lectures accompanying the book: Solid State Physics: An Introduction,by Philip Hofmann (1st edition, October 2008, ISBN-10: 3-527-40861-4, ISBN-13: 978-3-527-40861-0, Wiley-VCH Berlin. www.philiphofmann.net

README This is only the outline of a lecture, not a final product. Many “fun parts” in the form of pictures, movies and examples have been removed for copyright reasons. In some cases, www addresses are given for particularly good resources (but not always). I have left some ‘presenter notes’ in the lectures. They are probably of very limited use only.

Semiconductors what is a semiconductor? We are closer to this now because we could define what a metal is. Lot’s of good internet resources (such as the BS guide and the online book) ...because of very high commercial interest for this particular lecture, a very good source of images is the Britney Spears Guide to Semiconductor Physics (http://britneyspears.ac/lasers.htm)

Metals and insulators / semiconductors we have defined the difference between a metal and a non metal now we still have to define the difference between a semiconductor and an insulator

Can we predict this (here in 1D)? N unit cells -> N possible k values 2 possible states per band and k (spin) can we predict this? odd # electrons per unit cell is a metal even # electrons per unit cell CAN BE a semiconductor / insulator 2 valence electrons per unit cell fill one band An uneven number of valence electrons per unit cell results in a metal

Can a material with EF in a band gap conduct? valence band conduction explain valence band and conduction band first at 0 temperature

Can a material with EF in a band gap conduct? valence band conduction gap size (eV) InSb 0.18 InAs 0.36 Ge 0.67 Si 1.11 GaAs 1.43 SiC 2.3 diamond 5.5 MgF2 11 explain valence band and conduction band If the gap is sufficiently small against kT a material would not be a metal according to our definition but it would conduct! what is drawn here is greatly exagerated kT at room temperature was 100 meV most people would draw the line at around 3 eV most common: Si and Ge but also II/VI and III/V

Electrons and holes valence band conduction band The electrons in the conduction band can conduct, like in a metal. The remaining electrons in the valence band can be excited and also conduct. Alternative view: holes: instead of viewing all the electrons as moving, we can view one hole as moving. Later we put these ideas on a more formal basis

Intrinsic semiconductors Pure, i.e. not doped, semiconductors are called intrinsic. For the electronic properties of a semiconductor, “pure” means pure within 1 ppm to 1 ppb. as in the case of mech. prop. we need to include the notion of impurities to understand sc a VERY small amount of impurities has a major impact on the conductivity of a semiconductor. Not 99.9% or so. 10-6 to 10-9 It is almost impossible to make really pure samples (apart from Ge). But we start out like this here to understand the basic ideas.

The Fermi-Dirac distribution for a semiconductor For a metal, the Fermi energy is the highest occupied energy at 0 K. The chemical potential is temperature-dependent (but not much) and so the two are essentially the same. For a semiconductor, the definition of the Fermi energy is not so clear. We better use the chemical potential. Some (many) people also use the term “Fermi energy” for semiconductors but then it is temperature-dependent. For metals, it really does not make much of a difference. For semiconductors it does because the chem. pot. can change a lot with temperature. We use the chemical potential rather than a T-dep Fermi energy to emphasize this difference.

Where is the chemical potential? explain density of occupied states very elegant argument which is useful far beyond this simple case: charge conservation the solid cannot charge up when we warm it! mu just above VB no problem at 0K The chemical potential must be roughly in the middle of the gap. Otherwise one would get an imbalance between conduction electrons than vacant valence states.

Where is the chemical potential? explain density of occupied states very elegant argument which is useful far beyond this simple case: charge conservation the solid cannot charge up when we warm it! If I put the chemical potential just above the VB, no problem at T=0K mu is more or less in the middle of the gap. More precisely later on The chemical potential must be roughly in the middle of the gap. Otherwise one would get an imbalance between conduction electrons than vacant valence states. equal area!

