880:174 Physics of Modern Materials Chapter 6: Semiconductors

1. Intrinsic Semiconductors

As we saw in the previous chapter whether a material is categorized as an insulator or semiconductor depends on the band gap. For small band gaps, the material will be an insulator at T = 0 K but a fair conductor at room temperature. Such a material is a semiconductor. An intrinsic semiconductor is one without any impurities. Thus, its behavior is determined by intrinsic properties (e.g., lattice constant, band gap, etc.) alone. Silicon and germanium are the closest approximations to intrinsic semiconductors because of decades of improvement in processing technology. (GaAs is also produced with high purity for the IT industry.) However, even the purest Si and Ge have impurity concentrations of approximately 1 in 1010. Thus, an intrinsic semiconductor is an approximation, albeit a very good one.

We will assume (because of ease of calculation) that the carriers of current are electrons near the bottom of the conduction band (CB) and holes near the top of the valence band (VB). Under these conditions, the bands are nearly parabolic so that the electrons and holes act to good approximation like free particles with their masses taken to be the effective masses. (See Solymar & Walsh text for justification.) Thus, we can use the density of states expression previously derived. Let the top of the VB be at E= 0. Thus, the bottom of the CB will be at E = Eg. To find the number of electrons in the CB, we need to integrate Z(E)F(E)dE over all the states in the CB. However, we wish to assume that only electrons close to the bottom of the band are relevant, which would be in violation of the extended range of integration. We are rescued by the fact that the Fermi energy is usually near the midpoint of the energy gap at temperatures of interest, so that relevant energies in the conduction and valence bands are far from EF. Thus, at room temperature and below, the condition E ! EF >> kBT holds for electrons in the CB and

EF ! E >> kBT holds for holes in the VB. Under these conditions, the semiconductor is said to be non-degenerate. Under these conditions, the product Z(E)F(E) is only significant in the region near the band extrema because F(E) falls off exponentially with energy. The upper limit on the integral can be extended to infinity because at higher energies, F(E) ! 0.

With our chosen origin for the energy, for the CB, we have 4!(2m" )3/2 Z(E) = e (E # E )1/2 . (6.1) h3 g

For E ! EF >> kBT , the FD function approximates to

1 F(E) = e!(E ! EF )/kBT . (6.2) With all our approximations, we can now calculate the number of electrons in the CB: ! 4#(2m$ )3/2 ! N = Z(E)(F(E)dE = e (E % E )1/2 e%(E % EF )/kBT dE. (6.3) e " h3 " g Eg Eg

Using the substitutionu = (E ! Eg ) / kBT , one find finds that the integral is converted to the " form # x1/2e! xdx, which evaluates to ! / 2. Thus, we obtain 0

!(Eg ! EF )/kBT Ne = Nce , (6.4) where 3/2 # 2!m"k T & N = 2 e B . (6.5) c $% h2 '( One can employ similar tactics to calculate the number of holes in the VB. In this case, the number of holes in the VB equals the number of electrons removed from the VB. Thus, we need the probability that an electron state is not occupied, i.e., 1 – F(E). Now, 1 e(E ! EF )/kBT 1 1! F(E) = 1! = = " e(E ! EF )/kBT . (6.6) 1+ e(E ! EF )/kBT 1+ e(E ! EF )/kBT 1+ e(EF ! E)/kBT

The approximation at the end of Eq. (6.6) applies for EF ! E >> kBT. The density of states for the VB is given by 4!(2m" )3/2 Z(E) = h (#E)1/2 . (6.7) h3 Thus, the number of holes in the VB is given by 0 4$(2m% )3/2 0 N = Z(E)[1! F(E)]dE = e (!E)1/2 e(E ! EF )/kBT dE, (6.8) h # 3 # !" h !" which yields

!EF /kBT Nh = Nve , (6.9) where 3/2 # 2!m"k T & N = 2 h B . (6.10) v $% h2 '( For an intrinsic semiconductor, all electrons in the CB are excited there from the VB, i.e.,

Ne = Nh . (intrinsic semiconductor) (6.11) Equating Eqs. (6.4) and (6.9) yields

!(Eg ! EF )/kBT !EF /kBT Nce = Nve , (6.12) which can be easily solved for EF to give E 3 " m! % E = g + k T ln h . (6.13) F B $ ! ' 2 4 # me & Thus, at T = 0 K, the Fermi level is exactly in the middle of the band gap. As the temperature increases, EF may increase or decrease depending on the ratio of the effective masses. For example for Si, EF is slightly (<1% of Eg) below mid-gap at 300 K and for GaAs, it is slightly above mid-gap.

