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Shown here are three
graphical representations of the (fictitious) process upon contact. |
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Upon contact, the electrons from the conduction
band of the bulk will flow to the still empty states of the surface. This will
put an additional negative charge at the surface. |
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In response to the negative surface charge, the
electrostatic potential at the surface goes up and extends (decreasingly) into
the bulk - we have a built-in potential
V(x). The simplest way to visualize the
band bending going with this potential is to
remember that the electrons from the bulk are repelled by the negative surface
charge; work is needed to move them to the surface; this is an uphill
process |
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As long as energy can be gained by moving
electrons from the bulk to the surface, the process will continue. Equilibrium
is reached as soon as as it makes no difference any more if one more electron
is in the bulk or on the surface, i.e. if we change the electron count by
Dne. This means that
(¶G/¶ne) · Dne(bulk) = chemical potential
µ = Fermi energy EF in the bulk =
(¶G/¶ne) · Dne(surface), or in other
words: |
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This is a
crucial point; let's make it again: |
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| In thermal equilibrium, the Fermi energy
EF must be constant everywhere. |
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If you are not too sure about this,
read up about the concept of the chemical potential in the basic module. |
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This automatically leads to an
imbalance of charge, we now have local violations of
charge equilibrium. |
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The area with the band bending will
not contain free electrons, but still the ionized positively charged donor
atoms, it thus contains charges in space
and is called a space charge region
(SCR). |
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The SCR also contains an
electrical field E which is given directly (via the
Poisson
equation) by the derivative of the electrostatic potential V
by E = -
 V,
or for the one-dimensional case, Ex = -
dV/dx. |
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The easiest way to think about this
electrical field is to consider the field
lines, which must start at a positive charge and end at a negative
charge. |
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The negative charges
must be the surplus electrons sitting in the surface states (and thus also in
real space at the surface), the positive
charges must be the ionized donor atoms in the
bulk. |
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This view immediately leads to a
"quick and dirty" formula for the width
d of the space
charge region: We consider the capacitance
CSCR of the SCR. |
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Since the positive charges are spread
homogeneously through the volume of the SCR, we approximate the capacity
of the SCR by a plate capacitor with half the distance between the plates,
dcap = d/2. In other words, we sort of put all
the positive charge on on fictitious plate at half the width of the
SCR. |
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The capacitance then becomes |
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| CSCR = |
2 e ·
e0 · A
d |
= |
Q
U |
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With A = area of the
capacitor plates, Q = charge on the plates and U =
potential difference between the plates. |
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The charge on the plates is equal to
the number of ionized donors in the volume of the SCR. The volume is
d · A and the number of charges is just the concentration of
donors ND (assuming that all are ionized) times the
elementary charge e times the volume d · A, so we
have |
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Substituting this in
the equation from above we obtain |
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2 e · e0 · A
d |
= |
e · ND · d · A
U |
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This gives us one of the more
important semiconductors device equations in its simplest form - and this is
the correct equation despite the somewhat
questionable assumption of dcap = d/2. |
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| d |
= |
æ
ç
è |
2 ee0 U
e ND
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ö
÷
ø |
1/2 |
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The voltage
U is of course the potential difference of the extreme values of
the built-in potential which, in the case of the direct contact, is simply the
difference of the Fermi energies expressed as potential, i.e. |
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We immediately can generalize: If, in
addition to the "built-in" potential DEF/e an additional external
potential Uex is added from the outside by simply
connecting the material to a voltage source at Uex,
the total voltage becomes
U = DEF/e +
Uex and the width of the space charge region is |
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| d |
= |
1
e |
· |
æ
ç
è |
2 e · e0 · (DEF + e · Uex)
ND |
ö
÷
ø |
1/2 |
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It is easy to to obtain the
same equation by integrating
the Poisson equation for the case. This is done in
an illustration module. |
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This example illustrates nicely the
approach we take in this chapter: We start from the most simple consideration
of the case and try to deduce proper relations and formula by analogies -
cutting corners a little if necessary (but only as long the results are still
correct). |
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We now construct a pn-junction
exactly along the recipe given above. |
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Draw the band diagrams of both parts |
