 | 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. |
| First, we consider
the internal current (density) in a material with a gradient
of the carrier density (ne or nh). |
|  | Fick's first law then tells us that
the diffusion-driven particle current jp,diff
is given by |
| |
|
jp,diff = – De,h · Ñ
ne,h | | |
| | If the particles are carrying a charge q,
this particle current is also an electrical current (which
obviously is a diffusion current, then), given by |
| |
| je,h
| = | q · jp,diff
| = | – q ·
D e,h · Ñne,h |
| |
| | |
Considering only the one-dimensional case for electrons (i.e.
q = –e; holes behave in exactly the same way with q = +e),
we have |
| |
|
je (x) | = |
e · De · | dne(x)
dx | | |
| Since there can be no net
current in a piece of material just lying around (which nevertheless might still have a density
gradient in the carrier density, e.g. due to a gradient in the doping density), the carriers
displaced by diffusion always generate an electrical field that will drive the other carriers
back. |
| | Any
field E(x) (written
in mauve to avoid confusion with
energies) now will cause a (so far one-dimensional) current given by |
| |
| j
| = | s
· E(x) = q · n(x
) · µ · E(x) |
| |
| | With s = conductivity, µ
= mobility. |
| |
Note that the result is always the technical current density, which is positive for positive
charge carriers. Yet this equation also works for electrons because for them, effectively,
two minus signs cancel: one from their negative charge and the other from their direction
of movement opposite to the electric field. This means that in a strict sense, their mobility
should be negative. However, in this equation one only considers positive charges and positive
mobilities – also for electrons. Therefore, to use ths equation in full generality,
we write it as |
| |
|
| The total (one-dimensional)
current in full generality is then |
| | | jtotal(x)
= e · n(x) · µ · E(x
) – q · D · | dn(
x )
dx | | |
| | We will need this
equation later. |
| For
our case of no net current and only fields
caused by the diffusion current, both currents have to be equal in magnitude: |
| | | e ·
n(x) · µ · E(x) |
= | q · D
· | dn(x)
dx |
| |
| |
This is an equation that comes up repeatedly; we will encounter
it again later when we derive the Debye length. |
|
|
Note that in this equation, the sign on the right-hand side
depends on the type of charge carriers (since q = ±e). This is balanced
on the left-hand side by the direction of the electric field. |
| Now we are stuck. We need some
additional equation in order to find a correlation between D and µ.
|
| | This
equation is the Boltzmann distribution (here used as an
approximation to the Fermi distribution), because we have
equilibrium in our material. |
| | However,
we also know that, in this equilibrium situation, we have spatially varying charge carrier
densities and electric fields. We know such a situation from the p–n junction in equilibrium.
There, this was only possible due to the band bending, i.e. that the band edges were functions
of the lateral position. This we also consider here. |
|
|
Just to derive the relation to the electric field in the above
equation, for the moment we just consider the case of electrons as majority carriers. For
their local density it holds that |
| |
| n(x) | = | Neff · exp |
æ
è | – | EC(
x) – EF
kT |
ö
ø | | |
| | Differentiation
of the Boltzmann distribution gives us |
|
| dn
dx | = | – Neff ·
1/(kT) · | dEC(x)
dx | · exp |
æ
è | – | EC(
x) – EF
kT |
ö
ø | | |
| | |
| | dn
dx | = | –
n(x) · 1/(kT) · | dEC(x
)
dx | | |
| |
| |
The slope of the conduction band comes directly from the spatially
varying electric potential V(x); to convert the electric potential to
the absolute energy of a charge carrier, the elementary charge e is needed as an additional
factor. From the p–n junction we know that the sign (i.e., the direction) of the electric
field is identical to that of the slope of the conduction band. Thus, altogether we have |
