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Optoelectronics: Why Semiconductors?

  1. Mar 11, 2013 #1
    I want to confirm (or deny), and expand upon, why I think semiconductors are used in optoelectronic devices rather than conductors. For example, I THINK we use semiconductors in solar cells because they do not automatically re-emit the incident photons they absorb, which is what I THINK would happen if we used a conductor (metal).

    It seems to me that we are constantly trying to tune the bandgap of semiconductors (via doping, confinement, etc.) to REDUCE the rate of electron-hole recombination and therefore increase the conductivity and reduce (in the case of photovotaics, photocatalysts, etc.) or to INCREASE the rate of recombination (lasers, LEDs). In my mind, we are constantly trying to make semiconductors more like conductors or insulators to serve a specific purpose, rather than just using conductors or insulators.

    So, what exactly is the point? Is it just because they "absorb" without automatically emitting light? If I recall correctly, many narrow gap SCs do emit light quite quickly.
  2. jcsd
  3. Mar 11, 2013 #2
    I think the reason is that the electrical and optical properties can be more easily changed in a semiconductor than in a conductor or insulator.

    The conductivity of a crystal of silicon can be changed over 10 orders of magnitude by adding small amounts of impurities in the growth process. One can't change the conductivity of a piece of silver or a piece of quartz by nearly as much, even with large amounts of impurities. In fact, the silicon can be made a conductor or an insulator with small amount of impurity.

    One can change the electrical and optical properties of a semiconductor more easily with light than a conductor or an insulator. That is the basis of most photodetectors. A piece of high-resistivity silicon can be turned into a low-resistivity piece of silicon by shining light on it. A beam of light can never turn a piece of silver into an insulator, or a piece of quartz into a conductor.

    For technology, there is never such a thing as too much control. Every type of semiconductor has certain limitations on the properties that can be changed.Therefore, different semiconductors are being manufactured which have changeable properties in the range one wants.

    Think about vacuum tubes. Vacuum tubes used mostly conductors and an electron gas (i.e., vacuum with electrons moving in it). The properties of the conductor were changed by heat. The properties of the electron gas were controlled by applying different fields to the electron gas. Vacuum tubes couldn't be made as small as semiconductor devices, but they did the same type of work.

    Of course, there are always efforts to think outside the box. I know a physicist who worked with ionic solids (i.e., salt crystals) hoping to make electronic devices with them. The defects in ionic solids can carry electric charges, which can be changed in certain ways. However, the project didn't go to far. His team gave up. I haven't been able to get many details from him.
  4. Mar 12, 2013 #3
    It seems to me that this goes back to our endless effort to turn SCs into metals or insulators. We dope them one way or another to make them less like semiconductors and more like conductors or insulators, at least it seems, rather than just using insulators or conductors in the first place.
  5. Mar 12, 2013 #4


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    But that is not the point of doing that. What you get by using semiconductors and doping is functionality. A p-doped semiconductor region next to a n-doped one gives a p-n-junction, which is a very simple diode and therefore adds functionality. This is something you do not get by just using insulators or conductors. Two of these junctions already give you a simple transistor.

    To get back to the solar cell: These are basically huge p-n-junctions. The functionality of the p-n-junction lies in separating the carriers, do they do not recombine, and getting them away from the region of their creation. Just creating carriers in the material is rather useless unless you manage to get the carriers out of the solar cell.
  6. Mar 12, 2013 #5
    So, in other words, the only point of using semiconductors is pn junctions?
  7. Mar 12, 2013 #6
    no. you seem to know about band theory, so you should apply band theory.

    well, lets say you want an LED. you need a material that *actually has electricity flowing through it*. Otherwise its useless as an LED, since it'll just be a resistor. So you need something with a relatively small bandgap. However, it cannot be a metal, because there's no radiative transitions from one band to another involved when electrons are conducting. Therefore you must have a semiconductor and there are no alternatives.

    Lets say you want the LED to have different colors. Insulators and conductors have only 2 choices: its a resistor. it conducts with no radiative transitions. you can get them to glow with a big enough E field, but that's thermal radiation, and highly inefficient.

    Thus, the point then is to tailor the bandgaps of the material, through impurities, such that their bandgaps correspond to colors in the visible. Well, that seems to require semiconductors.

