2.2.4 Simple Junctions and Devices

In this section we will look at some junctions in a cursory manner with the goal to get a basic understanding for current flow and the driving forces behind it.
We will see that it is possible (with only a little cutting of corners) to come to a complete quantitative description of the current–voltage relationship of a p–n junction; even including the (usually omitted) generation current part, without doing a lot of math.
We will do this in three steps: First we look at a hypothetical junction, then at the classical p–n junction, then at a real p–n junction with recombination and generation currents in the space charge region.
But don't deceive yourself! These are difficult subjects, even if they are made to look relatively easy here.
We will delve somewhat deeper into these subjects later. However, within the scope of this course, there will not be enough time to go into junction theory thoroughly and in depth.
Here we only attempt to give enough background knowledge, so you can understand the rigid treatment of device physics that can be found in many books.
 

Equilibrium Between a Semiconductor and its Surface as a Model

First, we must realize that the surface is a defect – beyond it, there is no periodic lattice anymore; the bonds of the surface atoms have some other arrangements than the bonds in the bulk.
The dispersion relation of electrons in the two-dimensional surface "crystal" thus must be expected to be different from that of the bulk.
In the simplest approximation conceivable, we may picture the surface with the same band diagram as the bulk (but in only two spacial dimensions), but with plenty of additional states in the band gap.
We have then the following representation:
Surface and surface states
The surface introduces some unspecified states in the band gap of an n-type semiconductor, shown here arbitrarily as acceptor states. The Fermi energy will adjust itself as shown: Since states in the bandgap are available that are lower than the donor states (not shown for graphic simplicity), electrons will move down to these states – so the Fermi energy also comes down.
We will now consider what would be happening if we were to bring the bulk and the surface in close contact. Shown here are three graphical representations of this (fictitious!) process:
Upon contacting, 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. (Note that before that, the surface was not charged.)
Contact 1
 
 
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. Since work is needed to move them to the surface, this is an "uphill" process. (Remember that the band diagram gives the energy of the electrons.)
Contact 2
 
 
As long as the total energy can be reduced by moving electrons from the bulk to the surface, the process will continue. (Remember that energy minimization is the way towards equilibrium.) Equilibrium is reached as soon 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 amount by Dne. This means that equilibrium is given by 0 = dG = (G/ne)bulk · Dne(bulk) + (G/ne)surface · Dne(surface).
Contact 3
From the obvious relation Dne(bulk) = –Dne(surface) it follows that in equilibrium, we have (G/ne)bulk = (G/ne)surface. Since (G/ne) = chemical potential µ = Fermi energy EF, this means that:
EF(bulk)  =  EF(surface)
This is a crucial point; let's make it again:
In thermal equilibrium, the Fermi energy EF is the same everywhere.
If you are not too sure about this, read up about the concept of the chemical potential in the basic module.
This automatically leads to an imbalance of charge, we now have local violations of charge equilibrium.
The area with the band bending will not contain free electrons, but still the positively charged donor atoms (= "ionized"), it thus contains charges in space and is called a space charge region (SCR).
The SCR also contains an electrical field E which is given directly by the derivative of the electrostatic potential V by E = –ÑV (or, for the one-dimensional case, by Ex = –dV/dx), or it can be obtained via the Poisson equation.
 
The Space Charge Region
   
The easiest way to think about this electrical field is to consider the field lines, which start at a positive charge and end at a negative charge.
The negative charges are the surplus electrons sitting in the surface states (and thus also in real space at the surface), the positive charges are the "ionized" donor atoms in the bulk.
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.
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 d between the plates, i.e., dcap = d/2. In other words, we sort of put all the positive charge on a fictitious plate at half the width of the SCR.
The capacitance then becomes
CSCR =  2 er · e0 · A
d
 =  Q
U
With A = area of the capacitor plates, Q = charge on the plates and U = potential difference between the plates.
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 density of donors ND (assuming that all are ionized) times the elementary charge e times the volume d · A, so we have
Q  =  e · ND · d · A
Substituting this in the equation from above, we obtain
2 er · e0 · A
d
 =  e · ND · d · A
U
This gives us one of the more important semiconductor device equations in its simplest form – and this is the correct equation despite the somewhat questionable assumption of dcap = d/2:
d  =  æ
ç
è
2 er e0 Ubi
e ND
ö
÷
ø
1/2
The voltage Ubi 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 a potential, i.e.
Ubi  =  DEF
e
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
d  =  1
e
 ·  æ
ç
è
2 er · e0 · (DEF + e · Uex)
ND
ö
÷
ø
1/2
Note that, since the built-in potential Ubi is taken to be positive, a positive Uex will increase the width of the space charge region – by increasing the potential difference between bulk and surface.
It is easy to to obtain the same equation by integrating the Poisson equation for this case. This is done in an illustration module.
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).
 
