Hi professor, could you explain why the bands bend the way that they do i.e. p-type bends downwards and n-type upwards? I can't seem to follow any coherent logic for this with Schottky barriers. Thanks!
First, the conceptual explanation. A metal is full of electrons. The last place the semiconductor’s electrons want to go is into the metal. Thus, the electron energy increases close to the metal since that is an uncomfortable place for the semiconductor’s electrons to be. (Remember, high energy chases electrons away.) A p-doped semiconductor is inviting to the metal’s electrons. So, to lay out the welcome mat, the p-type semiconductor lowers the bands near the interface as if to invite the metal’s electrons in. The metal’s electrons still have a hill to climb to get into the semiconductor, but it isn’t as high at the junction as it is deeper inside. Now the theoretical explanation: When two materials come into contact, in this case a semiconductor and a metal, there will always be work function. This is the energy required for an electron to jump from the metal into the semiconductor. If it is vacuum instead of semiconductor, then the work function is the standard work function. However, between the metal and the semiconductor, the “work function” is the Schottky barrier potential. So, there are always two energies that need to be honored, the Schottky barrier potential and the Fermi energy, Ef. The conduction band has no choice by to be located at phi_Bn above Ef. But it is also required to be located KT*ln(n/Nc) above Ef. What can it do? At the interface, it makes the interface happy and is phi_Bn above Ef. But when it gets away from the interface, it has to follow the rule of the bulk semiconductor and be a different energy above Ef. For p-type semiconductor, the bands bend down because the carriers are holes, even though the bands depict electron energies. The valence band is required by law inside the semiconductor to be KT*ln(p/Nv) below Ef. But the junction requires the valence band to be fixed at q*phi_Bp below Ef. Again, what is it to do? Obey the law of the junction while in the junction’s territory and obey the law of the bulk while in the bulk’s territory.
Hi,MR! Now i am wondering about fermi pinning effect(fermi level pinning), would you please explain some about it ?😁
Hi professor, could you explain why the bands bend the way that they do i.e. p-type bends downwards and n-type upwards? I can't seem to follow any coherent logic for this with Schottky barriers. Thanks!
First, the conceptual explanation. A metal is full of electrons. The last place the semiconductor’s electrons want to go is into the metal. Thus, the electron energy increases close to the metal since that is an uncomfortable place for the semiconductor’s electrons to be. (Remember, high energy chases electrons away.) A p-doped semiconductor is inviting to the metal’s electrons. So, to lay out the welcome mat, the p-type semiconductor lowers the bands near the interface as if to invite the metal’s electrons in. The metal’s electrons still have a hill to climb to get into the semiconductor, but it isn’t as high at the junction as it is deeper inside.
Now the theoretical explanation: When two materials come into contact, in this case a semiconductor and a metal, there will always be work function. This is the energy required for an electron to jump from the metal into the semiconductor. If it is vacuum instead of semiconductor, then the work function is the standard work function. However, between the metal and the semiconductor, the “work function” is the Schottky barrier potential. So, there are always two energies that need to be honored, the Schottky barrier potential and the Fermi energy, Ef. The conduction band has no choice by to be located at phi_Bn above Ef. But it is also required to be located KT*ln(n/Nc) above Ef. What can it do? At the interface, it makes the interface happy and is phi_Bn above Ef. But when it gets away from the interface, it has to follow the rule of the bulk semiconductor and be a different energy above Ef.
For p-type semiconductor, the bands bend down because the carriers are holes, even though the bands depict electron energies. The valence band is required by law inside the semiconductor to be KT*ln(p/Nv) below Ef. But the junction requires the valence band to be fixed at q*phi_Bp below Ef. Again, what is it to do? Obey the law of the junction while in the junction’s territory and obey the law of the bulk while in the bulk’s territory.
@@stephenremillard1 much appreciated thank you
Thank you for that video.
You're very welcome.
the graph 1/C²: The V values at first quadrant are negative
Sir why schottky barrier act as a rectifier ?
See my video on Schottky diode rectification: th-cam.com/video/Q9TwNUgUkLE/w-d-xo.html