Physics of Semiconductor Devices - Exam March 2007, Problem 3

1. Exam question

In a Schottky contact, the Fermi energy is pinned to the middle of the gap by interface states. The semiconductor is p-doped.

(a) Draw the indicating the Fermi energy, the valence band, and conduction band.

(b) Explain the difference between a Schottky contact, an ohmic contact, and a tunnel contact.

(c) Why is the formed at the gate of a MESFET usually reverse biased during transistor operation?

2. Solution (a)

Figure 1: (left) Metal and Semiconductor separated. (right) Metal and Semiconductor in contact (Schottky contact).

As you can see in the left figure, the from the p-doped semiconductor is larger than the work function from the metal. When you bring the metal and the semiconductor in contact, electrons flow from the material with the lower work function towards to the material with the higher work function. When they do this, the metal will pick up a positive surface charge and a negative charged region occurs in the p-doped semiconductor. This negative charged region is the reason for the band bending in the p-doped semiconductor. Physics of Semiconductor Devices - Exam March 2007, Problem 3

(b)

Figure 2: Schottky contact with n-doped Semiconductor When the metal and the n-doped Semiconductor in figure 2 came in contact, the Fermi energy is the same on both sides. Electrons move from the semiconductor (lower Work function) to the metal (higher Work function). This causes the electric field E. To push electrons towards to the negative side (metal), you have to push them up. That is the reason why the bands bend up near the metal semiconductor interface.

Figure 3: Ohmic contact with n-doped Semiconductor The Fermi energy of the Semiconductor in figure 3 is lower than the Fermi energy of the Semiconductor in figure 2. When the metal and the semiconductor came in contact, the Fermi energy is once again the same on both sides. Electrons move from the metal to the Semiconductor. This causes the electric field E as marked in figure 3. In this case the electric field helps you to push the electrons towards to the metal. That is the reason why the bands bend down near the metal semiconductor interface.

At the Schottky contact the conduction band bends away from the Fermi energy and the semiconductor gets intrinsic in this region. That mean, that there are less mobile charge carriers and a depletion width occurs. At the Ohmic contact the conduction band is lower than the Fermi energy near the metal-semiconductor interface. The Physics of Semiconductor Devices - Exam March 2007, Problem 3

semiconductor behaves like a metal in this region and so an Ohmic contact with a linear resistance occurs.

In praxis when you bring a metal in contact with a semiconductor it is almost impossible to make an Ohmic contact. Because of the interface states (broken bands, impurities, etc.) you always get a Schottky contact. To avoid this you need a tunnel contact (figure 4).

Figure 4: Tunnel contact with n-doped Semiconductor To form a tunnel contact the semiconductor has to be heavily doped near the metal. If you degenerately dope the semiconductor, the is so thin that electrons can tunnel through it. The resistance of a tunnel contact is linear. The change from n++ to n is an Ohmic contact as explained in figure 3. But it works in this case because the n++ and the n region are on the same crystal and so there are no interface states which pin the Fermi energy to the middle of the gap.

(c)

The drain-source current in a MESFET is controlled by the depletion layer which is created by a Schottky contact. There are no free charge carriers in the depletion region and so the drain-source channel gets narrow if the depletion layer grows. The depletion width grows if you reverse bias the Schottky contact. This is the idea of a MESFET: There is a conducting channel and you increase the depletion layer in this channel to control the drain-source current.