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Chapter 10The 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 threedimensional, but lower dimensionality order is possible.)

Bonds in crystalline solids are more or less the same in energy, and crystalline solids havedistinct 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. Incrystalline B2O3, you can "look" in any direction and see the same environment, but not inamorphous 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.

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�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 ofatoms 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 impedeeach 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 incrystals.

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10.2 Ionic Crystals

An atom with a low ionization energy can give up an electron to another atom with a high electronaffinity. The result is an ionic crystal.

To calculate the stability of an ionic crystal, we need to consider all of the energies involved in itsformation.

�Positive energy is required to ionize an atom.

�Energy is released when a highly electronegative atom gains an electron (a negativecontribution to the total energy).

�There is a negative contribution to the energy from the Coulomb attraction between unlikecharged ions.

�There is a positive contribution to the energy from the overlap of core atomic electrons (thePauli exclusion principle at work).

Remember that negative energies mean stable systems. If you add up all the above energiesand get a more negative energy than for the separate, isolated atoms, then the ionic crystal isstable.

Ionic crystals are generally close-packed, because nature "wants" as many ions of different chargesqueezed 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 crystalrelative 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 itfor 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. Wewill add up all contributions to the Coulomb energy from ion-ion interactions.

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Each Na+ has 6 Cl- near neighbors at a near-neighbor distancer.

This figure shows four of them. There are two more in theplanes above and below the central atom. Can you visualizethem?

The contribution to the Coulomb potential from these sixnearest neighbors is

10

2

0V =

6(+e)(-e)

4 r = -

6e4 r

.π ε π ε

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, theinteraction is repulsive, and the contribution to the total energy is positive.

The figure to the right shows four of the next-nearestneighbors. There are four more in the plane above and fourmore in the plane below the one shown.

The contribution from the twelve next-nearest neighbors is

2

2

0

V = +12e

4 r 2.

π ε

We can do the same for additional ion "shells." The result is

V = -e

4 r6 -

12

2+... = -1.748

e4 r

= -e

4 r.

2

0

2

0

2

0π ε π εα

π ε

The constant � (which is constant only for a given type of structure) known as the Madelungconstant. Beiser gives the values of � for a couple of other structures.

Note that

( )α = 6 -12

2+... = 6 -8.485281...+... .

The convergence of this series is very poor. You have to find a clever way to do this series ifyou 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.

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Now we need to account for the repulsive forces that happen when electron shells start tooverlap.

Remember, overlapping shells might result in electrons in the same place with the samequantum number.

The Pauli exclusion principle says this can't happen. Electrons have to be promoted to higherenergies to allow atoms to come closer together. The result is a more positive energy, i.e., aless stable crystal.

We model this repulsive forcewith a potential of the form

repulsive nV = B

r,

where n is some exponent. Theexact value of n isn't too critical(see the figure to the right).

The potential energy of interaction ofthe Na+ ion with all the other ions is

V = -e

4 r+

B

r.

2

0n

απ ε

We can find B in terms of � and the equilibrium near-neighbor separation r0 by realizing that atequilibrium the energy is minimized.

0 = V

r =

e4 r

-nB

rr=r

2

0 02

0n+1

0

∂∂

απ ε

απ ε

2

0 02

0n+1

e4 r

= nB

r

B = e

4 nr .

2

0

n-1απ ε

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Plugging B back into the expression for V gives the lattice energy.

V = -e

4 r1-

1

n.

2

0 0

απ ε

The lattice energy is defined as the reduction in the energy of the ionic crystal relative to the ions atinfinite 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:

V = -e

4 r1-

1

n. = - (1.748)(9x10 )

(1.6x10 )

2.81x101-

1

9

2

0 0

9-19 2

-10

απ ε

= -1.27x10 Joules = - 7.96 eV.-18

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 tothe Cl. The increase is greater than the reduction:

net = (+5.14 -3.61)

2 = + 0.77 eV per ion.

The net reduction on bonding is -3.98 eV+0.77 eV=-3.21 eV, which compares quite well withthe 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.

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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'tattract charged particles.