    But wait, that's not all. Even some semiconductors cannot be used to emit light as they have indirect bandgaps, that is, energy minimum of the conduction band is not symmetric with the energy maximum of the valence band. since emitted photons result in lost energy, but not lost k (crystal momentum) you have to have a phonon in the process to carry away excess k from the transiting electron, which makes it less likely to occur, thus lowering emission efficiency. direct bandgap materials do not require a phonon for light emitting transitions.

    the only materials that you can actually change the band structure of conveniently are semiconductors.

    the take home message is that what distinguishes semiconductors from metals and insulators is not merely conduction; it is their band structure.
  8. Mar 12, 2013 #7
    Ok, I think I'm closing in on something here, at least with respect to LEDs. In this case, you're saying that conductors are ONLY capable of nonradiative transitions, correct? If this is true, I completely understand the LED part.

    On a side note, indirect bandgap semiconductors can still emit light, they just do so inefficiently, if my understanding is correct. The indirect transition is a vertical (photon) transition with accompanying phonon absorption or emission, depending on the band structure. In this respect, we still emit light, but at the cost of lattice vibration. Please correct me if I'm wrong.

    Now, tell me if I'm summing this up correctly. Conductors aren't used in optoelectronics because they are only capable of nonradiative transitions. We can't separate the charge carriers, so they all recombine and we're stuck with phonons.

    I think perhaps my ignorance is with regard to conductors. I've spent a great deal of time studying semiconductors, but I really know very little about metals.
  9. Mar 12, 2013 #8
    here's how I learned it:

    lets say you have a solid with a bandgap. A metal would be a solid that has electrons in the conduction band at T = 0. Therefore, there is no bandgap between occupied and unoccupied levels at T = 0, and the number of conduction electrons is roughly equal to the number of valence electrons (n ~ 10^28 m^-3).

    When an E field is applied to the metal, electrons flow in one direction since there's free carriers in the conduction band. There's no carrier separation here; it is just (to a good approximation) free electrons moving in response to an E field and slowed down by collisions (with phonons or impurities; see Bloch theorem).

    An insulator and a semiconductor both have no electrons in the conduction band at T = 0. The difference is that semiconductors have a bandgap small enough such that there will be a sizable amount of electrons (in practice, n ~ 10^20 m^-3) that can jump from the valence band to the conduction band at temperatures low enough such that the semiconductor is still a solid. This means a bandgap of 0.1 eV - 3 eV.

    When electrons in a semiconductor gets enough energy to jump to the conduction band, there's an effectively positively charged hole left behind in the valence band. This is thermal carrier separation. When they recombine, such as in an LED, they emit light, but only if they can recombine easily, which is what happens in direct bandgap materials.

    In an indirect bandgap material, lets say you have a band structure that was not symmetric. If you want the electron to jump down back into a hole in a radiative process, you need a phonon emitted or absorbed in order to change the momentum. However, this will also change the energy, since phonons also carry energy, which might make it incompatible with the bandgap at that k value (to move to that k value, you might need to raise the energy of the electron substantially). Thus, because the circumstances have to be "just right" for a radiative transition to happen, carrier recombination is slowed down greatly.

    In terms of picturing it on a K vs. E band diagram, a photon emission/absorption means either jumping straight up, or jumping straight down, while a phonon emission/absorption means moving in a diagonal.
  10. Mar 29, 2013 #9
    OK, so how exactly is photon absorption/emission explained in a metal as compared to a semiconductor? Metals easily absorb and rapidly emit photons, as opposed to semiconductors. With SCs, I understand the process as it's commonly explained, but how exactly do we treat absorption and emission in metals?
  11. Mar 29, 2013 #10
    Actually for optical frequency EM processes in metals (those that don't knock electrons out), its easier to think about it classically. Since an EM wave is just an AC signal, it can excite plasmon oscillations in metals which is basically the electrons moving back and forth in response to the AC signal.

    the plasmon frequency is given by ωp = sqrt( 4π∗ne2/m)

    the significance of the plasmon frequency is that it is the frequency above which the dielectric tensor ε(ω) is positive and admits oscillatory solutions; below this frequency the dielectric tensor is negative and only admits decaying wave solutions. Thus, the plasmon frequency is the frequency above which metals become transparent and below which they are reflective; because optics depends on reflection, anything above the plasmon frequency should not be called light.

    let us take the curl of Maxwell's 3rd equation ∇ x E = (-1/c) dH/dt:

    ∇x(∇xE) = ∇(∇∙E)-ΔE = (-1/c) ∇ x (dH/dt) = (-1/c) d/dt (∇xH)

    ∇xH = 4π/c * j + (1/c) dE/dt

    where Δ is the Laplacian operator. E is the electric field. In a metal we assume that there is no net charge so ∇∙E = 0. We also assume microscopic Ohm's law applies: j = σE where σ is the conductivity tensor.

