Carrier Action: Motion, Recombination
and Generation.
What happens after we figure out
how many electrons and holes are
in the semiconductor?
1
Carrier Motion I
Described
by 2 concepts:
• Conductivity: s
(or resistivity: r = 1/s)
• Mobility: m
Zero
Field movement:
Random – over all
Thermal – Energy
Distribution.
Motion
e-
Electrons
are
scattered by
impurities,
defects etc.
What happens when you apply a force?
2
Carrier Motion II


Apply a force: F = qE = qEx xˆ
Electrons
accelerate:
-n0qEx=dpx/dt {from F=ma=d(mv)/dt}
Electrons decelerate too.
• Approximated as a viscous damping force
(much like wind on your hand when driving)
dpx = -px dt/t {dt = time since last “randomizing
collision” and t = mean free time between
randomizing collisions.}
Net result: deceration = dpx/dt = -px/t
3
Carrier Motion III
Acceleration=Deceleration
in steady state.
• dpx/dt(accel) + dpx/dt(decel) = 0
• -n0qEx - px/t = 0.
Algebra:
• px/n0 = -qtEx = <px>
• But
qt

v
=

E   mn Ex
• <px> = mn*<vx> Therefore: x
* x
mn =
qt
mn*
, mp =
qt
mn
m*p
4
Currents
“Current
density” (J) is just the amount of
charge passing through a unit area per unit
time.
Jx = (-q)(n0)<vx> in C/(s m2) or A/m2
= +(qn0mn)Ex for e-’s acting alone.
= sn Ex (defining e- conductivity)
If both electrons and holes are present:



J = q(no mn  po m p ) E = sE =

E
r
5
Current, Resistance
How
do we find:
• current (I)? We integrate J.
I=
 J dydz
y = 0 to w
z = 0 to t
x
• resistance (R)?
L
r ( x ) dx
R =  w( x ) t ( x ) =
0
t
rL
wt
= s1 wtL
L
V
w
• Provided r, w, t are all
constants along the x-axis.
6
Mobility changes …
Although
it is far too simplistic we use:
t is the “mean free time.”
mn = qt/mn*
mn* is the “effective mass.”
(depends on material)
 t depends upon:
• # of scatter centers (impurities, defects etc.)
More doping => lower mobility (see Fig. in books)
More defects (worse crystal) => smaller mobility too.
• The lattice temperature (vibrations)
Increased temp => more lattice movement => more
scattering =>
m
smaller t and smaller m.
Increasing
Doping 7
Mobility Changes II
Mobility
is also a function of the electric field
strength (Ex) when Ex becomes large. (This leads
to an effect called “velocity saturation.”)
<vx>
Here m is constant (low fields). Note constant m => linear plot.
Vsat
107 cm/s
At ~107 cm/s, the carrier KE becomes
the same order of magnitude as kBT.
Therefore: added energy tends to warm
up the lattice rather than speed up the
carrier from here on out. The velocity
becomes constant, it “saturates.”
Ex (V/cm)
106 cm/s
105 cm/s
102
103
104
105
106
8
What does Ex do to our Energy Band Diagram?
Drift
currents depend upon the electric
field. What does an electric field do to our
energy band diagrams?
It “bends” them or causes slope in EC, EV
and Ei. We can show this. E
• Note:
Eelectron = Total E