�Something like hydrogen bonding (e.g., water), where the asymmetry of the molecule givesrise 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 inconstant 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 verylow temperatures.

It's a Physics 24 type problem to show that a dipole of moment p gives rise to an electric fieldE given by

�� � �

�E =

1

4

p

r-3(p r)

rr .

03 5π ε

•

This electric field can induce a dipole moment in a normally nonpolar molecule. The induceddipole moment is

� �′p = E,α

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.

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Things to note: the potential is negative (attractive, stable). The force is proportional to dV/drwhich 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 safelybe neglected.

Our brief derivation is a classical one, and, as you might suspect, not really correct at theatomic level. However, a full quantum mechanical derivation gives the same r-dependence.

The Lennard-Jones potential, which you may have encountered in chemistry (also sometimesknown as the Lennard-Jones 6-12 potential or the 6-12 potential) describes van der Waalsforces.

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 thischapter, 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 Heisenberguncertainty principle and the Pauli exclusion principle.

Metal atoms give up electrons (usually one or two, sometimes three). The result is a lattice ofpositive 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 inenergy 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 netelectron 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

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slightly altered. The energy differences are very small, but large enough so that a largenumber 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 butslightly split in energy. They are so close together that for all practical purposes we canconsider bands as a continuum of states, rather than discrete energy levels as we have inisolated 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 dothese bands come from?

We should first ask where the electronic energy levels in atoms come from.

Remember, we solved the Schr�dinger equation for an electron in the potential of thehydrogen nucleus. This gave us our energy levels and quantum numbers.

More complex atoms require more complex mathematics, but the idea is the same: the energylevels come from the solution of Schr�dinger's equation for electrons in the potential of thenucleus.

The same holds for metals, except now we have a periodic array of nuclei, and a periodicpotential. We still have to solve Schr�dinger's equation for an electron moving in this periodicpotential.

The problem can't be solved exactly, of course, but it can be solved with quite reasonableaccuracy. The energy bands result quite naturally from the solution, just as energy levels inatoms came naturally out of the solution to Schr�dinger'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 aretalking 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 toagree with experiment, but which doesn't come from any theory. Empirical equations are notvery 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.

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What happens when an electric field is applied?

Electrons accelerate and gain a net "drift" velocity in some direction. The drift velocity isactually 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 justthat?

Some force must eventually oppose the force due to the electric field and bring the driftvelocity 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 areoccupied).

If we define � as the mean free path of electrons between collisions, then the average timebetween collisions is

τλ

= v

.F

An applied electric field results in a force on the electron, and therefore an acceleration

a = F

m =

eE

m.

Since the electron is accelerated only during the time �, after which it undergoes a collisionand has its velocity randomized again, the average gain in velocity, or "drift velocity," is

dF

v = a = eE

mv.τ

λ

The total current I flowing through a conductor of length L, cross-sectional area A, with N freeelectrons per unit volume is

I = N A e v = NA e E

mv =

N emv

A

LV =

V

R,d

2

F

2

F

λ λ

where

R = mv

N e

L

A =

L

A.

F

2 λρ

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� is the resistivity of the metal sample. (We got this because E=V/L is the electric field insidethe 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

µτ

= e

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 carriervelocities. 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 onpage 349 says that electrons in copper at room temperature travel past an average of 150 + ionswithout "seeing" any of them. Isn't that strange?

Yes, but in fact, electron waves in a perfect periodic lattice interact with the lattice only undervery 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 severaltextbooks by itself.

What happens in crystalline solids when we bring atoms so close together that their valence electronsconstitute 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.] Theenergy levels of the overlapping electron shells are all slightly altered. The energy differencesare very small, but enough so that a large number of electrons can be in close proximity andstill satisfy the Pauli exclusion principle."

"We are describing the formation of energy bands, consisting of many states close togetherbut slightly split in energy. They are so close together that for all practical purposes we canconsider bands as a continuum of states, rather than discrete energy levels as we have inisolated atoms (and in the core electrons of atoms of metals)."