We should note that if we substitute for EF in the equations for Ne and Nh, we find that both quantities are proportional toe!Eg /2kBT . This is more easily seen if we calculate the product of the two quantities:

!(Eg ! EF )/kBT !EF /kBT !Eg /kBT NeNh = (Nce )(Nve ) = Nc Nve . (6.14) Since Ne = Nh for intrinsic semiconductors, defining each as Ni, the intrinsic carrier density, we have

!Eg /2kBT Ni = Ne = Nh = Nc Nv e . (6.15)

2. Extrinsic Semiconductors

An extrinsic semiconductor is one that has impurities in it, usually deliberately introduced to give the semiconductor specific properties. Let us consider silicon as our prototypical intrinsic semiconductor. Si has 4 valence electrons (due to sp3 hybridization). In a Si crystal, each Si atom is covalently bonded to 4 other Si atoms. The bonding requirements are satisfied for all Si atoms, with each essentially having an inert-gas configuration of 8 electrons (except at the surfaces of the crystal, which we do not consider). Suppose that we introduce a pentavalent impurity atom such as phosphorus (P). [Arsenic (As) and antimony (Sb) are others.] The P impurity will give up 4 valence electrons to bonding with 4 Si atoms. This leaves a lone electron without a partner to bond with. This extra electron is rather loosely bound to the parent P atom because of the screening of the nucleus caused by the 4 additional electrons from the Si atoms that form bonds with the P atom. Hence, this electron can be excited into the conduction band with much less energy than is necessary to excite an electron from the valence band across the band gap. Pentavalent impurities are called donors because they "donate" an electron to the CB. A donor atom that has donated an electron is ionized and positively charged. One can use a hydrogen-atom model to estimate the binding energy of an electron bound to the donor atom. The screening by the core electrons and the bonding electrons means that the effective nuclear charge seen by the extra electron is +e. We use the formula for the binding energy of the H atom, with electron mass replaced by the effective mass and the permittivity of free space !0 replaced by the permittivity of the semiconducting material. (The permittivity is equal to the dielectric constant K times !0 .) So, " 4 mee Ebinding ! 2 2 . (6.16) 2(4#$) ! Using parameters for Si, Eq. (6.16) yields 0.05 eV, which is in reasonable agreement with the experimentally measured value of 0.045 eV. The energy of the bound donor electron is usually given the symbol ED. Thus, the energy necessary to ionize a donor atom and promote the electron to the CB is Eg ! ED and so

3 Eg ! ED = Ebinding . (6.17)

Since Ebinding is small in comparison with the typical gap size (at least for Si, Ge, and GaAs), the donor levels are close to and just below the bottom of the CB.

If the impurity atom is trivalent, such as boron (B), aluminum (Al), or indium (In), then there is one fewer electron than is required to satisfy all the bonding requirements. Hence, one can consider this to be a hole in one of the covalent bonds. An electron near the top of the valence band can easily hop into this hole if it is supplied with a small amount of energy. Thus, the hole has hopped into the top of the valence band, where it can contribute to electric current. Trivalent impurities are called acceptors because they accept electrons from the valence band. (Alternatively, one can say they donate holes to the valence band.) An acceptor atom that has accepted an electron is ionized and negatively charged. The acceptor levels are close to the top of the VB and just above it. The acceptor levels are higher than the VB states because they represent the absence of a bound electron.

Real materials usually have both acceptors and donors. However, for electronics applications, extrinsic semiconductors are usually engineered with one type of impurity dominant, whether donor or acceptor. For example, to increase the number of electrons available for conduction at room temperature, Si is doped with donors.

The calculation the number of electrons and the number of holes that we performed for intrinsic semiconductors remains valid for extrinsic semiconductors. To calculate the Fermi level for extrinsic semiconductors, we must take into account all sources of electrons and holes. Now, when donors and acceptors are ionized, as a whole the material must remain neutral since charge is neither gained from nor lost to the surroundings. Thus, the total positive charge must be equal to the total negative charge: N N ! N N + , (6.18) e + A ! h + D ! + where N A is the number of ionized acceptors and N D is the number of ionized donors. Eq. (6.18) is a condition for dynamic equilibrium, whether either side may change (e.g., as the temperature is varied) but the equality of the LHS and the RHS is fixed. Now, the number of + ionized donors N D is equal to the number electrons that have been donated to the CB, i.e., the number of unoccupied donor-electron states. Thus, the number of ionized donors is equal to the total number of donors times the probability that a donor level is unoccupied: + N D = N D[1! F(ED )] . (6.19) Similarly, the number of ionized acceptors is equal to the total number of acceptors times the probability that an acceptor level is occupied by an electron: ! N A = N AF(EA ). (6.20)

Let us consider an n-type semiconductor, for which N D >> N A . Thus, the number of ionized donors will always be much greater than the number of ionized acceptors at temperatures of

4 interest. Let us also assume that at the temperatures of interest, Nh << Ne.Eq. (6.18) then becomes + Ne ! N D , (6.21)

Using Eqs. (6.4), (6.6), and (6.19), Eq. (6.21) gives

!(Eg ! EF )/kBT # 1 & Nce " N D % (E ! E )/k T ( . (6.22) $ 1+ e F D B '

EF /kBT Substituting x =e , one obtains a quadratic equation in x, which can be solved to give EF:

(Eg ! ED )/kBT !1+ 1+ 4(N D NC )e eEF /kBT = , (6.23) 2e!ED /kBT where we have discarded the solution that would give a negative value for the LHS. If

(Eg ! ED )/kBT 4(N D NC )e << 1 (i.e, ND<

Since N D NC << 1, the logarithm is negative and so EF is significantly smaller than Eg. If

(Eg ! ED )/kBT 4(N D NC )e >>1, i..e, N D NC >> 1 or low temperatures, we obtain E + E k T ! N $ g D B D . (6.25) EF = + ln# & 2 2 " Nc % Thus, for very low temperatures, the Fermi level is approximately halfway between the bottom of the CB and the donor level. At these temperatures, the number of electrons excited across the band gap from the VB is negligible. The only excitation of electrons is taking place from the donor states. Thus, material acts like an intrinsic semiconductor whose energy gap is only

Eg ! ED . Similar arguments can be made if acceptors significantly outnumber donors.

(In the figures above, 1, 2, 3, represent different impurity concentrations.)

5 3. Scattering

The solutions to the Schrodinger equation for an electron in a periodic potential such as a crystal lattice are Bloch functions, which have the periodicity of the potential. Because the electron is "aware" of the periodicity, it does not scatter from the positive ion cores at the lattice positions. (There are reflections, but the transmitted wave function continues untroubled. There is no transmission at k = ±n! / a ; only Bragg reflection of the electron wave function, which produces the energy gap.) Thus, a perfect crystal would have no resistivity. The resistivity in real crystals is due to imperfections in the crystals, such as deviations from periodicity due to thermal motion of the ion cores, defects, or impurity atoms (there are always impurities). Electrons and holes will scatter from these imperfections. In the following, we will consider electrons to be the current carriers for ease of discussion.

The mean free path ! is the average distance between scatterings of an electron. The MFP is temperature dependent, generally decreasing with increasing temperature. Suppose an electric field is applied to the metal or semiconductor. An electron will suffer many collisions per second, but there will be a slow overall motion, or drift, in the opposite direction to the field. The associated velocity is called the drift velocity vd. Note that the thermal velocity vth of the electron >> vd. The average time between collision, called the scattering time or relaxation time, is given by ! = " / vth . The drift velocity is the average velocity gained between collisions:

vd = a! = (eE / m)!. (6.26)

Now the current density (magnitude) is given by J = Nevd , where N is the number of electrons per unit volume. Further, the electric conductivity is defined as ! = J / E. Using the expressions for J and vd, we obtain Ne2" ! = . (6.27) m The scattering time can be due to several sources, e.g., scattering from lattice vibrations and scattering from impurities. Each of these sources has a different temperature dependence.