= PE + KE
How much is PE vs. KE???
electron
e-
Eg
EC
EV
h+
9
Energy Band Diagrams in electric fields
EC
is the lower edge for potential energy (the
energy required to break an electron out of a
bonding state.)
Everything above EC is KE then.
PE always has to have a
Eelectron
reference! We’ll choose
eKE
one arbitrarily for the
EC = PE
Eg
moment. (EREF = Constant)
EV = PE
PE
Then PE = EC-EREF
KE
h+
We also know: PE=-qV
E
REF
10
Energy Band Diagrams in electric fields II
Electric
fields and voltages are related by:
E = -V (or in 1-D E=-dV/dx)
• So: PE = EC-EREF = -qV or V = -(EC-EREF)/q
• Ex = -dV/dx = -d/dx{-(EC-EREF)/q} or
Ex = +(1/q) dEC/dx
 1
dEC
1
E = q EC or E x = q dx
 1
dEV
1
E = q EV or E x = q dx
 1
dEi
1
E = q Ei or E x = q dx
11
Energy Band Diagrams in electric fields III
The
Electric Field always points into the
rise in the Conduction Band, EC.
Eelectron
Ex
EC
Ei
Eg
EV
EREF
What
about the Fermi level? What
happens to it due to the Electric Field?
12
Another Fermi-Level Definition
The
Fermi level is a measure of the
average energy or “electro-chemical
potential energy” of the particles in the
semiconductor. THEREFORE:
The FERMI ENERGY has to be a constant
value at equilibrium. It can not have any
slope (gradients) or discontinuities at all.
The Fermi level is our real-life EREF!
13
Let’s examine this constant EF
+ V If current flows => it is
not equilibrium and EF must be
Semiconductor
changing.
Ex
 In this picture, we have no
Eelectron
connections. Therefore I=0 and
Ex
EC
it is still equilibrium!
 Brings us to a good question:
E
 Note:
i
• If electrons and holes are moved
by Ex, how can there be NO
CURRENT here??? Won’t Ex
move the electrons => current?
answer lies in the concept
of “Diffusion”. Next…
EF
EV
 The
Looks Looks
N-type P-type
14
Diffusion I
Examples:
• Perfume,
• Heater in the corner (neglecting convection),
• blue dye in the toilet bowl.
What
causes the motion of these particles?
• Random thermal motion coupled with a density
gradient. ( Slope in concentration.)
15
Green dye in a fishbowl …
 If
you placed green dye in a fishbowl, right in the center,
then let it diffuse, you would see it spread out in time until
it was evenly spread throughout the whole bowl. This
can be modeled using the simple-minded motion
described in the figure below. L-bar is the “mean
(average) free path between collisions” and t the mean
free time. Each time a particle collides, it’s new direction
is randomly determined. Consequently, half continue
going forward and half go backwards.
32
Dye Concentration
16
16
8
8
4
4
8
8
8
8
4
4
x
l
-3
-2
-1
0
1
2
3
16
Diffusion II
Over
a large scale, this would look more
like:
t=0
t1
Let’s look more in depth
at this section of the
curve.
t2
t3
tequilibrium
17
Diffusion III
 What
kind of a particle movement does Random Thermal
motion (and a concentration gradient) cause?
n(x)
nb0
nb1
nb2
Bin
(0)
Bin
(1)
Bin
(2)
Line with slope:
nb 2  nb1
l
Half of
e- go left
half go
right.
x0  l
x0
It causes net motion from
large concentration regions
to small concentration
regions.
x0  l
 dn
dx
x0
x-axis
18
Diffusion IV
 Net
•
•
•
•
number of electrons crossing x0 is:
Number going right: 0.5*nb1*ℓ*A
Minus Number going left: 0.5*nb2*ℓ*A
Net is = 0.5*ℓ*A*(nb1-nb2)
(note ℓ*A=volume of a bin.)
 Flux
= # of particles crossing a plane per unit time and
unit area. Symbol is: f
Or
f = 0.5*ℓ*A*(nb1-nb2)
t*A
f = 0.5*ℓ (nb1-nb2)
t
(t = mean free time.)
19
Diffusion V
 Using
the fact that slope (dn/dx) = -(nb1-nb2)/ℓ gives:
f = - 0.5*ℓ2 dn
t
dx
or f = -Dn*dn/dx (electrons)
or f = -Dp*dp/dx (holes)
 Now:
When charges move we get current. Consequently,
the current density is directly related to the particle flux.
The equations are:
• (electrons)
(holes)
J p =  qf p
J n = qfn
=
 qDn dn
dx
(1  D)
=  qDnn (3  D)
=
dp
qD p dx
(1  D)
= qD pp (3  D)
20
Diffusion VI
Let’s
n(x)
look at an example:
dn/dx = 0 here
x
J(x)
x
The electrons are diffusing
out of the center and
toward the edges.
21
Currents round-up
So
now we know that our total currents
have 2 components:
• DRIFT – due to any electric field we apply
• DIFFUSION – due to any (dp/dx, dn/dx) we
apply and thermal motion.


