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We have
encountered the Einstein relation
before. It is of such fundamental importance that we give two derivations: one in this paragraph,
another one in an advanced
module. |
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First,
we consider the internal current (density) in an homogeneous material with a
gradient of the carrier concentration
ne or nh. |
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Ficks first law than tells us that
the particle current j
pdiff is given by |
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If the particles are carrying a charge
q, the particle current is
also an electrical
diffusion current given by
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| jdiff |
= |
q · j pdiff |
= |
q · De,h · Ñne,h |
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Considering only the one-dimensional case for
electrons (i.e. q = e; holes behave in exactly the same way
with q = +e), we have |
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| jdiff (x) |
= |
e · De · |
dne(x)
dx |
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Since there can be no net current in a piece of material just lying around
(which nevertheless might still have a concentration gradient in the carrier
density, e.g. due to a gradient in the doping concentration), the
diffusion-driven movement of the carriers generates an electrical field that
always will drive the carriers back. |
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Any field E(x) (written
in mauve to avoid
confusion with energies) now will cause a (so far one-dimensional)
current given by |
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| jfield |
= |
s · E(x) = e ·
n(x) · µ · E(x) |
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With s =
conductivity, µ = mobility. |
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The total (one-dimensional) current in full generality
(even for fields not exclusively caused by
the diffusion current) is then |
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| jtotal(x) |
= |
e · ne(x) · µ ·
E(x) e ·
De · |
dne(x)
dx |
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We will need this equation later. |
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For our case of
no net current and only fields caused by the diffusion current, both
currents have to be equal in magnitude: |
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| e · ne(x) · µ ·
E(x) |
= |
e · De · |
dne(x)
dx |
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This is an equation that comes up repeatedly, we
had it, e.g., at the simple derivation of the Debye length. |
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Now we are stuck.
We need some additional equation in order to find a
correlation between D and µ. |
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This equation is the Boltzmann distribution (as an approximation to the
Fermi distribution) because we have
equilibrium in our material. |
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If we denote the electrostatic potential energy correlated to the electrical field
by V(x), we have |
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| E(x) |
= |
dV(x)
dx |
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n(x) |
= |
n0 · exp |
e · V(x)
kT |
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We now drop the index
"e" and continue in full generality. |
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Differentiation of the Boltzmann
distribution gives us |
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dn
dx |
= |
n0 · (e/kT) · |
dV(x)
dx |
· exp |
eV(x)
kT |
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dn
dx |
= |
n(x) · (e/kT) · |
dV(x)
dx |
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Using this equation, the current balance from
above becomes |
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| µ · n(x) · |
dV(x)
dx |
= |
D · (e/kT) · n(x) · |
dV(x)
dx |
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| D |
= |
µkT/e |
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In words: Equilibrium between diffusion currents
and electrical currents for charged particles demands a simple, but far
reaching relation between the diffusion constant D and the
mobility µ. |
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Distinguishing again between
electrons and holes gives as the final result the famous Einstein-Smoluchowski
relations. |
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| De |
= |
µe · kT
e |
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| Dh |
= |
µh · kT
e |
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You may want to have a look at a
different derivation in an
advanced module. |
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In the
consideration above we postulated that there is no net
current flow; in other words, we postulated total equilibrium. Now lets consider that there
is some net
current flow and see what we have to change to arrive at the
relevant equations. |
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In order to be close to applications,
we treat the extrinsic case and, since we
