2.3.5 Junction Reconsidered

	In this section we will give the p–n junction a new look using a somewhat more advanced point of view.
		A full treatment of p–n junctions in any kind of three-dimensional semiconductor, taking into account 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 section 2.2.4; this will clear the mind for the essentials.

Junction Without Contributions from the Space Charge Region

	Here we look at an advanced treatment of a p–n junction, but we still will have to make some simplifications.
		We consider a "narrow" "abrupt" 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 densities 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 the 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 diversity's sake, the p- and n-side reversed).
	In what follows, there will be a lot of shuffling formulas around – and somehow, like by magic, the I–V characteristics of a p–n junction will emerge. So let's be clear about what we want to do in the major steps (highlighted by the cyan background).
	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 density around the edge of the space charge region is somewhat larger than 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 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 (= p-side, raised index "p") and the right side (= n-side, raised index "n") of the junction is given by
		ECp – ECn  =  EVp – EVn  =  e · DV
		Here, DV is the difference of the (yet-to-be-determined) electrostatic potential between the n- and p-side, taken far away from the junction; the details (especially about its sign) will be given below.
	The band energy levels ECn,p(x) and the potential V(x) are functions of x, which makes all densities functions of x, too. From now on, however, we will omit the "(x)" when it's not considered necessary.
		As long as we discuss equilibrium, the Fermi energy is constant and the carrier densities 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. (Note that here the usual minus sign in the exponent was used to change the order of the terms in the differences.)
		nep  =  Neffe  · exp   EF  –  ECp kT    density of electrons on the p-side      nhp  =  Neffh  · exp  EVp  –  EF kT  density of holes on the p-side             nen  =  Neffe  · exp   EF  –  ECn kT  density of electrons on the n-side             nhn  =  Neffh  · exp   EVn  –  EF kT  density of holes on the n-side
		We also have the mass action law, here applicable everywhere since we are in full equilibrium:
		nep · nhp  =  nhn · nen  =    nmin · nmaj    =     ni2          =  Neffe · Neffh · exp æ è – Eg kT ö ø
	What we need to know to get on is the x-dependence of the energies or of the potential V(x) – this simply means we need the quantitative band diagram that so far we always just drew "by feeling". (This is one of the essential points why we reconsider the p–n junction here; the other one will be the usage of the quasi-Fermi energies.)
		For this we need to solve the Poisson equation and this demands to specify the total charge r(x) so that we can write down 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 (ne(x)), holes (nh(x)), ionized donors (ND+(x)), and ionized aceptors (NA–(x)). We have as before
		r(x)  =  e · æ è nh(x)  –  ne(x)  +  ND+(x) –  NA–(x) ö ø
		Inserting r(x) into 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 the situation is complicated: Since nh(x) and ne(x) depend on V(x) via the Fermi distribution, this is also an implicit equation for V(x).
		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 characteristic of a p–n 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. eV 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 density is constant up to the junction, and there is only one kind on each side of the junction, i.e.
		nhp(bulk)  =  NA  =  NA–           nen(bulk)  =  ND  =  ND+
		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 away from the junction, the Si extends into infinity (or at least to a distance much larger than several diffusion lengths) to both sides of the junction – in total we use the "abrupt" "large" junction approach
		This gives us for the carrier densities in equilibrium anywhere in the junction:
		nh(x)  =  NA · exp æ è –  e · V(x) kT  ö ø                 ne(x)  =  ND · exp æ è –  e · [V n  –  V(x)] kT ö ø
		Here, V n is the constant value of the potential deep in the n-type region. Note that, having chosen the zero point for V(x) at the p-side of the junction where there is the negative pole of the electric field, it holds inside the SCR that 0 £ V(x) £ V n.
		These equations mean that the carrier density is whatever you have in the undisturbed p- or n-part (i.e., the dopant density) times the Boltzmann factor of the energy shifts relative to this situation.
	V n is the built-in potential for equilibrium conditions, it is thus determined by the difference in the Fermi energies of the n- and the p-side before contact (relative to the band edges) – our simple view of a junction is totally correct on this point.
		With and without an external voltage Uext we have
		V n(Uext=0)  =  1 e  ·  (EFn  –  EFp)           V n(Uext)  =  1 e  ·  (EFn  –  EFp  +  e · Uext)
		Here, the sign of Uext is such that a positive external voltage increases the built-in potential difference. Note that this is just an interim choice; later on we will replace it by the usual standard.
		In the general case, the maximum potential at the n-side, V n(bulk), becomes
		V n(bulk)  =  V n(Uext=0) + Uext  =  1 e  ·  DEF + Uext  =  V n + Uext  =  DV
		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|SCR edge on the n-side, and V p|SCR edge = 0. This is an essential point, even so it is matter-of-course.
	We now can move towards our primary goal and find an expression for the carrier density at the edge of the SCR by considering the ratio of a carrier species on both sides of the junction. From the equations above, we obtain for the edge of the SCR:
		nep nen ÷ ÷ ÷ SCR edge   =    nhn nhp ÷ ÷ ÷ SCR edge   =  exp  æ è – e · DV  kT ö ø   =  exp  æ è – e · (V n  +  Uext) kT ö ø
		The minority carrier densities (always at the edge of the SCR without indicating it anymore) can now be written as
		nep(Uext)  =  nen(Uext) · exp  æ è – e · (V n + Uext) kT ö ø   electrons on the p-side             nhn(Uext)  =  nhp(Uext) · exp  æ è – e · (V n + Uext) kT ö ø 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(V n + Uext) 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 then changes monotonously.
		This means that for equilibrium we must obtain the same equations by computing the minority carrier density from the mass action law, i.e.
		nep(Uext=0)  =  ni2 nhp(Uext=0)
		We will see if this is true in a little exercise:
		Exercise 2.3.5-1 Show the equivalence of the two equations for the minority carrier density!

