EE 362 Electric and Magnetic Properties of Materials

1
EE 362 Electric and Magnetic Properties of
Materials

• Dr. Brian T. Hemmelman
• Chapter 6 Slides

2
Setting The Stage (2)

• We spent quite a bit of time talking in Chapter 2
  about the photoelectric effect and absorbing and
  emitting photons.
• In Chapter 4 we looked at changing the number of
  free charge carriers through the use of dopants.
• In Chapter 6 we will look at the creation and
  behavior of excess carriers in semiconductors.
• One of the ways to create excess carriers in
  semiconductors is through photon absorption.

3
Beyond Semiconductors (3)

• Photon absorption isnt limited to just
  semiconductors however.
• Jaundice of premature babies is caused by the
  retention of an excessive amount of pigment
  called bilirubin.
• Bilirubin derives from the breakdown of the
  oxygen-carrying chemical hemoglobin and is toxic
  to nerve cells.
• If it is present in excessive amounts for too
  long it can cause brain damage.

4
Beyond Semiconductors (4)

• Toxic bilirubin is soluble in fat but not water,
  therefore it cannot be excreted from the liver,
  which metabolizes only water-soluble chemicals.
• Bilirubin is ordinarily made water-soluble by the
  babys enzymes, but a premature baby may not have
  enough of the right enzyme in the liver to
  convert the bilirubin to a water-soluble form.
• When the baby is irradiated with blue light, the
  bilirubin is changed into compounds that are
  soluble in water and can be readily excreted by
  the liver.

5
Beyond Semiconductors (5)

• Photosensitive drugs have been investigated as an
  opportunity to enhance the effectiveness of
  ultraviolet light on cancer cells.
• Cancer cells are more likely to absorb these
  photosensitive materials (extremely high
  metabolism and reproduction or cell division
  rates), and ultraviolet light is known to be able
  to kill cells containing the photosensitive
  chemicals.
• If you can selectively get the photosensitive
  chemical into cancer cells and then irradiate
  with UV light you can selectively kill the cancer.

6
Porphyrins (6)

• Porphyrins are flat molecules with a hole in the
  middle (a molecular doughnut).
• The center of the molecule is a hole surrounded
  by nitrogen atoms.
• The center of the molecule, the hole, is 2 Å in
  radius. Several different metal atoms will fit
  in the hole.

7
Porphyrins (7)

• Magnesium in the middle turns it green.

8
Porphyrins (8)

• Magnesium in the middle turns it green.
• This is the building block of chlorophyll.

9
Porphyrins (9)

• Iron in the middle turns it red.

10
Porphyrins (10)

• Iron in the middle turns it red.
• This is the building block of hemoglobin.

11
Generation Recombination (11)

• Generation is the process whereby electrons and
  holes are created.
• Recombination is the process whereby electrons
  and holes are annihilated.
• In thermal equilibrium we have electrons breaking
  out of covalent bonds due to the acquisition of
  enough thermal energy to hop from the valence to
  the conduction band. This creates both a free
  electron and a free hole (generation).
• By the sake token there are free electrons that
  lose some of their thermal energy when they
  encounter a hole and fall back into covalent bond
  (recombination).