J n = qnmn E  qDnn
Drift
Diffusion


J p = qpm p E  qD pp
J total = J p  J n
22
Answering that old question
can we have an electric
Field and still have no current?
(Still have J = 0?)
+ V -
 How
Diffusion
must
balance Drift!
Semiconductor
Ex
Eelectron
Ex
Ei
EF
EV
Example:


J p = qpm p E  qD pp = 0
or

E=
Dp
1 dp
m p p dx
EC
Looks Looks
N-type P-type
23
Einstein Relationship
We
next remember: p=niexp((Ei-EF)/kBT)
Plugging this into our equation for the electric
field and noting that dEF/dx = 0 … we get
 The Einstein Relationships.
Dp
mp
=
k BT
q
and
Dn
mn
=
k BT
q
These
are very useful. You will never find a table
for both Dp and mp as a result of these. Once you
have m, you have D too, by this relationship.
24
A sanity check
Pretend
we have:
What
will be the fluxes
and currents?
x
Ex
Holes
Mechanism
Electrons
Diffusion
Flux (f)
Current Density (J)
n(x)
Drift
Flux (f)
p(x)
Current Density (J)
25
Recombination – Generation I
(G): How eand h+ are produced or
created.
Recombination (R):
How e- and h+ are
destroyed or removed
Generation
The
concepts are
visually seen in the
energy band diagram
below.
Ee
G
R
EC
At
equilibrium: r = g and
since the generation
rate is set by the
temperature, we write it
as: r = gthermal
hv
hv
EV
x
26
Recombination – Generation II
Recombination
must
depend upon
• the # of electrons: no
• the # of holes: po
(If no e- or h+, nothing can
recombine!)
From
the chemical reaction
• e- + h+ → Nothing
we can know that
• r = αrnopo = αrni2 = gthermal
The recombination
“rate coefficient”
When
the
temperature is
raised
• gthermal increases
Therefore
• ni must increase
too!
27
Recombination – Generation III
A variety
of recombination mechanisms exist:
Ee Direct, Band to Band
R
G
Auger
EC
hv
Ee
G
hv
R
EC
EV
x
Ee
EV
Indirect via R-G centers
R
G
x
EC
R-G Center Energy Level
EV
x
28
GaAs is a
Direct
Band Gap
Semiconductor
Eg – The
Band Gap
Energy
Direct
recombination
of electrons
with holes
occurs. The
electrons fall
from the bottom
of the CB to the
VB by giving
off a photon!
GaAs band structure produced by J. R. Chelikowsky and M. L. Cohen, Phys. Rev. B 14, 556 (1976)
using an empirical Pseudo-potential method see also: Cohen and Bergstrasser, Phys. Rev. 141, 789 (1966).
29
GaAs band structure produced by W. R. Frensley, Professor of EE @ UTD
using an empirical Pseudo-potential method see also: Cohen and Bergstrasser, Phys. Rev. 141, 789 (1966).
30
Si is an
Indirect
Band Gap
Semiconductor
Eg – The
Band Gap
Energy
Only indirect
recombination
of electrons
with holes
occurs. The
electrons fall
from the bottom
of the CB into
an R-G center
and from the
R-G center to the
VB. No photon!
Silicon band structure produced by J. R. Chelikowsky and M. L. Cohen, Phys. Rev. B 14, 556 (1976)
using an empirical Pseudo-potential method see also: Cohen and Bergstrasser, Phys. Rev. 141, 789 (1966).
31
Silicon band structure produced by W. R. Frensley, Professor of EE @ UTD
using an empirical Pseudo-potential method see also: Cohen and Bergstrasser, Phys. Rev. 141, 789 (1966).
32