Gallium arsenide (GaAs) is a compound of the elements gallium and arsenic. It is a III-V direct . GaAs diodes can be used for the detection of X-rays. Type I heterojunctions are quite common, the important GaAs – AlGaAs system is sometimes called an isotype junction, the pn-type a diode type junctions. The double-Schottky-diode model of Oldham and Milnes. The l-V characteristics of the GaAs-Ge isotype heterojunctions investigated by Anderson .

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Optoelectronics, as well as practically all other devices made from compound semiconductors, always contain heterojunctionsi. One possible reason gsas already mentioned: Transparency to light generated in some active part.

But there are many other advantages so compelling that extremely complicated heterostructures are now routinely produced despite a lot of problems that are encountered, too. Below the conduction band structure of an advanced Al x Ga 1-x As device is isotyp to illustrate that point.

There are many heterojunctions and we will not be able to delve very deep into the subject. The logic behind this structure will be made clear in a later chapter. In this sub-chapter we will look at some major properties of heterojunctions.

How do we construct a band diagram? Lets first look at the basic cases that we may encounter when considering heterojunctions. Naturally, the bandgaps are always different, but only specifying E g 1 and E g 2 and of course the Fermi energy in 1 and 2 is not sufficient to describe the heterosystem before the contact of the materials 1 and 2.

We also must specify the exact position on the energy scale of one of the band edges. This then gives rise to three distinct cases for heterojunctions as illustrated below together with the necessary definitions of the various energies isotpe.

You may want to consult a special modul in German dealing with some of the questions that may come up. Looking at the left hand Type I case, we first discuss the various energies encountered:.

In contrast to “simple” band diagrams in Sithe vacuum energy level is now included and defines the zero point of the energy axis. We also have the energy of the band edges, E C and E Vand from their difference the bandgap energy E g. Note that while c is a material constant, Q is not – it changes with the position of the Fermi energy. The discontinuities of the bands are such that both types gas carriers, electrons and holes, need energy D E Dioee and D E Vresp.

The isofype with respect to moving carriers from 1 to 2 or vice verse is no longer symmetrical. One kind of carrier gains energy dide the example if electrons move from right to leftthe other one needs energy the holes. The bandgaps do not overlap at all. The situation for carrier transfer is like type IIjust more pronounced. Yaas of Band Diagrams. What happens if we join the two materials? Exactly the same thing as for differently doped Si:. Carriers will flow across the junction, building space charges and now possibly also interface charges until the Fermi energy is the same everywhere in the material.

Far away from the junction, everything is unchanged. However, there are pronounced differences to the case of a pn -junction in Si. Lets imagine symmetrical junctions, i.

For homojunctionsthe number of electrons flowing into the p -type part is then the same as the number of electrons flowing into the n -part. However, at least in the type I case, only one kind of carrier will flow as is obvious and shown below. The space charge regions to the left and right of the junction thus might not be symmetric.

Still, some kind of carrier transfer will happen and the electrostatic potential far away from the junction will rise from a constant level on one side to a different, but constant level on the other side. The difference will be equal to the difference in Fermi energy before the contact divided by the elementary charge.

To the left and the right of the junction the bands are bend accordingly, and so is the vacuum energy. First, we align the Fermi energies. Then we bend the bands – in a smaller region on the more heavily doped side – but always identical for the band edges!

After all, the vacuum potential at some position x is fixed and so are the band edges relative to the vacuum potential. This leaves us with something new: We can not join the two materials!

## Gallium arsenide

If we adjust the band bending on both sides so that the conduction bands match, the valence band won’t match and vice verse. We must introduce a discontinuity right at the interface at one of the bands or at both. The next picture shows how to do that for type I heterojunctions for both possible doping cases. Some kind of cusp or notch must form in the conduction or valence band, depending on the details of the system. Exactly what happens and what the cusps look like depends on many details, you must solve the Poisson equation properly for a specific case.

Again, a discontinuity like this must happeneven at “ideal” interfaces. We don’t know exactly what it looks like, but we can now take this potential and plug it into a one-dimensional Poisson equations and see what it means for the charge distribution. Unavoidably, there must be a dipole layer right at the interface look at this basic module if you have problems figuring that out.

