is potential barrier in pn junction is depend upon doping level.if yes then how?

Answer 1

Yes..
In short.. the greater the doping, the narrower the depletion region.. so lesser potential barrier.

Answer 2

Yes . The built in pential or the potential barrier height depend on the doping level as follows.

Before the PNJunction is Formed :-

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).

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)


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.


After the Junction is Made:-.

A new thermal equilibrium is established when Fermi level of the p-side lines up with Fermi level of the n-side.
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.

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)
Hence the built in potential and hence the potential barrier is decided by the doping concentration level and the temperature.

Answer 3

in the pn diode the width of the barrier in forward bias is large under unbiased conditions. this is when the barrier potential will be equal to 0.7v. but when it is in forward bias the depletion region decreases and due to this the charge carriers flow from p side to n side. but when the width of the barrier decreases the potential should also decrease but it remains the same 0.7v. can anyone please tell me why i really want an answer .