PROFESSOR’S NOTES – SEMICONDUCTORS
7.1 SEMICONDUCTOR MATERIALS
When we consider solid–state materials in terms of their electrical properties, we generally group them according
to electrical conductivity as: conductors, semiconductors, and insulators. The distinction between these types of
materials is relative. All materials are conductive, some are just more conductive than others. But in the electronic
kingdom, the mechanisms of conduction (and non–conduction) are of considerable importance, and we will direct
our attention thereto.
The material that is the most interesting, semiconductors, is more than just a material whose conductivity falls
between that of insulators and that of conductors. It is a material with a highly variable conductivity. In its pure
state, a semiconductor might have a conductivity as much as 12 orders of magnitude less than that of copper which
makes it mighty like an insulator. But with trace amounts of impurities, the same material may be as much as 12
orders of magnitude more conductive than glass which make it reasonably conductive, although not as good as a
’real‘ conductor. Furthermore, this conductivity is highly sensitive to illumination, temperature, magnetic field
and electrical field, and maybe even to the presence of cognitive mental processes. It is this sensitivity to the wide
variety of factors, and the fact that impurities exercise considerable control over this sensitivity, that makes the
semiconductor one of the most important materials in the electrical engineering community.
A pure semiconductor is strongly temperature dependent, and serves well as a temperature–sensitive resistance.
Trace levels of impurities inhibit this thermistor behavior, and will strongly increase its conductive properties.
Although it might seem that a material with such sensitivity to impurities and temperature is limited in its usefulness, it is these ambivalent qualities that give the semiconductor its immense value. The advances in materials
technology have allowed the engineering community to gain a very fine control over the levels of impurity alloying
and an accurate patterning of this process. This technology is the way in which we have gained the many options for
control and channeling of electrical currents and fields and provided the large menu of transistor devices and integrated circuits.
Since a semiconductor is a poor conductor, it can accommodate separation(s) of charge and localized E–fields.
Localized E–fields can block or enhance current flow. Since a semiconductor is a conductor, it can sustain an electrically significant transport of mobile charge carriers. And since the electrical conductivity of a semiconductor is
strongly dependent on type and level of impurities, this feature can be patterned to provide channels, paths, and
junctions. It is this degree of control that allows us to create many variants of current and voltage characteristics,
and means for their control. Semiconductor devices may range from tiny semiconductor switches to huge power
transistors.
Semiconductors, whether silicon, germanium, gallium–arsenide, indium antimonide, or any one of a few dozen
others, generally exist in crystalline or polycrystalline form. When atoms join together in a group, as they do in the
case of molecules, their energy levels are shared, and they prefer to separate energies into slight differences. In the
crystalline state atoms share existence and organize into a highly regimented array, of a massive scale extending off
to infinity in all directions. Because of this huge degree of sharing, the energy states of the single atom become
considerably spread and smeared out, as atoms seek to slightly differentiate and yet remain a part of the regiment.
These energies tend to rearrange into parity groupings, which we define as ”energy bands”.
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When arranged in these crowded states of matter, particularly the highly organized crystalline state, materials share
their electrons, and elect, in accordance with the rules of parity, to form two types of states consistent with electron
structuring[7.1–1]:
(1) bonding states, which we identify as valence states, and
(2) anti–bonding states, which we identify as conducting states.
The anti–bonding states always are at the more excited, higher energy levels. The energy–level representation of
these states for a semiconductor material is shown by figure 7.1–1.
E
(crystalline)
(gaseous)
discrete energy states
anti–bonding
EC
EV
bonding
transition
atomic separation
5 nm
500 nm
Figure 7.1–1 Bonding and antibonding states. [McKelvey, p 258]
These states are defined by the parent type of atoms. It is the crowded proximity that causes the splitting and grouping of energies, as represented by figure 7.1–1. The grouping of these energies form into bands of energy states, as
represented by figure 7.1–1. Insulators will have these distinct energy bands, conductors will have have these
bands, and, of course, so will semiconductors. The bands in which the conductive and interactive action takes place
are called:
(1) the conduction band, which is a ”band” of anti–bonding energy states, and
(2) the valence band, which is the ”band” of chemical bonding energy states.
Insulators are the most respectable members of the solid–state community, since they have conduction and valence
bands widely separated, with an energy gap of value EG = EC – EV on the order of 5 eV or more. EC is the the lower
boundary of the conduction band of energy states and EV is the upper boundary of the valence band of energy states.
Conductors do not have this respectability, since their energy bands completely overlap, and the concept of a energy gap and separation of bonding and anti–bonding states is not a meaningful concept.
Electrons are like children. They are localized to an atomic site, their own little home, one among many houses
regularly spaced within the crystalline community. They are permitted a certain amount of local movement around
the neighborhood. Naturally these ”children” are influenced by the thermal events that take place within the crystalline community.
And thermal energy is like money. A material for which the energy band edges EC and EV are relatively close will
have electrons, who, like children, given a relatively small amount of this thermal energy money, will take leave of
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home (bonding site) in the valence band, take a leap, off into an antibonding state in the conduction band, and –
venture forth. Maybe after their thermal umbrage becomes depleted, they might sheepishly return, somewhat like
the prodigal son. But while they are in the conduction band they are only loosely attached to the lattice and can
wander about freely and easily.
In conductors, they overlap, so no extra thermal energy is needed for an electron to gain access to an antibonding
state. The parents do not provide a strongly–bound home situation for them, and they roam freely throughout the
streets of the crystal.
In insulators, the energy gap is large, and the electrons must acquire a great deal of thermal energy to leave home
and jump into one of the conducting states above the conduction band edge, EC .
But semiconductors have an energy gap that is ’just right’. It is about 1–3 eV. Since the band edges EV and EC are
relatively close, the statistical probability of an electron gaining enough energy from ambient thermal sources to
cross the gap is reasonable. And since the material is (usually) crystalline, impurities may be added to the lattice
which will dramatically influence the situation and statistics.
EC
EC
EV
EC
EV
semiconductor
(bandgap < 3 eV)
conductor
(overlapping bands)
EV
insulator
(bandgap > 5 eV)
Figure 7.1–2: Band structures for the different materials.
So, much to our surprise, it appears that much of our semiconductor electronics is nothing more than a gigantic
accounting process of thermal statistics associated with large numbers. Electrical behavior is determined by a
grand canonical ensemble of probability. The probability of molecular behavior is well documented, and the probability that an electron occupies a given energy state is given by the Fermi–Dirac distribution,
f (E)
1
1
e (E
(7.1–1)
EF) kT
This probability distribution is shown by figure 7.1–2. Note that the value of energy E = EF forms a statistical
”edge” that is relatively abrupt if the thermal energy kT is small. The edge is relatively broad if the thermal energy
is large. The parameter k is the Boltzmann energy–temperature constant, = 8.62 10–5 eV/k. At ambient temperature T = 300, kT .02585 eV. We define EF as the Fermi energy. Note that at E = EF, f(E ) = 0.5. This result tells
us that EF is an index about which there must be just as many empty states for E < EF as there are filled states for E
>EF.
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f(E)
1.0
0.5
E
EF
(rear)
(front)
Figure 7.1–2 The Fermi–Dirac distribution. Occupation probability of church pews being filled.
Fermi–Dirac electron statistics are not unlike the statistics of devout Baptists seated in church on Sunday morning.
The pews in the rear fill up first (these = lowest energy levels). As you look toward the front, more and more empty
seats are evident, eventually reaching a point where the occupancy level drops to about 50%. Beyond this point the
seats become exponentially more empty. Maybe Enrico Fermi was an Italian Baptist. Whatever his affiliation, he
was good at assessing combinational probability and the statistics thereto, on the scale of molecular and atomic
ensembles, in this case for numbers such as 1016 members being seated in 1022 seats. The result reduces to the
interesting and important probability distribution given so clearly and concisely by equation (7.1–1).
For an intrinsic (= pure) semiconductor, this 50% point, E = EF will be approximately at the midpoint of the energy
gap, since there must be just as many empty states in the valence band as there are free electrons above the energy
gap, just like the situation represented by Figure 7.1–2 or figure 7.1–3.
f(E)
EF
1.0
0.5
E
EC
EV
Figure 7.1–3 The Fermi–Dirac distribution overlaid upon energy bandgap for an intrinsic (= pure) semiconductor.
Note that when an electron leaves a valence state, a void is left behind. The void is the absence of an electron,
otherwise known as a hole. Since it is the absence of an electron it has net positive charge. The ”hole” can pass from
one site to another, much like the electrons in antibonding states do. Therefore ”holes” represent a form of conduction charges, just like conduction electrons are a form of conduction charges, except that ”holes” are of charge +q
each.
Even though holes are ”not there”, they are a more physically identifiable presence than are electrons. They are an
absence of matter, like the little ”bubblets” in a glass of ginger–ale. Holes are all about the same size, and they fall
in the opposite direction from that of the electrons when subjected to an E–field, just like the ”bubbles” fall upward
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instead of downward (where down is defined by the gravitational field). They can only exist in the presence of the
condensed environment, such as solid or liquid. These representations are illustrated by figure 7.1–4.
E – field
G – field
( holes and electrons )
( bubbles and pellets )
Figure 7.1–4 Holes and electrons
The flow of ”holes” within a solid–state material is, in all respects, equivalent to a flow of positive charge carriers.
They are identical to the concept of a flux of bubbles, which are a flow of negative (absence of) matter, but do not
exist unless the space is the ’filled’ solid or liquid molecular environment. Because of their mobility, the ”holes” in
the valence band are just like electrons in the conduction band, except they fall in the opposite direction of that of
the electrons when subject to the drift force of an an electric field.
Solid–state materials are characterized by a crystalline form based on a unit cell, which forms the basic ”stacking–
block” for an infinite lattice extending in all directions as far as the eye can see. For most semiconductor materials,
this is the ”diamond” lattice, shown by figure 7.1–5.
