# Chapter 10 The Solid State

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