do not assume equilibrium per se, we automatically do not assume that the
carrier concentrations have their equilibrium values ne(equ)
and nh(equ), but arbitrary values that we can express as some Delta
to the equilibrium value. We then start with |
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Since carriers above the equilibrium
concentration are often created in pairs we
have for this special, but rather common
case |
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| Dne |
= |
Dnh |
= |
Dn |
| |
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| Dne |
= |
ne
ne(equ) |
= |
nh
nh(equ) |
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This is a
crucial assumption! |
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This allows us to concentrate on
one kind of carrier, lets say we look at
n-type Si with electrons as the majority carriers. We now
focus on holes as the minority carriers
since we always can compute the electron density ne by
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| ne |
= |
ne(equ) + Dne |
= |
ne(equ) + nh
nh(equ) |
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We now must consider
Ficks second
law or the
continuity equation
(it is the same thing for special cases, but the continuity equation is more
general). |
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For the total (mobile) charge density
r (which is the difference of the electron and hole density
(r = ne
nh) in contrast to the
particle density, which is the sum!) we have |
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With jtotal = j
e + j h = sum of the electron and hole
current. |
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In the simplest form we have for the
holes |
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¶n
¶t |
= |
(1/e) · div (j h) |
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The factor 1/e is needed to
convert a particle current jpart to an electrical
current j via j = e ·
jpart. As always, we have to pick the right sign for the
elementary charge e (negative for electrons, positive for holes). |
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This is simply the statement that the charge is conserved. It would be sufficient that no
holes disappear or are created in any differential volume dV
considered, i.e. div j h = 0, to satisfy that
condition. |
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But this is, of course, a condition
that we know not to be true. |
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In all semiconductors, we have constant
generation and recombination of holes (and electrons) as
discussed before. In in equilibrium, of course, the generation rate
G and the recombination rate R are equal, so they
cancel each other in a balance equation and need not be considered - div
j h = 0 is correct on
average. |
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We are, however, considering non-equilibrium, so we must primarily consider the
recombination of the surplus minority
carriers given by |
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Why? Because, as
stated before, the generation
essentially does not change, so it still
balances against the recombination rate of the equilibrium concentration, and
only the recombination rate of the surplus minorities, RD = [nh
nh(equ)]/t needs to be
considered (t is the minority carrier life
time). |
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RD = [nh
nh(equ)]/t is the rate with
which carriers disappear by recombination, we thus must subtract it from the carrier balance as expressed in
the continuity equation, and obtain |
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dn
dt |
= |
nh nh(equ)
t |
(1/e) · div (j h) |
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The current j can
always be expressed as the sum of a field
current and a diffusion current
as we did above by |
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| j htotal(x) |
= |
e · n(x) · µ · Ex(x) e
· Dh · |
dnh(x)
dx |
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Inserting this
equation in our continuity equation yields |
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¶nh(x)
¶t |
= |
nh(x) nh(equ)
t |
nh(x) · µ ·
|
¶E(x)
¶x |
E(x) ·
µ · |
¶nh(x)
¶x |
+ D · |
¶2nh(x)
¶x2 |
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This is an important, if not so simple equation. It is not
so simple, because the electrical field strength E(x) at x is a function
of the carrier density ¶nh(x) at
x, which is what we want to calculate! We have used the symbols
for partial derivatives ("¶")
to emphasize that it is in reality a three-dimensional equation. |
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We will now look at some applications
of this equation. |
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Consider the minority carrier situation just outside of the
space charge region of a biased pn-junction. |
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If it is forwardly biased, a lot of majority
carriers are flowing to the respective other side where they become minority
carriers. They will eventually disappear by recombination, but the minority
carrier density right at the edge of the space charge region will be larger
