pn Junctino Electrostatics[edit | edit source]
Effect of an Electric Field - Conductivity and Ohm's Law[edit | edit source]
The Built-in Potential (Vbi)[edit | edit source]
The electric field is the derivative of the potential with position
Integrate across the depletion region
The sum of the drift and the diffusion at equilibrium is equal to zero
Use the Einstein relationship:
and are the n- and p-side doping concentrations.
Solution of charge density[edit | edit source]
Solution of electric field[edit | edit source]
Depletion Width with Step Junction and VA not equal to zero[edit | edit source]
pn Junction Diode: I-V Characteristics[edit | edit source]
- Diode operated in steady state
- Doping profile modeled by nondegenerately doped step junction
- Diode is one-dimensional
- In quasineutral region the low-level injection prevails
- The only processes are drift, diffusion, and thermal recombination-generation
Quasineutral Region Consideration[edit | edit source]
The quasineutral p-region and n-region are adjacent to the depletion region
The electric field is zero and the derivative of the electron and hole concentration is zero in the quasineutral regions.
Assume that the thermal recombination-generation is zero throughout depletion region. Sum the and solutions.
Strategy to find the minority carrier current density in the quasineutral regions:
- Evaluate current densities at the depletion region edges
- Add edge current densities
- Multiply by A to find the current
Establish boundary conditions at the edges of the depletion region.
Multiply defining equations of the electron quasi-Fermi level, , and the hole quasi-Fermi level, .
Monotonic variation in levels
"Law of Junction"
Evaluate the "law of junction" at the depletion region edges to find the boundary conditions.
- Solve minority carrier diffusion equations with regard to the boundary conditions to determine value of and in quasineutral region.
- Determine the minority carrier current densities in quasineutral region
- Evaluate quasineutral region solutions of and . Multiply result by area
Solve the equation below with regard to two boundary conditions.
Evaluate at the depletion region edges
Multiply the current density by the area:
Ideal diode equation:
The velocity of an electron in a one-dimensional lattice is in terms of the group velocity
Differentiate the equation of velocity
In the free electron model, the electronic wave function can be in the form of . For a wave packet the group velocity is given by:
In presence of an electric field E, the energy change is:
Now we can say:
where p is the electron's momentum. Just put previous results in this last equation and we get:
From this follows the definition of effective mass:
Strong disturbance when individual reflections add in phase
The potential energy is from the actual potential weighted by the probability function
Average over one period
The kinetic energy is the same in the case of both wave functions
The total energy is the kinetic energy plus the potential energy
The energy of an electron cannot be between the lower and higher value.