    -ΔE = (-1/c) d/dt (∇xH)

    Taking the Fourier transform of j, E and H to find their frequency dependence, and plugging back into Maxwell's 4th equation, we can get both sides to be in terms of frequency dependent electric fields:

    E(t)= ∫e^-iωt E(ω)dω
    dE(t)/dt = -iω ∫e^-iωt E(ω)dω
    j(t) = σE(t) = σ∫e^-iωt E(ω)dω
    H(t) = ∫e^-iωt H(ω)dω

    ∇xH = ∫e^-iωt H(ω)dω = (4πσ/c)∫e^-iωt E(ω)dω - iω/c ∫e^-iωt E(ω)dω = ∫E(ω)e^-iωt(4πσ/c-iω/c)dω
    Plug back into ΔE = ∫e^-iωt ΔE(ω)dω = (-1/c) d/dt (∇xH):

    -ΔE = ∫e^-iωt ΔE(ω)dω = d/dt (-1/c) ∫E(ω)e^-iωt(4πσ/c-iω/c)dω = (1/c^2)∫E(ω)e^-iωt(4πiωσ+ω^2)dω

    both sides have integrals with same limits, they can be removed. both sides have e^-iωt and that can be cancelled.

    -ΔE(ω) = (ω^2/c^2)*(4πiσ/ω+1)*E(ω)

    which has the form -ΔE(ω) = (ω^2/c^2)ε(ω)E(ω) where the dielectric tensor ε(ω) = (1+4πiσ/ω).

    This looks like a wave equation doesn't it?

    Let the conductivity tensor have the form σ = σ0 / (1-iωτ). This requires a separate derivation as to why it is this form.

    But if we make that assumption, then, at high frequencies:

    ε(ω) = (1-(ω_p)^2/ω^2).

    If ε(ω) is real and negative, then it means ω^2<(ω_p)^2 (and the frequency of the EM wave is lower than the plasmon frequency) and the equation admits only decaying exponential solutions. If ε(ω) is real and positive, then ω^2>(ω_p)^2 and the equation admits oscillatory solutions, which means the metal is transparent.

    The actual values come out to be about 200 nm wavelengths for typical metals.
  12. Apr 1, 2013 #11
    Is there a way to compare the two on a similar basis? If I'm explaining this to someone else, and I provide an explanation of insulator and semiconductor properties on the basis of band structure, people are likely to get confused if I change from band theory to classical EM if I explain conductors. Can't the concepts be explained using the same method?
  13. Apr 1, 2013 #12
    I like to explain plasmons by saying that the electrons in a metal move together in unison. The charge carriers in a metal move together like one solid block due to the Pauli exclusion principle. This is a visual image to explain the properties of a Fermi fluid.

    The carriers in a metal can be compared to a brick wall. If a photon with energy below the plasmon energy hits the surface of the metal, it is like a rubber ball hitting a brick wall. The photon bounces off. If a photon with an energy above the plasmon energy hits the surface of the metal, it is like a cannon ball hitting the brick wall. The photon does not reflect. The carriers come apart like the bricks in the wall come apart.
  14. Apr 2, 2013 #13
    I would think that a simple explanation would be that (all else constant) a wider bandgap would correspond to a longer recombination time, and vice versa. Therefore, for a conductor with a bandgap of ~0 eV, recombination times should be very short. Does this make sense, and is this a reasonable statement?
  15. Apr 2, 2013 #14


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    No, it is the other way round. Really keeping all else constant (also the number of carriers present), (optical) recombination times will be determined by the spontaneous emission rate and given by Fermi's golden rule. This is in turn proportional to the squared dipole moment of the transition and proportional to the third power of the photon to be emitted. As the band gap determines this quantity, spontaneous emission is way faster for large energy differences. This is also the reason, why creating terahertz photon sources is complicated.

    In reality, of course the available carrier densities and whether the band gap is direct or indirect are important factors determining carrier lifetimes.
  16. Apr 2, 2013 #15
    This does not make sense. It is not a reasonable statement.

    Recombination refers to the annihilation of charge carriers. The charge carriers can't carry charges after they annihilate. For example, the recombination can refer to the mutual annihilation between a conduction-electron and a valence-hole. The negative conduction-electron annihilates the positive valence-hole, producing a neutral something that can't carry electric charge.

    The charge carriers in a metal by definition aren't annihilated. There are sufficient carriers in the carrier band of a metal so that they can't relax to a lower energy state. Therefore, the charge carriers in a metal. The majority charge carriers have an infinite recombination time.