12
Generation Recombination (12)
Given that thermally generated free electrons and
holes must come in pairs
and they will also recombine in pairs so
In thermal equilibrium the total number of free
electrons and holes is constant so the rates at
which they are being generated must be equal to
the rates at which they are recombining.
13
Excess Carriers (13)
External events, such as incident photons, can
disrupt this equilibrium though and create
additional electron-hole pairs. These excess
charge carriers would be generated at equal rates
for electrons and holes, so
The number of actual excess electrons and holes
though are ?n and ?p. Thus the total number of
free electrons and holes in the semiconductor can
now be written as
The excess electrons and holes would also
recombine in pairs so we can write
14
Linking Recombination Rates To Quantities of
Excess Charge (14)
For the simple model of recombination we are
using (direct band-to-band recombination) the
probability of an electron-hole pair recombining
is constant with time. Moreover, the rate at
which electrons recombine must be proportional to
both the electron concentration and hole
concentration. We can describe this
mathematically with
where
The term is the thermal equilibrium generation
rate. Note then that the entire expression in the
parentheses will be less than (or equal to) zero
so that the derivative is negative. This should
make common sense as the value of n(t) is
decreasing due to recombination.
15
Linking Recombination Rates To Quantities of
Excess Charge (15)
Since n0 is constant with respect to time the
derivative can be taken with respect to ?n
instead of n. Also, since excess electrons and
holes are generated and recombine in pairs we
know that ?n(t) ?p(t). Making these
substitutions and expanding the terms out we find
16
Low-Level Injection (16)
The differential equation we have derived up to
this point isnt the easiest to solve at the
moment. However, if we restrict ourselves to the
case of low-level injection (a common
situation) it becomes much simpler. Low-level
injection simply means that the number of excess
carriers is much smaller than the thermal
equilibrium values of the majority carrier
concentration. That is (for p-type material),
?n(t) p0. For p-type material we also know
that n0 p0. Therefore, looking at our equation
we can see that the ?n(t)p0 term will dominate
the other two terms on the right-hand side of the
equation.
17
Low-Level Injection (17)
We can thus approximate this equation as
This is a simple first-order differential
equation with a solution of
where ?n0, the excess minority carrier lifetime,
is given by ?n0 (?rp0)-1.Note that the excess
minority carrier lifetime depends on the majority
carrier concentration.
The excess carrier recombination rate, Rn, is
the change in the number of excess carriers,
?n(t), so we can write
18
Low-Level Injection (18)
For direct band-to-band recombination, the excess
majority carrier holes recombine at the same rate
(if an electron has recombined, it obviously must
have recombined with a hole therefore subtract
BOTH one free electron and one free hole). Since
the two rates are equal we can write, for p-type
material,
A similar derivation can be done for low-level
injection in n-type material to yield
19
Continuity Equations (19)
Let us consider the flux of particles into and
out of a small box. Assume the flow of particles
only occurs in 1-D along the x-axis and the box
is a differential volume with dimension dx by dy
by dz.
Assume our particles are holes. Then using a
first-order Taylor expansion we can relate the
flux of particles into the box to the flux of
particles out of the box as
20
Continuity Equations (20)
The net increase in particles within the volume
would be the difference of the two fluxes
(multiplied by the surface area)
But the net increase in particles inside the box
could also be written
So
This only represents the buildup or decrease of
particles in the box due to different flow rates
however.
21
Continuity Equations (21)
We also need to account for the effects of
generation and recombination that may be
occurring within the box. These two phenomena
would also contribute to an increase or decrease
of the particle concentration within the box.
Including these terms we have
p is the density of holes and ?pt is the combined
hole lifetime (it includes both the thermal
equilibrium carrier lifetime and the excess
carrier lifetime). Dividing through by dxdydz
we reach the Continuity Equations.
22
Time-Dependent Diffusion Equations (22)
Our current densities are given by
By dividing the current density by the unit of
charge we obtain particle flux
23
Time-Dependent Diffusion Equations (23)
By substituting these expressions into the
Continuity Equations we get
Both p (or n) and E can be functions of position
so we need to use the chain rule
24
Time-Dependent Diffusion Equations (24)
Thus
If we assume we have a homogeneous semiconductor
(the doping concentration of electrons and holes
is uniform throughout the semiconductor), then
and our partial derivatives of p(x) and n(x) just
become partial derivatives of ?p(x) and ?n(x).
25
Time-Dependent Diffusion Equations (25)
Thus we get the time-dependent diffusion
equations for electrons and holes in a
homogeneous semiconductor
26
Ambipolar Transport Equations (26)
Since electrons and holes are generated in pairs
Similarly,
And since they are generated in pairs (assuming
charge neutrality), ?n ?p. Thus we can write
the time-dependent diffusion equations as
Multiplying the first equation by ?nn and the
second equation by ?pp we get
27
Ambipolar Transport Equations (27)
Adding these we eliminate the term
We can simplify/rewrite this by dividing through
by (?nn?pp).
28
Ambipolar Transport Equations (28)
The Ambipolar Transport Equation
However, since
29
Ambipolar Transport Equations (29)
For strongly p-type (or n-type) material under
low level injection this reduces considerably.
For p-type, low-level injection
For n-type, low-level injection
30
Ambipolar Transport Equations (30)
Also for low-level injection (small of excess
carriers) the probability of recombining will be
almost constant (the chance of hitting a majority
carrier wont change much), so ?nt ?n and ?pt
?p. For generation and recombination we have a
combination of thermal-equilibrium
generation/recombination and excess carrier
generation/recombination. Looking at just
electrons we have
However, for thermal equilibrium we know Gn0
Rn0 so
and we would have a similar expression for the
holes.
31
Ambipolar Transport Equations (31)
Combining all this gives our final Ambipolar
Transport Equations