The mobility of an electron is defined by v µ = d . (6.28) E From Eq. (6.26), we see that e! µ = . (6.29) m Thus, the smaller the relaxation time (i.e., higher scattering rate), the lower the mobility. With the definition of mobility, we can rewrite Eq. (6.27) as ! = Neµ. (6.30) In many metals and semiconductors, the electric current is carried by both electrons and holes. In this case, Eq. (6.27) becomes 2 2 Nee " e Nhe " h ! = # + # , (6.31) me mh where we have replaced the bare electron mass by the appropriate effective mass.

4. Relationship Between Hole and Electron Densities

We saw that for intrinsic semiconductors, Ne = Nh, and !Eg /kBT NeNh = Nc Nve . (6.32) Note that the carrier density product is independent of the Fermi energy and so is valid for extrinsic semiconductors as well. Eq. (6.32) is an example of the law of mass action. Thus, if donors cause more electrons to be added to the conduction band than in the intrinsic case, then the number of holes in the valence band must decrease. This seems counterintuitive since one might expect the number of thermally generated holes to be equal to the number of thermally generated electrons as seen in the intrinsic case. However, this expectation is incorrect because of dynamic equilibrium. Electrons and holes are constantly being generated and also constantly recombining to give bound electrons. The "reaction" is: electron + hole ! bound electron . The generation rate is constant at a given temperature. However, the recombination rate depends on the number of electrons and holes that are present (actually, their product). In equilibrium the two rates must be equal. Thus, if the number of electrons increases, then the number of holes must decrease. The reason is that as the concentration of electrons increases, the recombination rate initially increases as well. This increased recombination rate depletes the number of holes to a value below the intrinsic value until the recombination and generation rates are again in balance. (See Kittel for a good argument.)

5. III-V and II-VI Semiconductors

Silicon and germanium are covalently bonded semiconducting crystals with room temperature energy gaps of 1.11 eV and 0.67 eV, respectively. Silicon is the most technologically important semiconductor; however, there are other semiconductors that are becoming increasingly technologically important. These are compound semiconductors, e.g., gallium arsenide (GaAs). GaAs is an example of a III-V semiconductor – Ga is from Group IIIA of the periodic table and As is from Group VA. Because Ga has three valence electrons and As has five, the bonding between Ga and As in a crystal is partly ionic. Note that in the III-V materials, the greater the ionicity, the greater the energy gap, i.e., the more insulating the behavior. (This is not true in the II-VI materials; see the table.) More generally, larger atoms form

7 compounds with smaller energy gaps. The reason is that the valence electrons in larger atoms are less tightly bound because they are farther away from the nucleus. (Screening causes the effective nuclear charge seen by the outermost electrons to be about the same for atoms with the same number of valence electrons, e.g., Ga and In.) Also, in compounds with larger atoms, the lattice spacing tends to be greater. Thus the electrostatic bonding forces will be weaker, i.e., less energy is needed to break the bond and excite an electron to the conduction band. Hence, the band gap is smaller. In terms of bands, when the atoms are farther apart, the valence and conduction bands do not curve upward as much due to electron repulsion. This leads to smaller band gaps as can be seen in the figure to the right that illustrates the bands in the group IV elements. As in diamond, the bonding is tetrahedrally coordinated and the crystal lattice is diamond-like, except that there is a "double lattice" because there are two different atoms. This type of face-centered cubic lattice is the zinc blende lattice (after the mineral zinc blende, which is zinc sulfide and has the same crystal structure).

The II-VI compound semiconductors such as CdTe have similar characteristics to the III-V materials; however, their band gaps tend to be larger because of the greater ionicity of the II-VI compounds. These materials also tend to crystallize in the zinc blende lattice (ZB) or a close variant, the wurtzite lattice (which has a slight distortion relative to the ZB lattice, rendering it hexagonal).

One can tailor the band gap and other intrinsic properties of the III-V and II-VI materials by substituting for varying concentrations of one element with another element that has the same valence. For example, AlAs has a higher energy gap than GaAs. Thus, if one wishes to increase the energy gap of GaAs, one can produce Ga1-xAlxAs. The energy gap will vary approximately linearly with concentration x. The value of the energy gap is extremely important for optical applications – LEDs, semiconductor lasers, etc. (Question: You wish to reduce the energy gap of ZnSe for a particular optical application. How would you do it?)

6. Carrier Lifetime

We have, so far, only talked about the thermal excitation of electrons and holes for conduction. One can also use light. Light whose photons have a sufficiently high frequency can excite an electron from the valence band to the conduction band. In addition, if dopants are present (which they always are to some degree), electrons can be excited into the CB from donor centers and holes into the VB from acceptor centers. Thus, a semiconductor that is illuminated can become conductive if a

8 large enough number of carriers are excited. This phenomenon is called photoconduction. The excitation of carriers by photoconduction is a non-equilibrium process since the number of carriers is greater than would be predicted based upon the temperature alone. When the light is turned off, the number of carriers will relax back to the equilibrium value via diffusion from areas of high concentration to regions of lower concentration and by recombination of holes and electrons. The time taken for the number of excess carriers to fall to 1/e of its original value under illumination is called the lifetime ! of the carrier. (This symbol should not to be confused with collision time encountered in electrical transport in metals and semiconductors.)

If illumination changes the number carriers, then the number of minority carriers will undergo a much greater change than the number of majority carriers. Here is an example. Let us say the number of ionized donor atoms in a sample of material is 1022 m-3. At a given temperature, the number of intrinsic carriers (excited from VB to CB) is 1016 m-3. Then the number of holes is 2 10 -3 15 3 given by Ni / Ne = 10 m . Suppose illumination creates 10 electrons and holes per m . The number of holes has increased by 5 orders of magnitude whereas the number of electrons has increased by a factor of 10-7. Thus the change in the number of majority carriers is insignificant. It is the relaxation of the minority carriers back to equilibrium that can be more easily measured.

[This is Exercise 8.17 in the textbook.] Now, the recombination rate is proportional to the product of the number of electrons and the number of holes. At a given temperature, the carrier generation rate is constant (it is only dependent on temperature). If excess holes are created by, say, illumination, then the rate of change of the number of these holes is given by d!N h = G " #N (N + !N ), (6.33) dt e h h where G is the generation rate and ! is proportionality constant for the recombination rate.

Now, in equilibrium, the generation and recombination rates are equal, i.e., G = !NeNh . Then Eq. (6.33) becomes d!N h = "#N !N , (6.34) dt e h which can easily be solved to give "t /# !Nh = !Nh,0e , (6.35) #1 where the lifetime! = ("Ne ) . Thus, the excess carriers decay exponentially toward equilibrium values.

7. Real Semiconductors

The energy-band diagrams we have shown so far were obtained for idealized models. Real semiconductors have much more complex E vs. k diagrams because real lattices are more complex than the simple cases we have discussed. Schematic diagrams of the E-k diagrams for Si and GaAs are shown below. Note that in the case of Si, there are three valence bands: a heavy- hole band (h), a light-hole band (l), and a split-off band (s), which is a little lower in energy than the other two at k = 0. [Ask students explain the designations in terms of effective mass.] Note

9 that the light hole-band and the heavy-hole band are degenerate at k = 0. For properties involving the valence band edge, the heavy holes tend to dominate because their larger effective mass gives rise to a higher density of states.