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A detailed analysis of energy bands shows that there are as many separate energy levels in each bandas there are atoms in a crystal.

Remember that two electrons can occupy each energy level (spin), so there are 2N possiblestates in each band.

Figure 10.19 (show a transparency) shows how the sodium energy levels spread out into bands whenwe bring sodium atoms together.

Some atomic energy levels are shown to the right of the figure (the 3p level has alreadyformed a band in this figure). Sodium contains a single 3s electron, so the figure showsenergy levels which are not normally occupied.

As the sodium atoms are brought closer together than the 1.5 nm distance to the right of thefigure, notice how the 3p band spreads wider, and notice how the 3s state begins to form aband at a separation of about 1 nm.

Also notice how electron energies can be reduced uponformation of bands. The metal is stable, after all.

We schematically represent the energy bands in sodiumlike this:

This is highly schematic. Real bands aren't boxes or lines.

Sodium has a single 3s electron, so the 3s energy bandcontains twice as many states as there are electrons. Theband is half full. At T=0 the band is filled exactly halfwayup, and the Fermi level, �F, is right in the middle of theband.

Sodium is a metal because an applied field can easily giveenergy to and accelerate an electron.

Read how magnesium, which you might at first expect not to be a metal, is metallic because ofoverlapping 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, givingrise to semi-metallic behavior.

Now let's consider the energy bands in diamond. Carbon has two 2p electrons. A p shell has 6possible states, so a d-band should have 6N states, where N is the number of atoms in the crystal. 2N2p electrons filling a band with 6N states should produce a metal.

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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 8Nstates.

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 diamondlies within this range.

The 2N 2s and 2N 2p electrons occupy the lower of the bands with 4N states. The band isfull. There is a large energy gap (6 eV, a really large gap) between the full band and the nexthigher band.

Thus, in diamond,there is a full band, a 6eV gap, and an empty band. The 6 eV gap is a biggap. There is not enough thermal energy at normal temperatures to put any 2p/2s electronsinto the empty band.

An electric field can't give energy to electrons in the filled band, because there are nounoccupied energy states nearby. A huge electric field is required to get an electron across theband gap.

Thus, diamond is a very good insulator.

So far, we've covered conductors (odd numbers of valence electrons and partly filled bands) andinsulators (even numbers of electrons and all filled or all empty bands with large band gaps).

Remember that our band representationsshown in the last two figures are highlyschematic representations. Real bands aren'tsquare like this.

There's still another possibility for forming bands --bands which would normally be all filled and all emptybut with small gaps between them. Such materials aresemiconductors.

To the right is a schematic energy band diagram for asemiconductor.

The gap between bands is relatively small, e.g.,1 eV in silicon.

Room temperature is only 0.040 eV, but thereare lots of electrons in the filled (at T=0)valence band, and it doesn't take a very largeprobability to get quite a few of them across theband gap at moderate temperatures.

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Thus, at very low temperatures, silicon is an insulator, but at room temperature, it is a weakconductor (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'senergy in terms of its wavevector:

E = k

2m,

2 2�

where k is the wavevector and m is the electron'smass.

A plot of a free electron's energy looks like thefigure to the right.

What happens when you put the electron in a metal?

It moves in the periodic potential of the crystallattice.

Remember, an electron is a matter wave. Westudied (however briefly) diffraction of waves backin chapter 3. Waves can be diffracted by periodiclattices.

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 areenough wavevectors to essentially form a continuum.

The lattice diffracts electrons of a few select wavevectors. Exactly which wavevectors arediffracted 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 Schr�dinger's equation for these particular standing waves in the potential of thelattice, we find that there are two possible solutions. One solution has a lower energy than afree electron of the same wavevector, and the other solution has a higher energy than a freeelectron of the same wavevector.

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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-wavelengthrelationship is modified like this:

Electrons near the "zone boundary" (whichis determined by and related to theperiodicity of the lattice) are the ones mostaffected by the periodic potential of thelattice.

There are two possible solutions forelectron energy at the zone boundary.

Near the zone boundary, the energies aremodified by the lattice potential.