This dipole layer with its sharp wiggle in the charge distribution is therefore usually not included in drawings of the heterojunction. In reality, the matching of two lattice types with different atoms on both sides may well introduce some interface states in the bandgap, as discussed for the free Si surface.

Depending on the Fermi level which is of course influenced by the interface states, toothese interface states may be charged and introduce some band bending of their own.

### Ideal Heterojunctions

To make things even more complicated for pessimistsor to add more possibilities for engineering with heterojunction for optimistswe now can produce junctions with specific properties between materials of the same doping type – even for identical carrier concentrations. This type of heterojunction is sometimes called an isotype junctionthe pn -type a diode type junctions.

Other people use abbreviations, e. If we look ahead, we can gwas easily denote multi-junctions like PnPand so on. Isotype junctions must also have band discontinuities at the interface, the next picture shows examples.

### Gallium arsenide – Wikipedia

You may already wonder what properties to expect from this kind of junction and what it is good for; we will discuss that later. We now have a degree of freedom for all heterojunctions, which did not occur for homojunctions: How do we distribute the discontinuity? In the left hand diagrams above, e.

How can we find the real case? Nobody knows how to do that in some kind of comprehensive theory. The simplest model called the Anderson model assumes that D E C is equal to the difference in the electron affinities c. But that is only a rough estimate that may be quite wrong not to mention that bulk electron affinities can not be calculated with any precision and measurements always obtain the different surface electron affinities.

There is, however, one feature of the discontinuities that makes life somewhat easier: Whatever iostype value, it is determined by the interaction of the atoms at the interface and interatomic forces are responsible for its value.

This simply means that its value does not change much if we change properties of the materials on a scale much larger than the iwotype scale.

The band bending necessary for adjusting the potentials on both sides of the junction so that the Fermi energy is identical, may be seen as independent of the value of the discontinuities.

If we construct a band diagram, we simply always keep the same value for the discontinuities, no matter what else we do with the bands. How do we measure the values of the discontinuities? Make the heterojunction and measure the junction properties. In order to do that, we need to know how the discontinuities influence measurable quantities. How, for example, does the precise nature of the discontinuities influence the current – voltage characteristics across a heterojunction?

Or, if there is some radiative recombination, the quantum- or current efficiencies? We are now entering deep water.

After all, the equations for I-V -characteristics across isotupe junction without the space charge layer part in the simple or more complex form did not contain anything about the shape of the band bending – only the potential difference and bulk properties of materials to the left and right of the junction.

This tells us, that the basic diode characteristics assuming that nothing happens in the diods charge region must still be valid in its general form, but with one diove difference that transfers into a decisive property of heterojunctions:. The hole and electron part of the total current now are different even for a perfectly symmetric junction! Taking the values from the simple diode equation given beforewe obtain.

Rewriting this equation in terms of the carrier mobilities, and the doping concentration assuming fully ionized dopants with the relations given beforewe obtain.

For homojunctions, the intrinsic carrier concentration n i is the same on both sides, but not for heterojunctions! For the intrinsic carrier concentrations of any semiconductor we have the basic equations:.

But independent of the exact position of the Fermi energy, we always have. This is a very important equation for optoelectronics. If D E g is sufficiently large – and since it is in an exponential term, it does not have to be very large – it will always overwhelm the possible asymmetries in the pre-exponential term, e.

Getting all signs right, we have the following situation. In heterojunctions of the diode type, injection of the majority carriers from the material with the larger band gap almost always far surpasses the reverse process. To give a relevant example: Why is a large value of k so important? Because if we sandwich a small gap semiconductor between two large gap semiconductors, we should be able to inject a lot of electrons from one side and a lot of holes from the other side- with no means of escape.

The injected carriers must recombine in the small gap part which is our recombination zone – we have a large current efficiency h cu. If you think about that a minute and try to come up with some structure, you will realize, that there is a problem: You can not have injection via two pn -junctions – one junction must be an isotype junction. But luckily, isotype junctions have similar properties: