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Chapter 10 The Solid State 10.1 Crystalline and Amorphous Solids These are the two major categories into which solids are divided. Crystalline solids exhibit long-range order in their atomic arrangements. (The order is usually three dimensional, but lower dimensionality order is possible.) Bonds in crystalline solids are more or less the same in energy, and crystalline solids have distinct melting temperature. Amorphous solids exhibit only short-range order in their atomic arrangements. Their bonds vary in energy and are weaker; they have no distinct melting temperature. A good example is B2O3. See Figure 10.1b for crystalline B2O3, and Figure 10.1a for amorphous B2O3. Notice the difference. Every B is surrounded by 3 O's. That's the short-range order. In crystalline B2O3, you can "look" in any direction and see the same environment, but not in amorphous B2O3. Later on we will talk about several crystal structures. It's easy to think of real crystals as having these ideal structures. In fact, no crystals are perfect; all crystals have defects. Crystals can have: The "right" atoms in "wrong" places. "Wrong" atoms in "right" or "wrong" places. Etc. One type of defect is the point defect. There are three basic kinds of point defects. The vacancy. 117 The interstitial. The impurity, which can be either a substitutional impurity or an interstitial impurity. substitutional interstitial Point defects makes diffusion in solids possible. Either vacancies or interstitial atoms can migrate through a crystal. Diffusion is strongly temperature dependent. Higher dimensional defects include edge and screw dislocations. A dislocation occurs when a line of atoms is in the wrong place. Dislocations are important but more difficult to deal with than point defects. Edge dislocation are easy to draw and visualize. See Figure 10.3. Screw dislocation are more difficult to draw. See Figures 10.4 and 10.5. Work hardening. Work hardening occurs when so many dislocations are formed in a material that they impede each others' motion. Hard materials are often brittle. Annealing. Heating (annealing) any crystal can remove dislocations. Annealing makes metals more ductile, and can be used to remove secondary phases in crystals. 118 10.2 Ionic Crystals An atom with a low ionization energy can give up an electron to another atom with a high electron affinity. The result is an ionic crystal. To calculate the stability of an ionic crystal, we need to consider all of the energies involved in its formation. Positive energy is required to ionize an atom. Energy is released when a highly electronegative atom gains an electron (a negative contribution to the total energy). There is a negative contribution to the energy from the Coulomb attraction between unlike charged ions. There is a positive contribution to the energy from the overlap of core atomic electrons (the Pauli exclusion principle at work). Remember that negative energies mean stable systems. If you add up all the above energies and get a more negative energy than for the separate, isolated atoms, then the ionic crystal is stable. Ionic crystals are generally close-packed, because nature "wants" as many ions of different charge squeezed together as possible. Like-charged ions never come in contact. There are two primary structures for ionic crystals. Face-centered cubic (fcc). Draw a picture. Example: NaCl. Body-centered cubic (bcc). Draw a picture. Example: CsCl. It is not too difficult to calculate the energies involved in ionic bonding. We define the cohesive energy of an ionic crystal as the reduction in the energy of the ionic crystal relative to the neutral atoms. Later Beiser implicitly generalizes this definition to all crystals. Let's calculate the contribution to the cohesive energy from the Coulomb potential energy. Let's do it for an fcc structure. Let's take a Na+ ion as the reference ion. We get the same result if we take a Cl- as the reference. We will add up all contributions to the Coulomb energy from ion-ion interactions. 119 Each Na+ has 6 Cl- near neighbors at a near-neighbor distance r. This figure shows four of them. There are two more in the planes above and below the central atom. Can you visualize them? The contribution to the Coulomb potential from these six nearest neighbors is V1 = 6(+e)(-e) 6e2 = . 4 π ε0 r 4π ε0 r This represents a negative (more stable) contribution to the total energy. Each Na+ has 12 Na+ next-nearest neigh-bors at a distance of 21/2r. Since these are + ions, the interaction is repulsive, and the contribution to the total energy is positive. The figure to the right shows four of the next-nearest neighbors. There are four more in the plane above and four more in the plane below the one shown. The contribution from the twelve next-nearest neighbors is V2 = + 12 e2 . 