"Ideal" p–n Junction
   
We now construct a p–n junction exactly along the recipe given above:
Draw the band diagrams of both parts.
Join the two parts, move electrons to the materials with the lower Fermi energy, holes opposite.
Build up space charges and shift the potentials accordingly until the Fermi energy is the same everywhere.
These steps are illustrated below:
pn junction
As a graphical aid, which will be useful in cases to come, the carriers are schematically indicated as circles (electrons) and squares (holes). Many such symbols being present is meant as an indication for majority carriers, whereas few ones being present indicates the minorities.
The only differences to the first situation is that we now have a space charge region on both sides of the junction.
The field lines are now pointing from the positively charged donors on the n-type side to the negatively charged acceptors on the p-doped side.
The width of the space charge region is now a little more involved to calculate (but there is nothing new); it comes out to
d =   1 
e
· æ
ç
è
2 · er · e0 · æ
è
1
NA
 +  1
ND
ö
ø
 ·  æ
è
DEF + e · Uex ö
ø
ö
÷
ø
1/2
Now let's look at the various currents flowing in the conduction and valence bands without an external voltage.
We know that the net current is zero – we are in an equilibrium condition as long as we do not apply an external voltage (or shine some light on it).
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).
The same reasoning applies, of course, to the holes in the valence band.
The partial currents discerned above have several specific names. The current component flowing uphill in energy is called . . .
Diffusion current, because the driving force behind this current is the density gradient in the carrier density. This always leads to a current component given by Fick's first law to
jDiff = –D e Ñn
With n = density of the carriers in question. (The electric charge e has to be given explicitly in this equation because n only gives the number of particles present.)
It is also alternatively called recombination current because all the electrons (or holes) flowing to the other side become excess minority carriers there and must disappear by recombination, which can be depicted as a current flowing between the bands.
Looking a little ahead, a p–n junction is a diode and currents in diodes are classified as either forward or reverse current. 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.
The partial current flowing downhill in energy is called . . .
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!)
Drift current, because it is the current that results from a drift – caused by the electrical field – superimposed on random diffusion movements.
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 density constant.
Reverse current in the diode nomenclature explained above. We will use mostly this name and abbreviate this current component with jR.
 
