- #1
cybhunter
- 25
- 0
Hey everyone, got a quick question
Last semester, I ended up taking a solid state physics course and surprising myself I ended up getting a B- in the class (most of the other people ended up getting D's). For the most part much of this due to the 'memorize the equation' and 'plug and chug'
A friend of mine wants to go back to school to get a degree in Electrical Engineering. Thinking about how to explain a semiconductor junction, I may have come up with a simple explanation to him;
Electrons with increasing density have higher energy levels and go up, while holes (with increasing concentrations go down). When a biasing voltage is applied, the Fermi energy level must remain at the same potential energy difference with respect to the majority doping (ie valance band for P-types and Conduction band for n-Types). Since in a forward bias the concentration of holes increase on the P side and the electrons increase on the N side, the N side Fermi energy level increases upwards, while the P side Fermi energy level increase downward. In reverse biasing, the holes and electrons recombine, with the potential voltage of higher value taking precedent. The net result being the difference on the respective sides on the junction. Since the Fermi energy levels must remain the same with respect to the majority doping (since the P side will have a higher abundance of electrons, and the N side will have a higher abundance of holes), the band gaps thus look like the way they do
Last semester, I ended up taking a solid state physics course and surprising myself I ended up getting a B- in the class (most of the other people ended up getting D's). For the most part much of this due to the 'memorize the equation' and 'plug and chug'
A friend of mine wants to go back to school to get a degree in Electrical Engineering. Thinking about how to explain a semiconductor junction, I may have come up with a simple explanation to him;
Electrons with increasing density have higher energy levels and go up, while holes (with increasing concentrations go down). When a biasing voltage is applied, the Fermi energy level must remain at the same potential energy difference with respect to the majority doping (ie valance band for P-types and Conduction band for n-Types). Since in a forward bias the concentration of holes increase on the P side and the electrons increase on the N side, the N side Fermi energy level increases upwards, while the P side Fermi energy level increase downward. In reverse biasing, the holes and electrons recombine, with the potential voltage of higher value taking precedent. The net result being the difference on the respective sides on the junction. Since the Fermi energy levels must remain the same with respect to the majority doping (since the P side will have a higher abundance of electrons, and the N side will have a higher abundance of holes), the band gaps thus look like the way they do