Notice the energy gap at the zoneboundary. There are energies which aresimply not attainable by the electrons. Again, this is because they are wavesmoving in the periodic potential of thelattice. Not all waves can exist in theperiodic potential.

Effective Mass

I've switched from epsilon's to E's for electron energy here. No reason in particular, just did it anddidn'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 abovediscussion.

We know that

E = k

2m

2 2�

relates an electron's energy and wavevector.

Note that the coefficient of wavevector squared is proportional to 1/m, where m is the electronmass.

The above result holds in general. For an electron in a periodic potential, we write

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E(k) = k

2m,

2 2

*

�

where m* is the electron's effective mass.

In chapter 9, we found that the Fermi level for free electrons is

F

2 2/3

= h2m

3N

8 V.ε

π

Here, when we put electrons into a lattice, we have to account for the modification of theelectron 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 apotential from the lattice) is

F

2

*

2/3

= h

2m

3N

8 V.µ

π

The wavevector of an electron at the Fermi surface is just

F

2/3

k = 3N

8 V.

π

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 kcurve, 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 theyare 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, andinsulators have filled valence bands, empty conduction bands.

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Semiconductors (as described above) also have valence bands which are filled and conductionbands which are empty at T=0. For T>0 some electrons can get across the energy gap fromthe 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 themto 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 of4s24p3. After it shares 4 of its outerelectrons with neighboring silicons, theremaining electron is very loosely bound.

In fact, the arsenic impurity creates a donorimpurity level. Since the donor ionizationenergy is very small, the donor level sitsjust below (maybe a small fraction of aneV) the conduction band.

Electrons from the donor levels can easilyget to the conduction band, where they areavailable for conduction, just as normalconduction band electrons are.

It turns out that there don't have to be very many donor atoms around to result in a significantnumber of electrons in the conduction band. Beiser gives some numbers on this.

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The semiconductor we have been discussing are n-type, because conduction is by negativeelectrons.

Another type of semiconductor is the p-type semiconductor. As the name implies, conduction is bypositively charged objects, which we call "holes."

Here's how it works. Example: gallium impurities in silicon.

Ga has an outer electron configuration of4s24p1. It "wants" to borrow 5 electronsfrom neighboring silicon atoms. Four ofthe electrons are normally shared by siliconanyway. It turns out very little energy isrequired for the gallium to "borrow" theadditional electron.

The gallium impurity creates an acceptorimpurity level. Since the acceptorionization energy is very small, theacceptor level sits just above (maybe asmall fraction of an eV) the valence band.

Electrons from the valence band caneasily get to the acceptor bound states. That leaves holes in the valence band. The holes represent states into whichother electrons in the valence band canmove. Thus, electrons can easily movearound in the presence of an applied field. Alternatively, we can look at the holes,say they move around, and say thatconduction is due to holes.

Again, it turns out that there don't have tobe very many acceptor atoms around toresult 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 valenceband and sitting in an acceptor state.

Why don't we just talk about conduction by these electrons moving from state to state, insteadof worrying about holes.

Answer: holes really are more than just missing electrons. For one thing, electrons in thiscase are "stuck" on acceptor atoms, but the holes are free to move about in the valence band.

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For another thing, we can dope semiconductors so that there are excesses of holes andelectrons. 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 inthe previous edition, Beiser mentioned that a chip half a centimeter square can hold up to 50,000transistors. 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 goodenough contact that way.

Instead, we diffuse donors or acceptors into silicon. We define the regions into whichimpurities are diffused with masks. Sort of like stencils. We can build up a whole series oflayers in devices.

A p-n junction diode is a device which letscurrent flow in only one direction. Let'ssee how it works. Here's a junction.

Here's a plot of hole and electron concentrations(just examples, I've picked holeconcentration to be larger in thisparticular figure).

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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 sideand - ions on the acceptor side.

Coulomb repulsion will eventually stop the diffusion of holes and electrons, and we end upwith this:

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This charge double layer gives rise to a built-in electric field which inhibits diffusion of holesand electrons.

What happens to the Fermi level?