4π ε0 r 2 We can do the same for additional ion "shells." The result is e2 e2 e2 12 V = +... = -1.748 = -α . 6 4π ε0 r 4π ε0 r 4π ε0 r 2 The constant (which is constant only for a given type of structure) known as the Madelung constant. Beiser gives the values of for a couple of other structures. Note that 12 α = 6+... = ( 6 - 8.485281...+...). 2 The convergence of this series is very poor. You have to find a clever way to do this series if you want to calculate with a reasonable amount of effort. We've accounted for the Coulomb attraction. Note the - sign in the equation for V, so it is an attractive interaction. 120 Now we need to account for the repulsive forces that happen when electron shells start to overlap. Remember, overlapping shells might result in electrons in the same place with the same quantum number. The Pauli exclusion principle says this can't happen. Electrons have to be promoted to higher energies to allow atoms to come closer together. The result is a more positive energy, i.e., a less stable crystal. We model this repulsive force with a potential of the form Vrepulsive = B , rn where n is some exponent. The exact value of n isn't too critical (see the figure to the right). The potential energy of interaction of the Na+ ion with all the other ions is V = - α e2 B + n. 4π ε0 r r We can find B in terms of and the equilibrium near-neighbor separation r0 by realizing that at equilibrium the energy is minimized. α e2 nB ∂V 0 = = 2 - n+1 ∂r r=r0 4π ε0 r 0 r 0 α e2 nB 2 = 4π ε0 r 0 r 0n+1 B = α e2 n-1 r . 4 π ε0 n 121 Plugging B back into the expression for V gives the lattice energy. V = - α e2 1 1- . 4π ε0 r 0 n The lattice energy is defined as the reduction in the energy of the ionic crystal relative to the ions at infinite separation. Note the difference between lattice and cohesive energy. The lattice energy doesn't take into account the ionization energy and electronegativity. The cohesive energy also applies to non-ionic crystals. A "typical" value for n is 9. Beiser on pages 364-365 calculates the cohesive energy for NaCl. The lattice energy is: α e2 1 (1.6x10-19 ) 1 1- . = - (1.748)(9x109 ) 1- 2.81x10-10 9 4π ε0 r 0 n = -1.27x10-18 Joules = - 7.96 eV. 2 V = - This calculation counts pairs of ions. The lattice energy per ion is thus -3.98 eV. It takes energy to ionize an Na atom, and energy is reduced when an electron is transferred to the Cl. The increase is greater than the reduction: net = (+5.14 - 3.61) = + 0.77 eV per ion. 2 The net reduction on bonding is -3.98 eV+0.77 eV=-3.21 eV, which compares quite well with the measured value of -3.28 eV. Some properties of ionic crystals: They have moderately high melting points. They are brittle due to charge ordering in planes. They are soluble in polar fluids. 10.3 Covalent Crystals There are relatively few 100% covalent crystals in nature. Examples: diamond, silicon, germanium, graphite (in plane), silicon carbide. These materials are all characterized by tetrahedral bonding, which involves sp hybrid orbitals. 122 Some properties of covalent crystals: They are brittle due to their highly directional bonding. They have high melting points due to their high bond strengths. They have low impurity solubility due to their directional sp hybrid bonds. They are insoluble in polar fluids. Sometimes I show the diamond model. 10.4 Van der Waals Forces Even inert gases form crystals at low temperatures. What force holds them together? Ionic bonds -- no, because inert gases don't ionize. Covalent bonds -- no, because inert gases have no "desire" to share electrons. Some kind of Coulomb attraction? Inert gases are are electrically neutral, so they don't attract charged particles. Something like hydrogen bonding (e.g., water), where the asymmetry of the molecule gives rise to a nonuniform charge distribution and a polarity? Not if inert gas atoms are nonpolar. The answer is that, while on the average inert gas atoms are nonpolar, in fact their electrons are in constant motion and they can have instantaneous nonuniform charge distributions. The + and - portions of the atom can exert attractive force, which is enough to bond at very low temperatures. It's a Physics 24 type problem to show that a dipole of moment p gives rise to an electric field E given by 1 p 3(p • r) E = r . 4 π ε 0 r 3 r5 This electric field can induce a dipole moment in a normally nonpolar molecule. The induced dipole moment is p ′ = αE, where is the polarizability of the molecule or atom. The energy of the induced dipole in the electric field is V = - p ′ • E, and it is not too difficult to show that V is proportional to -1/r6. 