(1 0 0 )
a
(1 1 0 )
Figure 7.1–5 Diamond lattice structure of silicon, and Miller index notation.
For the diamond lattice, the bond length, or distance between nearest neighbors, is a 3 4 , where a is the lattice
constant. Note that the diamond unit cell is of the form of a face–centered–cubic lattice, if you are acquainted with
crystalline terminology. It has 8 atoms per unit cell. You might see if you can count these up, with attention to the
fact that some atoms are shared at the corners and at the faces with other cells. If we think about each site as being
encompassed by a sphere of a radius of half the bond length, then the packing fraction, which is the fraction of the
cube filled by the spherical zones, is
3 16 .
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If we apply this lattice geometry to silicon, which crystallizes in the form of a ’diamond’ lattice’ since it is tetrahedral, the lattice constant a = 0.543 nm, corresponding to a density of 5 1022 atoms/cm3.
7.2 CARRIER LEVELS IN SEMICONDUCTORS
The energy bandgap causes the semiconductor materials to have two types of charge carriers, (1) ”electrons” which
are charge carriers of magnitude –q, and (2) ”holes” which are carriers of magnitude +q. Because of the polarity,
these carriers are usually designated as ”n–type” and ”p–type”, respectively. Semiconductors in which ”n–type”
carriers dominate are called ”n–type semiconductors”, and semiconductors in which ”p–type” carriers dominate
are called ”p–type semiconductors”.
The conductive properties of a semiconductor are a function of the available conductive charge density, and therefore are proportional to the density of the charge carriers. The density of the mobile, conducting, n–type carriers
can be determined from the density of states that exist in the conduction band and the probability that they will be
occupied, as given by equation (7.1–1).
Density of energy states is a measure of how many possible energy levels exist between E and E + dE per volume.
These energy states are nothing more than energy harmonics ringing and vibrating within the crystalline lattice.
(They are at frequencies on the order of 1015 Hz, so don’t expect them to be audible). There are, of course, a virtually infinite number and density of these molecular harmonics. The energy density of these vibrational energy states
is defined by the number of possible wavenumber harmonics that can exist within the crystalline lattice, which
means that we consider 3–dimensions worth of harmonics. Like any kind of harmonics, they are only sustained if
they match the selection rules defined by the nodes (= atomic sites) of the crystalline lattice. It is these selection
rules, taken on a grand scale, which causes the grouping of these energy states into energy ”bands”.
Definition of these energy states is not as wicked an analysis as it sounds because molecular harmonic energy of
vibrating masses (electrons) of mass me is related to magnitude of the wavenumber k by[7.2–1]
2
E
1 h
2m e 2
k2
(7.2–2)
where k is the wavenumber, and where me is the effective mass of an electron within the semiconductor material.
Note that we need only to deal with the lattice energies in the conduction band, i.e. (E > Ec ), since these are the only
ones for which the electrons are free and mobile. Since we are dealing with atomic and molecular dimensions, we
find ourselves using h = Planck’s constant.
Do you remember Planck’s constant? Maybe not. It comes from thermal physics, equations of state and all that
stuff, and simply says that there is a minimum amount of momentum that an atomic or molecular structure of any
type can resolve. Planck’s constant has value of h = 4.135 10–15 eV.s (which is pretty durn small).
The density of states is the number of harmonics between k and k + dk in 3–dimensions, which is
dN(k)
1 (4 k 2dk)
3
8
2
(7.2–2)
If you elect to look more closely at this equation, you see that it is the volume of a spherical shell in ”k–space”, with
spherical area 4 k 2 and shell thickness dk, defined for the first octant. Equation (7.2–2) defines the density of
harmonics, where each harmonic = k/ . We have to include a factor 2 in equation (7.2–2), because electrons have a
measure called ”spin”,with two states, ”up” and ”down”, and these two states need to be included in the count density. As indicated by equation (7.2–1) the radius of the sphere represents energy, and so a ”shell in k–space” represents the energy density between E and E + dE.
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When equation (7.2–1) is applied to equation (7.2–2), we get the density of states in terms of energies rather than the
momentum harmonics:
dN(k)
4 (2m ) 3 2(E
e
h3
dN(E)
E c) 1 2dE
(7.2–3)
This information is sufficient to determine the equlibrium density, n0 , of mobile electrons having a statistical probability of existing in the conduction band. This density of negative charge–carriers is the sum of all the occupied
states for which E > EC , which is just the integral,
n0
n(E
EC
f (E) 4 3 (2m e) 3 2(E
h
f (E)dN(E)
E C)
E C) 1 2dE
EC
4 (2m ) 3 2
e
h3
(E E F) kT
e
(E
E C) 1 2dE
EC
(E C
N Ce EF) kT
(7.2–4)
where, in order to have a closed analytical form, we have made an approximation to the Fermi–Dirac distribution,
for the usual situation, in which E > EF,
f (E)
1
1
e (E EF) kT
e
(E EF) kT
This is called the the Boltzmann approximation, and makes the mathematics simpler with very little compromise in
accuracy. It allows us to have a nice, closed analytical form for density n of the conduction electrons.
If you want to check the mathematics of the steps in equation (7.2–4) we recommend the definite integral (which is
= gamma function):
x 1 2e 1
2a
ax
dx
a
0
Note that we have let the mess of coefficients in equation (7.2–4) be replaced either by NC , the effective density–of–
states for the conduction band, or by use of e , the thermal wavelength for electrons. The effective density of states
in the conduction band, NC , is given by
NC
and the thermal wavelength e
2
2 m ekT
h2 3 2
2
3e
(7.2–5)
is given by
e
h2
2 m ekT
(7.2–6)
At room temperature (T = 300K) the thermal wavelength for electrons in silicon is about 4.10 nm, assuming that the
density–of–states effective mass for electrons, me = 1.08m0 , m0 being the mass of a free electron (approx 9.11
10–31 kg). With this value for thermal wavelength, the effective density–of–states NC will be approximately 2.9
1019 #/cm3.
By a similar process, we can beat our way through an equal amount of mathematics to show that the equilibrium
density of holes, p0 , is given by
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Ev
p0
[1 23 e f (E)]dN(E)
(E F E v) kT
N Ve
(EF EV) kT
(7.2–7)
h
where
23
NV
(7.2–8)
h
is the effective density of states in the valence band.
h
is the thermal wavelength for holes, and is given by
h2
2 m hkT
h
(7.2–9)
The thermal wavelength for holes at room temperature is about 5.75 nm, assuming that the effective density–of–
states effective mass for holes mh = 0.56m0 . This thermal wavelength gives an effective density–of–states NV of
1.05 1019 #/cm3.
Note that h is different from that of e . These lengths are different because electrons and holes have distinctly
different effective masses, mn and mh , within the crystalline lattice. Keep in mind that an electron does not exist as a
little ’pellet’ ricocheting through the crowded solid–state array, since energy states, and therefore electrons, are
shared, often over many sites. Therefore the effective mass of a –q electron charge is not at all likely to be the same
as the free–electron mass m0 . The effective mass of a ”hole”, or ”bubblet” is a displacement mass, and is even less
likely to match the free–electron mass m0 .
Don’t be perturbed over the fudge–factor use of ’effective mass’. Just perceive an electron in the same manner as
you would a water droplet, where in free space it could be hurled from one point to another, with its translational
dynamics defined by the ’free’ mass of the droplet. But if we push it into the end of a hose, filled with water, a
droplet will pop from the other end of the hose, as if transmitted through the hose. The apparent velocity, and therefore the ’effective’ mass of the droplet, may appear to be greater or less than that of the ’free’ mass, depending on
whether the hose is a stiff hose or a flaccid hose. In like manner, if electrons are transmitted through a crystalline
lattice, they will have an apparent ’effective’ mass different from that of free electrons. They will even have different effective masses for different lattice directions! If we look at distinctly different lattice energy levels, they will
even have different effective masses according to the energy state in which they exist, so that we can have ”heavy
electrons” and ”lightweight electrons”, as well as ”heavy holes” and ”lightweight holes”.
If we take the product of (7.2–4) and (7.2–7), we see that the Fermi level (= equilibrium index), EF, is wiped out, and
the equilibrium equation can be put in the form
n 0p 0
N CN Ve
(EC
E V) kT
N CN Ve
EG kT
(7.2–10)
where EG = EC – EV is the value of the gap energy. EG is about 1.12 eV for silicon. This equation shows that the
product np is a constant for T = constant. This result is an important fact of equilibrium.
Our favorite semiconductor, silicon, is a tetravalent material. Its lattice is identical to the crystalline structure for
carbon that we know as diamond, represented by figure 7.1.5. Certain properties of the diamond lattice itself are
fascinating, but the primary feature of the lattice that we should recognize at the moment is that each lattice site is a
covalently–bonded node. It is a covalent community, and therefore impurities, particularly those which are pentavalent and trivalent, will be expected to share four of their electrons with their neighbors.
If a pentavalent impurity, such as phosphorus, is alloyed into a lattice node, it has the four electrons necessary for
sharing, but also has an extra electron which is loosely bound since it has no covalent assignment. With relatively
little effort, this extra electron can move right into an anti–bonding state, and therefore join the ranks of the rootless,
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free–roaming lifestyle of conduction electrons. The absence of the electron from the site of the impurity does not,
however, create a hole, since there are still four electrons left at the site to covalently bond with the neighbors. The
donor site does have a net (+) charge, so it will act as a strong deflection point to any incoming charges.
Such impurities are called ”donor” impurities, since they ”donate” an electron to the conduction band. We designate donor doping concentration as ND . Typically ND is on the order of 1015 #/ cm3, corresponding to an impurity
doping of only about five parts in 100 million!
