Question

A p-n junction is of area 1x10 cm? formed by doping at levels at Na = 1015 cm and Np = 101 cm and ni= 1.5 x1010cm. Calculate build in voltage of this p-n junction, zero bias space charge width and current at 0.5 volt forward bias conditions using parameters: Dielectric constant of the material = 11.8 F/cm, Mn =1500 cm?/V-sec, Hp =450 cm/V-sec, Tn = Tp = 2.5 x10- sce. (Note permittivity of free space is 8.85 x 10-14, = =

Answer 1

The build-in voltage of the p-n junction can be calculated using the formula:

Vbi = (k * T * ln(Na * Np / ni^2)) / q

where:

k = Boltzmann constant = 8.617 x 10^-5 eV/K

T = Temperature in Kelvin = 2.5 x 10^2 K

Na = Acceptor concentration = 10^15 cm^-3

Np = Donor concentration = 10^1 cm^-3

ni = Intrinsic carrier concentration = 1.5 x 10^10 cm^-3

q = Charge of an electron = 1.6 x 10^-19 C (Coulombs)

Vbi = (8.617 x 10^-5 eV/K * 2.5 x 10^2 K * ln(10^15 * 10^1 / (1.5 x 10^10)^2)) / (1.6 x 10^-19 C)

Vbi = (2.15425 eV * ln(10^16 / 2.25 x 10^20)) / (1.6 x 10^-19 C)

Vbi = (2.15425 eV * ln(4.44444 x 10^-5)) / (1.6 x 10^-19 C)

Vbi = (2.15425 eV * (-10.014)) / (1.6 x 10^-19 C)

Vbi = -0.1352 V

The build-in voltage of the p-n junction is approximately -0.1352 volts.

The zero-bias space charge width (W) of the depletion region can be calculated using the formula:

W = sqrt((2 * ε * (Vbi - V) / q) * (Na + Np) / (Na * Np))

where:

ε = Permittivity of the material = 11.8 F/cm = 11.8 x 10^-14 F/cm

V = Applied bias voltage = 0 volts

W = sqrt((2 * 11.8 x 10^-14 F/cm * (0.1352 V - 0 V) / (1.6 x 10^-19 C)) * (10^15 cm^-3 + 10^1 cm^-3) / (10^15 cm^-3 * 10^1 cm^-3))

W = sqrt((2 * 11.8 x 10^-14 F/cm * 0.1352 V / (1.6 x 10^-19 C)) * (10^16 cm^-3) / (10^16 cm^-6))

W = sqrt((2 * 11.8 x 10^-14 F/cm * 0.1352 V / (1.6 x 10^-19 C)) * (10^16 cm^-3) / (1 cm^-6))

W = sqrt((2 * 11.8 x 10^-14 F/cm * 0.1352 V / (1.6 x 10^-19 C)) * (10^22))

W = sqrt((3.19152 x 10^-15 F/cm) * (10^22))

W = sqrt(3.19152 x 10^7)

W = 5650.89 nm

The zero-bias space charge width (W) of the depletion region is approximately 5650.89 nanometers.

The forward bias current (I) at 0.5 volts can be calculated using the formula:

I = q * ni^2 * (exp(V / (k * T)) - 1) / (Na * Hp + Np * Mn)

I = 1.6 x 10^-19 C * (1.5 x 10^10 cm^-3)^2 * (exp(0.5 V / (8.617 x 10^-5 eV/K * 2.5 x 10^2 K)) - 1) / (10^15 cm^-3 * 450 cm/V-sec + 10^1 cm^-3 * 1500 cm/V-sec)

I = 1.6 x 10^-19 C * (1.5 x 10^10 cm^-3)^2 * (exp(0.5 V / 215.425) - 1) / (10^15 cm^-3 * 450 cm/V-sec + 10^1 cm^-3 * 1500 cm/V-sec)

I = 1.6 x 10^-19 C * (1.5 x 10^10 cm^-3)^2 * (exp(0.00232) - 1) / (450 x 10^15 cm^-2/V + 15 x 10^15 cm^-2/V)

I = 1.6 x 10^-19 C * (1.5 x 10^10 cm^-3)^2 * (1.00232 - 1) / (465 x 10^15 cm^-2/V)

I = 1.6 x 10^-19 C * (1.5 x 10^10 cm^-3)^2 * 0.00232 / (465 x 10^15 cm^-2/V)

I = 1.6 x 10^-19 C * 3.375 x 10^20 cm^-6 * 0.00232 / (465 x 10^15 cm^-2/V)

I = 1.6 x 10^-19 C * 7.83 x 10^17 / (465 x 10^15 cm^-2/V)

I = 1.2613 x 10^-1 C/V * 10^3 C/V

I = 0.1261 A

The forward bias current (I) at 0.5 volts is approximately 0.1261 Amperes.