The minimum in the CB for Si does not coincide with the minimum in the VB. Silicon is therefore called an indirect gap semiconductor. (Ge is an indirect gap semiconductor also.) The indirect gap makes excitation of carriers into the CB from the VB by light a bit more complex. When a photon a photon interacts with an electron in the VB, both energy and momentum have to be conserved. Note that photons have a relatively small amount of momentum compared to their energy (E = pc). If the CB and VB extrema coincide, a photon can boost an electron from the VB maximum to virtually directly to the CB minimum because the relatively small momentum of the photon makes Schematic bandstructure for Si no significant change in the crystal momentum, which is proportional to the k value. However, in the case of an indirect gap transition such as in Si, the momentum transfer necessary is relatively large because the crystal momentum has a non-zero value at the CB minimum. This extra momentum is provided by a phonon, which is a quantized lattice vibration. Phonons are a bit like photons except that they involve lattice vibrations and sound waves rather than electromagnetic-field oscillations and light waves. Because the speed of a sound wave is much smaller than the speed of light, for the same frequency, the wave vector value is much greater for a phonon. ( v = f !, k = 2" / ! ). Thus, phonons can easily supply the necessary momentum. The phonons also carry energy, so the energy of the photon that causes the excitation needs to be somewhat smaller than the energy gap. The indirect gap of Si makes it less desirable for use in optical applications involving band-to-band transitions. The probability of the transition depends on the population of phonons, which in turn depends on temperature. More phonons at every wave vector are available at higher temperatures. Schematic bandstructure for GaAs GaAs is a direct gap semiconductor, as seen from its band structure. Thus, GaAs and other direct-gap Ga-based materials are heavily used in optical applications. Note that GaAs has an additional "valley" in the conduction band above the CB minimum and displaced in k space. Also, the electrons at the bottom the CB have an effective mass ten times smaller than that of the holes at the bottom of the VB. Thus, the electrons at the CB band edge will have a greater mobility.

10 In amorphous semiconductors, there is no long-range crystal structure. Only very small (a few nm) crystallites are present and the large-scale structure is disordered. Recall that a diffraction grating that has many slits has very sharp fringes. A small number slits gives wide fringes that are less well defined. Similarly, small crystallites with relatively few atoms have less well defined band edges and more "fuzzy" energy gaps. Also the disorder produces dangling bonds that can serve as traps for carriers. Evidently, their properties are less desirable than those of single crystals. However, amorphous semiconductors are useful for applications where characteristics need not very stringent and large areas are needed (e.g., solar cells). Large areas can be produced relatively inexpensively, unlike single crystals.

8. Measurement of Semiconductor Properties

(a) Mobility

The mobility of the carriers of a semiconductor is an extremely important characteristic, which governs the electrical properties of the semiconductor. As seen before, the mobility is defined as µ = vd / E . Thus, to measure the mobility, one can measure the drift velocity and divide by the applied electric field. To measure the drift velocity of carriers, one can use an apparatus schematically shown in the figure to the right. The electric field in the n-type semiconductor bar is applied by the battery B1. Let us say a positive potential is applied at A when the switch S is closed. Electrons will leave the semiconductor and enter the electrode. Some of these electrons may come from the VB, leaving an excess of holes (minority carriers) in the semiconductor. To maintain charge neutrality, electrons will enter the bar at C, leading to an increase in the voltage across resistor R. This occurs at nearly the speed of light. The excess holes produced at A will drift in the direction of the electric field to C, at which time the voltage across R again increases. Knowing the distance between A and C and having measured the time taken between the two voltage increases, one can find the drift velocity and hence the mobility.

(b) Resistivity or Conductivity

The resistivity of a material is almost always measured by using a four-wire method. The use of a four-wire method eliminates the contribution of the resistance of the lead wires and contacts used to measure the voltage across the sample. Current is passed through the semiconductor

11 through contacts positioned closer to the ends of the specimen. The voltage is measured using a high-impedance voltmeter connected to contacts between those used to supply the current. The contacts can be sharp probes held in mechanical contact with sample, or, more usually in research, silver paint or silver epoxy is used to attach the leads to the sample. The sample is usually shaped as a rod or bar of uniform cross section in order to ensure homogeneous current flow (or as close to it as possible) through sample and enable the calculation of the resistivity. If the current leads are attached to the end faces of a thin sample, then one can use RA VA ! = = , (6.36) L IL where A is the cross sectional area and L is the length. V and I are the voltage across the sample and the current through it, respectively. Using the 4-probe set up shown in the diagram, if the leads are equally spaced and the spacing d is small in comparison with the thickness of the sample, then the resistivity can be calculated from 2"Vd ! = . (6.37) I If one measures the resistivity using a dc current, one also has to account for thermal voltages especially if the measured voltages are small. The thermal voltages are due to junctions between dissimilar metals being at different temperatures. One can eliminate the influence of thermal voltages by reversing the current and subtracting the two measurements (and dividing by two). This is because the thermal voltage does not change sign when the current is reversed. The resistivity can also be determined if the sample has an irregular shape using a method called the Van der Pauw method. [Picture from Singleton.] The method is as follows. (1) A current is driven from contact A to contact B and the voltage between contacts C and D is measured. The resistance RAB,CD is computed from the their ratio. (2) A current is driven from contact B to contact C and the voltage between contacts A and D is measured. The resistance RBC,AD is computed from the their ratio. (3) The resistivity is calculated from " R + R ! = d AB,CD BC,AD f , (6.38) ln2 2 where the function f depends on the ratio of the two resistances. The function is often plotted as a graph or typical values are given in a table.