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Join the two parts, move electrons to the
materials with the lower Fermi energy, holes opposite |
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Built up space charges and shift the potentials
accordingly until the Fermi energy is constant
everywhere. |
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These steps are
illustrates below: |
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As a graphical aid, which will be is useful in
cases to come, the carriers are schematically indicated as circles (electrons)
and squares (holes). Many symbols indicate majority carriers, a few ones the
minorities. |
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The only differences to the first
situation is that we now have a space charge region on both sides of the
junction. |
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The field lines are now pointing from the
positively charged donor ions on the n-type side to the negatively
charged acceptor ions on the p-doped side. |
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The width of the space charge region is now
a little more involved to
calculate (but there is nothing new); it comes out to |
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| d = |
1
e |
· |
æ
ç
è |
2 e ·
e0 |
æ
è |
1
NA
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+ |
1
ND
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ö
ø |
· |
æ
è |
DEF |
+ e · Uex) |
ö
ø |
ö
÷
ø |
1/2 |
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Now lets look at the various currents flowing in the conductance and valence
bands without an external voltage. |
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We know that the net current is zero - we are in
an equilibrium condition as long as we do not apply an external voltage. |
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We also know that we have a
dynamic equilibrium as in the case
of the recombination/generation business before. The net current is zero
because the local currents cancel each other. This implies that the electron
current flowing uphill (from left to right) must be identical in magnitude and
opposite in sign to the one flowing from right to left (downhill). |
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The same reasoning applies, of course, to the
holes in the valence band. |
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The partial currents discerned above
have several specific names. The current component flowing uphill in energy is called: |
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Diffusion current, because
the driving force behind this current is the concentration gradient in the
carrier density. This always leads to a current component given by
Fick`s first law
to |
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| jDiff = – eDn |
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With n = concentration of the
carrier in question. |
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It is also alternatively called
recombination
current because all the electrons (or holes) flowing into the other
side become excess minority carriers there and must disappear by recombination,
which can be depicted as a current flowing between the bands. |
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Looking a little ahead, a pn-junction is a
diode and currents in diodes are classified as forward- or reverse currents.
Well, the current component in question here, is responsible for the
forward current in the
diode and therefore is also addressed under that name. We will use mostly this
name and abbreviate this current component with
jF |
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The partial current
flowing downhill in energy is called. |
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Field current, because it is
the current that the electrical field in the junction produces; i.e. the
carriers flow in the general direction of the field lines with the respective
proper signs. The diffusion current, in contrast, flows against the force
exerted by the field (which will slow down these carriers!) |
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Drift current, because it is
the current that results from a drift - caused by the electical field -
superimposed on random diffusion movements. |
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Generation current,
because the (minority) carriers that were swept down the energy hill (or up in
the case of the holes) must be replaced by increased generation to keep the
concentration constant. |
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Reverse current in the
diode nomenclature explained above. We will use mostly this name and abbreviate
this current component with jR. |
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In equilibrium, without an external
voltage Uex, we know that the net current is zero,
or |
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jF = –
jR, or, to be more precise, |
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| jF(Uex = 0) |
= – |
jR(Uex = 0) |
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To make life a little easier, we now
drop some matter-of-course indices, and the
signs; i.e. we only look at the magnitudes
of the current components. It is easy enough to sort out the signs
in the end again; in case of doubt refer to the
link. In this shorthand we
have |
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If we draw these currents schematically into the
banddiagram from above, it looks like this: |
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Of course, the currents do not flow on both sides
of the band edge; this is simply a drawing device. |
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What happens for finite external
voltages? First lets look at the band diagrams: |
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An external potential eU is added
which will raise or lower the equilibrium potential, depending on its sign. For
sake of simplicity we only move the p-side band diagram. It goes
