2.3.5 Junction Reconsidered

Junction Without Contributions from the Space Charge Region

In this subchapter we will give the pn-junction a new look using a somewhat more advanced point of view.
A full treatment of pn-junctions in any kind of three-dimensional semiconductor, allowing for arbitrary doping profiles, finite size, and effects of the interfaces and surfaces, is one of the more difficult things to do in semiconductor theory; we will not attempt it here.
In all standard treatments of junctions, we always look at special (unrealistic) cases and use (lots of) approximations. This, admittedly, can become somewhat confusing.
It is thus very advisable, to become really well acquainted with the simple treatments given in subchapter 2.2, this will clear the mind for the essentials.
Here we look at an advanced treatment of a pn-junction, but we still will have to make some assumptions.
We consider a "narrow" one-dimensional junction in an infinitely long crystal, which is formed by a sudden change of doping, i.e. there are no gradients of the doping concentrations ND and NA to the left or right of the junction.
"Narrow" then means that the width of the space charge layer is much smaller than the diffusion length of minority carriers, but still much larger than the mean free path length needed for thermalization of carriers
In consequence, we do not consider recombination in the SCR, and we can assume thermal equilibrium in the bands, i.e. we can use the Quasi-Fermi energies.
First, we look at the junction in equilibrium, i.e. there is no net current and the Fermi energy is the same everywhere (we have the same situation as shown before; but for diversities sake, the p- and n-side reversed).
p-n-junction in equilibrium
In what follows, there will be a lot of shuffling formulas around - and somehow, like by magic, the I-V-characteristics of a pn-junction will emerge. So let's be clear about what we want to do in the major steps (highlighted in blue)
The first basic goal is to find an expression for the carrier density at the edge of the space charge region.
We know in a qualitative way from the consideration of pure diffusion currents that the minority carrier concentration around the edge of the space charge region is somewhat larger then in the bulk under equilibrium conditions - there is a Dnmin given by Dnmin = nmin(x) - nmin(bulk) and Dnmin will increase for forward bias, i.e. for non-equilibrium conditions.
We also know that Dnmin induces a diffusion current and that we therefore need a "real" current to maintain a constant Dnmin. Finding Dnmin and, if necessary, Dnmax, thus will automatically give us the necessary currents belonging to the non-equilibrium as defined by the voltage.
We first look at the various energies involved:
The energy difference between the left side (= n-side; index "n") and the right side (= p-side; index "p"; these indizes will appear "up" or "down" as dictated by the limitations of HTML) of the junction is given by
E CnECp  =  EVnEVp  =  e · (V nV p)  =  e · DV
With V n, V p = constant potentials at the n or p-side and far away from the junction, respectively.
The band energy levels ECn, p = ECn, p(x) and the potential V(x) are functions of x, which makes all concentrations functions of x, too. We will, however, write all these quantities without the "(x)" from now on.
As long as we discuss equilibrium, the Fermi energy is constant and the carrier concentrations are given by their usual expression. We consider them separately for the left- and right hand side of the junction i.e. for the n- and p-part.
npe  =  N eeff  · exp   EF  –  ECp
kT 
 =    concentration of
electrons on the p-side 
   
nph  =  N peff  · exp  EVp  –  EF
kT 
 =  concentration of
holes on the p-side
             
nne  =  N eeff  · exp   EF  –  ECn
kT 
 =  concentration of
electrons on the n-side
             
nnh  =  N peff  · exp   EVn  –  EF
kT 
 =  concentration of
holes on the n-side
We also have the mass action law applicable everywhere in equilibrium
npe · nph  =  nnh · nne  =  ni2  =  N eeff · N peff · exp – Eg
kT
Algebraic manipulation gives us any number of relations, some of which may be useful later on. Here some examples:
npe
nne
 =  nph
nnh
 =  exp e · (V nV p)
kT

