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?
Or is my answer correct?
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.

  • \$\begingroup\$ 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? \$\endgroup\$ – Naetmul Jul 26 '14 at 12:36
  • 1
    \$\begingroup\$ Yes. It seems the case. \$10^{17} \text{cm}^{-3}\$ includes \$ N_a\$ not only \$N_d\$ since scattering increases regardless of the type. \$\endgroup\$ – 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.

(I've been reading all about this, doing measurements, if I get something wrong or suspect, then please correct me.)

The mobility depends on the effective mass and scattering length. At high concentrations it's all impurity (dopant) scattering and mostly independent of what the impurity is. A disruption in the lattice/crystal. At lower concentrations phonon scattering dominates, I would expect that to change with temperature.

I was looking here, and got lost.. http://en.wikipedia.org/wiki/Electron_mobility#Relation_between_scattering_and_mobility

I never knew there was a saturation velocity.


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