up for the negative pole of the voltage
supply on the p-side (always think about if electrons are repelled or
attracted if you are unsure about how a potential moves bands), and down for the positive pole on the p-side |
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We no longer have an equilibrium
situation - the Fermi energy is no longer constant everywhere; we must leave it
undefined across the junction |
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Far away from the junction, however, nothing (or
very little) has changed. We still may consider these parts of the
semiconductor to be in equilibrium or at least very close to it. |
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The band diagram for the two possible
basic cases then look like this: |
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The two types of currents are shown increased
(thicker arrow) if the energy barrier was lowered, and decreased if it was
raised. |
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The situation now
is rather simple. The potential step is either increased or decreased. Lets
first look what happens to the forward
currents jF(U). |
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At zero volts the electron and
hole forward currents have a certain value that is certainly determined by the
height of the energy barrier that the carriers have to overcome, following
Boltzmann statistics (as an approximation
to the Fermi statistics, of course). |
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In other words, the equation for this current
includes an exp-E/kT term. Changing the energy barrier
E by DE simply means to
multiply the current at zero volts with exp –DE/kT. |
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Since the magnitude of
jF at zero Volts external voltage is just
jR(U = 0); the forward current current
jF(U) at any voltage U must be |
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| jF(U) |
= |
jR(U = 0) · exp – |
DE
kT |
= |
jR(U = 0) · exp – |
e · U
kT |
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and this is true for the electron and for the
hole forward current. |
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Imagine this as a lot of (drunken) bicyclists
with various random momentums driving around randomly at the foot of a hill.
Some of them on occasion will make it up the hill because their momentum was
large enough and they were heading in the right direction. The fraction of
bicyclist making it will simple change exponentially with the Boltzmann factor
relative to their "current" at some reference value if the hill is
raised or lowered. |
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Now to the reverse current
jR(U) |
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It corresponds to (drunken = randomly
moving) bicyclists that drive around for a certain time (corresponding to the
life time of the minority carriers) on the plateau on top of the hill before
falling off their bikes (recombining). |
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However, everybody who by accident makes it to
the edge of the hill, will invariably careen down, i.e. produce a reverse
current. |
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Clearly, it only matters how many carriers happen
to make it to the edge of the potential drop, not how deep it is. In other
words: The reverse current does not depend on the
external voltage or |
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The total
current is simply the difference
between forward and reverse current for the electrons and the holes so we
have |
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| j(U) |
= |
æ
è |
jF(U) – jR |
ö
ø |
e
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+ |
æ
è |
jF(U) – jR
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ö
ø |
h |
= |
jRe · |
æ
ç
è |
exp – |
e · U
kT |
– 1 |
ö
÷
ø |
+ jRh ·
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æ
ç
è |
exp – |
e · U
kT |
– 1 |
ö
÷
ø |
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This is the famous
diode equation - and this is all there is
to it for straight-forward pn-junctions. |
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Note that we did not assume that the forward or reverse current of
the holes must be identical to the forward or reverse current of the electrons,
respectively. Of course, if everything is symmetrical, we would have
jRe =
jRh, but we want to keep it a as general
as possible even at that level since many real devices employ wildly different
electron and hole currents across the junction. |
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All that is left to do is to consider
jR a bit more closely. |
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Thinking about drunken bicyclists, we might be
tempted to assume that jR should be proportional to their numbers, i.e. to the minority
carrier density on the plateau. |
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This, however, is wrong. |
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So lets sober up and think a bit harder: You only
can extract the bicyclists once! If you
want a constant current over time, the best you can do is to take all carriers
for the current that are generated per time unit and
make it to the edge. |
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In other words, you need
to refill the ranks of bicyclists and the current will be
proportional to the generation rate
G (to what comes out of the bars per time interval), which we
know is equal to the recombination rate
R in undisturbed semiconductors and
given by |
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| G |
= R = |
nmin
t |
= |
ni2
t · NDop |
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With t =
lifetime. |
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Thinking a bit
harder yet, we realize that minority carriers generated way back from the
junctions will not contribute to the current. They will fall off their bike
(recombine) long before the have a chance to come close to the drop-off.