    An electric insulator has a very short recombination time. If charge carriers are generated in an electric insulator, then they can recombine in a very short time. That is why they are insulators.

    This is why the photoconductivity in insulators is short lived. As an example, charge carriers can be generated in an insulator by a short pulse of light with photon energies greater than the band gap energy. Both conduction-electrons and valence-holes would be generated in equal numbers. Therefore, the electric insulator becomes an electric conductor immediately after the light pulse. However, conduction-electrons annihilate the valence-holes in a very short amount of time. Therefore, the electric conductivity disappears in a very short time.

    I conjecture that you are conflating mean collision times with recombination time. The charge carriers in a material can collide with each other over a very short time. The time between collisions can be very short in a metal, because the carriers are so dense. Thus, induced momentum has a very short lifetime associated with it. However, this is more like a momentum lifetime.

    If the momentum lifetime is short enough, then the carriers may have to move in unison because of momentum transfer. This picture may not be quite right, but it may be close enough for purposes of explaining to a nonscientific audience. The carriers in a metal tend to move in unison because momentum is transferred very rapidly in such a way that they act like a fluid. This fluid is referred to as a Fermi gas.

    The charge carriers in a metal behave like a very stiff fluid. The carriers in a metal are almost incompressible, with a very large "bulk compressibility". I put it in quotes because the relevant quantity is a close analog to bulk compressibility, but not quite the same. The bulk compressibility of the carriers and the density of carriers determine a plasmon frequency.

    Recombination times are not the same thing as collision times. The lifetime usually placed in expressions for conductivity in a metal is not the recombination time. It is sometimes called the dephasing time. The recombination time in a metal is effectively infinite.

    Semiconductors come in various types with varying conductivity. Low resistivity semiconductors are just like metals. The majority carrier has a large recombination time. High resistivity semiconductors are just like insulators. The carriers have a very short recombination time.
    Last edited: Apr 2, 2013
  17. Apr 3, 2013 #16
    Of course, I do know what recombination is and what carrier collisions are, but I don't know what makes the recombination rate short or long (which you and Cthugha have tried to explain to me).

    I've gathered that (all else constant) a larger band gap corresponds to faster recombination time, although I'd very much like to see this explained (mathematically) in terms of the golden rule as Cthugha has described.

    Returning to the original point, chill_factor mentioned that there are no radiative transitions in conductors. Is this the answer to the original question as to why we don't use conductors in optoelectronics?

    Also, as to my question about absorption and emission in conductors, I'm still not grasping this completely. Without thinking in terms of plasmons, what is a simple conceptual explanation as to the difference between semiconductors and conductors with respect to emission and absorption? Assuming no knowledge of plasmons.
  18. Apr 3, 2013 #17
    The electrical resistance of a semiconductor is highly sensitive to impurities. Hypothetically, a semiconductor crystal with no impurities, no light and at a low temperature is an insulator. It does not conduct electricity because it has no charge carriers. It does not absorb below band gap light because there are no empty mid gap states.

    A low resistivity semiconductor is basically the same as a metal. If a semiconductor crystals contains an excess of one type of shallow-impurity, at a temperature where that shallow-impurity will thermally ionize, then the semiconductor has an excess of charge carriers. The charge carriers in this low resistivity semiconductor do form plasmons.

    A high resistivity semiconductor is basically the same as an insulator. If a semiconductor crystal has no carriers, then it won't conduct electricity.

    If one visually compares two semiconductors with identical crystal structure (not including the impurities), the low resistivity semiconductor will look shinier than the high resistivity semiconductor. This is something one can see with the unaided eye. The extra reflection of light is due to charge carriers which are locked together to form plasmons. The reflectivity of the low resistivity semiconductor will resemble that of a metal.

    I am using the phrase charge carriers because it doesn't matter whether the excess charge carriers are valence-holes or conduction-electrons. The low resistivity semiconductor reflects light just like a metal.

    The majority carriers in a low resistivity semiconductor have basically an infinite recombination time. Just like a metal. If the excess carriers in the semiconductor are conduction-electrons, then the conduction-electrons have an infinite lifetime. If the excess carriers in the semiconductor are valence-holes, then the valence-holes have an infinite lifetime.

    What makes a low resistivity semiconductor different from a true metal is basically how the majority carriers were generated. A true metal has excess carriers even without impurities or light. Lead, for instance, has a low conductivity no matter what impurities you place in it. Silicon has no excess carriers unless it is grown with impurities. In a semiconductor, it is either impurities or light that provide the excess carriers.