123 Things to note: the potential is negative (attractive, stable). The force is proportional to dV/dr which is proportional to 1/r7, so the force drops off very rapidly with distance. Some general comments. Van der Waals forces are responsible for hydrogen bonding, such as occurs in water. Van der Waals forces are always present, but are often so relatively weak that they can safely be neglected. Our brief derivation is a classical one, and, as you might suspect, not really correct at the atomic level. However, a full quantum mechanical derivation gives the same r-dependence. The Lennard-Jones potential, which you may have encountered in chemistry (also sometimes known as the Lennard-Jones 6-12 potential or the 6-12 potential) describes van der Waals forces. The "6" part comes from the 1/r6 attractive term in the potential. There is also, of course, a repulsive term due to electron overlap. In the "6-12" potential, this is modeled with a 1/r12 dependence (recall we used 1/r9 earlier in this chapter, but that's not as serious a difference as it might look). 10.5 Metallic Bond Metallic bonding is caused by electrostatic forces acting in combination with the Heisenberg uncertainty principle and the Pauli exclusion principle. Metal atoms give up electrons (usually one or two, sometimes three). The result is a lattice of positive ion cores sitting in a "sea" or "gas" of electrons. The electrons in the gas repel each other. The electron gas and ion cores attract each other. Bonding occurs because the reduction in energy due to the attraction exceeds the increase in energy due to repulsion. Wait a minute. Don't we have a problem here? If we bring 1022 or so atoms together and try to combine that many electrons into a single "system," don't we have a problem with the Pauli exclusion principle? We sure do. It would seem that the energies of most of the electrons would need to be so high (remember, energies go up as we put electrons in successively further out shells) that the net electron energy would be so great that bonding could never occur. How do we get around this? The energy levels of the overlapping electron shells are all 124 slightly altered. The energy differences are very small, but large enough so that a large number of electrons can be in close proximity and still satisfy the Pauli exclusion principle. We are describing the formation of energy bands, consisting of many states close together but slightly split in energy. They are so close together that for all practical purposes we can consider bands as a continuum of states, rather than discrete energy levels as we have in isolated atoms (and in the core electrons of atoms of metals). This sounds kind of artificial. It sounds like an ad hoc explanation; maybe bad science. Where do these bands come from? We should first ask where the electronic energy levels in atoms come from. Remember, we solved the Schrdinger equation for an electron in the potential of the hydrogen nucleus. This gave us our energy levels and quantum numbers. More complex atoms require more complex mathematics, but the idea is the same: the energy levels come from the solution of Schrdinger's equation for electrons in the potential of the nucleus. The same holds for metals, except now we have a periodic array of nuclei, and a periodic potential. We still have to solve Schrdinger's equation for an electron moving in this periodic potential. The problem can't be solved exactly, of course, but it can be solved with quite reasonable accuracy. The energy bands result quite naturally from the solution, just as energy levels in atoms came naturally out of the solution to Schrdinger's equation. We will discuss band theory in more detail in later sections in this chapter. Ohm's Law Any reasonable theory of the solid state should be able to come up with Ohm's law. Since we are talking solids and metals, now is a good time to consider it. Ohm's law is an empirical relationship. I = V . R What does this really say? Remember, an empirical equation is something which seems to agree with experiment, but which doesn't come from any theory. Empirical equations are not very satisfying to physicists. Let's see if we can derive Ohm's law. What do electrons in a conductor do in the absence of an applied electric field? They move, very rapidly, but in random directions. There is no net current. 