than in equilibrium and will decrease as we go away from the junction. |
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This is now shown in the illustration
used before in the simple model of
the pn-junction, but the realistic
minority carrier situation is now included. |
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The region outside the space charge region,
while now showing a concentration gradient of the
minority carrier concentration, is essentially field free or at
least has only a small electrical field strength. |
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If we let Ex = 0 and consequently ¶Ex(x)/¶x = 0, too, the current equation from
above reduces to |
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¶nh
¶t |
= |
nh nh(equ)
t |
+ D · |
¶2nh(x)
¶x2 |
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Since ¶nh/¶t = ¶[nh(equ) + Dnh]/¶t = ¶Dnh/¶t, and correspondingly ¶2nh(x)/¶x2 = ¶2Dnh(x)/¶x2, we have |
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¶Dnh
¶t |
= |
Dnh
t |
+ D · |
¶2Dnh(x)
¶x2 |
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If we consider steady state, we have ¶Dnh/¶t =
0, and the solution of the differential equation is now mathematically
easy. |
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But how can steady state be achieved
in practice? How can we provide for a constant, non-changing concentration of minority carriers
above equilibrium? |
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For example by having a defined source of
(surplus) holes at x = 0. In the illustration this is the
(constant) hole current that makes it over the potential barrier of the
pn-junction. |
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But we could equally well imagine holes generated
by light a x = 0 at a constant rate. The surplus hole
concentration then will assume some distribution in space which will be
constant after a short initiation time - i.e. we have steady state and a simple
differential equation: |
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| D · |
¶2[Dnh(x)]
¶x2 |
|
Dnh(x)
t |
= |
0 |
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The solution (for a
one-dimensional bar extending from x = 0 to x =
¥) is |
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The length L is given by |
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L is simply the diffusion length of
the minority carriers (= holes in the example) as defined in the
"simple" (but in this
case accurate) introduction of life times and diffusion length. |
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This solution is already shown in the drawing
above which also shows the direct geometrical interpretation of
L. |
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The important point
to realize is that the steady state tied to
this solution can only be maintained if the hole current at x = 0
has a constant, time independent value resulting from
Ficks 1st law
since we have no electrical fields that could drive a current. |
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This gives us |
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| j h(x = 0) |
= e · D |
¶Dnh(x)
¶x |
÷
÷ |
x = 0 |
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By simple differentiation of our concentration
equation from above we obtain |
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¶Dnh(x)
¶x |
÷
÷ |
x = 0 |
= |
Dn0
L |
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Insertion into the current equation yields the
final result |
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| j h(x = 0) |
= |
e · Dh
Lh |
· Dnh(x =
0) |
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The physical meaning is that the
hole part of the current will decrease from
this value as x increases, while the
total current stays constant - the remainder is taken up by the
electron current. |
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The Debye
length and the dielectric
relaxation time are important quantities for majority carriers (corresponding to the diffusion length and the minority carrier life time for minority carriers). Let's see why this is so in this
paragraph. |
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Both quantities are rather general and come up
whenever concentration gradients cause currents that are counteracted by the
developing electrical field. |
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An alternative
simple treatment of the
Debye length can be found in a basic module. |
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Let us start with
the Poisson
equation for an arbitrary one-dimensional semiconductor with a varying
electrostatic potential V(x) caused by charges with a density
r(x) distributed somehow in the
material. We then have |
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| e · e0 · |
d2V(x)
dx2 |
= |
e · e0· |
dE(x)
dx |
= |
r(x) |
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E(x) is the electrical field strength;
always the derivative of the potential V. |
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The charge r(x) at any one point can only result from our
usual charged entities which are electrons, holes, and ionized doping atoms.
