2.3.4 Useful Relations

There is no way to cover all relevant semiconductor physics within the scope of this course. This subchapter provides some important or useful relations needed for the understanding of the topics.
It also serves as the "gate" to a number of modules providing additional information.
This subchapter therefore is more open than the other ones; it will fill out and sprout a network in connection with the lecture course that cannot be predicted by now.
 

Einstein Relation

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 · Ñ ne,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 · dn e(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
j   =  e · n(x) · µ · E( x)
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 –  EC(x) – EF
kT
Differentiation of the Boltzmann distribution gives us
dn
dx
 =  Neff · 1/(kT) · dEC(x)
dx
  · exp –  EC(x) – EF
kT
           
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  =  µkT/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
ne  =  ne (equ) + Dne

nh  =  nh(equ) + Dnh
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) + n hnh (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 · (nhne),  in contrast to the total particle density, which is the sum !) we have
r
t
 =  – div (jtotal)
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
Dnh  =  nhn h(equ)
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 = [nhnh(equ)]/t needs to be considered (t is the minority carrier life time).
R D = [nhnh(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 · D h · dnh(x)
dx
Inserting this equation in our continuity equation yields
nh (x)
t
 =  –  n h(x) – nh(equ)
t
 –  nh(x) · µ · E(x )
x
  –  E( x) · µ · n h(x)
x
 +  D · 2nh( x)
x2
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.
We will now look at some applications of this equation.
 
Pure Diffusion Currents
   
Consider the minority carrier situation just outside of the space charge region of a biased p–n junction.
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.
This is now shown in the illustration used before in the simple model of the p–n junction, but the realistic minority carrier situation is now included.
Realistic minority density in p-n-junction
The region outside the space charge region, while now showing a density gradient of the minority carrier density, is essentially field free or at least has only a small electrical field strength.
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)    =  Dn 0 · exp –  x 
L
The length L is given by
L  =  æ
è
Dh · t ö
ø
1/2
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 Dnh(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 · D h
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 · d2 V(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)
     
E V(x)  =  E V(V = 0) – e · V(x)
Thereby, the Poisson equation becomes
e · e 0 · d2V (x )
dx2
 =  e · e0
e
 ·  d2 EC (x)
dx2
 =  e æ
è
n h(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 ECEF 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 [– (ECEF )/(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 penerrate 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.
Band diagrams for fields
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.
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 a 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.
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 EDEF >> kT.
We then have only two charged entities:
ND +  =  ND  
     
n e   =  Neffe  · exp – ECEF
kT 
This means in what follows we only consider the majority carriers.
The Poisson equation then reduces to
e · e 0
e
  ·  d2EC (x)
dx2
  =    e æ
è
ND –  N effe  · exp –   EC( x) – E F
kT 
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 EC0 = EC(x = ¥ ). We thus write
EC(x)   =  EC0 + DE C(x)
The exponential term of the Poisson equation can now be rewritten, we obtain
Neffe  · exp – EC(x) – EF
kT 
 =  N effe  · exp – EC 0EF
kT 
· exp – DE C(x )
kT 
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 · N D
e · e0
 ·  æ
ç
è
1  –  exp æ
è
D EC
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 assume |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  ·  DEC
kT
That is easy now, the solution is
 
DEC(x)  =  DEC (x = 0) · exp – x 
LDb
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 Dr(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 · e 0
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 · e 0
 · 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 –  t 
td

t d   =   e · e0
µ · r0
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
td  =  e · e0
s 
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  =  LDb2
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; td and LD consequently are no longer small.
Moreover, whatever disturbance 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 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.
But this is exactly the process that we have neglected in this discussion (we had all density gradients in the continuity equation set to zero!).
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!
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 (Semiconductors - Script)