Solving the Poisson Equation for p–n Junctions

	We have the general expression for the space charge r(x) and the Poisson equation:
		r(x)  =  e · {nh(x) – ne(x) + ND+(x) – NA–(x)} e e0  ·   d2V(x) dx2  =  – r(x)
	V(x) is the internal electrostatic potential due to the space charge region, resulting from the flow of majority carriers to the other side.
		The zero point of this potential is set identical to the valence band edge in the bulk of the p-side of the junction.
		Since this side is negatively charged, the value of V(x) increases towards the n-side. There it reaches its maximum value, representing the increase in potential of a test charge that has been moved against the field across the whole junction.
	The internal electric field due to the space charge region (written in color to distinguish it from the energy) is related to the internal electrostatic potential according to the standard definition
		E(x)  =   –  dV(x) dx
		We consider the arrangement shown in the illustration: The p-side is on the left (where x < 0) and the n-side on the right (where x > 0). As a consequence, V(x) has a positive slope, and the electric field is negative. This corresponds to the polariy of the space charges.
	We consider a solution for the following approximations:
1.	All dopants are "ionized", i.e. NA– = NA = nh and ND+ = ND = ne.
		This is always valid as long as the Fermi level is not very close to a band edge.
		For the carrier density we have the general expression
		ne,h  =  NeffC,V  · exp æ è – DE kT ö ø
		Here, DE is EC – EF for electrons and EF – EV for holes.
		If EC,V are functions of x because the bands are bent (while EF stays constant), we may write the energy difference as DE = DE0 + e · V(x) for holes and DE = DE0 + e · {V n – V(x)} for electrons, with DE0 referring to the relevant situation without band bending (i.e., in the respective bulk parts) and V n being the constant value of the internal potential in the n-type bulk part, i.e. V n is the maximum value of V(x).
		The carrier densities then become
		nh(x)  =   NeffV · exp æ è – DE0 + eV(x) kT ö ø  =  NeffV · exp æ è – DE0 kT ö ø · exp æ è – eV(x) kT ö ø                      =   NA · exp æ è – e · V(x) kT ö ø                              ne(x)  =   NeffC · exp æ è – DE0 + e{V n – V(x)} kT ö ø  =  NeffC · exp æ è – DE0 kT ö ø · exp æ è – e{V n – V(x)} kT ö ø                      =   ND · exp æ è – e · {V n – V(x)} kT ö ø
		because the term with DE0 gives the relevant density in the bulk and that is the dopant density in our approximation.
2.	The density of the space charges is taken as given by a box profile.
		This is motivated by the fact that, as soon as V(x) deviates noticeably from its constant value of 0 (in the p-type bulk) or V n (in the n-type bulk) – in other words: inside the space charge region – the carrier densities decrease exponentially from their values NA or ND far outside of the SCR.
		The carrier densities then become
		nh  =  NA   for   x  <  – dA               nh  =  0   for   – dA  <  x  < 0               ne   =  ND   for   x  >  dD                ne  =  0   for   0  <  x  <  dD
		With dA, dD = boundaries of the space charge region with x = 0 at the geometrical junction.
	Thus, the space charge is only given by the density of the dopants. That's where we could have started right away, just plugging in the usual assumptions. We have
		r  =  – NA   for   – dA  <  x  <  0               r  =  ND   for   0  <  x  <  dD               r  =  0   for   everywhere else
	The Poisson equation then becomes
		d2V dx2   =   0 for – ¥  <  x  <  – dA                 d2V dx2   =  +  e  ee0 NA   for   – dA  <  x   <  0                 d2V dx2   =  –   e  ee0 ND   for   0  <  x  <  dD                 d2V dx2   =  0   for   dD  <  x  < ¥
		In addition, we have these boundary conditions:
		V  =  0 ö ÷ ÷ ÷ ø   for   x  =  – dA         dV dx   =  0                 V  =  V n ö ÷ ÷ ÷ ø    for   x  =  dD         dV dx  =  0               dA · NA   =  dD · ND   Charge neutrality