125 What happens when an electric field is applied? Electrons accelerate and gain a net "drift" velocity in some direction. The drift velocity is actually very small compared to the Fermi velocity. Since an electric field accelerates electrons, why don't they move faster and faster? Or do they do just that? Some force must eventually oppose the force due to the electric field and bring the drift velocity into a "steady state" condition. What is this "force"? Typically it is a result of collisions of electrons with impurities. Let's analyze this mathematically. Ohm's law had better come out of the analysis. Only electrons near the Fermi energy can be accelerated (remember, all states below F are occupied). If we define as the mean free path of electrons between collisions, then the average time between collisions is τ = λ . vF An applied electric field results in a force on the electron, and therefore an acceleration a = F eE = . m m Since the electron is accelerated only during the time , after which it undergoes a collision and has its velocity randomized again, the average gain in velocity, or "drift velocity," is vd = aτ = eEλ . m vF The total current I flowing through a conductor of length L, cross-sectional area A, with N free electrons per unit volume is I = N A e vd = N e2 λ A NA e2 Eλ V = , V = m vF m vF L R where ρL m vF L R = = . N e2 λ A A 126 is the resistivity of the metal sample. (We got this because E=V/L is the electric field inside the conductor.) What have we done? Simple drift velocity theory leads to Ohm's law. What are some consequences? Mobility of an electron is defined as µ = eτ . m The mass here is the electron effective mass, which we will investigate later. Some materials, such as GaAs, have very small effective masses, and thus very high mobilities and carrier velocities. They make good high frequency devices. I didn't see anything about the ion cores of the metal lattice in this theory anywhere. The example on page 349 says that electrons in copper at room temperature travel past an average of 150 + ions without "seeing" any of them. Isn't that strange? Yes, but in fact, electron waves in a perfect periodic lattice interact with the lattice only under very special circumstances. Resistivity (which comes from collisions) is due to imperfections in the lattice. 10.6 Band Theory of Solids This section in my notes will be very long, with many digressions. Band theory is worth several textbooks by itself. What happens in crystalline solids when we bring atoms so close together that their valence electrons constitute a single system of electrons? The discussion below comes directly from my lecture notes for section 10.5, and applies again here. "How do we get around this? [The fact that all the electrons constitute a single system.] The energy levels of the overlapping electron shells are all slightly altered. The energy differences are very small, but enough so that a large number of electrons can be in close proximity and still satisfy the Pauli exclusion principle." "We are describing the formation of energy bands, consisting of many states close together but slightly split in energy. They are so close together that for all practical purposes we can consider bands as a continuum of states, rather than discrete energy levels as we have in isolated atoms (and in the core electrons of atoms of metals)." 127 A detailed analysis of energy bands shows that there are as many separate energy levels in each band as there are atoms in a crystal. Remember that two electrons can occupy each energy level (spin), so there are 2N possible states in each band. Figure 10.19 (show a transparency) shows how the sodium energy levels spread out into bands when we bring sodium atoms together. Some atomic energy levels are shown to the right of the figure (the 3p level has already formed a band in this figure). Sodium contains a single 3s electron, so the figure shows energy levels which are not normally occupied. As the sodium atoms are brought closer together than the 1.5 nm distance to the right of the figure, notice how the 3p band spreads wider, and notice how the 3s state begins to form a band at a separation of about 1 nm. Also notice how electron energies can be reduced upon formation of bands. The metal is stable, after all. We schematically represent the energy bands in sodium like this: This is highly schematic. Real bands aren't boxes or lines. Sodium has a single 3s electron, so the 3s energy band contains twice as many states as there are electrons. The band is half full. At T=0 the band is filled exactly halfway up, and the Fermi level, F, is right in the middle of the band. Sodium is a metal because an applied field can easily give energy to and accelerate an electron. Read how magnesium, which you might at first expect not to be a metal, is metallic because of overlapping bands. This material is testable even if I don't lecture over it. When a filled and an empty band overlap, each of the two bands will be partially filled, giving rise to semi-metallic behavior. Now let's consider the energy bands in diamond. Carbon has two 2p electrons. A p shell has 6 possible states, so a d-band should have 6N states, where N is the number of atoms in the crystal. 2N 2p electrons filling a band with 6N states should produce a metal. 