Temperature dependence of the carrier density missing electrons (holes) in the valence band (VB) electrons in the conduction band (CB) carriers quite general for electrons and holes One thing clear: the higher T the more carriers! quite remarkable: can go from almost insulting to fairly conductive. the other way round than for metal How do we know how many carriers there are for conducting? we just have to know the density of states EXPLAIN the integration limits are the lowest energy in the CB and the highest energy in the VB

Band structures of real materials: Si and GaAs VBM CBM CB VB at the VBM and CBM the band structures are nearly parabolic, this has also been so in our simple model different gaps from different Fourier components in the potential. The chemical potential is in the middle of the gap and the Fermi function falls off quickly. So the only relevant places for electrons and holes have parabolic dispersion

Interpretation equ. of motion VB maximum as E=0 effective mass conduction band A negatively charged particle with a positive mass (”electron”) don’t be offended by the picture! valence band: negative charge and negative mass or: postive charge and positive mass: the hole effective masses for hole and electron are positive in definition what has the higher mass? valence band or “hole”

Simplified band structure free electrons VB maximum as E=0 conduction band let’s figure out the density of states need to calculate carrier densities! explain energy scale and conventions me and mh are effective masses for CB and VB and they are called so because of hole and electron so now we have simple expressions for the densities of states! valence band

Temperature dependence of the carrier density electrons in the conduction band (CB) we have a simple expression for g(E) but we cannot do the integral analytically we can do one more thing. We can simplify the Fermi distribution this can be dangerous: the parabolic approximation and g(e) calculation is a fairly good approximation but what comes now is not always. missing electrons (holes) in the valence band (VB)

Simplified Fermi-Dirac Distribution for the conduction band for the valence band Both are Boltzmann distributions! This is called the non-degenerate case. Assume for the moment that the gap is relatively large. Boltzmann distributions because E-mu is positive first and negative second it is called the non-degenerate case because the the values of the distribution are always much smaller than 1 therefore the problem with the Pauli principle does not occur. In other words, the number of electrons is so small that their de Broglie wavelength is short against their mean distance

The conduction band: occupation in the final expression, we have an effective number of states and a Boltzmann-like occupation. Neff also depends on T but not too much. Neff hides the dispersion through the effective mass substitution

The valence band: occupation in the final expression, we have an effective number of states and a Boltzmann-like occupation. Does not depend on the energy because I have defined the energy zero at the CBM Neff also depends on T but not too much.

The valence band: occupation so we can express things even simpler by just two states. its like a HOMO and LUMO in a molecule For practical calculation quite useless because we don’t really know mu but we know that p=n so we can figure it out CBM μ VBM

Law of mass action This does not depend on the position of μ. the law of mass action is useful because it does not depend on the position of the chemical potential (we shall see that this moves) see later why this is called like it is called finally we can calculate the carrier density! We can now calculate the carrier density more precisely! and finally with

Example gap size (eV) n in m-3 at 150 K at 300 K InSb 0.18 2x1022 1.11 4x106 2x1016 diamond 5.5 6x10-68 1x10-21 remember: metal is about 10^28 per m^3 InSb: high and fairly constant Si: strongly T-dep. From almost nothing to something relevant diamond: dramatic but irrelevant

Where is μ? These are the parameters one needs to know. where is the chemical potential as a function of temperature after lengthy but simple calculation: not always mid-gap! Discuss why it moves There are only three things to know Gap size is relatively simple. Use optical absorption for GaAs or ARUPS and IPES effective masses a bit less simple These are the parameters one needs to know. effective mass from conductivity?

The cyclotron effective mass centripetal force= Lorentz force so the hole mass is also different from the free electron mass. But it makes no sense to take about free holes! this only works if the mfp is long enough for the electron to go one circle. If it is not, what can you do? decrease T, increase purity, increase B in practice the B field is scanned because it is easier than scanning the microwave frequency which is given by the cavity dimensions and so on. Maybe show spectrum resonant absorption of the radio frequency signal at Absorption works because there are actually many (so called Landau) levels with an energy separation of quantized levels

The effective mass me*/me mh*/me InSb 0.014 0.4 InAs 0.022 Ge 0.6 0.28 Si 0.43 0.54 GaAs 0.065 0.5 Na 1.2 Cu 0.99 Sb 0.85 there is a much nicer version of this picture on the BS semiconductor site hole mass relative to electron mass. It makes little sense to talk about free holes! In any case, the electrons tend to be lighter. This is quite intuitive and it is the same in Britney’s case.

graphene real space reciprocal space

graphene band structure band structure around corner of BZ Energy insert picture of dispersion in graphene ky kx