• These equations simply say that there can be a
  change in our excess carrier concentation over
  time because
• They diffuse away.
• They drift away.
• More are generated.
• They recombine.

32
Ambipolar Transport Equations (32)
It is common in real situations that we can
simply these equations because of the specific
boundary conditions that apply.
Steady-state ? Uniform distribution of excess
carriers ? (uniform generation rate) Zero
electric field ? No excess carrier generation
? No excess carrier recombination ?
33
Example 6.1(33)
Question Assume we have been shining a light on
a thin semiconductor so that we have been
generating a uniform concentration of excess
carriers in a homogeneous n-type semiconductor.
At time t 0 we turn the light off and let the
semiconductor return to thermal equilibrium. (No
external electric field.) Calculate the excess
carriers as a function of time. Answer Since
this is n-type material we need to consider the
equation for the minority holes.
We have a uniform concentration of excess holes so
After t 0 we also have g 0 so all we have
left is
34
Example 6.1(34)
35
Example 6.2 (35)
We have an infinite, homogeneous n-type
semiconductor with zero applied bias. For t ? 0
we have a uniform generation rate. We start, of
course, with
As there is no concentration gradient (uniform
generation), we know
thus we are left with
Solution
36
Example 6.3 (36)
Here we have a p-type semiconductor that is
homogeneous and infinite with zero applied bias.
We will consider this to be a 1-D crystal.
Excess carriers are generated at x 0 only.
Calculate the steady-state concentrations of ?n
as a function of x.
For steady state there is no time-dependence
No E-field
and g 0 for all x ? 0.
So for -? lt x lt 0 and 0 lt x lt ? we have
37
Example 6.3 (37)
The characteristic equation is
Let where Ln ? Diffusion Length
Thus, for x gt 0, B must be 0, and, for x lt 0, A
must be 0.
38
Problem 6.18 (38)
p-type, NA 31015 cm-3 with no applied electric
field. At x 0, ?p(0) ?n(0) 1013 cm-3. ?n
1200 cm2/Vs, ?p 400 cm2/Vs, ?n0 510-7 sec,
?p0 10-7 sec.
Calculate the steady-state excess electron and
hole concentrations as a function of x.
For x ? 0 we again have
39
Problem 6.18 (39)
We just solved this as
For x gt 0 we must have B 0, and thus
Now apply our boundary conditions ?n(0) Ae0
1013 cm-3 ? A 1013 cm-3.
where
So
40
Problem 6.18 (40)
Now find the electron diffusion current density
as a function of x.
41
Quasi-Fermi Levels
The Fermi Level is constant in a sample under
equilibrium. However, with generation or current
flowing (even in steady-state) we are in
nonequilibrium so EF is no longer a
constant. However, we still want to know (or
relate) total electron and hole concentrations.
This can be done by considering a shift in EF
that represent the More or Less p or n nature
of the semiconductor. These new levels are
called EFn and EFp, the Quasi-Fermi Levels.