Obviously we only must consider carriers within a certain length of the
junction and this length is, of course, the
diffusion length L. |
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This gives us |
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| jR |
= |
const. · e · L · G |
= |
const. · e · L · nmin
t |
= |
const. · e · L · ni2
t · NDop |
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The elementary charge
e is needed to make an electrical current out of a particle
current. And as it turns out by more
involved calculations, the proportionality
constant is = 1. |
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This gives us the
complete diode equation exclusively in terms of primary materials properties: |
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| j(U) = |
æ
ç
è |
e · L · ni
2
t · NAcc
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æ
è |
exp - |
e · U
kT
|
– 1 |
ö
ø |
ö
÷
ø |
electrons |
+ |
æ
ç
è |
e ·L · ni
2
t · NDon
|
æ
è |
exp – |
e · U
kT
|
– 1 |
ö
ø |
ö
÷
ø |
holes |
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This is a remarkable achievement -
obtained without involved calculations and cutting corners only once
(proportionality const. = 1). But how good is
it? Only the experiment can tell. |
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If we measure the
current-voltage characteristics of an "ideal" pn-junction, we
will find the following curves |
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Generally, for Ge (or other
semiconductors with relatively small band gaps) the measured characteristics is
remarkably close to the one predicted by our formula. For Si, however
(and other semiconductors with large band gaps), it is
not a good formula, particularly for for the reverse current. We
have large deviations from theory, labeled with black lettering. Lets see what
went wrong (more
details in the link): |
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First, our theory has the usual (trivial)
omissions. We did not include any ohmic resistance which can be easily added by
putting a resistor in series to the junction. The result is the ohmic behavior
for larger voltages as seen at larger currents. |
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Second, everything will break down at high
field strength; this is true for a junction, too. So for large reverse voltages
(easily in the range 100V - 1000V), the junction will go up in smoke
while drawing a large current called "junction breakdown". |
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While these points would apply to a Ge
junction too (they are not shown in the ideal characteristics), the remaining
deviations for Si involve a major oversight in our present theory: |
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Third: We did not include carrier generation (and
recombination) in the space charge region! |
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In the bicyclist picture, we did not take into
account that there are bars along the slope of the hill, too, which will emit
bicyclists with a certain momentum and direction - they may either go up or
down the hill and thus add to the current of particles moving in either
direction. |
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This adds four more current components (forward
and reverse for holes and electrons) summarily called
generation currents from the
SCR. |
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How large are these SCR caused
current components? |
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The answer comes from one of the more involved
problems in semiconductor physics, it is not easy to obtain for real
semiconductors (we will do it
later). However, there is an easy way of thinking
about it that even comes up with the usually given formula resulting
from serious (but still approximate) computing. But we sure will have to cut a
few corners! |
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Here we go: |
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We know the maximum current
jmax that could emerge from the space charge region:
It is the generation rate of carriers times the width of the SCR in
complete analogy to the discussion above. More
than that cannot flow out (in one direction for the maximum) per time
unit. |
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Every generation event produces a hole and an
electron, so we have |
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With d = width of the SCR
and GSCR = generation rate inside the space charge
region. |
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How large is
GSCR? We cannot use the old equation G =
nmin/t because
nmin is not
constant across the SCR; it rather must look
very schematically somewhat like this: |
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This makes
GSCR a strong function of x - how are we
going to handle this? |
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Well, lets take a kind of average value for the carrier density, and what
suggests itself is the intrinsic carrier density
ni which must be the value at the point where the
two curves cross each other because of the
mass action law stating
ne · nh =
ni2. |
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Within this (questionable) approximation the
maximum current generated in the SCR then becomes |