“massless relativistic Dirac fermions” big story in graphene relativistic physics with “slow” electrons but what does this all mean? surely not that the effective mass is 0

relativistic energy m>0 m=0 EF E p

Doped semiconductors A very small amount of impurities can have a big influence on the conductivity of a semiconductor. Controlled addition of impurities is called doping. There are two types of doping: n doping (impurities increasing #electrons) and p doping (impurities increasing #of holes). Typical doping levels are in the order of 1019 to 1023 impurity atoms per m3. Remember: Si has a concentration of 5*1028 atoms per m3 and an intrinsic carrier concentration of 1016 electrons/holes per m3 at room temperature. actually it is not really possible to grow intrinsic semiconductors at all (apart of Ge). In practice both n and p doping at the same time but only numerically treatable. So if the intrinsic carrier concentration is only 10^16, and if we assume that every dopant atom contributes one electron, then an impurity concentration of 10^19 greatly matters! This concentration is really small. Very hard to achieve chemically otherwise. It is like one non-chinese person in the Chinese population making a great impact.

The case for high purity! (Nearly) intrinsic super-pure Si single crystals can be grown and quite big... this explains (partly) the bizarre, un-sexy and non-fashionable dress code in the industry.

n- and p-doping donor atom acceptor atom n-doping: penta-valent donor atoms, only four needed to do sp3, one left p-doping: tri-valent acceptor atoms donor atom acceptor atom

n-doping Estimate binding energy with Bohr model: phosphorus penta-valent, one electron too many using the modifications n-doping: one nuclear charge left on the P. The electron goes around. Very similar to H atom but as long as the electron is bound, it cannot participate in conduction donation corresponds to ionization, so we should calculate this for n=1 We know that the ionization energy is 13.6 eV but m has to be replaced with the correct effective mass (about 0.5 me) epsilon0 has to be replaced with the epsilon (about 12) epsilon0 and then the ionization energy becomes quite small the radius of this is quite big, 30 times the Bohr radius order of magnitude

the same applies for p doping only that I have to replace the word electron with hole! ionization of donor corresponds to freeing the electron, i.e. putting it in the lowest conduction band state we know ground state with respect to VB and CB CB and VB just treated as one energy level (CBM and VBM) for the acceptor the other way round chemical potential: at least at low T exactly between level and band, again charge neutrality. so quite close to CB and VB, 15 meV or so.

Carrier concentration Only numerical solution possible. Basis for calculation is charge neutrality. At very low temperature, μ must be between donor level and the conduction band minimum (n-doping). At very high temperature, μ must be in the middle of the gap because all donors are ionized. the trouble is that the simple models don’t work, i.e Boltzmann statistics really fails here.

this could be n doped Si very low temperature: no carriers, not even excited donors (kT much smaller than donor levels) mu is between donor state and cbm higher T, all donors get ionized. Then extrinsic region with saturation. Even higher T, intrinsic carriers get excited: unlimited increase, mu moves mid gap. In practice: there is a region in which the electron concentration is almost constant

Law of mass action at a given T come back to mass action and apply to doped case: this is like the equilibrium in a chemical reaction between decay of eh pairs and thermal generation of them. This means that there must be less holes in an n-doped semiconductor than in the intrinsic case. This is due to the higher recombination rate. at a given T This does not depend on the position of μ.

Majority and minority carriers the minority carriers are very much in the minority! (1) because of doping and (2) because of mass action but the minority carriers are important for devices. So we can’t ignore them. equal number of electrons and holes majority: electrons minority: holes

Measurement of carrier concentration: Hall effect electrons holes use the Hall effect again now the positive carriers are obviously holes if one carrier type is in the absolute majority, this works but in reality one often has both carrier types and then this is more difficult to figure out. exercise

The total conductivity just for electrons: conductivity We use the simple expression from the Drude model in the intrinsic case there are as many electrons as holes. But their contribution to the conductivity might be different because of the different effective mass, even if the relaxations times are the same. the concept of mobility really makes sense now! Figuring all this out is tricky: we have both donors and acceptors, we cannot use the approximation for the degenerate case anymore and so on. Complicated numerical solution. But we can learn some things here. mobility concentrations mobilities

