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I got this question in class:

An experiment is conducted as follows for n-Type semiconductor where \$N_D > 10^{16} [cm^{-3}]\$

The semiconductor (at room temperature) is placed in an electric field of \$E=50 [{V/cm}]\$. The current density is measured and found to be: \$J=4.5[A/{cm^2}]\$.

Then, the material is doped with \$N_a = 10^{14} [cm^{-3}]\$ and the test is repeated - the same electric field is applied. The final current density is measured to be: \$J=4.7[A/cm^2]\$.

What is the mobility of the minority charge carriers in the semiconductor?

In the answer it was stated that before the doping \$n=N_d,p=\frac{n_i^2}{N_d}\$, and afterwards \$n=N_d,p=\frac{n_i^2}{N_d} + N_a\$.(and then they subtract the two current densities...).

I don't understand why:

  1. We can use the equation \$np=n_i^2\$ for this case, because there is an electric field - this is non-equilibrium state.
  2. If we can, why after adding \$N_a\$, \$p\$ isn't equal to \$\frac{n_i^2}{n} = \frac{n_i^2}{N_d - N_a}\$
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  • \$\begingroup\$ 1) Can you quote your class question verbatim? 2) Do you know about the Haynes–Shockley experiment? The Haynes–Shockley experiment can be used to measure carrier mobility. As you write that you are ultimately asked to calculate the mobility of the minority charge carriers, it may help to teach you how one can measure carrier mobility, carrier lifetime, and diffusion coefficient. \$\endgroup\$ Commented May 4 at 8:56

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There are a couple of issues here.

While strictly speaking when doing a resistance measurement we are not really in equilibrium, we are usually approximately in equilibrium. For modest electric fields, very very close to equilibrium. So n*p=ni^2 WILL apply.

In any sensible way of thinking about the question the answer is wrong. If one set up an experiment where the resistance was measured with doping Nd, and then magically Na acceptors were introduced, then within picoseconds equilbrium would be re-established and the law of mass action (n*p=ni^2) would apply.

My guess is that this is an attempt to ask a question that is not one of the usual questions, for variety I guess, and the formulation was not completely thought out.

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