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# Haynes - Shockley experiment

In the Haynes - Shockley experiment, a piece of semiconductor gets a pulse of holes, induced by voltage or a short laser pulse.

To see the effect, we consider a n-type semiconductor with the length d. We are interested in determining the mobility of the carriers, diffusion constant and relaxation time. In the following, we reduce the problem to one dimension.

The equations for electron and hole currents are: $j_e=-\mu_n n E-D_n \frac{\partial n}{\partial x}$ $j_p=+\mu_p p E-D_p \frac{\partial p}{\partial x}$

where μ = eβD is mobility, Einstein relation states: v = μE.

We consider the continuity equation: $\frac{\partial n}{\partial t}=\frac{-(n-n_0)}{\tau_n}-\frac{\partial j_e}{\partial x}$ $\frac{\partial p}{\partial t}=\frac{-(p-p_0)}{\tau_p}-\frac{\partial j_p}{\partial x}$

The electrons and the holes recombine with the time τ.

We define p1 = pp0 and n1 = nn0 so the upper equations can be rewritten as: $\frac{\partial p_1}{\partial t}=D_p \frac{\partial^2 p_1}{\partial x^2}-\mu_p p \frac{\partial E}{\partial x}- \mu_p E \frac{\partial p_1}{\partial x}-\frac{p_1}{\tau_p}$ $\frac{\partial n_1}{\partial t}=D_n \frac{\partial^2 n_1}{\partial x^2}+\mu_n n \frac{\partial E}{\partial x}+ \mu_n E \frac{\partial n_1}{\partial x}-\frac{n_1}{\tau_n}$

Let us consider the part with the gradient of the electric field. Laplace equation states: $\rho=-\epsilon \epsilon_0 \frac{\partial^2 U}{\partial x^2}= \epsilon \epsilon_0 \frac{\partial E}{\partial x}$ $\frac{\partial E}{\partial x}= \frac{\rho}{\epsilon \epsilon_0}=\frac{e_0 ((p-p_0)-(n-n_0))}{\epsilon \epsilon_0}$

Introducing n2 = p1 + n1 and n3 = p1n1 < < n2, we rewrite the initial equations with the new parameters. $\frac{\partial n_2}{\partial t}=D_p \frac{\partial^2 n_2}{\partial x^2}-\mu_p p \frac{\partial E}{\partial x}- \mu_p E \frac{\partial n_2}{\partial x}-\frac{n_2}{\tau_p}$ $\frac{\partial n_2}{\partial t}=D_n \frac{\partial^2 n_2}{\partial x^2}+\mu_n n \frac{\partial E}{\partial x}+ \mu_n E \frac{\partial n_2}{\partial x}-\frac{n_2}{\tau_n}$

These two equations are coupled and can be combined: $\frac{\partial n_2}{\partial t}=D^* \frac{\partial^2 n_2}{\partial x^2}+ \mu^* E \frac{\partial n_2}{\partial x}-\frac{n_2}{\tau^*},$

kjer so $D^*=\frac{D_n D_p(p+n)}{p D_p+nD_n}$, $\mu^*=\frac{\mu_n\mu_p(p-n)}{p\mu_p+n\mu_n}$ in $\frac{1}{\tau^*}=\frac{p\mu_p\tau_p+n\mu_n\tau_n}{\tau_p\tau_n(p\mu_p+n\mu_n)}.$

Considering n>>p or $p\rightarrow 0$ (that is a fair approximation for a semiconductor with only few holes injected), we see that $D^*\rightarrow D_p$, $\mu^*\rightarrow \mu_p$ and $\frac{1}{\tau^*}\rightarrow \frac{1}{\tau_p}$. The semiconductor behaves as if there were only holes traveling in it.

The final equation for the carriers is: $n_2(x,t)=A \frac{1}{\sqrt{4\pi D^* t}} e^{-t/\tau^*} e^{-\frac{(x+\mu^*Et-x_0)^2}{4D^*t}}$

This can be interpreted as a delta function that is created immediately after the pulse. Holes then start to travel towards the electrode where we detect them. The signal then is Gaussian curve shaped.

Parameters μ,D and τ can be obtained from the shape of the signal. $\mu^*=\frac{d}{E t_0}$ $D^*=(\mu^* E)^2 \frac{(\delta t)^2}{16 t_0}$

## References

• Wang:Solid State Electronics