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Mathematical Analysis

Before the Junction is Formed

We have two bulk Si: p-type in red rectangle, and n-type in blue-rectangle. Underneath each is the equilibrium energy band diagram. The majority carrier concentration, or the impurity doping level (Nd and Na), is related to the separation of the Fermi level, ----------Ef, from the energy band edge (Ec or Ev). Click here to see an applet for a graphical illustration of Ef vs. n or p.

n-type: n = Nd = Nc exp[ -(Ec - Ef) / kT ].   -------- (1a)

p-type: p = Na = Nv exp[ -(Ef - Ev) / kT ].   -------- (1b)

Where Nc and Nv are the effective densities of states for conduction band and valence band, respectively, given by

Nc = 2 ( 2 p me* kT / h2 )3/2,   ------ (2a)

Nv = 2 ( 2 p mh* kT / h2 )3/2.   ------ (2b)

At room temperature (300K),

 Si GaAs Ge Nc (cm-3) 2.8x1019 4.7x1017 1.04x1019 Nv (cm-3) 1.04x1019 7.0x1018 6x1018

From eq.(1a) and (1b), the Fermi level relative to the band edge, Ec - Ef and Ef - Ev, is given by

n-type: Ec - Ef = kT ln(Nc/Nd), ------ (3a)

p-type: Ef - Ev = kT ln(Nv/Na)   ------ (3b)

Note that equations (1) and (3) apply to moderately-doped (nondegenerately-doped) semiconductors only. The upper limit for moderate doping level is approximately given by

n-type: Nd < 0.05 Nc, ------ (moderate doping)

p-type: Na < 0.05 Nv. ------ (moderate doping)

For materials more heavily doped than this, the accurate Fermi-Dirac statistics must be used. This applet implements the Maxwell-Boltzman approximation, eq.(1) or (3). Click here to see an applet for the heavily-doped (degenerately-doped) case.

A new thermal equilibrium is established when Fermi level of the p-side lines up with Fermi level of the n-side. The initial animation shows the diffusion of mobile carriers (due to the concentration gradient) which 'depletes' the carriers on either side of the junction. The depletion region is the white region in the sample schematic (top) and the curved region of the band diagram (bottom).

In order to understand the magnitude of concentration gradient: consider Na = 1015 cm-3 and Nd = 1015 cm-3 and a depletion thickness of W = 1 mm = 10-4 cm. The minority hole concentration in n-side is

pn = ni2/Nd = 1020 cm-6 / 1015 cm-3 = 105 cm-3.

Thus the hole concentration gradient in a region near the junction is

dp/dx = (pp - pn) / W = (Na - ni2/Nd)/W = (1015 - 105)cm-3 / 10-4 cm = 1019 cm-4.

Electron concentration gradient is the same for this example. This will induce a large diffusion flux for the mobile carriers. For hole diffusion coefficient, Dp = 10 cm2/s,

hole diffusion flux = Dp dp/dx = 10 cm2/s * 1019 /cm4 = 1020 holes/cm2.s

1020 holes move per cm2 per second by diffusion !

As soon as the junction is made, there is initially no mechanism to conuteract the diffusion process. This causes carriers to leave the regions of its own type, leaving behind the semiconductor which is depleted of mobile charges. The depleted semiconductor now has the ionized impurity charge not compensated by the carrier charges. Thus the depleted regions have net space charge.

As the new thermal equilibrium is made (Fermi level lined up!), the hole diffusion flux is exactly matched by an opposing hole flux. This opposing hole flux is due to the electric field forces holes in direction opposite to the diffusion direction. The electric field arises in the depletion region due to the uncompensated ionized impurity charge (the space charge). Thus the net hole flux is now zero. The same thing is true for electrons.

I. The built-in potential, V0.

Once the new equilibrium is reached, the built-in potential (the potential energy difference between the Ec of p-side and the Ec of n-side), V0, can be found from

pn = pp exp(-qV0/kT) [that is, ni2/Nd = Na exp(-qV0/kT)]   ---- (4a)

np = nn exp(-qV0/kT) [that is, ni2/Na = Nd exp(-qV0/kT)] ---- (4b)

Equation (4a) relates the equilibrium hole concentrations between the n-side (minority hole) and p-side (majority hole), and equation (4b) relates the equilibrium electron concentrations between the p-side (minority electron) and the n-side (majority electron).  Therefore, the built-in potential is found if the doping levels are known:

V0 = (kT/q) ln(NaNd/ni2)   ----- (5)

II. The depletion region thickness xn, xp, and W = xn + xp.

Two equations are needed. Both are obtained by solving the Poisson's equation:

e dE/dx = q (p - n + Nd - Na) --- (6a)

where E = electric field. If we assume that

p » 0 and n » 0 for - xp < x < xn   [depletion approximation]  --- (7)

Then the Poisson equation in the depletion region [in depletion approximation] becomes

e dE/dx = - q Na for -xp < x < 0   ---- (6b).

e dE/dx = q Nd for 0 < x < xn    ----- (6c)

The first equation needed is the solution of (6b) and (6c) with the boundary condition [E(x) is continuous at x=0]:

E0 = - (q/e) Naxp = - (q/e) Ndxn  ---- (8)  [charge balance equation]

Using E = - dV(x)/dx, the Poisson eq. becomes

e d2V/dx2 = - q (Nd - Na) for - xp < x < xn    ---- (9)

The second equation needed is obtained by two successive integrations of (8) over the depletion region:

V0 = - 1/2 E0 (xp + xn)   ----- (10)   [built-in potential]

Where, V0 is found from (5). From (8) and (10) we find:

W = xp + xn = [ (2 eV0/q ) (1/Na + 1/Nd) ]1/2, ------ (11a)

xp = W Nd /(Na + Nd), ------ (11b)

xn = W Na/(Na + Nd), ------ (11c)

Equations (5) and (11) are used in the calculation of numbers in theis applet.