e = 1.602 \times 10^{-19}\ \text{C}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $N_D = 3.0 \times 10^{21}\ \text{m}^{-3}$ and $N_A = 1.0 \times 10^{21}\ \text{m}^{-3}$ .

Plan. Neutrality says $n_h + N_D = n_e + N_A$ . Because the doping is much larger than $n_i$ , the larger of the two kinds sets the majority carriers, and the difference gives their number.

Step 1. The type and the electrons. $N_D > N_A$ , so the wafer is n-type. $n_e \approx N_D - N_A = 2.0 \times 10^{21}\ \text{m}^{-3}$ .

Step 2. The holes. $n_h = \dfrac{n_i^2}{n_e} = \dfrac{(1.0 \times 10^{16}\ \text{m}^{-3})^2}{2.0 \times 10^{21}\ \text{m}^{-3}} = 5.0 \times 10^{10}\ \text{m}^{-3}$ .

Step 3. The conductivity. $\sigma \approx e\, n_e \mu_e = 1.602 \times 10^{-19}\ \text{C} \times 2.0 \times 10^{21}\ \text{m}^{-3} \times 0.135\ \text{m}^2/\text{V s} = 43.3$ S/m.

Step 4. The resistivity. $\rho = \dfrac{1}{\sigma} = \dfrac{1}{43.3\ \text{S/m}} = 0.0231\ \Omega\ \text{m} = 2.31\ \Omega\ \text{cm}$ .

Step 5. After more acceptors. Now $N_A = 5.0 \times 10^{21}\ \text{m}^{-3}$ is larger than $N_D$ . The wafer is p-type, with $n_h \approx 5.0 \times 10^{21} - 3.0 \times 10^{21} = 2.0 \times 10^{21}\ \text{m}^{-3}$ . So $\sigma \approx e\, n_h \mu_h = 1.602 \times 10^{-19}\ \text{C} \times 2.0 \times 10^{21}\ \text{m}^{-3} \times 0.048\ \text{m}^2/\text{V s} = 15.4$ S/m.

Answer: $2.0 \times 10^{21}\ \text{m}^{-3}$ electrons, $5.0 \times 10^{10}\ \text{m}^{-3}$ holes, $43.3$ S/m, $2.31\ \Omega$ cm, and $15.4$ S/m after the change, with the wafer now p-type.

Watch out. Doping both ways does not give $3 + 1$ carriers. The two kinds cancel. The conductivity falls even though more atoms were added, because the holes are slower than the electrons.

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