ECpEVn  =  – kT · ln  nne
N eeff
  – kT · ln  nph
N heff

e · (V nV p)  =  ECpECn  = Eg + kT   · nne · nph
N eeff · N heff
This is about all we can do in full generality.
What we need to know to get on is the x-dependence of the energies or of the potential V(x) - this simply measn we need the (quantitative) banddiagram that so far we always just drew "by feeling".
For this we need to solve the Poisson equation and this demands to specify the total charge r(x) so we can write dow the charge as a function of x. This is easy in principle:
The total (space) charge r(x) at any point along the junction is the sum of all charges: Electrons, holes, ionized donors (N+D(x)), and ionized aceptors (NA(x)). We have as before (with the necessary "minus" sign being part of the elementary charge e)
r(x)  =  e · æ
è
nh(x)  –  ne(x)  +  ND+(x ) +  NA(x) ö
ø
Inserting r in the Poisson equation gives
e e0
e 
  ·   d2V(x)
dx2
 =  nh(x)  –  ne(x)  +  ND+(x)  +  NA(x)
Solving this equation with the proper boundary conditions will yield V(x) and everything else - but not so easily because nh(x) and ne(x) are complicated functions of V(x) via the Fermi distribution.
It is, however, not too difficult to find good approximations for "normal", i.e. highly idealized junctions; this is shown in an advanced module accessible through the link.
For our final goal, which is to describe the current-voltage characteristics of a pn-junction, we use the same approximations and conventions, namely
1. The zero point of the electrostatic potential is identical to the valence band edge in the p-side of the junction , i.e. e · V p = EVp = 0 as shown in the complete illustration to the situation shown in the picture above. This is a simple convention without any physical meaning.
2. All dopants are ionized, their concentration is constant up to the junction, and there is only one kind on both sides of the junction, i.e.
nhp(bulk)  =  NA  =  N A
         
nen(bulk)  =  ND  =  N +D
This is a crucial assumption. Note that while nh, ep, n(bulk) are constant, this is not required for nh, ep, n(x) around the junction.
3. We also assume that the Si extends into infinity (or at least to a distance much larger than a diffusion length) to both sides of the junction - in total we use the "abrupt" "large" junction approach
This gives us for the carrier concentrations in equilibrium anywhere in the junction:
nh(x)  =  NA · exp –   e · V(x)
kT  
       
       
ne(x)  =  ND · exp –   e · [V(x)  –  V n]
kT
With V n = constant value of the potential deep in the n-type region.
These equations are already a special case of the more general equations above. What they mean is that the carrier concentration is whatever you have in the undisturbed p- or n-part (i.e the dopant concentration) times the Boltzmann factor of the energy shifts relative to this situation.
V n is the difference of the built-in potentials for equilibrium conditions, it is thus determined by the difference in the Fermi energies of the two semiconductors before contact - our simple view of a junction is totally correct on this point
With and without out an external voltage U we have
V n(U = 0)  =   1
e
 ·   (E nF  –  E pF)
         
V n(U)   =   1
e
 ·   (E nF  –  E pF  +  e · U)
The maximum potential at the n-side, V n(bulk) becomes
V n(bulk)  =  V n(U = 0) + U  =  1
e
 ·   DEF + U  :=  V n     for brevity
Looking at the proper solution of the Poisson equation for our case, we realize that the space charge region was defined as the part of the Si where the potential was not yet constant. This means that V n(bulk) = V n = V n|edge SCR on the n-side, and V p|edge SCR = 0. This is an essential point, even so it is matter-of-course.
We know can move towards our primary goal and find an expression for the carrier density at the edge of the SCR by inserting the proper potential in the n pe and n ne equation from above. We obtain for the edge of the SCR:
npe
nne
÷
÷
÷
edge
SCR
  =    nnh
nph
÷
÷
÷
edge
SCR
  =   exp – e · (V n  +  U)
kT
The minority carrier concentrations (always at the edge of the SCR without indicating it anymore) can now be written as
npe(U)  =  nne(U) · exp – e · (V n + U)
kT
  electrons on
the p-side
           
nnh(U)  =  n ph(U) · exp – e · (V n + U)
kT
holes on
the n-side
These equations are nothing but the Boltzmann distribution giving the number of particles (nmin) that make it to the energy e(Vn + U) out of a total number nmaj - in thermal equilibrium. We used essentially the same equation before, but now we know the kind of approximations that were necessary and that means we also know what we would have to do for "better" solutions of the problem.
Since this is important, let's review the approximations we made:
Besides the "abrupt" "large" junction, we used the approximations from the simple solution to the Poisson equation which implies that the potential stays constant right up to the edge of the SCR and than changes monotously.
This means that for equilibrium we must obtain the same equations by computing the minority carrier concentration from the mass action law, i.e.
npe(U = 0)  =  ni2
n ph(U = 0)
We will see if this is true in a little exercise
Exercise 2.3.5-1
Show the equality of the two equations for the minority carrier density
Now comes a crucial point: We are looking at non-equilibrium. We first review the starting point again:
At equilibrium (U = 0), the majority carrier concentrations nne |edge SCR and nph |edge SCR are given by .
nph  =  Npeff · exp –   EF
kT  
       