    The optical properties of a metal are largely determined by its conductivity. When you take electromagnetic theory, you will learn how to calculate reflectivity from the conductivity of the metal. The same formulas for metal work in low resistivity semiconductors. From the standpoint of electrical properties, a metal and a low resistivity semiconductor are the same thing.

    The conductivity of a material is determined by the product of the carrier density and the mobility. The mobility is proportional to the time between collisions. The carrier density is determined differently in metals and semiconductors. Basically, the carrier density of semiconductors is the density of the impurities that contribute carriers. The carrier density in the true metal is basically the density of atoms times the number of free carriers in the outermost band.

    Recombination time is important when it comes to photoconductivity. Photoconductivity can be important in high resistivity semiconductors. If a semiconductor crystal has no impurities, or is doped with impurities that absorb free carriers, then the carrier density will be very low. The unilluminated crystal may effectively have no free carriers. However, a small amount of light can make that crystal become a "conductor".

    There is no unique answer to the question, "What causes recombination." There is more than one process that can cause recombination of free carriers in a semiconductor. There are journals devoted to that sort of stuff. Band gaps are sometimes involved but not always.

    I happen to be doing some research now on the recombination time in a high resistivity semiconductor. The research happens to be rather specific, not general. However, a short description may be helpful to you.

    The material that I am working on is a high resistivity semiconductor crystal. The material has free carriers only under certain illumination conditions. If kept in the dark at room temperature, it can not conduct electricity. I did not know at the beginning of the research anything about the impurities in the material. In fact, I was told that the crystal was "absolutely pure". The other students called it "intrinsic material". Their logic was that a high resistivity semiconductor must be pure. My research showed not only that it wasn't pure, but the acceptor destroyed the free carriers made by other impurities.

    I did not know the mechanism of recombination when I started the research. However, my research has revealed that the recombination lifetime is determined by the concentration of acceptor.

    In the dark, it has no free carriers. Exposed to bright light, it has lots of impurities. In the specific situation that I am working on, the recombination time is determined by impurities.

    The semiconductor crystal that I am working on is "compensated". It has a nearly equal amount of shallow-donors and shallow-acceptors. Therefore, the conduction-electrons and the valence-holes annihilate each other. A few deep-level impurities are present which destroy any free-carriers caused by the small imbalance between donors and acceptors. Because the unilluminated material has no free-carriers, the unilluminated material is an insulator.

    When I write the article, I will refer to it as semi-insulating. This way, new students won't have to ask me whether it has semi-conductivity. The semiconductor is acting like an insulator because that is how the impurities in it were added.

    What I am finding is that the recombination time of the free carriers is determined by the shallow-acceptor concentration. The acceptor is a recombination center. When the crystal is illuminated, an equal amount of conduction-electrons and valence-holes are generated. Supposedly, the conductivity should go down though I am using other measurements to detect these carriers.

    I didn't know that to begin with. I initially thought that the recombination time was determined by something called an "exciton". An exciton is a conduction-electron in orbit around a valence-hole. The exciton doesn't require any impurities. I thought that the lifetimes I was measuring were determined by some property of the exciton. Maybe in some other semiconductor under different conditions, the exciton does determine the lifetime! There are many things that can affect recombination time. Excitons are sometimes important and sometimes not.

    The acceptor first captures a conduction-electron, then it captures a valence-hole, and then it forces then to combine. A photon is given off (which I am detecting). However, the rate at which these carriers are destroyed is determined by the concentration of acceptors.

    Note that the band gap does not uniquely determine the recombination time of the material. The recombination time in this material is highly sensitive to what impurities are in the crystal. There are apparently a large number of different types of impurity in the crystal. However, I managed to determined that the acceptor (which may itself be a mixture) mostly determines the recombination time of the carriers.

    I think what I am trying to say is that you can't ignore the impurities. You can't determine the properties of a semiconductor from the band gap alone.

    In any case, the extra reflectivity in metals and low resistivity semiconductors is caused by plasmons. The carriers in these materials move in unison under low frequency light. The carriers absorb high frequency light. High frequency light causes the plasmons to break up.
    Last edited: Apr 3, 2013
  19. May 3, 2013 #18
    OK, let me see if I've got this straight. I've done a little reading into plasmons. How about this:

    We think of a metal as a plasma, because our model is that of an electron gas with ionized atoms. For (certain) doped semiconductors, we think of plasmons as we do in metals. In metals, the significance of the plasmon frequency (ωp) is that for ωpphoton, the metal is reflective, and for ωpphoton, the metal is transparent. Semiconductors have a lower ωp than metals, and are therefore less reflective (metals have a higher tendency to re-emit photons).

    Am I getting warm?
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