In like manner, if a trivalent impurity, such as boron, takes up residence at a lattice node, it only has three electrons,
and therefore must ”borrow” one from a neighbor. This creates a ”hole”, and the neighbor must borrow an electron
from another neighbor to fill his hole. In this way, electron theft on a frantic scale takes place, and the hole is frantically passed around, creating the (+) mobile charge carrier that we call a ”hole”.
These impurities are called ”acceptor” impurities, since they ”accept” an electron, leaving a hole in the valence
band. We designate the acceptor doping concentration as NA . The acceptor site does have a net (–) charge since it
has an extra electron, any therefore it can strongly deflect any incoming charges.
Note that for a pure semiconductor, thermal statistics yield a hole for every electron created, and n0 = p0 = same
ni . We define ni = the intrinsic carrier density. Then equation (7.2–10) can be expressed in the more compact form
n 0p 0
n 2i
(7.2–11)
This equation is an important and convenient relationship, and we will make extensive use of it in other calculations. Before we do make use of it, however, we should take note of some of the features of ni . In particular, we note
that ni is strongly affected by the temperature, since, according to equation (7.2–10),
n 0p 0
n 2i
N CN Ve
(E C
EV) kT
N CN Ve
E G kT
(7.2–12a)
ni
or,
N CN V e
EG 2kT
(7.2–12b)
where EG is the magnitude of the energy gap. Whenever we consider temperature behavior of our semiconductor
materials, then this equation is where we begin.
Since EG is on the order of about 40 kT for normal ambient temperatures, then ni is on the order of 108 to 1010 #/cm3.
This level, although relatively weak, is electrically significant. For our favorite semiconductor, silicon, the benchmark value for ni is usually accepted as:
ni
n i0
1.5
10 10# cm 2
at T
300K
The other thing of which we should take note is that equation (7.2–12b) is an equation derived from molecular
quantum statistics, and therefore is somewhat ideal. In this respect, equation (7.2–12b) is primarily a disciplinary
equation, but is useful in that it identifies the strong temperature sensitivity (thermistor behavior) of an intrinsic
semiconductor. If we use the values for e and h , which are 4.103 nm and 5.7508 nm, respectively, at T = 300K, and
EG = 1.1245 eV (for Si at room temperature), we get ni = 6.255 109 #/cm3. This result is about half the benchmark value. In solid–state physics vernacular, this discrepancy is ’OK’, but may not be satisfactory for the engineering analysis. Engineers should use the equation,
3 2
ni
n i0
T
T0
exp
EG
1
2kT 0
78
T0 T
(7.2–13)
This equation is the one that you should program into your calculator, not equation (7.2–12). We assume T0 = 300K
(typical semiconductor ambient temperature), and ni0 = ni at T = To . We usually assume silicon, and therefore take
the benchmark value ni0 = 1.5 1010 #/cm3.
Since silicon has about 5 1022 atoms/cm3, then ni0 represents about 1 ionization in 1012 sites!
EG (eV)
ni0 (#/cm3)
Si*
1.12
1.5
GaAs*
1.43
2.25
SiC–
2.9
3.0
SiC–
1010
2.2
106
10–5
20
Table 7.2–1 Intrinsic carrier concentrations at T=300K for various semiconductors
(* Includes the presence of both heavy holes and light holes in the valence band.)
For the temperature approximately constant, we note that the complete mess of equilibrium statistics is concisely
represented by equation (7.2–11). It tells us that if we increase the number of electrons without changing temperature, then the statistics of equilibrium will decrease the number of holes. We can easily increase the number of
electrons without changing the physical temperature, simply by alloying with donor impurities. Likewise, we can
very simply increase the number of holes by alloying the semiconductor with acceptor impurities.
Equation (7.2–11) therefore is usually called the ‘mass–action’ law, since the increase of one quantity results in
decrease of the other.
When we add a donor impurity, we increase the number of free electrons, since the extra electron of the impurity
represents energy levels which are close to the conduction band edge, EC . These donor energy levels are within
about .05 eV of EC , which is a very small ionization energy difference, and means that virtually all of these sites will
donate an electron, as indicated by figure 7.2–1.
EC
ED
EV
Excess electrons escaping ’homesite’ responsibilities
Figure 7.2–1 Donor doping and ionization energy ED
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Electrons contributed by donor impurities
Note that when a site ’donates’ an electron, it becomes ionized with net polarity +q. The material itself stays electrically neutral, and therefore the charge count will balance:
n0
ND
n0
ND
p0
If we apply the mass–action law, then
n 2i n 0
which is a quadratic equation with solution
1 N
2 D
n0
N 2D
4n 2i
(7.2–14)
Only the positive solution is used since a negative density would be meaningless. Note that if ND >> ni ,
n0
ND
for which
p0
n 2i
Nd
within a good approximation. It does not take much of an impurity doping to establish this condition. A donor
doping of only 1 part in 107, which is a lot more pure than most of the water we drink, corresponds to ND = 5 1015
#/cm3. This concentration is approximately five orders of magnitude greater than that of ni !
Similarly, if the semiconductor is doped with acceptor impurities, of acceptor concentration NA ,
p0
1 N
2 A
N 2A
4n 2i
(7.2–15)
which, if NA >> ni , then
p0
NA
for which
n0
n 2i
NA
If both types of doping are applied, with ND > NA , then we expect that
n0
ND
NA
or, if, instead, NA > ND , then
p0
NA
ND
since, in order to make the neighbors happy, the donors will contribute electrons to acceptor sites, and vice versa.
This case, which we call ’counter–doping’ is represented by figure 7.2–2
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EC
ED
EA
EV
Band diagram: Counter–doping
Some excess electrons find homes at acceptor sites
Figure 7.2–2: Counter–doping with ND > NA . Note that the donor electrons merely fill up the acceptor sites, so that n ND – NA .
Majority–carrier density n0 = ND – NA if ND > NA >> ni , and minority–carrier density p0 is found by the mass–action law (equation 7.2–11) p0 = ni 2/n0 . If NA >> ND , then p0 = NA – ND , and the minority carrier density n is
determined by n0 = ni 2/p.
DON’T BE FOOLED into thinking that both types of charge carriers can be increased by doping. The mass–action
law, equation (7.2–11), which is our statement of equilibrium, clearly forbids this option.
Whatever the case, it is clear that we can considerably shift the magnitude of the levels of charge–carrier by means
of the doping. We therefore usually identify a semiconductor according to the type of doping as n–type if doped
with donors, and p–type if doped with acceptors. Since both types of charge carriers, n and p, are always present, we
can further designate them according to the relative densities, as majority carriers and minority carriers.
For example, in a material which is n–type, the majority carrier level might more properly be designated by nn =
ND , for which the minority carrier level will be designated by np = ni 2/ND.
********************************************************************************
EXAMPLE: A monolithic (= single–crystal) wafer of n–type silicon of doping ND = 1014 #/cm3 is implanted by boron (= acceptor) at 2 parts in 107 to form a localized ’p–well’ region. What is the minority
carrier–level density in the p–well?
SOLUTION: Since there are
NA
(5
10 22)
then p = NA – ND = 1016 – 1014
np
n 2i
NA
2
10 7
10 16# cm 3
1016#/cm3 = pp , and
2.25 10 20
10 16
10 4# cm 3
2.25
********************************************************************************
A semiconductor to which a dopant has been added is called an extrinsic semiconductor. If the dopant is a donor,
the majority and minority carrier densities are given by equations (7.2–14) and (7.2–11), which yield the complementary pair of equations:
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nn
pn
1
2
1
2
N 2D
4n 2i
ND
(7.2–16a)
N 2D
4n 2i
ND
(7.2–16b)
As identified earlier, nn
ND for ND >> ni . However, should we have elevated temperatures, e.g. T = 250oC, (T =
527 K) we now see, via equation (7.2–13), that ni will increase, to become approximately the same order of magnitude as ND , for which we must turn to the more complicated equation, (7.2–16). Behavior of an extrinsic semiconductor with respect to temperature is shown by figure 7.2–3. Note how its thermistor properties are corrupted..!
Over a large part of the temperature range we see that nn
ND . But at temperatures in excess of 300 oC, n ni $,
and the semiconductor is said to have ”become” intrinsic, (i.e. ni > ND ).
n
intrinsic
ND = 1016
ni
T
80 K
550 K
Figure 7.2–3 Electron density as a function of temperature in n–type silicon. An ND = 1016 #/cm3 is assumed.
Note that at lower temperatures, the carrier density falls off. This roll–off is a result of thermal statistics affecting
even the ionization density. Typically these temperatures are on the order of 60–90 K, which are temperatures
where maybe another semiconductor material would serve better.
Similarly, if the dopant is an acceptor, the majority and minority carrier densities are given by
pp
np
1
2
1
2
N 2A
4n 2i
NA
(7.2–17a)
N 2A
4n 2i
NA
(7.2–17b)
In general, a semiconductor which is acceptor–doped will have doping density NA >> ni , and therefore pp
NA .
7.3 CONDUCTION IN SEMICONDUCTORS
In semiconductors, the density levels of charge carriers can be varied with respect to position due to impurity doping, or as result of non–equilibrium situations. Because of this gradient and because a finite E–field can exist within
the material, the conduction processes in semiconductors will have two different driving ”forces”. These driving
forces are called the ”drift” component and the ”diffusion” component, respectively, of the current. The drift component is due to the carriers being ”drifted” by the E–field force. The diffusion component is due to the thermodynamic ”force” resulting from a gradient in the carrier density.