r(x) is always the net sum of this charges, i.e. |
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| r(x) |
= e · |
æ
è |
nh(x) + ND+(x)
[ne(x) +
NA(x)] |
ö
ø |
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(The sign in this
formulation is negative if more negative charges are present than positive ones
- which takes care of the minus sign usually attached to Poissons
equation). |
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The electrostatic potential
V needed for the Poisson equation is now a function of
x and shifts the conduction and valence band up or down by the
amount eV relative to some zero point at x = 0. We
thus may write |
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| EC(x) |
= |
EC(V = 0) + e · V(x) |
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| EV(x) |
= |
EV(V = 0) + e · V(x) |
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The Poisson equation becomes |
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| e · e0 · |
d2V(x)
dx2 |
= |
e · e0
e |
· |
d2EC(x)
dx2 |
= e |
æ
è |
nh(x) +
ND+(x) [ne(x) +
NA(x)] |
ö
ø |
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If we now insert
the proper equations for the four
concentrations, we obtain a formidable differential equation that is not
easy to solve, but of prime importance for semiconductor physics and
devices. |
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However, even if we could solve the differential
equation (which we most certainly cannot), it would not be of much help,
because we also a need a "gut-feeling" of what is going on. |
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The best way to
visualize the basic situation is to imagine a homogeneously doped semiconductor
with a fixed charge density at its surface and no net currents (imagine a
fictive insulating layer with infinitesimal thickness
that contains some charge on its outer surface). |
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Carriers of the semiconducor thus can not neutralize the charge, and the surface charge
will cause an electrical field which will penetrate into the semiconductor to a
certain depth. |
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This is the most general case for disturbing the
carrier concentration in a surface-near region and thus to induce some band-bending. |
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There are two distinct major situations: |
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1. The surface charge has the
same polarity as the majority carriers in
the semiconductor, thus pushing them into the interior of the material. |
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This exposes the ionized dopant atoms with
opposite charge and a large space charge
layer (SCR) will built up. This is also called the depletion case. |
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The SCR is large because the dopant
density is low and the dopant atoms cannot move to the
interface. Many dopant atoms have to be "exposed" to be
able to compensate for the surface charge; the field can penetrate for a
considerable distance. |
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However: In contrast to what we learned about SCRs in
pn-junctions, even for large fields (corresponding to large reverse
voltages at a junction), the Fermi energy is EF
still constant (currents are not possible).
The bands are still bent, however, this means that EC
EF incrases in the direction toards the
surface. |
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If the majority carrier concentration then is
becoming very small in surface near regions (it scales with exp
(EC EF) after all), the minority
carrier concentration increases due to the mass action law until minority
carriers become the majority - we have the case of inversion |
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2. The surface charge has the
opposite polarity as the majority carriers
in the semiconductor, thus accumulating them at the surface-near region of the
material. |
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Then majority carriers can move to the surface
near region and compensate the external charge. The field cannot penerrate
deeply into the material. |
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This case is called accumulation. |
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The situation is best visualized by
simple band diagrams, we chose the case for n-type materials. The
surface charge is symbolized by the green spheres or blue squares on the
left. |
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Between depletion and accumulation must be the
flat-band case as another prominent special
case. This is not necessarily tied to a surface charge of zero (as shown in the
drawing where a blue square symbolizes some positive surface charge), but for
the external charge that compensates the
charge due to intrinsic surface
states. |
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We have
some idea about the width of the
space charge region that comes with the depletion
case. But how wide is the region of appreciable band bending in the
case of accumulation? |
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Qualitatively, we know that it can be small - at
least in comparison to a SCR - because the charges in the semiconductor
compensating the surface charges are mobile and can, in principle, pile up at
the interface |
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For the quantitative answer for all
cases, we have to solve the Poisson equation from
above. However, because we cannot do that in full generality, we look at
some special cases. |
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First we restrict ourselves to the
usual case of one kind of doping -
n-type for the following example - and temperatures where the donors are
fully ionized, which means that the Fermi energy is well below the donor level
or ED EF >>
kT. |
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We then have only two charged entities: |
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| ND |
= |
ND |
| |
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| ne |
= |
Neeff · exp
|
EC EF
kT |
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This means in what follows we
only consider the majority carriers. |
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The Poisson equation than reduces
to |
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e · e0
e |
· |
d2EC(x)
dx2 |
= e |
æ
è |
ND Neeff
· exp |
EC(x) EF
kT |
ö
ø |
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And
this, while special but still fairly general, is still not easy to
solve. |
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We will have to specialize even more.