	The solutions are easily obtained, they are
		VA(x)  =  e  2ee0  · NA · (dA + x)2     for      – dA  <  x  <  0   VD(x)   =  V n  –   e  2ee0 · ND · (dD  –  x)2   for         0  <  x  <  dD     V n     =    e 2ee0 (NA · dA2  +  ND · dD2)
		The last equation comes from the condition of continuity at x = 0, i.e. VD(x = 0)  =  VA(x = 0).
	The two limits of the space charge region, dA and dD, as well as the field strength E = – dV/dx in the SCR thus could be calculated if we knew V n.
		V n, of course, is the built-in potential Ubi, i.e. it is the difference of the potential across the SCR and thus identical to 1/e times the difference of the Fermi energies before contact in thermal equilibrium; hence, we have
		V n   =  EFn  –   EFp e   =:  DEF e   =   Ubi
		If we superimpose an external voltage Uext, V n becomes (watch out for the correct sign!)
		V n   =   DEF e    ±  Uext
	Now that we know (at least in principle) the internal potential V(x), we can draw a band diagram givging the total energy of the electrons – and the latter is most important for the correct understanding of the band bending (i.e., the dependence of EC,V on x). Here is why:
		In general, the potential energy of an arbitrary charge q in an arbitrary electrostatic potential V(x) is just q · V(x).
		In the band structure of the p–n junction, we need to include the potential energy of the electrons (remember that the band structures of our semiconductors always give the energies of the electrons). And since an electron carries a negative elementary charge, we have to add the term q · V(x) = –e · V(x) to the standard energy diagram coming from the quantum-mechanical properties of the electrons.
		Since this electrostatic term is negative, the band bending shows the opposite behavior compared to that of the internal potential: Where the latter goes up, the former goes down.
	The following illustration shows the whole situation in one drawing (without external voltage):
	However, we still need to find the width dSCR of the space charge region.
		For this, we first consider the p–n junction in equilibrium and only later add an external voltage.
		The electric field is the strongest at the inflection point of the internal potential, i.e., at x = 0; the relevant absolute value is
		Emax    =     | E(x = 0) |    =     e ee0 · NA · dA    =     e ee0 · ND · dD
		With this, the above explicit result for V n can be written as
		V n   =   e 2ee0 (NA · dA2  +  ND · dD2)   =   1 2 Emax (dA  +  dD)
		Combining this with the relation expressing charge neutrality, dA · NA = dD · ND, and re-inserting the relevant expression for the max. field strength, one finds for the boundaries of the space charge region
		V n   =   1 2 Emax dA  æ è 1  +  NA ND ö ø     Þ     dA   =   2 V n Emax   ND NA + ND   =   2 ee0 V n e dA   ND  NA (NA + ND)   Þ     dA   =   æ ç è 2 ee0 V n e    ND  NA (NA + ND) ö ÷ ø 1/2   V n   =   1 2 Emax dD  æ è 1  +  ND NA ö ø     Þ     dD   =   2 V n Emax   NA NA + ND   =   2 ee0 V n e dD   NA  ND (NA + ND)   Þ     dD   =   æ ç è 2 ee0 V n e    NA  ND (NA + ND) ö ÷ ø 1/2
	We now can express the width dSCR of the space charge region as
		dSCR  =  dA  +  dD            =   æ ç è 2 ee0 V n e (NA + ND) ö ÷ ø 1/2 ·  æ ç è æ è ND NA ö ø 1/2  +  æ è NA ND ö ø 1/2 ö ÷ ø              =   æ ç è 2 ee0 V n e (NA + ND) ö ÷ ø 1/2 ·  ND + NA (NA ND)1/2            =   æ ç è 2 ee0 V n e  ND + NA NA ND ö ÷ ø 1/2   =     æ ç è 2 ee0 V n e  æ è 1 NA  +  1 ND ö ø ö ÷ ø 1/2
	Replacing the internal potential by the Fermi energy difference and including an external voltage, we obtain the final result as
		dSCR  =   1  e æ ç è 2ee0 · æ è 1 NA  +  1 ND ö ø · æ è DEF + e · Uext ö ø ö ÷ ø 1/2

With frame

3.4.1 Junction Diodes

2.2.4 Simple Junctions and Devices

2.3.4 Useful Relations

2.3.5 Junction Reconsidered

Basic Equations

Potential Discontinuities and Dipole Layers

Solution to 2.3.5-1

© H. Föll (Semiconductors - Script)