The energy gap of a semiconductor can be measured electrically or optically. Electrically, one measures the resistivity and calculates the conductivity as its reciprocal. Now, the conductivity is given by

! = (Neµe + Nhµh )e. (6.39) In the intrinsic regime,

3/2 "(Eg /2kBT ) Ne = Nh = Ni = constant ! T e . (6.40)

In the intrinsic regime (high temperatures), The exponential dependence will dominate. Thus, if one plots a graph of ln! vs. 1/T, at high temperatures, the graph will be nearly linear with slope of –Eg/2kB. Note that at lower temperatures when the donors (or acceptors) are just beginning to ionize, the Fermi level is nearly midway between the CB minimum and the donor energy level (or VB maximum and the acceptor level). Here again, the graph of ln! vs. 1/T will be nearly linear with slope of –(Eg – ED)/ 2kB. [See graph to the right.] Optically, one can measure Eg by measuring the transmission of electromagnetic radiation through a thin slice of the semiconductor. Above a certain frequency given by hf = Eg, the photons will be energetic enough to cause transitions from the VB to the CB and absorption will increase rapidly. Thus, one can obtain the value of Eg by measuring the location of the absorption edge. Note that this is only true for direct gap semiconductors. For indirect gap semiconductors, the frequency of the absorption edge will be different than f = Eg/h.

(d) Hall Coefficient

Consider a homogeneous sample of material in the shape of a rectangular parallelepiped ("Hall bar") as shown in Figure 1.

I z - w y VH t + x

Figure 1: Hall effect configuration for a rectangular parallelepiped sample. The sign of the Hall voltage shown pertains to positive charge carriers. The bold lines simply emphasize the circuit for the measurement of the Hall voltage. ! The sample is immersed in a uniform magnetic field B directed along the z axis. If a current I flows in the x direction, the magnetic force will deflect the charge carriers in the y direction. For positive charge carriers moving in the direction of I, the deflection will be in the negative y ! direction. An electric field EH directed in the positive y direction will therefore build up. For negative charge carriers moving in the opposite direction to I, the deflection is also in the ! negative y direction. An electric field EH directed in the negative y direction will arise in this case. Hence, from the sign of the electric field and accompanying voltage, the sign of the charge carriers can be deduced. The electric field will eventually strengthen until repulsion prevents additional carriers from accumulating. At this point, the Lorentz force is zero:

! ! ! q(EH + vd ! B) = 0, i.e., ! ! ! EH = !vd " B . (i) ! In the foregoing, vd is the drift velocity of the carriers, which is in general different for holes and electrons. In terms of magnitudes, Eq. (1) gives

EH = vd B . (ii)

Now the magnitude of the current density is given by J = pqvd, where p is the number of charge carriers per unit volume and e is the charge of each carrier. Further, I = JA, where A is the cross sectional area along the direction of the current, i.e., A = wt. Thus,

I = pqvdwt. (iii)

In addition, the voltage VH (the Hall voltage) produced by the electric field EH is given by

VH = EHw. (iv)

Eliminating vd and EH from Eq. (ii) using Eqs. (iii) and (iv) yields

IB VH = . (v) pqt

If one defines the Hall coefficient by

E V t R ! H = H , (vi) H JB IB

Eq. (vi) then becomes

1 RH = . (vii) pq

Note that RH depends only on the characteristics of the charge carriers. Further, if q is negative, then RH is negative and if q is positive, then RH is positive. Hence, the sign of RH indicates the sign of the dominant charge carriers. From Eq. (7), one finds that

1 p = . (viii) R q H

Thus, the number of charge carriers per unit volume can be calculated from the measured value of RH.

