Answer:

Given data:

- Critical electric field at breakdown, Ecrit = 4 x 10^5 V/cm
- Breakdown voltage, Vbr = 25 V

To calculate:

- Maximum n-type doping concentration in an abrupt p-n junction

Explanation:

The maximum electric field that can be sustained in a p-n junction without causing avalanche breakdown is given by:

E = Vbr / W

Where W is the depletion width of the p-n junction.

The depletion width W is given by:

W = sqrt((2 e ε Vbi / q) (1 / Na + 1 / Nd))

Where:

- e is the electronic charge
- ε is the permittivity of silicon
- Vbi is the built-in potential of the p-n junction
- q is the electronic charge
- Na is the acceptor doping concentration (p-type)
- Nd is the donor doping concentration (n-type)

The built-in potential Vbi is given by:

Vbi = kT / q ln(Na Nd / ni^2)

Where:

- k is the Boltzmann constant
- T is the temperature in Kelvin
- ni is the intrinsic carrier concentration of silicon

Substituting the expressions for W and Vbi in the expression for E, we get:

E = sqrt((2 e ε kT / q) / (Na Nd (1 / Na + 1 / Nd)) ln(Na Nd / ni^2))

The critical electric field at breakdown Ecrit is defined as the electric field at which the impact ionization rate equals the recombination rate in the depletion region. This can be written as:

Ecrit = q μn n

Where μn is the electron mobility and n is the electron concentration in the depletion region.

Substituting the expressions for E and n, we get:

n = Ecrit / (q μn)

The electron mobility μn is given by:

μn = μn0 / (1 + αn / N)

Where μn0 is the electron mobility at low doping concentrations, αn is the electron scattering coefficient, and N is the doping concentration.

Substituting the expression for μn in the expression for n, we get:

n = Ecrit (1 + αn / N) / (q μn0)

Substituting the values of the constants, we get:

n = 1.5 x 10^10 Na Nd / sqrt(Na + Nd) exp(-0.687 eV / kT) (1 + 0.8 / Nd) / Na

Using the given values of Vbr = 25 V and Ecrit = 4 x 10^5 V/cm, we can calculate the maximum doping concentration Na:

W = Vbr / E = 25 / (4 x 10^5) = 6.25 x 10^-5 cm

Na = Nd = (2 ε Vbi / q W^2) (1 / sqrt(1 + 2 ε Vbi / qW^2 Ecrit) - 1)

Substituting the expressions for Vbi and W, we get:

Na = Nd = (ni^2 / (2 (1 + sqrt(1 + 2 ε kT / q Na W^2 / ni^2))))

Substituting the given values, we get:

Na = Nd = 2 x 10^16 cm