82
When we treat current as a velocity of carriers, then we are assessing their drift current flux:
Jn
(7.3–1)
qnv
The drift velocity v of the carriers is highly dependent upon the extent to which they are scattered as they travel
through the material. Assuming that this scattering is approximately a statistical collision process, then the drift
velocity will be an averaged velocity proportional to the electric field, given by
v
(7.3–2)
E
where we define the coefficient of proportionality, as the mobility of the charge carriers. We might note that this
definition gives us that old familiar law of conduction that we know as Ohm’s law(!):
Jn
q nnE
n
(7.3–3)
E
mobilities (in cm2/V–s)
Bandgap EG
n
p
1.12
1340
460
1.43
8500
400
2.9
400
50
–SiC
2.2
1000
40
Ge
0.66
3900
1900
InP
1.35
4600
150
InAs
0.36
33000
460
GaP
2.26
110
75
Si
GaAs
SiC
Table 7.3–1 Intrinsic mobility values at T=300K for various semiconductors
as written for electron current density. In equation (7.3–3) n is the conductivity of the n–type charge carriers and is their mobility. From these equations we see that conductivity of n–type charge–carriers will then be
qn
n
n
(7.3–4a)
n
We have been careful to make a distinction about the type of charge carriers, since, in like manner, for semiconductors we will have a complementary form of equation (7.3–4a) for p–type charge carriers:
p
qp
(7.3–4b)
p
The mobility of the p–type charge carriers p may be considerably different than that of the n–type charge carriers.
In silicon p is a factor of about 2.5 smaller than n .
From a purely conductive standpoint, the current through a uniform semiconductor is then
J
E
83
just like any other kind of material. But for semiconductors,
p
(7.3–5)
n
Given that semiconductors are dominated by either n–type or by p–type carriers, we may safely assume that
n if the material is n–type or
p if the material is p–type.
********************************************************************************
EXAMPLE: A monolithic substrate of n–type silicon of doping ND = 1014 #/cm3 is implanted by boron
at doping level of 1016 #/cm3 to form a localized ”p–well” region. Assuming that the mobility of the n–
and p–types of carriers is n = 1209 cm2/Vs and p = 437 cm2/Vs, find the conductivity of the p–well.
SOLUTION:
p
and
np
then
n
p
NA
n 2i
NA
10 16# cm 3
(1.602
2.25
10
0.70( cm)
10 4# cm 3
)
1209 19
(2.25
10 4)
437
(10 16)
1
Note that it was not truly necessary, in this case, to include conductivity due to minority carriers since its
value is approximately 12 orders–of–magnitude smaller than that due to majority carriers.
********************************************************************************
The mobility of the charge carriers is closely associated with scattering effects. Therefore the analytical measure of
mobility can be a mess, since scattering is dependent on the crystalline lattice structure and everything therein.
Mobility is particularly affected by the presence of the donor/acceptor impurities, since these impurities produce
ionized sites, and thereby constitute a set of very powerful scattering nodes.
We prefer a direction within the crystalline diamond lattice for which the scattering cross–section is minimal. For
this reason, the transistors are usually oriented so that the current paths are along the (110) paths, shown by figure
7.3–1. Note that when we sight along the (110) direction, the diamond lattice shows a relatively open cross–section, with lattice spacings aligned to form hexagonal ”microchannels”.
84
Figure 7.3–1 Structure of silicon as viewed along the (110) direction of (diamond) lattice.
The scattering due to ionized impurities is independent of the type of impurity since a charge carrier will be deflected by either a (+) site or a (–) site. All impurity doping of a semiconductor, whether ND or NA , represents a
density of ionized impurities. The scattering is also strongly temperature dependent since carrier velocities are
associated with thermal energy. A semi–empirical relationship of mobility that accommodates both ionized impurity scattering and lattice scattering is given by the form[7.3–2]
1 (N N ) max
min
min
I
R
has been developed for silicon[7.3–2] and the forms for
and
p
0.57
0.57
(88T n
n
(54T n
1
)
7.4
[N I (1.26
n
and
p
10 8T 2.33
0.88
10 17T 2.4
n )]
are given as follows:
T
n
1.36 10 T 1 N (2.35
10 T ) T 2.23
8
)
17
I
0.146
2.4
n
0.88
n
(7.3–6a)
(7.3–6b)
0.146
where Tn = T/300. NI represents the total density of ionized impurities, whether ND , or NA , or ND + NA , in measure
of #/cm3
At T = 300 K, which is the usual temperature for benchmarks, these relationships reduce to
n
88
1
1252
[N I (1.26 10 17)] 0.88
p
54
1
407
[N I (2.35
and
10 17)] 0.88
(7.3–7a)
(7.3–7b)
A plot of mobility vs impurity doping, using these equations is shown by figure 7.3–2. Note that the mobility for
both holes and electrons rolls off dramatically for NI > 1017 #/cm3.
85
(in cm2/Vs)
1330
n
p
461
NI (in #/cm3)
1017
Figure 7.3–1 Empirical behavior of mobility in silicon along the [110] direction, with respect to impurity
doping.
Take note again that any ionized impurity, NI , will affect the mobility
and
n
p
for both type charge carriers.
******************************************************************************
EXAMPLE: A monolithic (= single–crystal) substrate having a thick p–type layer of silicon of doping
NA = 1016 #/cm3 is implanted by phosphorus at ND = 1.5 1016 #/cm3 to form a shallow layer. Find the
densities and mobilities of the n– and p–type charge carriers in this shallow implant layer.
SOLUTION:
n
p
NI
ND
NA
1.5
n 2i
(N D
NA
10 16
1.0
2.25 10 20
5 10 15
4.5
N A)
ND
10 16
5
10 15
10 4# cm 3
1.5
10 16
1.0
10 16
2.5
10 16# cm 3
applying this value of NI to equations (7.3–7a) and (7.3–7b) gives
1088cm 2 Vs
n
p
412cm 2 Vs
An ambient temperature T = 300 K has been assumed. Note that, for counter–doping, BOTH types of
impurity dopings have contributed ionized scattering sites, and hence NI = NA + ND . We recommend that
you program equations (7.3–6) or (7.3–7) into your calculator, rather than manually evaluating these
equations every time you need a value of silicon mobility.
******************************************************************************
The diffusion component of current results from the presence of a gradient of the charge–carriers. Such a gradient
is usually a result of a non–equilibrium situation in which the carriers are ”injected” across an interface. This occurs whenever we forward bias a pn junction or when radiation falls upon a semiconductor surface. In the case of
the forward bias of the pn junction, an excess of n–type carriers flow into the p–type region from the n–side, and
86
conversely, an excess of p–type carriers flow into the n–region from the p–side. This flux of carriers may be identified entirely by the negative gradient of the carriers, as follows:
Jp
qD p
dp
dx
(7.3–8a)
and
Jn
qD n dn
dx
(7.3–8b)
where Dp and Dn are the diffusion coefficients of the p– and n–type charge–carriers, respectively, representing the
ability of the material to diffuse from one point in the material to another, as governed by transport properties. Note
that direction of flow of positive carriers relates to a negative gradient, so the sign of equation (7.3–8a) is negative;
whereas the diffusion flow of negative carriers (electrons), as represented by equation (7.3–8b), takes a (+) sign
since the polarity of these charge carriers is negative.
It should not be any surprise that diffusion coefficient D and mobility, , are linearly related to each other, since they
both are defined by scattering processes. This relationship is
Dn
n
VT
and D p
p
VT
(7.3–9)
where VT = kT/q is defined as the thermal voltage, and is of value .02585 V at T = 300K. Equation (7.3–9) is called
Einstein’s law. It may be derived from molecular thermodynamics as a disciplinary exercise, but otherwise we will
assume it to be gospel. But it also may be straightforwardly derived from the analysis of thermal equilibrium, as we
will see in chapter 8 when we develop a few more concepts.
7.4 THE FERMI ENERGY LEVEL: AN INDEX OF EQUILIBRIUM
In section 7.2 we showed that Fermi–Dirac statistics define the equilibrium distribution of charge–carriers. The
result of applying this probability distribution to the semiconductor energy situation was the thermodynamic relationships for active charge–carrier densities n and p, as follows:
n
N Ce (EF
p
N Ve
E C) kT
(7.4–1)
(E F E V) kT
(7.4–2)
where EF is the Fermi energy, and NC and NV are the effective density–of–states for the conduction and valence
bands, respectively.
Note that EF is not a real, existing energy level, but is only an index energy level. If we look back at its point of
origin, we see that it is a level marking the ”fill level”, of available electrons into energy states of the crystalline
lattice. It is also a representation of the average equilibrium thermodynamic energy, since at E = EF, the probability
of occupation of energy states is f(E) = 0.5.
In this respect, if more electrons are supplied to energy slots in the conduction band, by way of donor doping or
other means, the index EF will ”rise” closer to the conduction band. If electrons are taken away from the valence
band, which is another way of saying that more (hole) charge carriers are supplied to the valence band, usually as
result of acceptor doping, then EF will ”sink” closer to the valence band. If there are just as many carriers in the
conduction band as in the valence band, this index energy EF should be approximately at (EC + EV )/2.
87
We can confirm that EF is nearly at mid–gap energy for the intrinsic case, in which n = p = ni , by equating (7.4–1)
and (7.4–2). This equality gives intrinsic Fermi energy,
EF
Ei
1 (E
2 C
1 (E
2 C
E V)
1 kT ln
2
3 kT ln
4
E V)
NV
NC
me
mh
(7.4–3)
We see that Ei , the Fermi index for the intrinsic case, is almost at gap center, as we expected. It is not exactly at gap
center, since equilibrium statistics also depend on the effective densities of states, NV and NC , which differ from one
another as a consequence of the different effective masses me and mh . Equation (7.4–3) shows that it is slightly
lower than the midpoint by approximately .0131 eV, for ambient temperature T = 300 K.
EG
(eV)
me/m0
mp/m0
Si
1.12
1.08
0.56*
GaAs
1.43
.067
0.48*
SiC–
2.9
Table 7.4–1 Density–of–states effective masses of various semiconductor materials.
* includes the presence of both heavy holes and light holes in the valence band.