Measuring the Hall voltage can be tricky because if the voltage leads are not directly across from each other, there will be an additional voltage due to the longitudinal resistivity of the sample. This voltage can be eliminated by reversing the magnetic field, subtracting the two readings, and dividing by two.

(e) Carrier Lifetime

As discussed before, one can measure the lifetime of minority carriers by exciting a non- equilibrium population, e.g., by illumination, removing the exciting agent and then measuring the decay in excess carrier population (by measuring the current). The exponential decay constant is the minority carrier lifetime. A schematic set up is shown below.

9. Preparation of Single-Crystal Semiconductors

(a) Growth from the Melt

The purified material is placed in a crucible and then heated in a furnace until it melts. The molten material is then slowly cooled. The greater surface area of the conical end of the crucible causes it to cool a bit faster than the rest of the material. Thus, crystal growth begins there, i.e., it is "seeded". The growth of the crystal then progresses through the melt. A single crystal grown this way will have a concentration gradient of impurities, with the purest material (smallest impurity concentration) at the bottom.

One can understand the concentration gradient by examining the phase diagram for the semiconductor and the impurity. Temperature is on the vertical axis and impurity concentration is on the horizontal axis. Each concentration represents a different alloy of the semiconductor and the impurity. For temperatures above the liquidus line, the material is entirely molten (liquid); for temperatures below the solidus line, the alloy is completely solid. Let us assume that the molten material crystal has concentration Ca of the impurity. The material will remain molten until temperature Ta is reached. At temperature, liquid of with concentration Ca can coexist with solid of concentration Ca. Thus, material of this composition will crystallize from the melt. Note that this impurity concentration is smaller than the initial concentration in the melt. With the crystallization of this low-impurity material, the remaining molten material has a higher impurity concentration than Ca, e.g., Cc. No further crystallization takes place until temperature Tc is reached. Then solid of impurity concentration Cd crystallizes, etc. Thus, each successive crystallization results in an alloy of higher impurity content. For slow cooling, one has a continuous distribution of impurity concentration in the boule (crystal), with the material that crystallized first containing the smallest concentration of the impurity. Zone refining methods take advantage of this. One can melt a section or zone of the material and cause the molten zone to move through the entire length of the crystal. The end of the crystal where the

16 zone started will have the smallest level of impurity. This process can then be repeated several times to increase purity.

(b) Epitaxial Growth

Epitaxial growth entails the growth of crystals in a layer-by-layer fashion. The first layer is deposited on a substrate, which is a high-quality single-crystal that serves as the template for the growth of the crystalline layers. Thus, usually the crystal structure and lattice spacing of the substrate matches as well as possible the crystal structure and lattice spacing of the crystal to be grown. In fact, the substrate is often the same material as the semiconductor being deposited. A buffer layer is also frequently deposited on the substrate to provide an atomically smooth surface. The buffer layer is often the same material as the substrate.

In molecular beam epitaxy (MBE), each layer is grown by depositing the constituent elements of the compound in the correct proportions. (See figure to the right.) Each element has its own source (Knudsen cell), from which a beam of evaporated atoms is directed toward the substrate. The beams are controlled precisely by using shutters and temperature control; the growth takes place in high vacuum. The growth of the crystal is monitored by electron-beam diffraction from the surface. MBE can often be used to prepare materials that cannot be fabricated by other means because MBE can be used as a non- equilibrium method of growth. This

17 away-from-equilibrium growth process makes it possible to fabricate, e.g., materials with crystal structures that are not possible under equilibrium conditions.

Another type of epitaxial growth method is MOCVD – metal-organic chemical vapor deposition. Chemicals bound to organic radicals are transported by hydrogen gas towards the heated substrate. (See figure below.) The chemicals react close to the substrate, depositing the desired semiconductor on the substrate. An example of a reaction is Ga(CH3)3 + AsH3 ! 3CH4 + GaAs.