| | | E(x) | = –
| dV(x)
dx |
= | 1/e | · |
dEC(x)
dx | |
|
| | Using
this relation, the current balance from above becomes |
|
| | | e · n(x)
· µ · E(x) | =
| q · D · | dn(x
)
dx | | |
| | | |
| e · n(x) · µ ·
E(x) | = | –
q · n(x) · D · e/(kT) · |
E(x) | | |
| | |
| | D |
= | µ
· kT/e | | |
| | 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. |
| Distinguishing again between electrons and holes gives
as the final result the famous Einstein–Smoluchowski
relations: |
|
|
| De | =
| µe · kT
e |
| | | |
| Dh | = |
µh · kT
e |
| |
| You
may want to have a look at a different derivation
in an advanced module. |
| |
|
Non-Equilibrium
Currents |
| | |
| In the consideration above we postulated that there is no net current flow; in other words, we postulated
total equilibrium. Now let's consider that there is
some net current flow and see what we have to change to
arrive at the relevant equations. |
| 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 densities have their equilibrium values n
e(equ) and nh(equ), but arbitrary
values that we can express by some Delta to the equilibrium value. We thus start with |
| | |
| |
Since carriers above the equilibrium density are often created
in pairs we have for this special,
but rather common case |
| | | Dne |
= | D
nh | = |
Dn | | |
| | | | | D
n | = | ne
– ne(equ) | = | nh – nh(equ) |
| |
| |
This is a crucial assumption!
|
| This allows us to concentrate on one kind of carrier, let's 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 |
| |
|
ne | = | n
e(equ) + Dne | =
| ne(equ) + nh – nh
(equ) | | |
|
|
We now must consider Fick's second law or the continuity equation (it is the same thing for special cases, but the continuity
equation is more general). |
| For
the net (mobile) charge density r (which is the
difference of the electron and hole density, r
= e · (nh – ne), in
contrast to the total particle density, which is the sum
!) we have |
| |
|
| | With jtotal
= je + jh = sum of the electron and hole currents. |
| In the simplest form we have for the holes
|
| |
¶n
¶t
| = | –
(1/e) · div (jh) | | |
| The factor 1/e is needed to convert
an electrical current j to a particle current jpart
via j = q · jpart, with q = ±e.
Here, as always, we have to pick the right sign for the elementary charge e (negative
for electrons, positive for holes). |
| |
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 jh = 0,
to satisfy that condition. |
| But
this is, of course, a condition that we know not to be
true. |
| | 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 – since div jh = 0 is correct on
average. |
|
| We are, however, considering non-equilibrium,
so we must primarily consider the recombination of the surplus
minority carriers given by |
| | |
| | Why? Because, as stated before,
the generation essentially does not change, so it still
balances against the recombination rate of the equilibrium density, 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). |
| | R
D = [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 |
| |
dn
dt | = – |
nh – nh(equ)
t | –
(1/e) · div (jh) | | |
| The current j can always be
expressed as the sum of a field current and a diffusion current as we did above
by |
| |
|
jh,total(x) | = |
e · n(x) · µ · Ex(
x) – e · Dh · |
dnh(x)
dx | |
|
| If we let Ex
= 0 and consequently ¶Ex
(x)/¶x = 0, too, the current equation
from above reduces to |
| |
¶nh
¶t | = – |
nh – nh(equ)
t | + D
· | ¶2nh(
x)
¶x2 |
| |
| | Since ¶nh/¶t = ¶[nh(equ)
+ Dnh]/¶t
= ¶Dnh/
¶t, and correspondingly ¶2n
h(x)/¶x2 = ¶
2Dnh(x)/¶
x2, we have |
| | ¶D
nh
¶t | = – | D
nh
t | +
D · | ¶2D nh(x)
¶
x2 | | |
| If we consider steady state
, we have ¶Dnh/¶ t = 0, and the solution of the differential equation is now
mathematically easy. |
| But
how can steady state be achieved in practice? How can we provide for a constant
, non-changing density of minority carriers above equilibrium? |
|
|
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 p–n junction. |