But before we do this, we will rewrite the equation somewhat. |
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For what follows, it is convenient to express the
band bending of the conduction band in terms of its deviation from the
field-free situation, i.e. from EC0 =
EC(x = ¥). We thus
write |
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The exponential term of the Poisson equation can
now be rewritten, we obtain |
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| e · Neeff · exp |
EC(x) EF
kT |
= |
e · Neeff · exp |
EC0 EF
kT |
· exp |
DEC(x)
kT |
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The first part of the right hand side gives just
the electron (charge) density in a field free part of the semiconductor, which
- in our approximations - is identical to the density
ND of donor atoms. This leaves us
with a usable form of the Poisson equation for the case of accumulation: |
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d2EC
dx2 |
= |
d2DEC
dx2 |
= |
e · ND
e · e0 |
· |
æ
è |
1 exp |
DEC
kT |
ö
ø |
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DEC characterizes the amount of band
bending. We can now proceed to simplify and solve the differential equation by
considering different cases for the sign and magnitude of DEC. |
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Unfortunately, this is one of
the more tedious (and boring) exercises in fiddling around with the Poisson
equation. The results, however, are of prime importance - they contain the very
basics of all semiconductor devices. |
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We will do one approximative solution here for the most simple case of
quasi-neutrality which will give us the all-important Debye length. |
|
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The other cases can be found in advanced modules:
|
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Quasi-neutrality is the mathematically most
simple case; it treats only small
deviations from equilibrium and thus from charge neutrality. |
|
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The condition for quasi-neutrality is simple: We
assume DEC <<
kT. |
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We then can develop the
exponential function in a Taylor series and
stop after the second term. This yields |
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d2DEC
dx2 |
= |
e2 ·ND
e · e0
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· |
DEC
kT |
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That is easy now, the solution is |
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| DEC(x) |
= |
DEC(x = 0) · exp
|
x
LDb |
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The solution defines
LDb =
Debye length for n-type semiconductors = Debye length
for electrons, we have |
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| LDb |
= |
Ö |
e · e0
· kT
e2 · ND |
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Obviously the Debye length
LDb for holes in
p-type semiconductors is given by |
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| LDb |
= |
Ö |
e · e0
· kT
e2 · NA |
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For added value, our solution also
gives the field strength of the electrical field extending from the surface
charges into the depth of the sample. |
|
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Since the field strength E(x) is the derivative of the
electrostatic potential, which in turn is is simply V(x) = e
· EC(x), we have |
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| E(x) |
= |
dV(x)
dx |
= |
1
e · LDb |
· DEC(x) |
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The Debye length gives the typical
length within which a small deviation from
equilibrium in the total charge density -
which for doped semiconductors is always dominated by the majority carriers - is relaxed or screened; in
other words is no longer felt. |
|
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LDb
is a direct material parameter - its
definition contains nothing but prime material parameters (including the
doping). |
|
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For medium to high doping densities, it becomes
rather small. The dependence of
the Debye length on material parameters is shown in an illustration. |
|
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The Debye length is also a prime material
quantity in materials other than semiconductors - especially in ionic
conductors and electrolytes (for which it was originally introduced). It also
applies to metals, but there it is so small that it rarely matters. |
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The Debye length comes up in all
kinds of equations. Some examples are given in the advanced modules dealing
with the other cases of field-induced band
bending |
|
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The Debye length is to
majority carriers what the diffusion length
is to minorities. And just as the diffusion
lenght is linked to the minority carrier lifetime t, the Debye
length correlates to a specific time too, called the dielectric
relaxation time td. |
|
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This will be the subject of the next
paragraph. |
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|
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Lets start from the same assumption
that lead to the Debye length: A doped semiconductor, all dopants ionized, and
some small disturbance in the charge equilibrium expressed as some small
Dr(x,
t) somewhere, starting at some time t0;
i.e. we still assume quasi-neutrality. |
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The Poisson equation now is extremely simple, we
write it directly for the electrical field strength and have |
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dE(x, t)
dx |
= |
Dr(x,
t)
e · e0
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We now want to
find out about how long it takes to
establish a steady state, so we need some expression for dr/dt. The Poisson equation won't help because
it does not explicitely contain the time dependence. |
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But simply using the continuity equation from above, and replacing
nh by Dr (because this is the relevant density of charged
carries now), provides a dDr/dt term. Moreover, we can hugely simplify
this equation, because we
- Look only at majority carriers; i.e. we may neglect the first term
[nh nh(equ)]/ t.