Note that this is consistent with use of either (7.4–1) or (7.4–2) as a means of defining intrinsic Fermi energy, Ei , in
terms of ni . Or we can alternatively define ni in terms of Ei , i.e.
ni
N Ce (Ei
EC) kT
N Ve
(E i E V) kT
(7.4–4)
The Fermi level is therefore not only an index for equilibrium thermodynamic energy, it is an index for the charge–
carrier density. If we apply equation (7.4–4) to (7.4–1) and (7.4–2) then we have a convenient means of relating
extrinsic carrier levels to intrinsic carrier level through their Fermi energies:
p
E i) kT
(7.4–5a)
(EF Ei) kT
(7.4–5a)
n ie (EF
n
n ie
This behavior is indicated by figure 7.4–1. These equations are important and convenient relationships, and will be
used extensively to deal with equilibrium thermodynamic statistics particularly where we have both n– and p–type
semiconductor material present.
88
EC
EC
EF
Ei
Ei
EF
EV
EV
doped with ND type impurities
doped with NA type impurities
Figure 7.4–1 Donor/acceptor impurity doping and effect on the Fermi level
The figure shows that we may use the Fermi level EF as an index for carrier levels, as well as for the equilibrium
thermodynamics.
******************************************************************************
EXAMPLE: A monolithic (= single–crystal) wafer of p–type silicon, doped at NA = 1014 #/cm3 is implanted by phosphorus at ND = 2 x 1017 #/cm3. Where is the Fermi level relative to (a) Ei (b) EC ?
SOLUTION:
(a)
n
ND
n ie (EF
E i) kT
Inverting this equation, we get
EF
Ei
N
kT ln nD
i
.02585eV
NC
ND
.02585eV
ln 2
1.5
10 17
10 10
0.425eV
Similarly:
(b)
EC
EF
kT ln
ln 2.9
2
10 19
10 17
0.129eV
where we have assumed that NC = 2.9 x 1019 #/cm3.
******************************************************************************
7.5 THE HALL GALVANOMAGNETIC BALANCE
If a semiconductor carrying a current I is placed in a magnetic field such that current I is transverse to the magnetic
field, the flow of charges through the semiconductor will develop an electromagnetic condition in which the magnetic force is balanced by an induced electrostatic force. This ”galvanomagnetic balance” is a useful and interesting application of semiconductor materials, and is also appropriate to the measurement of semiconductor properties. The galvanomagnetic balance is usually identified as the transverse Hall effect.
To interpret the transverse Hall effect, we assume that a slab of semiconductor material carrying current in the x–
direction, Ix , is immersed in a magnetic field in the z–direction, Bz , as shown by figure 7.5–1.
89
Figure 7.5–1 Transverse Hall–effect geometry. The reference frame is chosen so that current flow is in
the x–direction and the magnetic field is in the z–direction.
For the transverse Hall effect, charges with velocity vx transverse to the magnetic field will experience the Lorentz
force, transverse to both Ix and Bz ,
Fy
qv xB z
For a galvanomagnetic balance, this Lorentz force will be balanced by an induced E–field force, Fy = +qEy, so that
qE y
(7.5–1)
qv xB z
If the slab is is of uniform, rectangular geometry, as represented by the figure, then current density Jx is
Jx
(7.5–2)
qpv x
where p represents the density of +q charges and vx represents the average velocity of the carriers. Combining
equations (7.5–1) and (7.5–2),
Ey
1
qp J xB z
R H J xB z
EH
(7.5–3)
where EH is usually called the Hall electric field. and RH is called the Hall coefficient. Since we have a uniform
rectangular geometry we may define
VH
wE y
where the induced voltage VH is called the Hall voltage.
If we solve equation (7.5–3) for RH , then the Hall coefficient will be
RH
Ey
J xB z
(7.5–4)
The fields Bz , Ey and current Jx are all readily measurable quantities, and therefore provides us a straightforward
means for measurement of the Hall coefficient, RH . From equation (7.5–3), we expect that:
RH
RH
1
qp ,
1
qn ,
p
n
type charge charriers
type charge charriers
90
NO!
Oops! Why not? All of our mathematics seemed to be OK!?
Trouble is, equation (7.5–3) is not complete, because charge carriers in a crystalline lattice are not free to assume
just any velocity. At best, vx is an averaged velocity, due to scattering. And the scattering effects will alter direction
of carrier velocities, and thereby not be confined to the nice transverse Hall geometry. In semiconductors, charge
carriers also obey a natural distribution of velocities due to the distribution of energies in conduction and valence
bands. Carriers will be scattered by collisions, either from ”acoustic” collision processes with the crystalline lattice, or from scattering by ionized impurity sites.
1 mv 2 and the probability statistical distribu2
tion of carriers in energy states. The distribution of occupied states is defined by Fermi–Dirac statistics, with mathematics simplified by use of Boltzmann distribution. Effective mass m will differ for different crystalline directions, and this wrinkle must be considered in analyzing the velocity distribution. When these effects are integrated
into the galvanomagnetic balance condition, the coefficient that relates Ey to Jx and Bz is changed: the Hall coefficient RH will take the form:
The natural velocity distribution is defined by the relationship, E
rH
qp
RH
or
RH
rH
qn
(7.5–5)
We still find RH by use of the field equation (7.5–4). We see that the mess of effects due to the velocity distribution
nicely fall into nothing more than a extra coefficient, rH , called the Hall factor. The Hall factor itself also separates
cleanly into crystalline and velocity–distribution factors, as given by[7.5–1]
3K(K 2)
(2K 1) 2
rH
2
m
2
m
(7.5–6)
The process of developing equation (7.5–6) does takes a serious amount of mathematical physics, the finer details
r
of which we will politely skip. In this equation, m represents an energy–dependent scattering time m
0 (E/kT) ,
where 0 represents the collision relaxation time, typically on the order of 0.1 ps. Exponent r is a consequence of the
type of collision process. Parameter K is the ”mass ratio” of the effective–mass tensor for the crystalline lattice, and
represents a ratio of the longitudinal and transverse components of the ”mass–tensor ellipsoid”. The mass tensor
ratio K = 5.2 for silicon[7.5–1]. F represents the averaging process over the velocity distribution,
F(t)
f (v)v x
f 3
dv
vx
f 0d 3v
v 2F
v2
(7.5–7)
where f0 represents the equilibrium velocity distribution function and F is the average over the equilibrium distribution function,
F
0
F(v)f 0(v)d 3v
f 0(v)d 3v
(7.5–8)
0
vx is the velocity of the carriers in the x–direction, and f0 is the equilibrium Boltzmann velocity distribution function. The v 2 in the numerator comes from relationship between velocity and energy for an electron gas of free
particles, E
1 mv 2 . Each of the integrals in equation (7.5–8) is of the form
2
91
2
3
v F(v)f 0(v)d v
4 v 2F(v)e 3
e
0
mv 2 2kT
dv
0
Which are called Maxwell–Boltzmann integrals, and are little more than a modified form of the gamma function.
If we take the function m = 0 (E/kT) r and follow the mathematics to completion over the Boltzmann integrals, the
result is
4
3
m 2
Similarly, the average of the quantity
0
(r
5 2)
over the velocity distribution will give
m
2
0
4
2
m 3
(2r
5 2)
where 0 is the relaxation time constant for collisions. Applying this analysis to equation (7.5–6) gives a Hall factor
of[7.5–1,7.5–3]
rH
3K(K 2) 3 (2K 1) 2 4
(2r
[
(r 5 2)
5 2)] 2
kK
3
(2r
4 [
(r 5 2)
5 2)] 2
(7.5–9)
where (x) is an old friend, the gamma function. If the collision scattering is due to acoustic deformation of the
lattice, r = –1/2. If the scattering is due to ionized impurities, such as donor and acceptor impurities, then r = 3/2.
The Hall factor has value rH = (3 /8) kK if the scattering is due to acoustic deformation (= lattice) scattering, and rH
= (315 /512) kK if the process is due to ionized impurity scattering.
We also may identify conductive properties of a semiconductor along the direction of applied current by:
Jx
pE x qp pE x
(7.5–10)
where, for convenience, we have assumed that only p–type charge–carriers are present. When equations (7.5–4),
(7.5–5) and (7.5–10) are combined, we have a means by which we may measure mobility:
p Ey 1
Ex Bz
1
qpR H
(7.5–11a)
For n–type charge carriers, everything is much the same:
n 1
qnR H
Ey 1
Ex Bz
(7.5–11b)
For convenience, we usually define the mobility measured by the Hall effect as the the Hall mobility:
H RH
(7.5–12)
for which
H Ey 1
Ex Bz
(7.5–13)
92
From equation (7.5–11) and (7.5–13), and using equation (7.5–5) we see that the relationship between Hall mobility H and drift mobility, , will be:
H
rH
(7.5–14)
The Hall mobility is sometimes cited as being virtually the same as the drift mobility, or that rH 1. Whereas this
is a reasonably good approximation for metals, it is a lousy approximation for crystalline semiconductors, since will overestimate the value of by factor rH .
Otherwise the Hall effect is a simple and forthright means for measuring the mobility. It is useful in other ways, in
which a galvanomagnetic transducer may have an advantage for either measuring either magnetic or electric fields.
At one time the Hall effect served a circuit function, – that of an analog multiplier [7.5–2], since Ey = RH Bz Jx ,
represents the product of two independent electromagnetic quantities.
********************************************************************************
EXAMPLE: A sample of silicon is doped with 1017#/cm3 phosphorus. (a) Determine the conductivity of
the sample. (b) If the sample is of 100 m thickness and carries a current Ix = 1 mA, what Hall voltage
should be expected if it is subjected to magnetic field Bz = 1 kGauss = 10–5 Wb/cm2? Assume that the Hall
factor rH = 1.65.