Simple Current–Voltage Characteristics
   
In equilibrium, without an external voltage Uex and without illumination, we know that the net current is zero, or
jF = –jR, or, to be more precise,
jF(Uex=0)  =  – jR(Uex=0)
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 notation we have
jF(0)  =  jR(0)
If we draw these currents schematically into the band diagram from above, it looks like this:
Junction currents in equilibrium
Of course, the currents do not flow on both sides of the band edge; this is simply a drawing means.
What happens for finite external voltages? First let's look at the band diagrams:
An external potential Uex 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. But do we have to move it up or down?
It depends: We have two possibilities for choosing the polarity of the external voltage, either increasing the already existing electrical field or weakening it. (Remember that the field lines are pointing from the n-side to the p-side; if you have forgotten why this is so, look again at the explanation given above.)
Thus, the potential on the p-side 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.
We no longer have an equilibrium situation – the Fermi energy is no longer the same everywhere; we must leave it undefined across the junction. (Later on we will have a look at how the Fermi energy behaves inside the SCR.)
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).
The band diagram for the two possible basic cases then look like this:
Biased p–n junction
p–n junction
The diffusion currents are shown increased (thicker arrow) if the energy barrier was lowered, and decreased if it was raised. The drift currents are unchanged.
The situation now is rather simple. The potential step is either increased or decreased. Let's first look what happens to the forward currents jF(Uex).
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).
In other words, the equation for this current includes an exp[–E/(kT)] term. Changing the energy barrier E by D E simply means to multiply the current at zero volts with exp[–DE/(kT)].
Since the magnitude of jF for zero external voltage is just jR(Uex=0), the forward current current jF(Uex) at any voltage Uex must be (and this is true both for the electron and the hole forward current)
jF(Uex)  =  jR(Uex=0) · exp æ
è
DE
kT 
ö
ø
 =  jR(Uex=0) · exp æ
è
e · Uex
kT 
ö
ø
Note that, due to the minus sign in the exponent (which we inherited from the Boltzmann distribution), increasing the energy barrier leads to a decrease of the forward current; we will come back to this later.
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 bicyclists making it will simply change exponentially with the Boltzmann factor relative to their "current" at some reference value if the hill is raised or lowered.
Now to the reverse current jR(Uex).
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).
However, everybody who by accident makes it to the edge of the hill will invariably careen down, i.e., produce a reverse current.
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
jR(Uex)  =   jR(Uex=0)  =   jR
The total current is simply the difference between forward and reverse current for the electrons and the holes, so we have
   
j(Uex)  =  æ
è
jF(Uex)  –  jR ö
ø
e
 +  æ
è
jF(Uex)  –  jR ö
ø
h
                     
j(Uex)  =  æ
è
jRe  +   jRh ö
ø
· æ
ç
è
exp æ
è
–  e · Uex
kT
ö
ø
 – 1 ö
÷
ø
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 were 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.
This is the famous diode equation, and this is all there is to it for straight-forward p–n junctions – except that the technically relevant diode voltage UD, being related to the technical current flow direction, is taken as positive for the current flowing in forward direction; thus, we have UD = –Uex .
This just means that, in order to make the forward current flowing, the external voltage must be applied such that it effectively reduces the built-in potential difference.
Combining the reverse currents of electrons and holes in a single constant j0 := jRe + jRh, we arrive at the final form of the ideal diode equation:
   
j(UD)  =  j0 · æ
ç
è
exp æ
è
e · UD
kT
ö
ø
 – 1 ö
÷
ø
All that is left to do is to consider the reverse currents j0 a bit more closely.
Thinking again about drunken bicyclists, but now driving around on the top of the hill (and thus representing electrons), we might be tempted to assume that jRe should be proportional to their numbers, i.e., to the minority carrier density on the plateau.
This, however, is wrong.
So let's sober up and think a bit harder: You only can extract the bicyclists once! Yet if you want a constant current over time, the best you can do is to take all carriers for the current that are (i) generated per time unit and (ii) making it to the edge.
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
G  =  R =  nmin
t
 =  ni2
t · NDop
With t = lifetime.
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 they have a chance to come close to the drop-off. Obviously we only must consider carriers within a certain distance of the junction, and this distance is, of course, the diffusion length L.
This gives us
jR  =  const. · e · L· G  =  const. · e · L · nmin
t
 =  const. · e · L · ni2
t · NDop
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.
This gives us the complete diode equation exclusively in terms of primary material properties:
 
j(UD)  =  æ
ç
è
e · L · ni2
t · NA
æ
è
exp ( e · UD
kT
)  –  1 ö
ø
ö
÷
ø
electrons   +   æ
ç
è
e · L · ni2
t · ND
æ
è
exp ( e · UD
kT
)  –  1 ö
ø
ö
÷
ø
holes
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.
If we measure the current–voltage characteristic of an "ideal" p–n junction, we will find the following curves:
I-V-characteristics in theOry and experiment
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. So, let's see what went wrong (more details in the link) in each case.
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.
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 100 V . . . 1,000 V), the junction will go up in smoke while drawing a large current called "junction breakdown".
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:
Third: We did not include carrier generation (and recombination) in the space charge region!
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.
This adds four more current components (forward and reverse for holes and electrons) summarily called generation currents from the SCR.
How large are these SCR-caused current components?
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!
Here we go:
 