128 But diamond is an insulator. Figure 10.22 shows why. The 2s band (2N states) and 2p band (6N states overlap when carbons are brought close together, and form a single band with 8N states. As carbon atoms are brought still closer together, the single band splits into two hybrid bands, each having 4N available states. The equilibrium separation of the carbon atoms in diamond lies within this range. The 2N 2s and 2N 2p electrons occupy the lower of the bands with 4N states. The band is full. There is a large energy gap (6 eV, a really large gap) between the full band and the next higher band. Thus, in diamond,there is a full band, a 6eV gap, and an empty band. The 6 eV gap is a big gap. There is not enough thermal energy at normal temperatures to put any 2p/2s electrons into the empty band. An electric field can't give energy to electrons in the filled band, because there are no unoccupied energy states nearby. A huge electric field is required to get an electron across the band gap. Thus, diamond is a very good insulator. So far, we've covered conductors (odd numbers of valence electrons and partly filled bands) and insulators (even numbers of electrons and all filled or all empty bands with large band gaps). Remember that our band representations shown in the last two figures are highly schematic representations. Real bands aren't square like this. There's still another possibility for forming bands -bands which would normally be all filled and all empty but with small gaps between them. Such materials are semiconductors. To the right is a schematic energy band diagram for a semiconductor. The gap between bands is relatively small, e.g., 1 eV in silicon. Room temperature is only 0.040 eV, but there are lots of electrons in the filled (at T=0) valence band, and it doesn't take a very large probability to get quite a few of them across the band gap at moderate temperatures. 129 Thus, at very low temperatures, silicon is an insulator, but at room temperature, it is a weak conductor (intermediate between conductor and insulator, hence semiconductor). A rule of thumb: band gap of less than 3 eV gives rise to semiconductor. We will skip section 10.8 (Energy Bands: Alternative Analysis) which tries (with only limited success, in my opinion) to explain the origin of forbidden bands. I will try to justify forbidden bands here. Remember that I can calculate a free electron's energy in terms of its wavevector: 2 k 2 E = , 2m where k is the wavevector and m is the electron's mass. A plot of a free electron's energy looks like the figure to the right. What happens when you put the electron in a metal? It moves in the periodic potential of the crystal lattice. Remember, an electron is a matter wave. We studied (however briefly) diffraction of waves back in chapter 3. Waves can be diffracted by periodic lattices. Because the electron exists in the "box" of the crystal, only certain wavevectors are allowed. For a free electron in a large enough box (such as a macroscopic piece of a metal), there are enough wavevectors to essentially form a continuum. The lattice diffracts electrons of a few select wavevectors. Exactly which wavevectors are diffracted depends on the details of the lattice, but are just a generalization of Bragg's law. Electrons of wavevector appropriate for diffraction "bounce back and forth" inside the metal, constantly being diffracted. Such electrons, going nowhere, form standing waves. If we solve Schrdinger's equation for these particular standing waves in the potential of the lattice, we find that there are two possible solutions. One solution has a lower energy than a free electron of the same wavevector, and the other solution has a higher energy than a free electron of the same wavevector. 130 The stronger the lattice potential, the wider a range of wavevectors near the "standing wave" vector are affected, and the larger the gap between the two allowed solutions. The free electron energy-wavelength relationship is modified like this: Electrons near the "zone boundary" (which is determined by and related to the periodicity of the lattice) are the ones most affected by the periodic potential of the lattice. There are two possible solutions for electron energy at the zone boundary. Near the zone boundary, the energies are modified by the lattice potential. Notice the energy gap at the zone boundary. There are energies which are simply not attainable by the electrons. Again, this is because they are waves moving in the periodic potential of the lattice. Not all waves can exist in the periodic potential. Effective Mass I've switched from epsilon's to E's for electron energy here. No reason in particular, just did it and didn't want to go back and change everything. This discussion of effective mass is not in the assigned material in the text, but it fits in after the above discussion. We know that 2 k 2 E = 2m relates an electron's energy and wavevector. Note