SOLUTION: Phosphorus doping represents n–type carriers, and the majority–carrier mobility for NI =
1017 #/cm3in Si at T = 300K is n = 771.6 cm2/Vs, using equation (7.3.7a). Therefore
since p
q nn 0 (1.602 10 19
) 771.6 10 17 12.36(
cm) 1
is negligible.
The Hall coefficient is
rH
1.65
3
qn 0 (1.602 10 19) 10 17 103cm C
Then the Hall voltage is
3
5
I BZ
103 10 210 0.103mV
VH RH x
d
10 RH ********************************************************************************
All of this treatment of the transverse Hall effect makes assumptions that the fields are sufficiently low so that linear
relationships exist between fields and currents, and that the relaxation time 0 is independent of field strength.
Equation (7.5–3) is generally identified as the low–field Hall effect, and is usually appropriate only for B–fields
that are less than 10 kGauss.
7.6 INDUCED ELECTRIC FIELDS:
The fact that there are two mechanisms for conduction in semiconductors,
J J(drift) J(diffusion)
(7.6–1)
tells us that non–uniformly–doped semiconductors will contain a built–in electric field. This statement may be
confirmed by considering a n–type semiconductor for which ND = ND (x) varies uniformly with respect to position.
93
Then, from equations (7.3–3) and (7.3–8b):
J
J n(drift)
where we are recognizing that n(x)
q nn(x)E
J n(diff)
qD n dn
dx
(7.6–2)
ND (x).
The condition of n = n(x), by definition of equilibrium, show that the energy bands EC and EV will also vary with
respect to position, as represented by figure 7.6–1.
EC
EF
Ei
EV
Figure 7.6–1 Energy bands for a semiconductor with impurity gradient ND (x), at equilibrium. Note that,
at equilibrium, EF must be a constant throughout the material.
Since n = n(x), then
dn(x)
dx
d N e (EF
dx C
N Ce (EF
E C) kT
EC) kT
1 dE C
kT dx
dE C
n(x) 1
kT dx
(7.6–3)
The variation of (potential) energy with respect to position is the definition of electric field,
E
1 q dEdx ddx C
(7.6–4)
provided we include the basic unit of charge associated with these energy levels, which is the charge –q of the electron.
Applying equation (7.6–3) to equation (7.6–4), the electric field is then
E
kT 1 dn
q n(x) dx
V T 1 dn
n(x) dx
(7.6–5)
where VT is defined as the thermal voltage, and is (approximately) .02585 V at T = 300K.
If equation (7.6–5) is applied to equation (7.6–2), and equilibrium is assumed, (for which J = 0), then
0
q
n
n(x)V T 1 dn
n(x) dx
qD n dn
dx
(7.6–6)
Eliminating common terms in equation (7.6–6),
Dn
n
VT
(7.6–7)
which confirms equation (7.3–9), which was defined by argument rather than by analysis.
In like manner we can carry through the process of equations (7.6–2) through (7.6–7) for p–type charge carriers
with like result:
Dp
p
VT
94
(7.6–8)
7.7 EXCESS CARRIERS IN SEMICONDUCTORS
Whenever non–equilibrium conditions are created in a semiconductor, the natural transition process will drive the
situation back toward a state of equilibrium. The equilibrium state is represented by the mass–action law, equation
(7.2.11). Non–equilibrium is typically a result of excess charge carriers being injected into the semiconductor material.
In most cases, excess carriers are a result of the following situations:
(1) Injection of carriers across a surface of the semiconductor as result of conduction
(2) Photoresponsive injection, due to transitions from radiation incident upon a surface.
These cases are represented by figure 7.7–1.
carrier
density
n(x)
Ix
x
Figure 7.7–1a Carrier injection due to a conduction process
carrier
density
n(x)
x
Figure 7.7–1b Carrier injection due to a photoelectric process
It should be noted that photoresponsive processes are a result of photons of energy greater than EG being absorbed
by the semiconductor. This energy causes an increase in the transition probability, as represented by figure 7.7–2.
95
EC
EC
EV
EV
photoresponse
photoemission
Figure 7.7–2 Photoresponsive processes
Note that photoresponsive processes involve both absorption and emission. This tells us that an excess of charges
may also yield an emission of radiation. Although this is usually not the case for silicon devices, some very unusual
circuit glitches have occurred as result of photoreactive situations.
For this reason, it is important that working circuits live in a sealed package, protected from external photogenerative radiation levels. Also, the measurement of device characteristics must take place in the dark. Normal room
lighting generally is of radiation of wavelength on the order of 0.6 m, or energy Eph = 2.07eV. The silicon energy
gap EG = 1.12 eV, corresponds to radiation of wavelength 1.1 m, and therefore room lighting is more than adequate to cause transitions across the gap and create a flux of excess carriers away from the surface and into the
semiconductor.
The concept of excess carriers makes it necessary to identify equilibrium levels by an additional subscript, such that
the mass–action equation must be rewritten in the form
n 0p 0
n 2i
(7.7–1)
If we elect to be more specific, then we can identify the equilibrium level of p–type carriers in an n–type material as
pn0 , or equilibrium n–type carriers in an n–type material as nn0 .
In the majority of cases that need to be considered, excess carriers will exceed the level of minority carriers, but not
the level of majority carriers. This situation is referred to as low–level injection. The rate at which this excess will
return to equilibrium is proportional to the excess density of minority carriers, p = pn – pn0 , or
d p
dt
B p
p
p
where the constant of proportionality B may be identified in terms of a recombination time
equivalent to an exponential decay
p n(t)
p n0
p(t)
p(0)e
t p
(7.7–2)
p.
Equation (7.7–2) is
(7.7–3)
similar to most natural decay processes. Recombination time constants are a function of the transition probability
within the semiconductor, and, in most semiconductors, ranges from ns to s.
The excess of minority carriers at a surface, regardless of the mechanism that originated the excess, will result in a
flow of carriers into the material. This flux is governed by the continuity equation,
dp
dt
dF
dx
(G
R p)
96
(7.7–4)
where F = flux, G = rate at which the carriers are generated, and Rp = rate at which the carriers are annihilated by
recombination. This equation merely says that the rate of increase of carriers, dp/dt, depends on how much the flux
F will increase across a slice dx and by the rate at which are generated in excess of their death rate.
Equation (7.7–4) often is clarified by considering a volume V = Adx as represented by figure 7.7–3. F represents the
flux of carriers in from the left and F + dF represents the flow of carriers out to the right. Within this volume generation and recombination may take place. The rate at which these carriers increase within this volume is
dN
dt
F
where N = p Adx = (carrier density)
(F
dF)
(G
R p)Adx
volume = total number of carriers in the slice of thickness dx.
F
F + dF
area = A
dx
Figure 7.7–3 Carrier accretion in a one–dimensional slice.
Note that if we divide out the volume dV = Adx we end up with the continuity equation (7.7–4)
Equation (7.7–4) is the continuity equation in one–dimension. If we extend this concept to three dimensions then
the continuity equation will be
F dp
dt
(G
R p)
(7.7–5)
The continuity equation occurs in almost every situation in which we have to consider the balance between flux and
the rate of generation or recombination. It is likely that you have seen it elsewhere, in similar form. The explanation elsewhere may have taken more of a mathematical development, but the result is the same.
When an excess of carriers exists at a surface, then the flux away from the surface is due to the diffusion process,
which occurs as result of a gradient in the density of (excess p–type) carriers, i.e.
F
dp
dx
Dp
and the recombination rate Rp is given by equation (7.7–2).
Rp
p
p
p
p
For the case where no carriers are being generated G = 0. At steady–state conditions, for which the dp/dt = 0, then
0
d 2p
Dp 2
dx
Assuming that p
0 at x
d 2p
Dp 2
dx
Rp
p
d2 p
Dp
dx 2
p
(7.7–6)
, this differential equation has solution
p(x)
p n(x)
p n0
p(0)e
97
x Lp
(7.7–7)
where
Lp
D
(7.7–8)
p p
Lp is defined as the recombination length or diffusion length of the carriers as they flow from the surface into the
semiconductor and are annihilated by the recombination processes. It is not unlike the famous march of the Spanish
army into the jungle from the coast of Nicaragua in 1839 with 10,000 men. Twelve days and 120 miles later, the
strength of the army was only 2,000 men, as result of the deaths due to malaria, snakes and snipers, as they marched
forward through the steaming jungle. The death rate S translated into a ”death distance” LS as the soldiers marched
(diffused) forward through the hostile territory.
In like manner, when the p–type carriers are injected into the hostile territory, their numbers will also decline exponentially as a function of distance, according to how fast they diffuse forward Dp and what the characteristic
”death” time constant p may be. This concept is represented by figure 7.7–4.
carrier
density
p n(x)
p n0
p n(0)e
Lp
Ix
pn0
x
Figure 7.7–3 Carrier recombination as a function of recombination distance, as the carriers diffuse forward into the material.
Equation (7.7–7) and all of the analysis preceding it has defined the recombination distance Lp in terms of an excess
of p–type carriers injected into an n–type region. Similarly we will have a recombination distance (or diffusion
length) for n–type carriers with lifetime n , injected into a p–type region, of
Ln
D
(7.7–9)
n n
The flow of electrons and holes, of course, also corresponds to a flow of current, for which F = Jp /q, if the carriers
are p–type, and F = – Jn /q, if the carriers are n–type.
********************************************************************************
EXAMPLE: Determine the diffusion length for electrons injected into a p–type material doped with
5 1016 #/cm3 of Boron, assuming recombination time for the electrons n = 200 ns. Assume T=300K.