       
nne  =  Neeff  · exp +   (EF  –  e · V n)
kT  
- our first equations from above, but expressed now with the normalized potential V. This is fully correct as implicitely shown in the exercise. However, we are now going to look at non-equilibrium.
The essential point for the majority carrier concentration at the edge of the space charge region for non-equilibrium is that it remains practically unchanged (approximately at its bulk value) if we now apply a voltage U, i.e
nn, he, h(U) ÷
÷
edge
SCR
 =  nn, pe, h(equ) ÷
÷
edge
SCR
 =  nn, pe, h ÷
÷
bulk
This can be seen from the equations above since neither EF nor [EF  –  e · (V n + U)]|edge SCR depend on the voltage since the Fermi energy in the bulk moves up or down with the band edges!
The minority carrier concentrations npe |edge SCR and nnh |edge SCR, however, depend very much on the applied voltage as expressed in the formula above if we replace Vn by Vn + U.
In other words: While the quasi-Fermi energy EmajF for majority carriers remains at the equilibrium value EF near (and even for some distance in) the SCR, the quasi-Fermi energy for the minority carriers, EminF, branches off early (the illustration below shows this, albeit in a slightly different connection).
We now ask for the difference of the minority carrier concentration relative to equilibrium, i.e. for
D np, ne, h ÷
÷
edge
SCR
 =  n p, ne, h(U)  –  n p, ne, h(U = 0)  
It comes out as
Dn pe  =  n ne  · æ
ç
è
exp –  e · (V n + U)
kT
  –   exp –  eVn
kT
ö
÷
ø
                 
Dnnh  =  n ph æ
ç
è
exp +  e · U
kT
      –  1 ö
÷
ø
   
Inserting the general expressions for the minority carrier concentration from above yields the final formula for our first goal:
Dnpe ÷
÷
edge
SCR
 =  npe(equ) · æ
ç
è
exp   eU
kT
  –  1 ö
÷
ø
                   
Dnnh ÷
÷
edge
SCR
 =  nnh(equ) · æ
ç
è
exp   eU
kT
  –  1 ö
÷
ø
In other words: The concentration of minority carriers at the edge of the SCR will be changed by an external voltage. In steady state conditions (which does not imply equilibrium, just that nothing changes) this concentration must remain constant as a function of time.
Since deep in the material the minority carrier concentration is unchanged and has its equilibrium value, we now must have a current, driven by the concentration gradient alone, and this current must be maintained by the voltage/current source if we want steady state.
Physically speaking, the excess concentration of minority carriers will diffuse around and disappear after some diffusion lengths - deep in the material they are not noticeable any more.
This is exactly the situation treated under "useful relations" for pure diffusion currents.
We can take the formula derived there with Dn e, hp, n(x = 0) given by the equation from above and obtain immediately for the current-voltage relationship of a pn-junction
j e(U)  =  e · De
Le
 · Dne ÷
÷
edge
SCR
  or  

j e(U)  =  e · De
Le
 · nep(equ) ·   æ
ç
è
exp   e · U
kT
  –  1   ö
÷
ø

j h(U)  =  e · Dh
Lh
 · nhn(equ) ·   æ
ç
è
exp   e · U
kT
  –  1   ö
÷
ø
In terms of sign conventions we assumed that positive voltages lower the barrier.
For the final result we add the electron and hole currents, drop suffixes and functional arguments now unnecessary, and obtain the diode equation:
jtotal(U)  =  æ
ç
è
e · n ep · De
Le
  +   e · n hn · Dh
Lh
ö
÷
ø
 ·   æ
ç
è
exp e · U
kT 
  –  1 ö
÷
ø
This is the same equation as before, if we take into account that the pre-exponential factor can be written in many ways. To see that we use the following identies:
From the Einstein relation we have
L e, h  =  æ
è
De, h · te, h ) ö
ø
 
