Accumulation

	This is the case where an electrical field of arbitrary origin attracts the majority carriers .
		Starting with the Poisson equation for doped semiconductors and all dopants ionized, we have seen that we can approximate the situation by
		d2DEC dx2  =  –   e2 · NDee0 · æ ç è 1   –  exp – DEC kT ö ÷ ø
		In contrast to the case of quasi-neutrality, we now have (and the sign is important)
		– DEC  >  kT
		This allows the approximation
		1  –  exp – DEC kT  »  – exp – DEC kT
		and the Poisson equation reduces to
		d2DEC dx2  =  –  e2 · ND ee0 · exp – DEC kT
	Using the Debye length LD= {(ee0kT)/(e2ND)}1/2, or ND = ee0kT/e2L2D , the Poisson equation for accumulation can be rewritten as
		d2DEC dx2  =  –  kT   L2D  · exp – DEC kT
		While this looks like a simple differential equation, it is not all that easy to solve it.

	What we would need first, are defined boundary conditions so we can tackle the differential equation. There are no obvious candidates, so we have to think a little harder now.
		Accumulation means that we have some surface charge r that we put on the surface of the semiconductor (with our fictive thin insulating layer in between).
		We thus need to refomulate the differential equation so that surface charge can be included. the (not overly obvious) way to do this is to introduce the electrical field strength E(x) as a new variable besides DEC.
		For that we use the relation
		d2DEC dx2  =  d dx æ ç è dDEC dx ö ÷ ø  =  d dx [e · E(x)]  =  e · dE(x) dDEC  ·   dDEC dx  =  e2 · E(x) ·   dE(x) dDEC
		We also made use of the equaltity dDEC/dx = e · E(x) with E(x) = field strength.
	Inserting and separating the variables (and omitting the "(x)" for clarity) gives
		e2 · E · dE  =  kT L2D  · exp – DEC kT   · dDEC
		Tricky, but worth it. Now we can integrate both sides. The integrations run from far inside the bulk , i.e. from E = 0, to some value of E, and that means from dDEC = 0 to some corresponding value dDEC.
	Omitting the integration (which is trivial), we obtain
		E2 2  =  æ ç è kT e · LD ö ÷ ø 2   ·   æ ç è exp – DEC kT   –   1 ö ÷ ø
		i.e. an equation relating the amount of band bending at some position x to the electrical field strenght at this point, which is
		E(x)  =  ±   kT e · LD æ ç è 2exp – DEC(x) kT   –   1 ö ÷ ø 1/2
		For n-type semiconductors, which we are considering, DEC is negative and large (i.e. DEC >> kT) - and we may neglect the – 1, obtaining
		E(x)  »  ±   kT e · LD  ·   æ ç è 2exp – DEC(x) kT ö ÷ ø 1/2
	While this is fine, we still don't have the solution we want. We must now remember that there is a simple relation tying surface charge to volume charge.
		This is Gauss law, stating that the flux of the electrical field through a surface S is the integral over the components of E perpendicular to the surface.
		The charge is usually expressed in terms of charge density r(x,y,z). Gauss law then states:
		ó õ ó õ  E · n · da  =  1  ee  ·  ó õ ó õ ó õ  r(x,y,z) · dV S   V
		With n = normal vector of the surface S, da = surface increment, dV = volume increment. For more details use the link.
	For our case it means that we could replace the total charge r contained in a slice between x = ¥ (where there is no charge and the field strength is E = E bulk = 0) and x, by a surface (or better areal) charge s area(x) at x given by
		s area(x)   =  ee0 · (E(x)  –  E bulk)  =   ee0 · E(x) s area(x)  =  ±   ee · kT e · LD  ·   æ ç è 2exp – DEC(x) kT ö ÷ ø 1/2
	The total amount of band-bending induced by a real external surface charge sex is simply DEC(x = 0) which we call DEC0:
		DEC0  =  ±  2kT · ln   sex · e · LD 21/2 · ee0 · kT
		So we have all we need. The +/- sign came from the two solutions of the square root; we have to pick the correct one depending on the situation (holes or electrons considered).

To index

Gauss law or integral theorem

© H. Föll