	Now comes a crucial point: We are looking at stationary non-equilibrium. We first review the starting point again:
		At equilibrium (Uext = 0), the majority carrier densities nen |SCR edge and nhp |SCR edge are given by
		nhp  =  Neffp · exp æ è –  EF kT  ö ø             nen  =  Neffe · exp æ è +  EF  –  ECn kT   ö ø
		Do you remember them? These are two of our first equations from above, but given here for the choice of EVp = 0.
	The essential point for the majority carrier density 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 Uext, i.e.
		nen(Uext) ÷ ÷ SCR edge  =  nen(equ) ÷ ÷ SCR edge  =  nen(bulk)   nhp(Uext) ÷ ÷ SCR edge  =  nhp(equ) ÷ ÷ SCR edge  =  nhp(bulk)
		The trick here is that we consider the majority carrier density at the SCR edge – and the position of the latter may vary with the applied voltage!
		Nevertheless, beyond that point we have the bulk behaviour of the majorities – because that's how we have defined the SCR edge: The bulk potential stays constant right up to the edge, and this is only possible for a constant density of majority carriers.
	The minority carrier densities nep |SCR edge and nhn |SCR edge, however, depend very much on the applied voltage as expressed in the formulae above.
		Thus, we have to adjust the minority carrier density independent of the majority density, which means we have to use the quasi-Fermi energies.
		In other words: While the quasi-Fermi energy EFmaj for majority carriers remains at the equilibrium value EF near the SCR, the quasi-Fermi energy for the minority carriers, EFmin, branches off early; the details will be shown below.
	We now ask about the difference of the minority carrier density relative to equilibrium, i.e. we look at
		Dnep ÷ ÷ SCR edge  =  nep(Uext)  –  nep(Uext=0)   Dnhn ÷ ÷ SCR edge  =  nhn(Uext)  –  nhn(Uext=0)
		For this, we can only use the relationship to the corresponding majority carrier density at the other side of the junction, because only then we include the effect of the applied voltage.
		For both types of carriers it comes out as
		Dnmin  =  nmaj  · æ ç è exp æ è –  e · (V n+ Uext) kT ö ø   –  exp æ è –  eV n kT ö ø ö ÷ ø                    =  nmaj ·  exp æ è – eV n kT ö ø ·  æ ç è   exp æ è – eUext kT ö ø – 1    ö ÷ ø
	Inserting the general expressions for the minority carrier density from above for the case Uext = 0 yields the final formula for our first goal:
		Dnmin ÷ ÷ SCR edge  =  nmin(equ) · æ ç è exp æ è – eUext kT ö ø  –  1 ö ÷ ø
		In other words: The density 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 density must remain constant as a function of time.
		Since deep in the material the minority carrier density is unchanged and has its equilibrium value, we now must have a current, driven by the density gradient alone, and this current must be maintained by the voltage/current source if we want steady state.
		Physically speaking, the excess density 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 Dnp,ne,h(x=0) given by the equation from above and obtain immediately for the current–voltage relationship of a p–n junction (just considering the absolute magnitudes):
		| je(Uext) |  =  e · De Le  · Dne ÷ ÷ SCR edge   or   | je(Uext) |  =  e · De Le  · nep(equ) ·   æ ç è exp æ è – eUext kT ö ø  –  1   ö ÷ ø | jh(Uext) |  =  e · Dh Lh  · nhn(equ) ·   æ ç è exp æ è – eUext kT ö ø  –  1   ö ÷ ø
	We now see that the external voltage, as we have introduced it, raises the potential barrier and therefore decreases the minority carrier density – and, thus, also the current flow.
		This means that, in order to enhance the current flow over the p–n junction, we have to apply the external voltage in a way that it lowers the barrier.
		Therefore, the forward voltage is UD := –Uext, and since it is the forward voltage, it is also the one which is taken as positive; the subscript "D" refers to the p–n junction functioning as a diode.
	For the final result we add the electron and hole currents, drop suffixes and functional arguments now unnecessary, and obtain the diode equation (giving the total current density, including the reverse current, counted in the standard way):
		jD(UD)  =  æ ç è e · nep · De Le   +   e · nhn · Dh Lh ö ÷ ø  ·   æ ç è exp eUD 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:
		For the diffusion length we have
		Le,h  =  æ è De,h · te,h ö ø   1/2   De,h  =  Le,h2 te,h   te,h  =  Le,h2 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) we get
		nep  =  ni2 NA       nhn  =  ni2 ND
	Shuffling everything around with these identities gives us – among many other equivalent formulations – . . .
		jD(UD)  =  æ ç è e · Le· ni2 te· NA  +   e · Lh· ni2 th · ND ö ÷ ø  · æ ç è exp eUD 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! (Hint: Start from UD = 0.)
	What is left is just to consider the quasi-Fermi energies relevant for the forward direction; not only was the relevant drawing promised already above, it will also show explicitly what is meant by "surplus of minorities, having to be moved away from the junction" – because it will show us where those minorities end up.
		To cut a long story short, here it is:
		That the quasi-Fermi energies of the majorities remain constant throughout the SCR corresponds to the expressions giving the ratio of each carrier type on both sides of the junction.
	Note that in the dawing, deliberately there are more minority carriers close to the SCR edges than deeper in the bulk. Yes, that's where the surplus minorities go. But that's not the end of the story:
		That the quasi-Fermi energies of the minorities outside the SCR linearly merge towards the majorities' ones corresponds to the exponential decay of the surplus minority density away from the SCR, with the decay constant given by the diffusion length – as already discussed for the case of pure diffusion currents.
		Think for yourself about why all this is the case! And think about the possible consequences of the surplus minorities' presence in the case of a direct semiconductor.