1/2

  D e, h  =  (L e, h ) 2 · te, h  

te,h  =  (Le,h ) 2
De,h
   
From the mass action law which is still valid for the bulk, and the general approximation for the majority carrier density (that is already contained in our equations) anyway) we get
ne,hp,n  =  ni2
nh, ep, n
     
n ep  =  ni2
NA
     
n hn  =  ni2
ND
Shuffling everything around with these identities gives us - among many other equivalent formulations
jtotal(U)  =  æ
ç
è
e · L e · ni2
te · NA
 +   e · L h · ni2
th · ND
ö
÷
ø
 · æ
ç
è
exp e · U
kT 
  –  1 ö
÷
ø
and that is exactly the equation we got before! However, we did not have to "cut corners" this time and we did not have to assume that some proportionality constant equals 1!
More important, however: The interpretation of what happens may now be different. Different in the sense of looking at one and the same situation from a different point of view, not different in the sense that it is something else. The two points of view are complementary and not mutually exclusive; neither one is wrong!
In the simple picture we looked at the minority carriers that had to be generated to account for the loss of carriers accounting for the reverse current and running down the energy slope.
Here we looked at the surplus of minorities accounting for the forward current and which has to be moved away from the junction.
Think about, why this is the same thing (start with U = 0)!
 
Contributions from the Space Charge Region
   
We now should include the generation currents from the space charge region, as we did (in a somewhat fishy way) in our simple consideration of a junction.
This, however, is not so easy to do in a correct (albeit still very approximate) fashion.
For the reverse part of the generation current from the SCR, we can obtain an equation directly from the Shockley-Read-Hall theory. All we have to do is to consider the Quasi-Fermi energies of a junction in reverse bias. This is shown in the picture.
Reversely biased juction and Quasi-Fermi energies
The Quasi-Fermi levels must behave in the way shown (the details do not matter), because otherwise the concentration of carriers in the junction would be too high (it must be close to zero!). The decisive point is that in any given thin slice of the SCR, we may consider to be in local equilibrium, and that the Quasi-Fermi energy of the electrons is lower than that of the holes.
This the exact opposite of the situation that we have considered so far in the recombination business, where we looked at an increased concentration of carriers, e.g. produced by irradiation with light. Then recombination outweighs generation and UDL, the difference between recombination and generation was positive.
In the case we are considering here, UDL is negative, i.e. there is more generation than recombination. And this means that the space charge region is busily producing carriers, always in pairs because of neutrality, which will run down the energy barrier producing an additional reverse current.
Pair production means that a deep level first emits a hole to the valence band, and than an electron to the conduction band.
Lets look at this using the formula for UDL:
UDL  =  

v · se · NDL · (ne · nh  –  ni2)

nenh  +  2ni · cosh EDLEMB
kT
For making estimates easier, we assume a mid-band level (i.e. cosh[(EDLEMB)/kT = 1), and ne, nh << ni. This leaves us with
UDL  =  –   v · se · NDL · ni
2
For these assumptions we have seen that 1/v ·se · NDL = te, or since we do have holes and electrons now on an equal footing, we drop the suffixes and write 1/v ·s · NDL = t.
However, because we now have more generation than recombination, t is now called the generation life time tG for this case. (More to that topic in the link).
This leaves us with a net generation of one kind of carrier of
UDL  =  G  =   ni
2tG
The current density from the net generation of carriers in the SCR is then given by the product of the net generation rate with the width d of the SCR; adding up the holes and the electrons, yielding
jR(SCR)  =  e · ni · d
tG
With d = width of the SCR.
and this is exactly the same formula (give or take a factor of 2) as in our "quick and dirty" estimate from before. The physical reasoning wasn't so different either, if you think about it.
How about the contribution to the forward current from the SCR?
The proper treatment is much more complicated and physically different from our simple explanation. The physical reasoning is as follows:
We have seen that we need to sustain a certain concentration of surplus minority carriers, Dne, hp, n at the edges of the SCR to maintain local equilibrium. The surplus carriers needed were injected from the other side of the junction and crossed the junction without losses - in our running approximation.
In reality, however, some injected holes from the p-side will recombine with the injected electrons from the n-side. Recombination in the SCR thus reduces the current needed to maintain Dne, hp, n and an additional current has to be produced which exactly compensates the losses.
The necessary calculations are shown in an advanced module, suffice it to state here that the final result for the forward current from the SCR is (in a rather crude approximation)
jR  =  e · ni · d
2tG
· exp – eU
2kT
Again, besides the factor 2 (and the new kind of life time), the same formula as before. But this time it was a kind of lucky coincidence, not really very well justified.
Or was it? Think about it!

With frame Back Forward as PDF

© H. Föll (Semiconductor - Script)