The Fermi level is like the level of a liquid in a glass. It's thesame everywhere in the material.

The Fermi level is equalized because of the electric potentialdue to the charge double layer, which raises the Fermi level ofthe p-region relative to the n-region until the Fermi level is thesame everywhere.

Now let's re-do the above analysis by looking at the band structure ofthe junction. Here are the bands of the p- and n-type semiconductorsnot 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 whathappens:

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 eachother.

3. Eventually, the built-in field (which came about from the flow of holes and electrons) shutsoff further net flow of holes and electrons. To move any more, electrons must now flow uphilland holes "float" downhill.

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4. The result is a steady state, where there arecontinuous small flows of holes and electrons. These flows cancel each other, so that the netflow is zero. (Where do these flows comefrom?)

5. Electrons flow downhill, holes float uphill,until we get a band structure that looks like theone 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 358of Beiser.

Next, let's reverse bias the junction, like this:

The reverse bias results in a net electron flowto the right and hole flow to the left. Both n-and p-regions are quickly depleted of chargecarriers, and current stops almost immediately.

Also, the reverse bias increases the potentialdifference, 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 theelectrons, so now electrons can flow.

Electrons and holes crossing the junction meetand annihilate, but a steady flow of electronsand holes is "pumped" into the device by theexternal emf, so current flows.

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The voltage-current characteristic of a p-njunction 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 theholes.

Also, holes from the p region have gotten into the n region and recombined with some of theelectrons.

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 torecombine.

Light is emitted in some semiconductors when electrons and holes recombine.

A voltage applied across such a semiconductor can result in light. Advantage: very lowcurrent 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.

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Application: photovoltaic detectors.

The idea is the same as for solar cells, except they are used to detect photons rather thangenerate useful power.

Another kind of p-n junction device: the tunnel diode.

We can heavily dope the semiconductorswhich make up the pn junction, so that thebands overlap like this:

Electrons and holes can tunnel both ways across the very narrow depletion layer between thesemiconductors. In the absence of an applied voltage, the net current is zero.

A small forward voltage does this to the bandstructure:

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 isfrom n to p only. The result is an "excess" current for this voltage.

Additional forward bias raises the n-regionconduction band too high for tunnelling tooccur, and the device turns into an ordinarydiode:

This is the current-voltage characteristic (seefigure 10.2:

The advantage to a tunnel diode is that it is adevice which changes current very rapidly inresponse to a voltage.

Zener diode, avalanche multiplication, and Zener breakdown. Read about these. I will skip them, notbecause 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, reversevoltage normally shuts off the flow of electrons), electrons can be accelerated to large enough

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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 alsomake 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 reachingequilibrium, they can't diffuse any further because ofthe built-in voltage.

Now forward bias the first np junction:

Electrons ("majority carriers" in the n-region) willflow from the n-region (called the "emitter") into thep-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...

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But suppose we make the base thin, so that electrons don'trecombine before they reach the other n-region, and supposewe reverse bias the second pn junction. Remember, thereverse bias sweeps electrons from the p-region into the n-region.

Above was a crude schematic showing the resulting energybands. To the right is a picture of the connections.

In the circuit we have drawn, the output (reverse) current is avery high fraction of the input current.

But remember the base-collector junction is a diodeconnected in reverse bias.

And also remember, this is what the IV curve lookslike for a diode.

A very small reverse current corresponds to a verylarge reverse bias.

In other words, the voltage on the output side can be made very much larger than the voltageon the input side.

We have made a voltage (or power, since both currents are about equal) amplifier. Note thatthe current doesn't get amplified.

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We can also operate a pnp junction transistor as a current amplfiier.

A small input current into the base results in a veryhigh output current.

The reverse current is very sensitive to the forward voltage. In typical cases, we can getincreases of factors of 1000 in reverse current when we apply a forward bias across the emitterand the base.

These npn transistors are "junction transistors" and are called "minority carrier devices" because theirbehavior 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 isundesirable for a number of reasons.

Field effect transistors--read pages 364-6 in Beiser. The main idea is that voltages can be amplifiedwith very little current flowing.