# Relation between drift velocity & conductivity of a doped semiconductor

Here is the problem, and I am stuck.

When an electric field with strength $1 \times 10^3 \text{V/cm}$ is applied to a p-type uncompensated Si sample at room temperature, the electron drift velocity, $v_d$ is $1 \times 10^6 \text{cm/s}$.

Calculate the conductivity of this sample. ($q = 1.6 \times 10^{-19} \text{C}$, $m_0 = 9.11 \times 10^{-31} \text{kg}$, $m_{n,\text{Si}}^{*} = 0.26 m_0$, $n_{i, \text{Si}} = 1.5 \times 10^{10} \text{cm}^{-3}$. The graph was from Solid State Electronic Devices, 6th Ed., by B. G. Streetman & S. K. Banerjee.)

I used the following equations : $\sigma = q n \mu_n$, $\mu_n = - \dfrac{\langle v_x \rangle}{\mathcal{E}_x}$.

I can find $\mu_n = \dfrac{1 \times 10^6 \text{cm/s}}{1 \times 10^3 \text{V/cm}} = 1 \times 10^3 \text{cm}^2/\text{V s}$.
Using the graph, $n \approx 1 \times 10^{17} \text{cm}^{-3}$.
Therefore, $\sigma = q n \mu_n\ = 1.6 \times 10^{-19} \text{C} \times 1 \times 10^{17} \text{cm}^{-3} \times 1 \times 10^3 \text{cm}^2/\text{V s} \\ = 16 \text{C} / \text{cm V s} \\ = 16 \text{A s} / \text{cm A } \Omega \text{ s} \\ = 16 \text{cm}^{-1}\Omega^{-1}.$

However, I don't know where the condition "p-type" should be used.
It is also strange that $n \gg n_i$ even though it is p-type.
I thought the impurity concentration from $\mu_n$ leads to the $n$ since it is not $\mu_p$, but is it wrong?
How to solve this problem correctly?

Taking a stab at ths- and it's been a while- I'm thinking that your supposed to divine the dopant concentration from the minority carrier mobility, then use $\mu_p$ at that concentration for the majority carrier mobility $\mu_p$, so around $3\times 10^2$, so you'd get a considerably smaller conductivity number.

• So $n \approx N_d$ and $p \approx N_a$, isn't it? You mean it can be deduced that $N_a \approx 1 \times 10^{17} \text{cm}^{-3}$ from the figure, not $N_d$? But why? Mobility is dependent on the concentration of dopants, but not dependent on the type of those dopants? – Naetmul Jul 26 '14 at 12:36
• Yes. It seems the case. $10^{17} \text{cm}^{-3}$ includes $N_a$ not only $N_d$ since scattering increases regardless of the type. – Naetmul Jul 26 '14 at 13:41

This is mostly in answer to Naetmul's comments to Spehro. (correct answer IMHO) It's a bit of a weird question, they give you the minority drift velocity. (I think there are ways to measure minority properties of a sample, but it's hard.. so.. a made up problem.)

It's interesting that the mobility doesn't depend on the type of dopants. And below a concentration of ~10^16 it's independent of the concentration.