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We know more. If the external voltage
is zero, the total current is zero and that means that half of the SCR
current must be forward and the other half reverse. We have |
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| jF(0) |
= |
jR(0) |
= |
e · ni · d
t |
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If we now change the barrier height
by eU, the forward current will change
as before. However, we cannot simply multiply
by exp –(eU/kT) as before! |
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Whereas a carrier generated at the bottom of the
hill experiences the full added potential, a carrier generated further up sees
less or even no potential if it originates all the way uphill. |
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So again, lets be sloppy and assume an
average additional energy
barrier, average between everything and nothing - and this will
be eU/2. |
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Again assuming that
jR(U) does not depend on the
barrier height, but somewhat on
U because the width d(U) of the SCR
is voltage dependent, this gives us |
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| jF(U)
= |
e ·ni ·
d(U)
t |
· exp – |
eU
2kT |
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The two components no longer add up to
jmax, but we don´t have to worry about this. We
only would be very wrong for large forward
currents, but in this case the bulk forward
current is always much larger anyway - so it does not really matter much for
the diode behavior. |
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The total current
from the space charge region then becomes jSCR(U) =
jF(U) – jR(U), and written
out we have |
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| jSCR(U) |
= |
e · ni · d(U)
t |
· |
æ
ç
è |
exp – |
eU
2kT |
– 1 |
ö
÷
ø |
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This happens to be exactly the same formula (give or take a factor
of 2) that we would obtain with
the "proper"
theory. |
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We now can write down the diode
equation in all its splendor: |
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| jtotal(U) =
|
e · L · ni
2
t · NAcc
|
æ
ç
è |
exp – |
eU
kT
|
– 1 |
ö
÷
ø |
+ |
e · L · ni
2
t · NDon
|
æ
ç
è |
exp – |
eU
kT
|
– 1 |
ö
÷
ø |
+ |
e · ni ·
d(U)
t |
æ
ç
è |
exp – |
eU
2kT
|
– 1 |
ö
÷
ø |
| |
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e– bulk |
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h+ bulk |
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h+,
e– SCR |
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Lets see what it means in
forward direction: |
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We may neglect the –1 in the
SCR part of the forward current; it then adds a component that increases
with exp(eU/2kT) (for neg.
U), i.e. with half the slope
of the bulk current (in an Arrhenius diagram). The half-slope component will
always "win" at small voltages but pale to insignificance at higher
voltages as shown in the illustration |
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The combined characteristics looks very much like
the measured behavior shown above (The sharp drop of the current close to
U = 0 V is due to the factor –jR in
the diode equation which we neglect for larger voltages). |
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We also see that the exact value of the
SCR forward current indeed only matters for small values, as claimed
above. |
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What do we get
in reverse directions? |
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First, the reverse
current now is voltage dependent because the width of the space
charge region and thus jR increases with a
U1/2 law. |
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Second, the total
reverse current now is larger. Assuming a symmetric junction
(ND = NA = NDop)
and identical life times, and so on, for electrons and holes, it is easy to
calculate the relation
jR(bulk)/jR(SCR); we have |
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jR(bulk)
jR(SCR)
|
= |
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= |
L · ni
N · d |
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The decisive factor is
ni. It decreases exponentially with increasing band
gap Eg. |
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This answers the question why Ge junctions
follow the simple theory, while Si junctions are far off: If
ni >> NDop,
jR(bulk) >> jR(SCR) and the
SCR contribution will not be felt. |
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The generation current from the SCR thus
is much more important in semiconductors with larger band gaps. The
characteristics from above show this rather clearly. |
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Whereas the SCR part may be safely
neglected for Ge (Eg = 0,6 eV), it is about
(102 – 103) times larger than the bulk
diffusion current in Si. |
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The SCR currents should be absolutely dominating in large band-gap
semiconductors. Not only is ni rather small, but
t is very small, too, since these materials
are often direct semiconductors. |
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© H. Föll