- We are treating quasi neutrality, ie. we neglect all terms with gradients in the carrier concentration
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This leaves us with the following continuity equation |
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¶Dr
¶t |
= e · r ·
µ · |
¶E(x)
¶x |
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Inserting dE/dx from the
Poisson equation gives |
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¶Dr
¶t |
= |
e · r · µ
ee0 |
· Dr
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r is the total
carrier concentration, we can write it as r =
r0 + Dr » r0 since we have quasi neutrality;
µ, as always, is the mobility of the carrier in question. |
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This is a differential equation for
Dr(x, t)
with the simple solution |
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| Dr(x, t) |
= |
Dr(x, 0) · exp |
t
td |
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With
td =
dielectric relaxation time =
another basic material constant for the same
reason as the Debye length. |
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The dielectric
relaxation time tells us exactly what we wanted to know: How long
does it take the majority carriers to respond to a disturbance in the charge
density. |
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While this definition of some special
time is of some interest, but not overwhelmingly so, the situation gets more
exciting when we consider relations between our basic material constants
obtained so far: |
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Since e · µ ·
r = s, the
conductivity of the
material (for the carriers in question), we have the simple and fundamental
relation |
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Now let's see if there is a correlation to the
Debye length: |
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We use the
Einstein relation D =
µ(kT/e), the Debye length
definition (LDb = {(ee0 ·
kT)/(e2 · r)}1/2, pluck it into the definition of the
dielectric relaxation time (again replacing ND by
r) and obtain |
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| td |
= |
LDb2
D |
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| LDb |
= |
Ö |
D · td |
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This is exactly the
same relation for the majority
carriers between a characteristic time
constant and a length as in the
case of the minority carriers where we had the minority lifetime t and the correlated diffusion length
L. |
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The physical meaning is the same, too. In both
cases the times and lengths give the numbers for how fast a deviation from the
carrier equilibrium will be equalized and over which distances small deviations
are felt. |
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This merits a few more thoughts.
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If the carrier concentration is high, td is in the order of pico seconds and LDb
extends over nano meters. Any deviation
from equilibrium is thus almost instantaneously wiped out, or, if that is not
possible, contained within a very small scale. |
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And this is the regular situation for majority carriers. The few minority carriers always
present in the semiconductor, too, can be safely neglected. |
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For minority carriers, however, the situation is
entirely different. |
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Their concentration is very small; td and LD consequently
are no longer small. |
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Moreover, whatever disturbance occurs in the
concentration of minorities, there are
plenty of majorities that can react very quickly (with their td) to the electrical field always tied to
a Dr. |
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The majority carriers are always
attracted to the minorities and thus will quickly surround any excess minority
charge with a "cloud" of majority carriers, essentially compensating
the electrical field of the excess minorities to zero. |
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They will, of course, eventually remove the
excess charge by recombination, but that takes far
longer than the time needed to do the screening. |
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Since the electrical field is now zero, the
excess charge cannot disappear or spread out by field currents - only spreading
by diffusion in the concentration gradient (which is automatically introduced,
too), is possible. |
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But this is exactly the process that
we have neglected in this discussion (we had all concentration gradients in the
continuity equation set to zero!). |
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Dielectric
relaxation (i.e. the disappearance of charge surpluses driven by
electrical fields) is thus not applicable to minority carriers. Charge
equilibration there is driven by diffusion - which is a much slower process!
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This then justifies the simple approach we took before, where we
only considered the diffusion of minorities and did not take into account the
majority carriers. |
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© H. Föll (Semiconductor - Script)