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Physics 361 Principles of Modern Physics Lecture 25 Bonds, Bands, and Semiconductors This Lecture • Hybridization • Solid State Physics -- Lattices and Bands • Fermi-Dirac Statistics • Semiconductors, Impurities, and Doping • Law of Mass Action and Doping Next Lecture • Semiconductor Devices • Bose-Einstein Statistics • Lasers Hybridization – hybrid sp orbitals For the atoms in the third and fourth group – they contain at least one empty p orbital. For such atoms, it does not cost much energy to take an electron in an s orbital and move it to the vacant p orbital. The physical reason for this is that the two s electrons will be feeling a relatively strong electrostatic repulsion because they occupy the same orbital. By moving one of these electrons to an empty p orbital relieves this electrostatic repulsion, so costs very little energy. After this promotion, we now have two additional unpaired orbitals for making molecular bonds. When both s and p orbitals form bonds together, a bit of the s and p orbitals are present in each atomic orbital that creates the molecular orbitals. Hybrid sp orbitals Each hybrid orbital contains a linear combination of the s and p orbitals. For example, for the sp2 case we have three orbitals: Hybrid sp orbitals and bonding with carbon Ethene is H2C=CH2 It has three sigma bonds in order to connect all the atoms. The remaining pz orbitals form pi bonds, which makes the carbon atoms double-bonded. In this way, all four electrons in each carbon is used in bonding. Other examples are methane (CH4) which has four single sp3 bonds to hydrogen atoms. Also, Ethyne (HC≡CH): Combining large numbers of atoms together As we combine atoms together in large groups, their coupling forces new orbitals to form, which break degeneracies. As we form new orbitals, we always conserve the number of orbitals. As the number of orbitals approaches a very large number for a macroscopic piece of matter, the new mixed orbitals form a nearly continuous band of states. Band Theory For example, consider atoms containing s and p orbitals. As the atoms are brought together, they form bands. At the top of the bands are the highest energy antibonding states. At the bottom, are the lowest energy bonding states. There can be energy gaps where there are no states. Filling of Bands To have electrical conduction, we require electrons to be in states near to ones which are empty. Thus, we only have conduction when bands are only partially filled. To determine conducting properties, we just need to count the electrons in the material! As electrons are added to the material, the fall to the lowest energy level. (Just like a ball rolling down a hill.) Conductors, semiconductors, and insulators For a material to be a good conductor, we need a partially filled band. First, there need to be electrons in a band for them to conduct a current. Second we need adjacent empty states for the energized electrons to inhabit. This gives the electrons some wiggle room in order to accelerate. Compare to a classical charged particle in an electric field. As the particle experiences a force, the energy state of the particle changes. If there are no energy states available for the particle to inhabit – it can’t accelerate!! Force Insulators and semiconductors only have filled bands and empty bands – neither of which contribute conductivity at low fields or temperature. Conductors For the material at right, this could represent the bands for a group 1 material like Na. In that case, we have one valence electron for each 3s orbital. If we have N atoms in a chunk of material, there will be N 3s electrons that can contribute to the s-band. The s-band must have N orbitals in the band, since orbital number is conserved. So we conclude that the s-band can hold 2N electrons (because each of its orbitals can hold a spin up and a spin down electron). We conclude that the top s-band for a typical group-1 metal must be only half filled. This makes such a material a conductor. Conductors – and complicated bands We might think that a typical group 2 material like Be or Mg would be an insulator. However, bands can mix in complicated ways as the atoms approach each other. When bands overlap with each other in energy, we can obtain a partially filled band. For Be and Mg, we have p and s bands which overlap in energy, so neither band is completely filled. Therefore, they are both metals. Insulators and Semiconductors These are materials have completely filled bands. Good examples of these are the group 4 materials. Carbon, Silicon, and Germanium. They all form the “diamond” lattice structure and have the same characteristic bands made from sp3 orbitals – shown at right. As the atoms are brought close together, the bonding and antibonding bands split and a gap opens. For Si and Ge, the band gap is a little larger than 1 eV, but for C (diamond) with a closer lattice spacing, the bandgap is larger (~7 eV). All these materials are what are called band insulators – having filled bands. However, as an insulators band is reduced to close to 1 eV, one typically calls such insulators semiconductors. Semiconductors To understand why a small bandgap materials is a semiconductor, we need to understand how to fill states occupied by fermions (like electrons) at a finite temperature. Fermions fill states according to the Fermi-Dirac distribution function fFD (E) = 1 (E-EF )/kBT e What this function tells +1 us is the probability to find an energy state at that energy filled. This function is plotted at right for two different temperatures. The Fermi-Dirac Function & Orbital States Energy Let’s understand what the FD function tells us. EF Let’s first consider very low temperature near absolute zero. In this case the FD function is a step function. If we have a bunch of discrete orbitals, the FD function tells us the probability they will be filled. At low temperature, the electrons will fill the lowest energy states possible, without the multiplefilling of states (except for an electron spin degeneracy) – because they are fermions. Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 The Fermi-Dirac Function & Orbital States for T>0 Energy At larger temperatures there are a larger number of statistical fluctuations which can cause electrons to be thermally excited to higher-energy states. EF This only happens close to the Fermi energy because low lying states require too much energy compared to the thermal fluctuations (kBT) to reach an unoccupied level. Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 Density of states larger chunk band The widths of the bands are determined by the proximity of the atomic orbitals. small chunk The number of actual orbitals in a band is proportional to the total number of the orbitals – that is, the amount of material in the chunk of matter. Over a small relevant range of energy, we will be interested in knowing the number of orbitals per unit volume of material. This is the density of states . g(E) The density of electrons per unit volume per unit energy is given by the product g(E) fFD (E). fFD (E) = 1 e(E-EF )/kBT +1 The Fermi-Dirac Function & Insulators For insulators with large band gaps, the FD function is almost completely 1 in the highest occupied band (“valence” band) and zero in the lowest unoccupied (“conduction” band). So the number of electrons per unit volume in the conduction band is approximately zero, and the number of empty states in the valence band is approximately zero as well. Thus for a large band gap material, there is no wiggle room for electrons to accelerate within any of the bands. fFD (E) = 1 e(E-EF )/kBT +1 The Fermi-Dirac Function & Semiconductors For semiconductors, as the band narrows we get some excited electrons from the valence band to the conduction band. Thus for a small band-gap material, there is some wiggle room for electrons to accelerate within the bands. The excited electron in the conduction band can now be accelerated and contribute to the conductivity. The excited electron also leaves behind an empty state in the valence band – this is called a hole – and it also contributes to the conductivity. e- fFD (E) = 1 e(E-EF )/kBT +1 The Fermi-Dirac Function & Semiconductors As the temperature increases (to lowest order), we expect more excited charge carriers and thus more electrical conductivity. e- fFD (E) = 1 e(E-EF )/kBT +1 Holes Conduction The empty states in the valence band act like positively charged particles with an effective mass. Energy They will contribute to the conduction very much like the excited electron and we must consider both their contributions. EF h+ Valence Let’s now look at conductivity as a function of carrier density. e- Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 Charge Carrier Motion in a Conductor The zig-zag black line represents the motion of a charge carrier in a conductor The net drift speed is small The sharp changes in direction are due to collisions The net motion of electrons is opposite the direction of the electric field Charge Carrier Motion in a Conductor The drift speed Vd is small (~1meter/hour) The fermi speed Vf is very fast! Due to quantum mechanics of electrons. (~106 meters/second) Charge rearrangements and signals travel near speed of light (~108 meters/second) vf 23 Written as a vector and a current density we have j = nqvd 24 Ohms law Ohm’s law just says that there is a proportional relation between the applied electric field in a conductor and the current that flows. j =sE Using Newton’s second law, we have F = -eE = me a Assume electrons accelerate for an average time t . So, at = vd = -eEt / me Inserting this into the current density, ne 2t je = nqvd = E me which is Ohm’s law. For holes we get a similar expression. pe 2t jh = E mh E -e F Relaxation time The relaxation time t is present in the Ohm’s law relations, ne 2t je = nqvd = E me pe 2t jh = E mh This time is the average time between collisions and is given by the Fermi velocity (v f) and the mean free path ( l ) between collisions as, t= l vf (Note: The relaxation time can be different for electrons and holes.) E -e F Conduction in Semiconductors Energy The carriers near the band edges are the most important for understanding conduction within a semiconductor. Thus, we can estimate the density of DE carriers (per unit volume) in the bands within a small energy range given by DE = kBT is e- EF n » g(EC ) fFD (EC )DE » N(me ) fFD (EC ) DE EC h+ EV p » g(EV ) fFD (EV )DE » N(mh ) fFD (EV ) where EC and EV are the band edges and N(mh ) and N(me ) are the orbitals/volume. Materials with roughly the same effective mass have roughly the same density of states near the band edges. So the number of states near band edges are roughly the same for most materials, and the FD function dominates the determination of the number of carriers. Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 Ohms law for semiconductors For semiconductors, the total current is the sum due to the electrons and the holes jTotal = je + jh The two forms we derived earlier for these are: 2 ne t je = E me pe 2t jh = E mh The carrier densities are typically the most sensitive to temperature and band gap sizes. The carrier densities can vary by many orders of magnitude. Thus, a great deal of focus in semiconductors is on the carrier densities. E -e F Intrinsic Semiconductors Generally, EF lies close to the middle of a pure semiconductor. The reason for this is due to the fact that for every excited electron we DE need to have a left-over hole – so ni = pi (where the subscript denotes intrinsic). This is considered the intrinsic behavior of semiconductors. Before moving on to extrinsic behavior, we will look at another way to excite electron-hole pairs with lower energy – these pairs are called “excitons”. DE Energy e- EC EF h+ EV Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 Excitons Since electrons and holes have opposite charge, they can be bound into a hydrogen-like state. As with hydrogen, this energy is negative, so it reduces the energy barrier for forming a hole-electron pair – and thus increases the probability for their formation. Energy ee- bound h+ The binding energy is much reduced (to less than 1 eV) compared to the 13.6 eV ground state of hydrogen due to dielectric screening and the effective reduced mass of the system. Since the hole-electron pair is bound (and must move together), it does not contribute to the net electrical current since it has a total charge of zero. Thus, we can ignore excitons when considering electrical conduction. EC h+ EF EV Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 Impurities in Semiconductors Intrinsic An intrinsic semiconductor will have zero net charge. This is even the case when we have thermally excited electron and hole carriers, because n = p . If we insert a neutral impurity to a semiconducting material, the overall charge of the material will be constant, but the material can disrupt the carrier charge balance. Well engineered impurities for semiconductors in technological applications are designed to easily ionize when they are imbedded within the semiconductor. Impurity atom with extra electron Impurities in Semiconductors -2 Since the impurity can ionize easily in the semiconductor, the impurity will either give up an electron to the rest of the semiconductor (called “ndoping” or “electron doping”) or take in an electron from the rest of the semiconductor (called “pdoping” or “hole doping”). Impurity atom with extra electron Impurities which give up electrons Extra electron excited to conduction band (n-doping) (like the one at right) are called “donors”. They give the electron to the conduction band. Impurities which take in an electron are “acceptors”. They take electrons from the valence band and thus create a hole. + Impurities in Semiconductors -3 An example of an acceptor is shown at right. For silicon, the most common acceptor impurity used is boron. This is a group 3 element that can also form sp3 hybrid orbitals and bond in the diamond structure with silicon – so it can easily replace a silicon atom in the lattice. However boron has one fewer electron than Si, so it cannot fill all of its sigma bonding orbitals. Thus it likes to take one from the valence band of the Si lattice to form a hole. In a similar way, donor impurities (ndopants) are taken from the group 5 elements (such as arsenic or phosphorus). Impurity acceptor atom Electron taken from valence band (p-doping) - Doped Semiconductors Impurity states can be represented on the band diagram as located within the band. If an impurity state is close to the conduction band, it can be easily ionized to form an electron in the conduction band This will increase the conductivity. It will also increase the electron density such that n > p. This will also increase the Fermi level, EF . All of this holds for hole doping if the impurities have levels located here. In that case EF is driven downwards, Energy e- e- e- EC EF EV Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 Doped Semiconductors Impurity states can be represented on the band diagram as located within the band. If an impurity state is close to the conduction band, it can be easily ionized to form an electron in the conduction band This will increase the conductivity. It will also increase the electron density such that n > p. This will also increase the Fermi level, EF . All of this holds for hole doping if the impurities have levels located here. In that case EF is driven downwards. Energy e- e- e- EC EF EV Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 Physical argument for decrease in hole density As we dope with donors, the electron density increases while the hole density must decrease – if Fermi energy is to increase. Energy e- e- e- EC EF The decrease in holes is due to recombination (“hole-electron annihilation”). Increased density of electrons means probability for holes to recombine increases – so their density decreases. This is the “Law of Mass Action”. The Law of Mass Action makes doping consistent with a shift up and down of the Fermi energy. EV Probability 1 fFD (E) = 1 e(E-EF )/kBT +1 Law of Mass Action When doping, the product of the densities of electrons and holes is a constant in equilibrium. np = n 2 i So in equilibrium, we can change the balance between carriers, but not their product. For a material, the product is a function of only temperature. Energy e- e- e- EC EF EV Probability 1 fFD (E) = 1 e(E-EF )/kBT +1