that the coefficient of wavevector squared is proportional to 1/m, where m is the electron mass. The above result holds in general. For an electron in a periodic potential, we write 131 2 k 2 E(k) = , 2 m* where m* is the electron's effective mass. In chapter 9, we found that the Fermi level for free electrons is εF h2 3N = 2m 8πV 2/3 . Here, when we put electrons into a lattice, we have to account for the modification of the electron behavior by the periodic potential of the lattice. This is where equation 10.24 comes from. The Fermi energy in a metal (where there is a potential from the lattice) is µF h2 3N = 2 m* 8πV 2/3 . The wavevector of an electron at the Fermi surface is just kF 3N = 8 πV 2/3 . Notice that 1/m* has something to do with the slope of the E versus k curve. "Flat is fat." The less the E versus k curve changes over some k, the flatter the E versus k curve, and the bigger ("fatter") the effective mass m*. An electron with a very big effective mass is basically being held in place by the lattice. An electron with a very small effective mass is basically getting a kick from the lattice. What about an electron with a negative effective mass? Some electrons almost have the appropriate wave vector for diffraction. If I give them just a little more energy, their wavevector satisfies the Bragg condition, and they are diffracted. I push them in one direction and they bounce back in another. Hence a negative "mass." Impurity Semiconductors Let's move back towards devices. We saw above how metals have partly filled conduction bands, and insulators have filled valence bands, empty conduction bands. 132 Semiconductors (as described above) also have valence bands which are filled and conduction bands which are empty at T=0. For T>0 some electrons can get across the energy gap from the valence band to the conduction band and conduct electricity, but not very many. In semiconductor devices, we want materials which are good conductors when we want them to conduct, and not conductors at all when we don't want them to conduct. The (idealized) materials we've talked about don't sound like this. What do we do to make semiconductors more effective? We put in "junk" -- impurities. Here's how it works. Example: arsenic impurities in silicon. As has an outer electron configuration of 4s24p3. After it shares 4 of its outer electrons with neighboring silicons, the remaining electron is very loosely bound. In fact, the arsenic impurity creates a donor impurity level. Since the donor ionization energy is very small, the donor level sits just below (maybe a small fraction of an eV) the conduction band. Electrons from the donor levels can easily get to the conduction band, where they are available for conduction, just as normal conduction band electrons are. It turns out that there don't have to be very many donor atoms around to result in a significant number of electrons in the conduction band. Beiser gives some numbers on this. 133 The semiconductor we have been discussing are n-type, because conduction is by negative electrons. Another type of semiconductor is the p-type semiconductor. As the name implies, conduction is by positively charged objects, which we call "holes." Here's how it works. Example: gallium impurities in silicon. Ga has an outer electron configuration of 4s24p1. It "wants" to borrow 5 electrons from neighboring silicon atoms. Four of the electrons are normally shared by silicon anyway. It turns out very little energy is required for the gallium to "borrow" the additional electron. The gallium impurity creates an acceptor impurity level. Since the acceptor ionization energy is very small, the acceptor level sits just above (maybe a small fraction of an eV) the valence band. Electrons from the valence band can easily get to the acceptor bound states. That leaves holes in the valence band. The holes represent states into which other electrons in the valence band can move. Thus, electrons can easily move around in the presence of an applied field. Alternatively, we can look at the holes, say they move around, and say that conduction is due to holes. Again, it turns out that there don't have to be very many acceptor atoms around to result in a significant number of holes in the valence band. What's all this fuss about holes. It seems like a hole is really just an electron missing from the valence band and sitting in an acceptor state. Why don't we just talk about conduction by these electrons moving from state to state, instead of worrying about holes. Answer: holes really are more than just missing electrons. For one thing, electrons in this case are "stuck" on acceptor atoms, but the holes are free to move about in the valence band. 134 For another thing, we can dope semiconductors so that there are excesses of holes and electrons. A "hole" really is more than just a missing electron which is somewhere else. 10.7 Semiconductor Devices We now consider how we make useful things out of semiconductors. In the introductory paragraph in the previous edition, Beiser mentioned that a chip half a centimeter square can hold up to 50,000 transistors. That edition of Beiser is dated 1988. How many transistors are in an Intel 486 CPU? (About 1.25 million; about 3 million in an original pentium; how many now?.) Let's consider the simplest device, a junction diode. A junction is just an n-type semiconductor in contact with a p-type semiconductor. How do we make a junction? Not by shoving two semiconductors together. Can't make good enough contact that way. Instead, we diffuse donors or acceptors into silicon. We define the regions into which impurities are diffused with masks. Sort of like stencils. We can build up a whole series of layers in devices. A p-n junction diode is a device which lets current flow in only one direction. Let's see how it works. Here's a junction. Here's a plot of hole and electron concentrations (just examples, I've picked hole concentration to be larger in this particular figure). 135 Now I'll put the last two figures together so you can see how they correlate. Now, nature doesn't like this. It's like trying to fill only the right half of a glass with water. What happens next? Holes and electrons try to diffuse (holes to the n-region, electrons to the p-region). What, if anything, will stop them? Remember, the crystal as a whole is electrically neutral. There are + ions on the donor side and - ions on the acceptor side. Coulomb repulsion will eventually stop the diffusion of holes and electrons, and we end up with this: 136 This charge double layer gives rise to a built-in electric field which inhibits diffusion of holes and electrons. What happens to the Fermi level? The Fermi level is like the level of a liquid in a glass. It's the same everywhere in the material. The Fermi level is equalized because of the electric potential due to the charge double layer, which raises the Fermi level of the p-region relative to the n-region until the Fermi level is the same everywhere. Now let's re-do the above analysis by looking at the band structure of the junction. Here are the bands of the p- and n-type semiconductors not in contact. Here are the bands when we put the above two semiconductors in contact: This last figure ought to look wrong. What's wrong is that there are two Fermi levels. Here's what happens: 1. Electrons and holes flow to shift the Fermi levels, until the F's are equalized. 2. The bands, being fixed relative to the Fermi levels, are shifted up or down relative to each other. 3. Eventually, the built-in field (which came about from the flow of holes and electrons) shuts off further net flow of holes and electrons. To move any more, electrons must now flow uphill and holes "float" downhill. 137 4. The result is a steady state, where there are continuous small flows of holes and electrons. These flows cancel each other, so that the net flow is zero. (Where do these flows come from?) 5. Electrons flow downhill, holes float uphill, until we get a band structure that looks like the one shown to the right. Example of an application: a junction diode. The above discussion was basically a detailed explanation of the "No bias" case on page 358 of Beiser. Next, let's reverse bias the junction, like this: The reverse bias results in a net electron flow to the right and hole flow to the left. Both nand p-regions are quickly depleted of charge carriers, and current stops almost immediately. Also, the reverse bias increases the potential difference, discouraging the flow of charges. Finally, let's forward bias the junction, like this: Now the potential energy barrier is lowered. But remember, the barrier stopped the electrons, so now electrons can flow. Electrons and holes crossing the junction meet and annihilate, but a steady flow of electrons and holes is "pumped" into the device by the external emf, so current flows. 138 The voltage-current characteristic of a p-n junction diode looks like this: When a pn junction is formed, a depletion region occurs between the two. In this region, electrons from the n region have gotten into the p region and filled some of the holes. Also, holes from the p region have gotten into the n region and recombined with some of the electrons. The net result is a deficiency of both types of charge carriers in this narrow region. Application of pn junctions: light-emitting diodes. As we saw above, the proper application of potentials can cause electrons and holes to recombine. Light is emitted in some semiconductors when electrons and holes recombine. A voltage applied across such a semiconductor can result in light. Advantage: very low current or very low power lamps. Population inversions can be created in some pn junctions, which lead to solid-state lasers. Application: solar cells (just light-emitting diodes operating in reverse). Shine light on a pn junction. The light creates extra electrons and holes. These extra carriers diffuse to the energy barrier. This gives a forward voltage across the barrier. The forward voltage can deliver power to an external circuit. 