1
\$\begingroup\$

I'm studying the metal-semiconductor junction on Muller-Kamins book. I have two questions.

  1. This is the band diagram of the metal-semiconductor junction in thermal equilibrium:

enter image description here

In order to get this plot, the book gives some rules. One of these rulese says that the vacuum level E0 must be continuous because if it were not then "one could conceive of means to extract work from an equilibrium situation by emitting electrons and then reabsorbing them an infinitesimal distance away where E0 had changed value". What does this mean?

  1. This is a picture taken from the book representing the situation that occurs when a voltage is applied to the system:

enter image description here

Now, when you apply a voltage, you go out of thermal equilibrium and you enter a quasi-equilibrium condition. In the first chapter of the book, the Authors say that in this case, in order to represent n and p (electron and hole concentration in the silicon, respectively), you need to define two quasi-Fermi levels: one for the electrons (Efn) and one for the holes (Efp). Why is only one generic Fermi level for silicon (called Efs) represented in the picture?

\$\endgroup\$

1 Answer 1

Reset to default
1
\$\begingroup\$
  1. "one could conceive of means to extract work from an equilibrium situation by emitting electrons and then reabsorbing them an infinitesimal distance away where E0 had changed value". What does this mean?

In physics, the vacuum level refers to the energy of a free stationary electron that is outside of any material (it is in a perfect vacuum). It may be taken as infinitely far away from a solid, or, defined to be near a surface. (a quote from Wikipedia)

In the source you quote, the vacuum level E0 is defined to be near a surface separating a vacuum from materials.

Notice than that a) the energy of a free stationary electron in a vacuum is an electrostatic potential times the electron charge, and b) at the discontinuity of electrostatic potential, the electric field strength is infinite.

Imagine a setup with a horizontal interface separating materials from a vacuum and vertically oriented metal-semiconductor junction. Now, consider a closed rectangular contour of an infinitesimal (very narrow) width and of a commensurate height. Let its vertical sides cross a material-vacuum interface, and one of its horizontal side, which lies in a vacuum, crosses the electrostatic potential discontinuity. We calculate an electric field circulation, which in this case is expected to be zero (inferred from energy conservation in static fields). Since there is no discontinuity of electrostatic potential in a material (metal and semiconductor pieces are in contact), the electric field strength along the horizontal side in a material is finite and the circulation contribution of the horizontal side lying in a material is proportional to the rectangular width. Since there is no infinite electric field components orthogonal to a material-vacuum interface, the contribution of the vertical sides is proportional to the rectangular height. Decreasing the rectangular dimensions, the contributions of the three sides of the closed rectangular contour can be made arbitrarily small, and therefore they cannot compensate the contribution of the horizontal side lying in a vacuum which tends to a value of the discontinuity divided by the elementary charge value. We arrive at a contradiction, therefore, there cannot exist a vacuum level discontinuity in a setup under consideration.

  1. Why is only one generic Fermi level for silicon (called Efs) represented in the picture?

see, for example, https://www.iue.tuwien.ac.at/phd/ayalew/node53.html : "Obviously, when the excess carrier concentration is small compared to the equilibrium carrier concentration, the quasi-Fermi level must be very close to the Fermi level. For device operation, we often use a low-level injection condition, meaning that while the minority carrier concentration is changed, the majority carrier concentration remains un-affected. Thus the quasi-Fermi level of the majority carrier is the same as the Fermi level."

Notice also that the Schottky diode is a majority carrier device.

\$\endgroup\$
3
  • \$\begingroup\$ Thank you for your answer. Point 1 is clear. I have some doubts on the second point. I accept the fact that, under Low Level Injection, majorities are basically not modified while minorities are significantly affected (like in the base of a BJT), and thus Efs represents electrons (it should be called Efn). But what if I also consider holes to be more precise? Should I also consider a quasi-Fermi level for holes inside the metal (this sounds strange to me, since you don't have holes in a metal) coming from the silicon? \$\endgroup\$
    – Stefanino
    Jul 9, 2020 at 7:58
  • \$\begingroup\$ Maybe you shouldn't, because, as you noticed, there are no minority carriers in metals and therefore a quasi Fermi level for holes in semiconductor (if there were one) cannot participate in a band bending diagram (we have nothing to bend from the metal side of a junction). \$\endgroup\$
    – V.V.T
    Jul 9, 2020 at 9:08
  • \$\begingroup\$ Anyway, minority carriers from a semiconductor side may happen to reach a quasi equilibrium state, but we have no counterpart from a metal side to equalize (quasi) Fermi levels on both sides of the junction \$\endgroup\$
    – V.V.T
    Jul 9, 2020 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.