	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 schematically shown in the following picture:
		The quasi-Fermi energies must behave in the way shown (the details do not matter), because otherwise the density of charge carriers (especially minority carriers!) in the junction would be too high.
		Note that in the dawing there are no minority carriers close to the SCR edges (deliberately!); only in those regions away from the SCR, where there is a single Fermi energy (shown in red), minority carriers are depicted. There, the standard full-equilibrium mass action law holds.
	The decisive point is that we may consider any given thin slice of the SCR to be in local equilibrium, and that the quasi-Fermi energy of the electrons is lower than that of the holes throughout the SCR.
		The latter is a direct consequence of the applied reverse bias, increasing and steepening the potential barrier in the SCR, in combination with the diffusion length of the minorities being larger than the width of the SCR (remember the narrow junction approximation from above).
		This ordering of the quasi-Fermi energies is the exact opposite of the situation that we have considered so far in the recombination business, where we looked at an increased density of carriers, e.g. produced by irradiation with light. Then recombination outweighs generation and UDL, the difference between recombination and generation, was positive.
	Hence, 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 at a deep level means that, while the hole is created in the valence band, the corresponding electron at first just occupies the deep level; only after some time it gets emitted to the conduction band.
		Let's look at this using the formula for UDL:
		UDL  =  v · s · NDL · (ne · nh  –  ni2) ne+  nh +  2ni · cosh EDL – EMB kT
		For making estimates easier, we assume a mid-band level (i.e., cosh[(EDL – EMB)/(kT)] = 1) and ne, nh << ni. This leaves us with
		UDL  =  –   v · s · NDL · ni 2
		For these assumptions we have seen that, treating holes and electrons on equal footing, 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 yields
		jR(SCR)  =  e · ni · d tG
		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 of the SCR to the forward current?
		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 density of surplus minority carriers, Dnep and Dnhn, 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 – at least in our present 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 Dnmin, 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)
		jF(SCR)  =  e · ni · d 2tG · exp eUD 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!

© H. Föll (Semiconductors - Script)