SOLUTION: The mobility for n–type carriers in a material of ionized impurity density 5 1016 #/cm3,
according to equation (7.3–7a) is:
n
88
1
Then the diffusion coefficient,
[(5
Dn
VT
905
1252
10 16) (1.26
n
98
10 17)] 0.88
.02585
905cm 2 Vs
23.4cm 2 s
And the diffusion length for the dying electrons will then be:
Ln
D
n n
23.4
(200
10 9)
21.6 m
********************************************************************************
Note that it is unnecessary to evaluate the majority carrier mobility p, even though we sometimes do so for disciplinary purposes. The only relevant mobility (and diffusion length) for a recombination process is that of the minority–carriers. It is NOT necessary to consider the diffusion length for majority carriers.
7.7 NON–EQUILIBRIUM THERMAL STATISTICS
Since it is convenient to think of carrier levels n and p relative to equilibrium, as represented by Fermi level EF, then
equations (7.4.5) and (7.4.6) give us an equilibrium reference. These can be related to electrical potential energies,
since the relationship between energy and potential for electrons, of charge –q, is given by
qE
(7.7–1)
Equations (7.4.5a) and (7.4.5b) can then be rewritten as
n
n ie (
p
n ie (
and
where
= –qEi and
F
F
F
) VT
(7.7–2a)
(7.7–2b)
= –qEF , and VT is the thermal voltage, given by
kT
q
VT
F) V T
is usually defined as the Fermi potential and
(7.7–3)
is identified as the intrinsic potential.
Equations (7.7–2a) and (7.7–2b) represent equilibrium conditions. If we should have a non–equilibrium condition,
we prefer to keep the simplicity of these equations and their relationships to the semiconductor potentials. This is
accomplished by defining quasi–Fermi potentials Fn and Fp , such that
n
n ie (
p
n ie (
and
Fn) V T
Fp
) VT
(7.7–4a)
(7.7–4b)
which is equivalent to ”splitting” the equilibrium reference Fermi level EF. This is represented by figure 7.7–1.
99
EC
EFn
Ei
EFp
EV
Figure 7.7–1 Non–equilibrium: The Fermi level EF is ”split” into two ”imrefs”, EFn and EFp .
The ”quasi–Fermi levels” are also called ”imrefs”, since they are associated with imaginary reference levels used
outside of the framework of the equilibrium statistics that define Fermi level EF.
As the semiconductor approaches equilibrium, for which Note that the product np no longer satisfies equation (7.2.11) since the semiconductor is no longer in equilibrium
np
n 2i e (
Fp
Fn) VT
Fp
(7.7–5)
Fn ,
then np
ni 2.
7.8 RECOMBINATION PROCESSES AND THE RECOMBINATION RATE
The recombination process is one in which an electron an electron gives up energy and falls into a hole Energy,
either in the form of radiation or thermal energy, is released. The process is continuous, since recombination and
emission, wherein an electron gains energy and jumps into the conduction band. are driven by thermodynamics.
The thermodynamic process is governed by transition probability. When in non–equilibrium, the transition probability and the recombination time constant, pulls the system back toward an equilibrium state.
The transition probability is dependent on the magnitude of the energy gap, EG , loosely defined as the difference
between the upper edge of the valence band, EV, and the lower edge of the conduction band EC . Unfortunately, in
the three dimensional crystalline lattice, these ”band edges” vary with respect to position in momentum space, and
the ”peak” of the valence band does not necessarily line up with the ”valley” of the conduction band. Since, for a
conducting electron the free energy is E = h 2k 2/2me , we see that the concept of varying energy is one for which E =
E(k), where k is the wavenumber. This is the ”plot space”, otherwise known as ”k–space” which is represented by
figure 7.8–1.
100
E
EC
EG
EV
k
Figure 7.8–1 Transition probabilities and the energy–gap in the crystalline lattice. Note that for Silicon,
the band–to–band transition must take an indirect path, with a given momentum value necessary to make
the transition.
An indirect band–to–band transition, as represented by this figure, requires that an energy transition must be accompanied by a particular value of momentum. This complementary requirement drastically reduces the transition
probability, and therefore may make the process of regaining equilibrium relatively slow, much too slow for the
quick electrical responses demanded of modern semiconductor circuits. Pure silicon may have a recombination
time on the order of s, whereas ns are usually needed.
The recombination rate is a transition rate. For recombination of electrons, it will then be directly proportional to
the excess density of electrons, n, and the density of available recombination sites (holes), p0 . Therefore it will be
of the form
Rn
c np 0 n
(7.8–1)
This principle is illustrated by figure 7.8–2.
(excess n–type carriers
in a p–type semiconductor)
EC
n
Ei
EF
EV
p0
Figure 7.8–2 Recombination process in a semiconductor material.
Equation (7.8–1) also has a similar form for recombination of holes as excess carriers.
Rp
c pn 0 p
101
(7.8–2)
Impurities and surface defects will considerably improve the speed of the recombination process. If intermediate
energy states are introduced within the bandgap, an alternate recombination process, indirect recombination, will
also occur. This process is illustrated by figure 7.8–3.
EC
R2
R1
trap levels
Et
R4
R3
EV
Figure 7.8–3 Indirect recombination using intermediate ”trap”centers.
Both recombination and emission processes are involved. In reference to the figure, the four processes are:
(1)
(2)
(3)
(4)
capture of an electron from the conduction band by an empty trap center, R1 .
emission of an electron to the conduction band from an occupied trap center, R2 .
capture of a hole from the valence band by a filled trap center, R3 .
emission of a hole to the valence band from an empty trap center, R4 .
The probability that a center is occupied is given nicely by the Fermi–Dirac distribution, assuming that the trap
centers fall at energy level E = Et ,
1
ft
f (E
E t)
1
e (Et
E F) kT
(7.8–3)
As noted by equation (7.8–1) the transition rate is proportional to the number of carriers available and the number of
sites available to them. Using the occupation probability given by equation (7.8–2), the transition rates associated
with the indirect recombination process can be defined in terms of the densities as:
R1
c nnN t (1
R2
e nN t f t
(7.8–4b)
R3
c ppN t f t
(7.8–4c)
R4
e pN t (1
f t)
f t)
(7.8–4a)
(7.8–4d)
where cn , en , cp , ep , represent capture and emission coefficients, for the electrons and holes, respectively. Note that
the emission rates R2 and R4 are proportional only to the density of the emitting population, although the idea of
emitting a ”hole” does seem somewhat strange. Nt represents the density of the trap centers within the semiconductor.
At equilibrium, the number of electrons lost and gained from the conduction band should be equal, and therefore R1
= R2 , for which:
102
en
c nn
1
ft
ft
c n(n i(EF
E i) kT
)(e (Et
E F) kT
c nn ie (Et
)
E i) kT
c n n ie
(7.8–5a)
Similarly R3 = R4 is also necessary for equilibrium, which gives
ep
c pp
ft
f
1
c pn ie
t
(7.8–5b)
Under a non–equilibrium process, the rate at which electron are lost from the conduction band must be equal to the
rate at which they are gained by the valence band (same as the rate as which holes are lost from the valence band).
This condition is called the principle of detailed balance, and requires that
R1
R2
R3
(7.8–6)
R4
If equations (7.8–3) and (7.8–4) are applied to (7.8–5) then
c nnN t(1
f t)
c n n ie N tf t
which gives ft as
ft
c nn
c ppN tf t
c n n c n n ie
c p p n ic n e
c pn ie
n ic p e
N (1 f t)
t
(7.8–7)
(7.8–8)
The net recombination rate is the same as the loss rate from the conduction band, given by R1 – R2 . Applying equation (7.8–7) to this definition we get
R
R1
R2
c nn
c nc pN t(pn n 2i )
c pp n ic ne
n ic p e
(7.8–9)
If we assume that the capture cross–sections for electrons and holes are about the same, then cn = cp , and equation
(7.8–8) will be somewhat simplified,
R
c nN t(pn n 2i )
n p 2n i cosh
(7.8–10)
Under low–level injection, p = pn >> p0 , and n = nn = n0 >> p. Since ni 2 = n0 p0 , then equation (7.8–9) will reduce
to
R
Since pn – p0 =
n0
c nN t(p n p 0)n 0
p n 2n i cosh
(1
c nN t(p n p 0)
(2n i n 0) cosh )
p
p
(7.8–11)
p, then the recombination lifetime is of the (approximate) form
p
1 1
c nN t
2n i
n 0 cosh
(7.8–12)
Note the similarity of equation (7.8–10) to equation (7.8–1).
This analysis tells us that the recombination rate is dependent on the density Nt of the intermediate states, the density of carriers n0 + pn , and the location of the intermediate energy level Et within the energy gap EG . The recombination rate is a minimum for = 0, for which Et
Ei , and tells us that impurities such as Au or Cu are the most
effective for speeding up the recombination time.
103
Homework, Chapter 7: Semiconductor materials
Vers 2.2
(* Indicates advanced problems)
7–1. Determine the temperature at which there is a 10–6 probability that an energy state 0.55 eV above the Fermi
level will be occupied by an electron.
7–2. Consider the energy gap and band edges as shown for Si. (a) If EC – EF = 0.3 eV, determine the probability that
an energy state at E = EC is occupied by an electron, and the probability that a state at E = EV is empty.
(a)
EC
0.3 eV
EF
EG = 1.12 eV
EV
7–3. Repeat problem 7–2 for the case when the gap energy is 1.43 eV (as in the case of GaAs ).
*7–4. What is the maximum allowed temperature that assures that the intrinsic carrier concentration in Si will be
less than 1.5 1012#/cm3. Hint: Use your calculator to set up the equation for ni . You will have to iterate. Assume
that EG = 1.12 eV for Si.
n i(T
n i(T
7–5. Find the ratio of intrinsic carrier concentrations
330K)
for:
300K)
(a) Si
(b) GaAs.
7–6. If the density–of–states function in the conduction band is
g = K(E – EC ) for E
EC ,
derive an expression for the thermal equilibrium concentration of electrons in the conduction band, assuming that
Fermi–Dirac statistics can be replaced by the Boltzmann approximation.