139 Application: photovoltaic detectors. The idea is the same as for solar cells, except they are used to detect photons rather than generate useful power. Another kind of p-n junction device: the tunnel diode. We can heavily dope the semiconductors which make up the pn junction, so that the bands overlap like this: Electrons and holes can tunnel both ways across the very narrow depletion layer between the semiconductors. In the absence of an applied voltage, the net current is zero. A small forward voltage does this to the band structure: Now more of the electrons in the n-region want "want" to tunnel "downhill" into the p-region, and fewer of the holes in the p-region "want" to tunnel into the n-region. The tunnelling is from n to p only. The result is an "excess" current for this voltage. Additional forward bias raises the n-region conduction band too high for tunnelling to occur, and the device turns into an ordinary diode: This is the current-voltage characteristic (see figure 10.2: The advantage to a tunnel diode is that it is a device which changes current very rapidly in response to a voltage. Zener diode, avalanche multiplication, and Zener breakdown. Read about these. I will skip them, not because they are not important, but because of lack of time. The basic idea behind the Zener diode is that at very high reverse voltages (remember, reverse voltage normally shuts off the flow of electrons), electrons can be accelerated to large enough 140 energies to ionize atoms, which creates an avalanche of electrons, and current can flow in the "wrong" direction, or else electrons can tunnel in the "wrong" direction. Transistors. We will discuss an npn junction transistor here first. Electrons are the charge carriers. You can also make pnp junction transistors, with holes as the charge carriers. Look at two np junctions back-to-back: Holes and electrons want to diffuse, but after reaching equilibrium, they can't diffuse any further because of the built-in voltage. Now forward bias the first np junction: Electrons ("majority carriers" in the n-region) will flow from the n-region (called the "emitter") into the p-region (called the "base"). In other words, electrons are injected into the base, where they are minority carriers. What happens to these electrons? They would recombine with holes in the base... 141 But suppose we make the base thin, so that electrons don't recombine before they reach the other n-region, and suppose we reverse bias the second pn junction. Remember, the reverse bias sweeps electrons from the p-region into the nregion. Above was a crude schematic showing the resulting energy bands. To the right is a picture of the connections. In the circuit we have drawn, the output (reverse) current is a very high fraction of the input current. But remember the base-collector junction is a diode connected in reverse bias. And also remember, this is what the IV curve looks like for a diode. A very small reverse current corresponds to a very large reverse bias. In other words, the voltage on the output side can be made very much larger than the voltage on the input side. We have made a voltage (or power, since both currents are about equal) amplifier. Note that the current doesn't get amplified. 142 We can also operate a pnp junction transistor as a current amplfiier. A small input current into the base results in a very high output current. The reverse current is very sensitive to the forward voltage. In typical cases, we can get increases of factors of 1000 in reverse current when we apply a forward bias across the emitter and the base. These npn transistors are "junction transistors" and are called "minority carrier devices" because their behavior is dominated by minority carrier diffusion (through the base). Both majority and minority carriers may participate in the current, so these are bipolar devices. We typically dope the semiconductors to reduce this effect. We can also construct pnp transistors. The ideas are the same. The conduction is by holes. Disadvantages of junction transistors: It is hard to incorporate many of them into miniature circuits. They are low input impedance devices. Low impedance means large currents flow. This is undesirable for a number of reasons. Field effect transistors--read pages 364-6 in Beiser. The main idea is that voltages can be amplified with very little current flowing. 143