# Kris's Research Notes

## February 4, 2011

### Model Parameters — Part 2.

Filed under: GaAs Simulations — Kris Reyes @ 9:30 pm

This is a follow-up to this post.

I am currently in the process of fitting the surface termination phase diagram (see this post) to experiments. While doing this, I have learned more about the model. Specifically, I more clearly understand how the $\gamma_{2,4}, \gamma_{2,3}, \gamma_{4,2}, \gamma_{3,2}$ and $\mu_{As}$ parameters affect crystal growth — where $\gamma_{i,j} = \gamma(Ga(i), As(j))$.

## Step-flow growth

Initially, I had used the following $\gamma_{i,j}$ energies in re-simulating crystal growth:

$\left(\gamma_{i,j}\right) = \begin{pmatrix} 0.19 & 0.19 & 0.19 & 0.19 \\ 0.19 & 0.19 & 0.95 & 0.95 \\ 0.19 & 0.19 & 0.95 & 0.95 \\ 0.19 & 0.19 & 0.95 & 0.95 \end{pmatrix}$,

which is identical to the energies used in the previous post except $\gamma(Ga(2), As(3)) = 0.95 eV$. I fixed the temperature $T = 833K$. According to experiments, the transition from a $Ga$-terminated surface to an $As$ one occurs between $r = r_{\downarrow As}/r_{\downarrow Ga} = 1$ and $r=2$ at this temperature. I also varied $\gamma={3,2} = \gamma_{4,2}$ — the bond energy of an $As$ atom on the surface of the $GaAs$ substrate — between 0.19 eV and 0.95 eV. In the 0.95 eV case (i.e surface As atoms are strongly bonded to the substrate), the surface of the growing crystal is much rougher than the 0.19 eV case. For example, at $r = 20$, here are two snap shots for the 0.95 eV and 0.19 eV case, respectively:

This is easily explained. Consider the case of a step-edge:
If $\gamma_{3,2}=\gamma_{4,2}= 0.19 eV$ the energy gain in starting a new level (the top option) is $2\times\gamma_{3,2} = 0.38$ eV, while the gain in continuing on the current level (the bottom option) is $2\times\gamma_{2,4} = 1.9$ eV. In this case, the system favors building on the current level, and so encourages step-flow growth. If $\gamma_{3,2} = \gamma_{4,2} = 0.95$ eV, the both transitions gain $1.9 eV$ energy, and so both transitions may occur with equal probability. This explains the situation pictured above; there are both level ares and pointed hills/valleys in the $\gamma_{4,2}=0.95$ eV example.

It is by this that the parameters $\gamma_{3,2}$ and analogously $\gamma_{2,3}$ control step-flow growth. When they are close to $\gamma_{4,2}, \gamma_{2,4}$ — i.e. surface bond strength, we do not observe step-flow growth. We can address this by setting the energies to be

$\left(\gamma_{i,j}\right) = \begin{pmatrix} 0.19 & 0.19 & 0.19 & 0.19 \\ 0.19 & 0.19 & 0.19 & 0.95 \\ 0.19 & 0.19 & 0.95 & 0.95 \\ 0.19 & 0.95 & 0.95 & 0.95 \end{pmatrix}$,

Here the difference between $\gamma_{2,3}, \gamma_{3,2}$ and the surface bonding energies are large. Note that the matrix is symmetric. I believe this effectively creates a $GaAs$ species whose bonding energies correspond to the bottom right corner of the matrix. The smaller bonds are not considered part of a $GaAs$ crystal. In particular, we do not regard the $Ga(2)-As(3)$ and $Ga(3)-As(2)$ bonds (i.e. those on the edge of a step) as crystal bonds — although we do so for the surface $Ga(2)-As(4)$ and $Ga(4)-As(2)$ bonds. Note further this will encourage desorption at the ends of the islands.

## As vs. Ga – terminated surfaces

Using these energies, we attempt to fit the surface termination phase diagram to experiments. We fix $T=833K$ and consider

$\mu_{As} \in \left\{ 0.2, 0.3, \hdots, 1.0 \right\}$ eV.

We vary $r = 1, 2$. Again, the transition between a Ga-terminated surface and an As-terminated one occurs between these two ratios.
Here is a plot of percent surface Ga as a function of $\mu_{As}$ and $r$:

We see that the correct transition from Ga-terminated to As-terminated between ratio $r=1$ and $2$ occurs for $\mu_{As} \geq 0.5$, approximately. For example, here is the $\mu_{As}=0.5eV$ for the $r = 1, 2$ cases, respectively:

Here is the $\mu_{As} = 0.4$ eV and $r = 2$ case:

It is clearly not As terminated.

However, the $\mu_{As} = 0.50$ eV case (the smallest possible $\mu_{As}$ for which we get the correct point of transition for $T=833$K) is too large a desorption barrier in the low temperature/high $As$ overpressure regime and so we have excess $As$. For example, when $T=435K$ and $r = 80$, we observe the following:

This is perhaps undesirable.

For mid-temperature regime, we can consider $T = 573 K$. Here are the results when $r = 80$ and $\mu_{As}$ varies in $\left\{0.4, 0.5, 0.6\right\}$:

While these are taken at different times, it is clear that for $\mu_{As} = 0.4$ eV, we do not get excess As, while at $\mu_{As} = 0.6$ eV we do.

We can address this in two ways. First we can lower the $As-As$ bond strength and implicitly encourage $As$ desorption. Second, we can give $As(0)$ a separate, and lower desorption potential, which would explicitly allow for more $As$ desorption. The first option would require more computational time as $As$ on $As$ diffusion would occur at a high rate since the energy to diffuse would be that much less than the energy to desorb. More diffusion would take place before an $As$ desorption in this case. The second case would state that in our model, for sufficiently low $\mu_{As(0)}$, we effectively disallow any excess $As$ except for low temperatures/high $As$ overpressure.

We shall address this in a future post.