Current–Voltage Characteristics with Generation Currents from the Space Charge Region
   
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 unit time.
Every generation event produces a hole and an electron, so we have
jmax  =  2e · GSCR · d
With d = width of the SCR, and GSCR = generation rate inside the space charge region.
How large is GSCR? This needs to be considered in detail since it wouldn't help to simply use the known equation G = nmin/t, because nmin is not constant across the SCR; very schematically, it rather looks like this:
Carrier cONCentraiOn acrOss the space charge regiOn
Inside the SCR, this makes G a strong function of x – how are we going to handle this?
Well, let's take a kind of average value for the carrier density, and what suggests itself is the intrinsic carrier density ni, which is the value at the point where the two curves cross each other because of the mass action law stating ne · nh = ni2.
Within this (questionable!) approximation the maximum current generated in the SCR then becomes
jmax  =  2e · ni · d
t
And we know more: If the external voltage is zero, the total current is zero and this means that half of the SCR current must be forward and the other half reverse. We have
jF(0)  =  jR(0)  =  e · ni · d
t
If we now change the barrier height by eUex, the forward current will change as before. However, we cannot simply multiply by exp[–eUex/(kT)] as before!
While 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.
So again, let's be sloppy and assume an average additional energy barrier, average between everything and nothing – and this will be eUex/2.
Again assuming that jR does not depend on the barrier height (but slightly on U since the width d of the SCR is voltage dependent), this gives us:
jF(Uex) =  e · ni · d
t
· exp æ
è
–  eUex
2kT
ö
ø

jR =  e · ni · d
t
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.
The total current from the space charge region then becomes jSCR = jF – jR. Written out and, as above, using UD = –Uex, we have
jSCR(UD)  =  e · ni · d
t
 ·  æ
ç
è
exp æ
è
eUD
2kT
ö
ø
  –  1 ö
÷
ø
This happens to be exactly the same formula (give or take a factor of 2) that we would obtain with the "proper" theory.
We now can write down the diode equation in all its splendor:
  jtotal(UD) =  e · L · ni2
t · NA
æ
ç
è
exp æ
è
eUD
kT
ö
ø
  –  1 ö
÷
ø
  +    e · L · ni2
t · ND
æ
ç
è
exp æ
è
eUD
kT
ö
ø
  –  1 ö
÷
ø
   +    e · ni · d
t
æ
ç
è
exp æ
è
eUD
2kT
ö
ø
  –  1 ö
÷
ø
                             
      e bulk         h+ bulk         h+, e SCR    
We may use the same abbreviation for the bulk reverse current as above and additionally abbreviate the SCR one in a similar manner, then we see more clearly that we effectively have two diodes parallel, the first (prefactor j01) representing the bulk currents and the second (prefactor j02) representing the SCR currents:
   
j(UD)  =  j01 · æ
ç
è
exp æ
è
e · UD
kT
ö
ø
 – 1 ö
÷
ø
  +   j02 · æ
ç
è
exp æ
è
e · UD
2kT
ö
ø
 – 1 ö
÷
ø
Let's see what it means in forward direction:
We may neglect the –1 in the SCR part of the forward current; it then adds a component that increases with exp[eUD /(2kT)], 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:
COmparison bulk- and SCR fOrward current
The combined characteristic looks very much like the measured behavior shown above; the sharp drop of the current close to U = 0 V is due to the –1 in the diode equation (i.e., the reverse current jR) which we neglected for larger voltages.
We also see that the exact value of the SCR forward current indeed only matters for small voltages, as claimed above.
What do we get in reverse direction?
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.
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
jR(bulk)
jR(SCR)
 = 
e · L · ni 2
t · NDop

 =  L · ni
NDop · d
e · d · ni
t
The decisive factor is ni. It decreases exponentially with increasing band gap Eg.
This answers the question why Ge junctions follow the simple theory, while Si junctions are far off: If ni >> NDop, then jR(bulk) >> jR(SCR) and the SCR contribution will not be felt.
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.
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.
The SCR currents should be absolutely dominating in large band-gap semiconductors. Not only is ni rather small, but L = (Dt)1/2 is very small, too, since these materials are often direct semiconductors.

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