| | But we could equally well imagine holes generated by light
a x = 0 at a constant rate. The surplus hole density 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: |
| | | D · |
¶2[D nh(x)]
¶x2 | –
| D nh(x)
t | =
| 0 | | |
|
The solution (for a one-dimensional
bar extending from x = 0 to x = ¥)
is |
| |
|
Dn(x) | =
| Dn0 · exp |
æ
è | – |
x
L |
ö
ø | | |
| | The
length L is given by |
| | |
| |
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. |
| | This solution
is already shown in the drawing above which also shows the direct geometrical interpretation
of L. |
| 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 Fick's 1st law since we have no electrical fields that could drive a current. |
| | This
gives us |
| |
|
jh(x = 0) | = – e
· D | ¶D
nh(x)
¶x |
÷
÷ |
x = 0 | |
|
| | By
simple differentiation of our density equation from above we obtain |
| |
¶
Dnh(x)
¶
x | ÷
÷
| x = 0 | =
– | Dn0
L | | |
| | Insertion into the current
equation yields the final result |
| |
| jh(x = 0) | =
| e · Dh
Lh
| · Dnh(x
= 0) | | |
|
|
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. |
| | |
| General Band-Bending and Debye Length |
| | |
| 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. |
| | Both quantities are rather general and come up whenever
density gradients cause currents that are counteracted by the developing electrical field. |
| | | An
alternative simple treatment of the Debye
length can be found in a basic module. |
| 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 |
| |
|
– e · e0 ·
| d2V(x)
dx2 |
= | e ·
e0· | d
E(x)
dx | =
| r(x) |
| |
| | E(x) is the electrical
field strength; always minus the derivative of the potential V. |
| | 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. |
| |
| r(x) | = e · | æ
è | nh(x) + ND
+(x) – [ne(x) + NA–(
x)] | ö
ø |
| |
| | | The electrostatic potential V
needed for the Poisson equation is now a function of x and shifts the conduction
and valence band by the potential energy qV relative to some
reference point for which one has V = 0. Since
the band structure refers to the energy of electrons, we have that q = –e
and thus may write |
|
|
| EC(x) | =
| EC(V = 0) – e · V(x) |
| | | |
| EV(x) | = |
EV(V = 0) – e · V(x) |
| |
| |
Thereby, the Poisson equation becomes |
| |
| – e ·
e0 · | d2V(x
)
dx2 | = |
e · e0
e | · | d2
EC(x)
dx2 | =
e | æ
è | nh(x) + ND+(x)
– [ne(x) + NA–(x)]
| ö
ø |
| |
| | If we now insert the proper
equations for the four densities, we obtain a formidable differential equation that is
of prime importance for semiconductor physics and devices, but not easy to solve. |
| | 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. |
| 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 (think of a fictional insulating
layer with infinitesimal thickness that contains some charge on its outer surface).
|
| | 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. |
|
| This is the most general case for disturbing the carrier density in a surface-near
region and thus to induce some band-bending. |
| There are two
distinct major situations: |
|
1. The surface charge has the same polarity as the
majority carriers in the semiconductor, thus pushing them into the interior of the material. |
| | 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. |
| | 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 the surface charge; the field
can penetrate for a considerable distance. |
|
|
However: In contrast
to what we learned about SCRs in p–n 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. |
|
|
If the majority carrier density then is becoming very small
in surface-near regions (it scales with exp[– (EC – EF
)/(kT)] after all), the minority carrier density increases due to the mass action
law until minority carriers become the majority – we have the case of inversion
. |
| 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. |
| | Then
majority carriers can move to the surface near region and compensate the external charge.