Temperature-dependent conductivity General trend of the conductivity ws temperature. This is a very central figure but it is also dangerous because that’s not what happens in your ipod. In fact, “normal” doped semiconductors behave around room temperature like metals: the conductivity is almost constant but drops slightly with temperature. Can you explain it? What happens at even higher temperatures? Here it emerges that the concept of the mobility is a very useful one!

visible spectrum: 1.7 to 3.1 eV Optical properties light absorption light emission absorption: photoconductivity, photo-sensors, solar-cells emission: solid-state light, LED, semiconductor lasers for real insulator such as MgF or diamond, band gap 11 eV and 5 eV. No excitation, light just passes through, transparent visible spectrum: 1.7 to 3.1 eV

Optical properties this is because omega=ck. c is VERY high. High energy for small k. Optical photons carry energy but almost no momentum. A transition with a change in k can therefore not be achieved.

Semiconductor devices Current application: mainly electronics, many billion Euros turnover each year, major economic factor. Future: optical applications such as solid state lightning, optical communication, solar cells The conductivity can be controlled by an electric field. Turn electricity into light and vice versa.

(of identical semiconductor material) The p-n junction (of identical semiconductor material) E two identical semiconductors: gap is the same, made by diffusion of dopands p-type: some acceptors charged, holes and some minority electrons, chem. pot close to bottom n-type: some donors ionized, electrons and some minority holes, chem. pot close to top join them by local doping unstable: electrons diffuse into p-type and recomb. with holes holes diffuse into n-type and recombining with electrons. DEPLETION LAYER very poor conductivity here electric field across the depletion layer / potential difference: process can’t go on electrons have higher potential energy on p side due to dipole layer thermodynamical equilibrium: the chemical potential must be constant bands are potential energy for electrons: they want to move downhill, holes want to move up hill.

What is the potential difference? What is the length of the depletion zones? quantitative treatment needed to answer these questions. equivalently, how big is the electric field?

The p-n junction: quantitative solution assume: no free carriers in depletion region and Poisson’s equation abrupt change in the dopant density p/n go through all definitions depletion instantaneous treat the + and - regions separately! poisson macroscopic: change on a bigger scale than atomic dimensions net charge density outside is zero. So no field get an expression for the contact potential but we don’t know d boundary conditions: U and dU/dx continuous result for the total potential change

The p-n junction: quantitative solution the length of the depletion layer by combining these, we get also expressions for the lengths. But this contains the contact potential. This is a little bit of a trap But at very low T, the mu for the p is close to the VBM and for the n close to the CBM and so deltaU =Eg Length macroscopic, it was ok to use Poisson. come back to an alternative description in a minute note: sign convention. The potential energy is -e times the potential. This explains why the U and band curves go the other way round. An electron has a higher E_kin, lower E_pot on the right hand side so it must have a higher U. at very low T The depletion layer length is of the order 0.1 μm to 1μm

(very approximate) quantitative solution From these and the law of mass-action one gets simple but lengthy derivation assume all concentrations assume that all acceptors and donors ionized assume law of mass action which is more useful than and because the parameters are known. typical values for Si diodes: contact potential of 0.7 V and depletion layer width of 0.1 to 1 μm.

The p-n junction with an applied voltage consider electrons only no voltage, look only at the electrons In the p part only few minority electrons, they “drift” with the field when the come to the depletion zone. In the n part many majority electrons. Some of them have enough energy to diffuse into empty state in the CB of the n semiconductor drift and diffusion are of course equal. Indeed the number of electrons above the CB in the p-side is equal on both sides, at least in our Boltzmann picture and when the DOS in both bands is independent of the energy drift of minority electrons (p) and diffusing of majority electrons (n) equal.

The p-n junction with an applied voltage consider electrons only Chemical potential deep in the semiconductors as before. Change at the contact due to the additional voltage, voltage drop over depletion zone since the resistance is high there. negative terminal on n side, positive terminal on p side. At the contact not well defined because of non-equilibrium Drift current the same as before Diffusion current much higher, exponential increase with external voltage exponential because this is like a Boltzmann tail The important point is that the chemical potential in the n-part rises with respect to the CB in the p part. So if the temperature is the same, there are many more majority electrons which can contribute to the p area. Voltage drop only over depletion zone. Increased diffusion current, drift current unaffected.

