Question

The p-type region of a silicon p-n junction is doped with 1016 boron atoms per cubic centimeter,
and the n-type region is doped with 1018 phosphorus atoms per cubic centimeter. Assume a step p-n
junction and that all doping atoms are ionized. The intrinsic carrier concentration in silicon at 300K is
1010cm−3. Calculate the width of the depletion region at 300 K (the relative permittivity of Si is
r = 11.7).

Answer 1

Any choices

Answer 2

To calculate the width of the depletion region at 300K, we can use the following equation:

W = √(2εrε0(Vbi + Vapplied)/q)(1/Nd + 1/Na)

where:
- W is the width of the depletion region
- εr is the relative permittivity of Si
- ε0 is the permittivity of free space
- Vbi is the built-in potential of the p-n junction
- Vapplied is the applied voltage across the junction
- q is the elementary charge
- Nd is the doping concentration of the n-type region
- Na is the doping concentration of the p-type region

At 300K, the built-in potential of the p-n junction can be calculated as follows:

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

where:
- k is Boltzmann's constant
- T is the temperature in Kelvin
- n_i is the intrinsic carrier concentration in silicon at 300K

Plugging in the given values, we get:

Vbi = 0.026V * ln(1018*1016/10^20)
= 0.76V

Assuming no applied voltage, Vapplied = 0, we can simplify the equation for the depletion region width as:

W = √(2εrε0Vbi/q)(1/Nd + 1/Na)

Plugging in the given values, we get:

W = √(2*11.7*8.854*10^-12*0.76/1.6*10^-19)(1/1018 + 1/1016)
= 1.14*10^-5 meters

Therefore, the width of the depletion region at 300K is approximately 11.4 micrometers.