*7–7. If the probability of occupation of states in the conduction band is f(E ) = 1.0 determine the upper limit E1 =
EC + E to which these states would fill, assuming that the lowest fill first, and n is limited to n = 2 n 3 = NC .
2 3
Answer:
E
E1
EC
3
4
kT
1.21kT
7–8. Determine the value of n0 and p0 for Si at T = 300K if EF – Ei = 0.22 eV
7–9. The value of p0 in Si at T = 300K is 1015 #/cm3. Determine: (a) n0 and (b) EC – EF.
7–10. A Si substrate doped with ND = 1015 #/cm3. Determine n0 , p0 , and (EC – EF ).
7–11. Assume that a semiconductor is doped with concentrations ND = 2 1015 #/cm3 and NA =1.5 1015 #/cm3.
Determine: (a) n0 and p0 at T = 300K (b) determine ni , n0 , and p0 at T = 500K.
7–12. A 10mm x 10mm x 10mm cube of Si at T = 300K has 1014 #/cm3 of Ga and 1.5 1014 #/cm3 of As, as trace
impurity concentrations, respectively. Determine the resistance between opposite faces of the cube. Assume n =
1330 cm2/Vs and p = 460 cm2/Vs
104
7–13. A strip of Si having cross–sectional area 2500 m 20 m and length 2 mm has 25 mA of current flowing
through it. If we desire a voltage drop of not more than 2 V, determine: (a) maximum allowed resistance (b) minimum allowed conductivity (c) Assuming light doping so that n = 1330 cm2/Vs, determine concentration of donor
atoms needed to achieve this goal. (d) What concentration of acceptor atoms must be added to counter–dope the
material such that the same conductivity will be achieved for a p–type material?
*7–14. In the year 2009 engineers have been tasked to design a series of probes that will assess some disturbing
activity in and around the Jovian red spot. They have discovered that, as well as surface temperatures on the order
of 450K, there are often boron and phosphorus storms. In order to monitor the extent and duration of these storms, a
young engineer has proposed that sensor strips made up of Si be plated on the hull of the probe, (an excellent idea,
since the hull is to be made of pure quartz (SiO2)) and that the impurity doping be assessed by monitoring the resistance of these strips. The resistance of a strip of thickness 25 m doped at ND = 1014 #/cm3 is observed to decrease
from 12.5k to 10k . Assuming that diffused impurities have affected the top 5 m of the strip, identify the possible type(s) and density(s) that have caused of this change. Assume that mobilities of Si at this temperature, for
low levels of doping, are:
n
= 551 cm2/Vs
p
= 207 cm2/Vs
Figure P7–14: Jovian surface probe
7–15. (a) Determine the electron mobility at T = 300K for a Si substrate when it is doped at ND = 5 1016 #/cm3.
(b) If it is counter–doped with an acceptor of concentration NA = 1017#/cm3, what is the hole mobility? (c) What is
the conductivity of the counter–doped material?
7–16. Engineers at the Thames River power complex have decided to monitor the #4 reactor with a network of Si
thermistor sensor strips doped at NA = 1014 #/cm3 plated on each fuel rod. They will be used to control and inhibit
temperature fluctuations within the core. At T = 500K, which is the best operating temperature for a non–pressurized reactor, sensor conductance is G = 0.2 –1. However, there is some concern, that over a period of time, phosphorus by–products from the reaction may contaminate the sensors and give false readings that indicate lower temperatures than truly exist. Assuming that the Si does become contaminated with phosphorus at 1 part in 107,
determine the conductance at T = 500 K. Measurements of lightly–doped Si have shown that at T = 500 Si has n =
443 cm2/Vs and p = 171 cm2/Vs.
7–17. (a) Determine the electron mobility at T = 300K in Si for doping concentration NA = 5 1016 #/cm3. (b) If it
is counter–doped with a donor material of concentration ND = 6 1016 #/cm3, what is the electron mobility? (c)
what is the hole mobility for case (b)? and (d) what is the conductivity?
105
7–18. An 10 m thick stripe of Si doped at NA = 1015#/cm3 is implanted with NA = 1017#/cm3 to form a more
heavily–doped layer of thickness 1 m, as shown by figure P7–18. If width and length of the stripe are 15 mm and 5
cm, respectively, determine (a) the resistance in the long direction of the stripe before implant (b) after implant (c)
after a long annealing process in which the heavily doped layer diffuses throughout the Si to form a uniformly
doped stripe. Assume T = 300K. As part of your assessment, calculate the mobilities for the various doping densities.
implant NA = 1017 #/cm3
thickness = 1 m
thickness = 10 m
NA = 1015 #/cm3
Figure P7–18: silicon strip
7–19. Determine the average drift velocity of holes in a strip of Si having cross–sectional area 500 m 200 m
that contains a hole concentration of 4 1015 #/cm3 and carrying a current of 20 mA.
*7–20. In the year 2004, research in medical electronics determined that nerve impulses may be enhanced by several orders of magnitude if the nerve filament is accompanied in parallel by a 40 m dia. strand of gold–doped
silicon. Research being conducted on a $6 million mouse (codename: Mighty Mouse) discovered that the mouse
developed a strange weakness as result of exposure to Krypton gas. It was since determined that the Krypton acted
as a catalyst that causes alloying of trace elements of phosphorus (=donor) with the silicon filaments, thus creating
an implanted cylindrical outer layer, of thickness approximately 5 m and doping concentration ND =
2 1015#/cm3. If the filament was originally doped with Boron (= acceptor), of concentration 1015#/cm3, determine the resistance per length of the fibers: (a) before the phosphorous infusion layer is formed, and (b) after phosphorus infusion layer is formed. Assume T = 300 K and evaluate n and p thereto as part of your analysis.
implant ND = 2 1015 #/cm3
layer thickness = 5 m
Figure 7–20: Silicon filament
7–21. In a p–type GaAs semiconductor, the conductivity is measured to be = 5 ( –cm)–1. Determine the equilibrium values of electron and hole concentrations, assuming that T = 300K.
7–22. Determine the intrinsic conductivity
i
of (a) Si and (b) GaAs, assuming T = 300K.
*7–23. A semiconductor material has electron and hole mobilities
mum value of conductivity is:
n
and
p,
respectively. Show (a) that the mini-
2
min
i
n
n
p
p
where
i is the
intrinsic conductivity. (b) Show that the corresponding hole concentration is:
106
p0
ni
n
p
.
7–24. The electron concentration in a modulation–doped Si substrate is formed such that
n(x) 10 16e x a
where a = 15 m, and where 0 < x < 25 m. Assume that Dn = 25 cm2/s. The total electron current density through
the semiconductor is Jn = 10 A/cm2. The electron current has both diffusion and drift components. Determine the
electric field as a function of x that exists in the semiconductor.
7–25. A constant E–field of magnitude Ex = 12 V/cm exists in a GaAs material for 0 < x < 50 m. The total current
density Jn = 100A/cm2. At x = 0 the drift and diffusion currents are equal. Assume that T = 300K and that n =
8000 cm2/Vs. Determine: (a) the expression for the electron concentration n(x), (b) the electron concentration at x
=0
(c) electron concentration at x = 50 m. (d) drift and diffusion components of Jn at x = 50 m.
7–26. A sample of Si is doped with 1016#/cm3 of Boron. A strip of dimensions as shown and oriented so that it is
subject to field Bz = .04T. Assume that the Hall factor rH = 1.10. If current Ix = 1 mA, determine (a) Ey and (b) VH .
*7–27. A Si sample of dimensions as shown is evaluated using the transverse Hall effect. If Bz = 0.1T, Ix = 25 mA,
Vx = 16V and VH = 30mV determine (a) carrier type, (b) majority carrier concentration, (c) Hall mobility, (d) Hall
factor. Note that an iteration is necessary since rH is an unknown.
d = 50 m,
w = 500 m,
L = 0.5 cm,
7–28. A GaAs sample at T = 300K is evaluated by a Hall effect test strip of dimensions d = 100 m, w = 0.5mm, L =
0.5cm. For Ix = 2.5mA, Vx = 2.2V, and Bz = 25 mT the Hall voltage is –4.5V. Find (a) type carrier (b) carrier
concentration (c) Hall mobility (d) resistivity.
7–29. In a p–type substrate material with NA = 5 1016 #/cm3, excess carriers (electrons) are injected into the
surface such that np (0) = 1014 #/cm3. (a) Determine n , and Dn . Assuming that ! n = ! p = 0.1 s determine (b) Ln and
(c) the steady–state excess electron concentration n(x) at: (c) x = 5 m and at x = 20 m. (d) Determine p(x) at x =
5 m and x = 20 m.
107
7–30. Power transistors are constructed with a relatively thick, lightly–doped layer anode capable of withstanding
reasonably high levels of voltage in reverse bias. The drawback is that, in forward bias, these layers have a significant series resistance to current flow. If we have a Si power transistor as shown has layer doping ND = 5 1014
#cm3, determine (a) the current that the transistor will conduct in forward saturation, when 1.6 V falls across this
layer, and (b) the power that will be dissipated in heat when the transistor is in this conducting state.
cm
20 m
Figure P7–30: Power transistor
*7–31. The typical cross–section and surface pattern of a power MOSFET ( HEXFET) is shown, with current
flow lines as indicated by the arrows. Assuming that the bulk layer of the FET is made of lightly–doped Si at ND =
1014 #/cm3, (a) determine the voltage drop that will fall across it when conducting 12 kA, assuming that the device
pattern shown is replicated many thousands of times to cover a circular wafer of diameter 10 cm with this pattern.
(b) Determine the power dissipated in the lower 40 m of the conducting layer. (c) Determine the power/area dissipated in the upper 4 m of the conducting layer. Hint: think triangles.
4 m
40 m
linear model of current flow
distance between hexagon pattern centers = 20 m
20 m
5 m
Figure P7–31: Power MOSFET (HEXFET) device.
108