The p-n junction with an applied voltage consider electrons only Drift current the same as before Diffusion current much lower. The important point is that the chemical potential in the n-part lowers with respect to the CB in the p part. So if the temperature is the same, there are less majority electrons which can contribute to the p area. current saturates with drift current Voltage drop only over depletion zone. decreased diffusion current, drift current unaffected.

The p-n junction with an applied voltage exponential increase in forward direction decrease and eventual small saturation current in reverse bias nice example of violation of Ohm’s law!

(very approximate) quantitative solution calculate the carrier densities outside the depletion region Approximation of the degenerate semiconductor. Certainly not good quantitatively at low T! Never mind now! we use for electrons for holes but this is a bit too much of an approximation....

(very approximate) quantitative solution calculate the carrier densities outside the depletion region. The VB maximum of the n-side is the energy zero. then this is really quite simple (but not really correct). We use MB distribution and we only have to find expressions for the energies. we use

(very approximate) quantitative solution calculate the carrier densities outside the depletion region we use

(very approximate) quantitative solution calculate the carrier densities outside the depletion region we use

(very approximate) quantitative solution calculate the carrier densities outside the depletion region we use

The p-n junction with an applied voltage we say that both drift and diffusion are given by the number of carriers. easy to understand this: (1) drift and diffusion currents must be equal because of the equilibrium. (2) they must also formally be equal because the diffusion current applies only for electrons with energies higher than the CB minimum on the p side. first consider no bias voltage

The p-n junction with an applied voltage so if we only increase eV, this is getting quite big, exponentially or if eV is negative it goes to I_0

The p-n junction with an applied voltage exponential increase in forward direction decrease and eventual small saturation current in reverse bias this is a good example of a solid which does not follow ohm’s law!

For the holes, we can construct exactly the same arguments.

2kbit magnetic core memory Transistors Bardeen and Brattain 1947 Two types: bipolar transistors and field-effect transistors. Bipolar transistors can be found as separate devices and in integrated circuits. Field-effect transistors can only be found in integrated circuits. Both can be used as amplifiers and switches. vacuum tube replaced vacuum tubes (as amplifiers) and later magnetic core memories (for storing information) amplifiers and switches: both transistors can be used for both purposes. But here we discuss only the FET. 2kbit magnetic core memory

The Metal Oxide Field Effect Transistor (MOSFET) general idea: control the conductivity of a semiconductor by a voltage current from source to drain, controlled by the voltage on the gate.

The field effect transistor: principle of operation the positive voltage repels the majority holes and attracts the minority electrons finally, an inversion layer is formed which conducts. inversion layer: high e-density. Explain more clearly in a minute bit asymmetric because superposition of applied E-field and intrinsic E-field of the junctions This happens above the so-called threshold voltage so this works like a switch Ugate>Uthreshold

The field effect transistor: principle of operation inversion layer clearer in band picture deeply inside the p material, the Fermi level is just above the VB. BUT: the positive voltage repels the majority holes and attracts the minority electrons. It lowers the energy of the bands. eventually CB closer to mu than VB: inversion The gate voltage induces a band-bending close to the interface. For a sufficiently high voltage, the CB is closer to the chemical potential than the VB and the semiconductor shows n-type behavour.

The field effect transistor It is not good to have heavy carriers because this affects the mobility and thereby the switching speed of the device. For silicon, it does not make much of a difference but for GaAs it does (almost a factor of 10!) pnp MOSFETs are also possible but they have the disadvantage that the current is carried by the holes. Holes tend to be heavier than electrons in most semiconductors. Remember Britney!

Optical properties Optical photons carry energy but almost no momentum. A transition with a change in k can therefore not be achieved.

Solar cell light induced voltage (current) for reverse-biased diode light-induced electron-hole pair generation in the depletion layer. light induced voltage (current) for reverse-biased diode

Light emitting diodes (LED) GaAsP p-n junction of direct band gap semiconductor operated in forward bias. recombination leads to light emission, colour given by band gap.

Light emitting diodes (LED) efficiency: > 70 lumens / watt a “normal” incandescent lamp around 10 lumens / watt

LED colours Just use the right material blue has been a problem until recently but it isn’t any more. this can be done by GaAsP for almost the whole visible spectrum just by changing the P contents. Blue has been a problem but it is achieved now.