The field cannot penetrate deeply into the material. |
|
|
This case is called accumulation. |
| 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. |
|
| |
| | 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, but for the external charge that compensates the charge
due to intrinsic surface states (as shown in the drawing where a blue square symbolizes
some positive surface charge). |
| 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? |
| | Qualitatively, we know that it can be small – at least in comparison
to an SCR – because the charges in the semiconductor compensating the surface
charges are mobile and can, in principle, pile up at the interface |
| 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: We restrict ourselves to one
kind of doping – namely n-type for the following example – and to
temperatures where the donors are fully ionized, which means that the Fermi energy is well
below the donor level (i.e., ED – EF >>
kT). |
| |
We then have only two charged entities: |
| |
| ND
+ | = |
N D | | | |
| | n
e | = |
Neffe · exp |
æ
è | – | EC
– EF
kT |
ö
ø | |
|
| |
This means in what follows we only consider the majority
carriers. |
| The
Poisson equation then reduces to |
| |
e · e0
e | · | d
2 EC(x)
dx2 | = e |
æ
ç
è | ND –
Neffe · exp | æ
è | – | EC
(x) – EF
kT |
ö
ø | ö
÷
ø | |
|
| |
And this, while special but still fairly general, is still not easy
to solve. |
| We
will have to specialize even more. But before we do this, we will rewrite the equation somewhat.
|
| | 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 E C0
= EC(x = ¥). We thus write |
| | |
| |
The exponential term of the Poisson equation can now be rewritten,
we obtain |
| |
|
Neffe · exp |
æ
è | – | EC(x
) – EF
kT |
ö
ø | = |
Neffe · exp | æ
è | – | EC
0 – EF
kT |
ö
ø | ·
exp | æ
è | –
| DEC(x)
kT | ö
ø
| | |
|
|
The first part of the right hand side gives just the electron
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: |
| |
d2EC
dx2
| = | d2(
DEC)
dx2 |
= | e2 · ND
e · e0 |
· | æ
ç
è | 1 – exp | æ
è | – |
DEC
kT |
ö
ø |
ö
÷
ø | | |
| 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 D
EC. |
| | 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. |
| We
will do one approximative solution here
for the most simple case of quasi-neutrality which will give us the all-important
Debye length. |
|
|
The other cases can be found in advanced modules:
|
| Quasi-neutrality
is the mathematically most simple case; it treats only small
deviations from equilibrium and thus from charge neutrality. |
| | The condition for quasi-neutrality
is simple: We require |DEC| << kT. |
| | We then can approximate the exponential function by its Taylor
series and stop after the second term. This yields |
| |
d2(DEC)
dx2 |
= | e2 · ND
e · e0
| · | D
EC
kT | |
|
| | That
is easy now, the solution is |
| | | DEC(x)
| = | DE
C(x = 0) · exp | æ
è | – | x
LDb | ö
ø |
| |
| |
The solution defines LDb
= Debye length for n-type
semiconductors = Debye length for electrons, we have |
| |
| LDb
| = | Ö
|
e ·
e0 · kT
e2 · ND |
| |
| |
Obviously the Debye length LDb for holes in p-type semiconductors is given by |
| | | LDb
| = | Ö
|
e ·
e0 · kT
e2 · NA |
| |
| 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. |
| | Setting it to zero at the top of the valence band in the
p-type material (as it is conventionally
done), the electrostatic potential is related to the conduction band edge by EC
(x) = Eg – e · V(x). As discussed already
above, the minus sign stems from the negative charge of an
electron. |
| | Since
the field strength E(x) is minus the derivative
of the electrostatic potential, we now have |
|
| | E(x) |
= – | dV(x)
dx | = | 1/e |
· | dEC(x)
dx | = – | 1
e · LDb | ·
DEC(x) | |
|
| | Note
that in the case of accumulation at the surface of an n-type semiconductor, D
EC(x) is negative, so the electric field comes out positive –
in full agreement with the surface (at x = 0) being positively charged in this
case. |
| 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, it is no longer felt. |
| | LDb
is a direct material parameter – its definition
contains nothing but prime material parameters (including the doping). |
| | For medium to high doping
densities, it becomes rather small. The dependence
of the Debye length on material parameters is shown in an illustration. |
| | 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. |
| 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 |
|
|
The Debye length is to majority carriers what the diffusion length is to minorities. And
just as the diffusion length is linked to the minority
carrier lifetime t, the Debye
