Anisotropy of III-V Compounds

Anisotropy of the substrate has a paramount
impact on the nucleation and stable growth of pores.

3.2.1 Anisotropy of III-V Compounds

Anisotropy of the substrate has a paramount impact on the nucleation and stable growth of pores. A certain threshold difference in electrochemical behaviour of different crystallographic planes is necessary for pore formation, otherwise only uniform dissolution will take place. In spite of the fact that Si and III-V compounds have the same diamond structure, the anisotropy is totally different. Due to the difference in anisotropic properties of Si and III-V compounds the direction of pore growth in the two type of semiconductors are different as well.

In order to understand how anisotropy of the substrate can influence the direction of pore growth let us have a look at the crystal structure of III-V compounds:

  • With the exception of nitrides, III-V compounds crystallize in the so called cubic zinc blend lattice;
  • The crystal consists of two cubic, surface-centered sublattices being shifted relative to each other by the translational vector [1/4, 1/4, 1/4];
  • Each sublattice is occupied by one kind of atoms, i.e. from III-rd or V-th group (see Figure 3.24a).

3.2.1  Anisotropy of III-V Compounds

Figure 3.1: a) GaAs cubic zinc blende crystal structure. b) The double layer structure along <111B> directions; c) The equidistant layers along <100> directions.

The most investigated low indexed crystal surfaces are {111}, {110} and {100}. Each set of planes has its distinct dissolution behavior. One of the reasons for the dissolution differences of different crystallographic planes could be the density of atoms within these crystallographic planes. It is reasonable to think that a high density of atoms in a crystallographic plane will result in a lower dissolution rate. However, this is not the only factor determining the anisotropy. In III-V compounds an additional factor contributing to the anisotropy is related to different chemical properties of the atoms from the third and the fifth groups of the periodic table of elements, which can influence the passivation behaviour of the crystallographic planes by the species from the electrolyte.

For example, along <111> directions, atomic planes are occupied alternatively by atoms from the third and fifth groups forming double layers, i.e. alternately short and long distances exist between {111} planes (Figure 3.1b). Each atom has 3 bonds within the same double layer and one bond in the exterior of the double layer. The surfaces terminated by Ga or In atoms are called {111}A, whereas those terminated by As or P atoms are called {111}B. In nearly all kinds of electrolytes, {111}A facets show the slowest rate of dissolution [35]. This can be explained assuming that the three bonds (within the same double layer) are much more difficult to break from the A side than from the B side, due to the fact that the electronic cloud of one bond is polarized towards the B atom, which is more electronegative. Thus the electrolyte species attacking the substrate from the B side will be able to come in closer contact with the electronic clouds of the bonds. On the other hand, the distance between the electronic clouds and electrolyte species will be larger when the species approach the substrate from the A side. Thus, the electron exchange reaction will be slightly easier from the B as compared to the A side.

Let's assume that in order to brake a bond from the B side an average time t1 is required, whereas from the A side the time t2. Thus, in order to remove a double layer from the B side we have to spend the time 3t1+t2, whereas from the A side the total time 3t2+t1. It is easy to realize that 3t1+t2>3t2+t1. As a consequence, it is much faster to etch from the B side (along <111B> directions) than from the A side (along <111A> directions).

On the other hand, along <100> directions the atomic planes are also occupied alternatively by atoms from the third and fifth groups. However, the planes are equidistant and each atom is symmetrically bonded with neighboring layers, i.e. two and two bonds. Thus, the time needed to etch two layers from the A side is 2t1+2t2, whereas from the B side 2t2+2t1. As a consequence, in this case there is no difference from what surface the etching will take place, because the average time is equal (Figure 3.24c) .

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