length correlates to a specific time, too, called the dielectric relaxation time
td. |
| | This will be the
subject of the next paragraph. |
| | |
| Dielectric Relaxation Time |
| | |
| Let's start from the same situation that lead
to the Debye length: A doped semiconductor, all dopants ionized, and some small disturbance
in the charge equilibrium expressed as some small D
r(x, t) somewhere, starting at some time t0;
i.e. we still assume quasi-neutrality. |
| | The Poisson equation now
is extremely simple, we write it directly for the electrical field strength and have |
| | dE(x, t)
dx | = | Dr(x, t) e
· e0 | |
|
| We now want to find out
about how long it takes to establish a steady state, so
we need some expression for d(Dr)/dt. The Poisson
equation won't help because it does not explicitly contain the time dependence. |
| | But simply using
the continuity equation for the relevant
charge density Dr provides a d( Dr)/dt term. Since we are treating quasi-neutrality, we neglect all
terms with gradients in the carrier density (this will
turn out to be fully justified). |
| | Since the only relevant current is the drift curent j(x)
= s E(x) =
r · µ · E(x), this leaves
us with the following continuity equation |
| |
¶(Dr)
¶
t | = – r ·
µ · | ¶E
(x)
¶x |
| |
| Inserting dE/dx from the Poisson equation gives |
|
| ¶(Dr
)
¶t | =
– | r · µ
e · e0 |
· Dr |
| |
| | r is the total carrier density,
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. |
|
| This is a differential equation for Dr(x, t) with the simple solution |
| |
| Dr(x, t) | =
| Dr
(x, 0) · exp | æ
è
| – | t
td | ö
ø |
| |
| | With td
= dielectric relaxation time = another basic
material constant for the same reason as the Debye length. |
| | The
dielectric relaxation time tells us exactly what we wanted
to know: How long does it take for the majority carriers to respond to a disturbance in the
charge density. |
| 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: |
| | Since
µ · r = s, the conductivity of the material (for the carriers in question),
we have the simple and fundamental relation |
|
| |
|
| Now let's see if there is a correlation to the Debye length: |
| | We use the Einstein relation
D = µ kT/e, the Debye length definition
(LDb = [e · e0
· kT/(e · r)]1/2), pluck it into
the definition of the dielectric relaxation time (again replacing e · ND
by r) and obtain |
|
| | td
| = | LDb
2
D | | |
| | | LDb | = |
Ö |
D · td |
| |
| | 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 . |
|
|
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. |
| This merits a few more thoughts. |
| | If the carrier density is
high, td is in the order of picoseconds
and LDb extends over nanometers.
Any deviation from equilibrium is thus almost instantaneously wiped out, or, if that is not
possible, contained within a very small scale. |
|
|
And this is the regular situation for majority
carriers. The few minority carriers always present in the semiconductor, too, can be safely
neglected. |
| For minority carriers, however, the situation is entirely
different. |
| | Their density is very small; consequently, their t
d and LDb would not be small. |
| | Moreover, whatever disturbance
Drmin occurs in the density of
minorities, there are plenty of majorities that can react very quickly (with their
td) to the electrical field always tied to such
a Drmin. |
| The majority carriers are always attracted to the minorities and thus will quickly
surround any excess minority charge with a "cloud" of majority carriers (which is
called screening), essentially compensating the electrical
field of the excess minorities to zero. |
|
|
They will, of course, eventually remove the excess charge by
recombination, but that takes far longer than the time
needed to do the screening. |
| | Since the electrical field is now zero, the excess charge cannot
disappear or spread out by field currents – only spreading by diffusion in the density
gradient (which is automatically introduced, too) is possible. |
| And this is exactly the process that we have neglected
in this discussion (we had all density gradients in the continuity equation set to zero!); now we know why this is justified. |
| | Dielectric relaxation
(i.e. the disappearance of charge surpluses driven by electrical fields) is just not applicable
to minority carriers. Charge equilibration there is driven by diffusion – which is a
much slower process! |
|
| This also justifies the simple approach
we took before, where we only considered the diffusion of minorities and did not take into
account the majority carriers. |
| | |
© H. Föll (Semiconductors - Script)
