Why is the conduction band lowered in Schottky diodes?

In summary, the conduction band bends down for a metal against p-type semiconductor contact due to the nature of the metal/SC interface, while it bends up for n-type. This band bending is caused by the matching of chemical potentials across the interface and is not affected by donor atoms. The band bending can also be seen as a reduction in the bandgap. The potential V(x) falls parabolically as one moves away from the metal contact and into the SC, with the barrier width L_0 related to the built-in potential and the Debye length. The electron density also changes exponentially in this region. Additionally, the Fermi energy of the metal matches the chemical potential of the SC. In most cases, the SC band bends
  • #1
jasum
10
0
Here is my question in the formation of Schottky Diode:

"Why does conduction band is lower than the original value in the semiconductor side?"

And I think it is lower because the postitve charged donor ions provide a attractive force ,which makes the electrons in semiconductor more easy to escape from bonding and go to conduction band.

Am I right?
 
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  • #2
The conduction band bends down (lower energy) for a metal against p-type semiconductor (SC) contact. It bends up, naturally, for n-type.

The band bending does not arise from donor atoms--why would it be different at the interface than in the bulk material? Instead it is the nature of the metal/SC interface. Imagine that a planar metal sheet is brought up to the surface of the SC. The Fermi energies must become equal, so Ef of the metal decreases to match Ef of the SC. In the process electrons leave the metal and penetrate the SC, depressing its conduction band. The charge density decays exponentially in a distance characterized by the Debye depth, so the band bends back up to the bulk value.

Most SC books cover this. For an exhaustive treatment, starting from a simple description in ch. 1, see
Henisch, "Semiconductor Contacts", Oxford U Press
 
  • #3
I've read about schottky diode that , such a metal-semiconductor combination is taken in which Ef of metal lies between the bottom of conduction band and top of valence band of the semiconductor. Why is it so?
 
  • #4
marcus, your post is a little confusing to me! It seems like you are describing band bending on the metal side of the interface! There isn't a Debye length for a semiconductor (you need lots of free charges for debye screening, so the Debye depth applies to the metal side of the interface), and the charge density on the semiconductor side doesn't have an exponential decay profile - in fact, the profile is pretty close to the depletion profile you see in a regular PN junction.

jasum, you're partially correct in that if you have an n-type Schottky interface, the field on the SC side is primarily due to the ionized donors within the "depletion region". However, this additional potential doesn't make it "more easy to escape from bonding". What you describe would be seen as a reduction in the bandgap, not a bending of the bands. The bending is caused, as marcus explained, by the matching of chemical potentials across the interface. The bending profile, however, is a function of the total charge density, but typically, it goes approximately quadratically within the depletion region (assuming a nearly uniform density of ionized donors here).

savi, you want any "diode" to have the ability to rectify (conduct in one direction only). With a Schottky diode, you do that by putting the conduction band (of the SC) just above the Fermi energy (ofnthe metal). This way, a small positive bias voltage (exceeding the Schottky barrier height) results in a large conductivity, but any negative bias voltage (short of breakdown) leaves you with an essentially insulating barrier. That's what a diode needs to be able to do.
 
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  • #5
Oh you're right, my wording was confusing. :blushing: Only the SC (semiconductor) bands bend, downwards for p-type and upwards for n-type.

I'm not a semiconductor guy, Gokul, so please correct any errors in the following. As I understand it, the potential V(x) falls parabolically as one moves away from the metal contact and into the SC of a classic Schottky barrier. The barrier width L_0 through which the potential is nonzero is related to the built-in potential across the contact and to the Debye length L_D, where (a) The built-in potential depends on properties of the metal and of the doped SC and (b) The Debye length exists (by definition) for any material with free charge carriers, including a doped semiconductor. Typical numbers listed in the books for a silicon device are L_D = 100 nm and L_0 = 4 * L_D.

The electron density n(x) for a parabolic potential in n-type material rises from a low value (the barrier is depleted, as you say) to the bulk value according to the Boltzmann distribution exp(-eV(x) / kT). Thus log(n(x)) rises parabolically as one moves away from the contact, but the density n(x) itself changes exponentially.

ZapperZ notes the difference between Fermi energy in a metal and chemical potential in a SC:
http://https://www.physicsforums.com/showthread.php?t=133914" [Broken]
Accordingly I should have said that Ef of the metal matches the chemical potential of the SC in my earlier post.
 
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  • #6
but I thought ,Marcus, that SC band bends downwards for N type and upwards for Ptype ?!
 
  • #7
Nope. Here's an illustration from wikipedia
http://en.wikipedia.org/wiki/Image:Ohmic2.png" [Broken]
and you should see the same thing in any book on semiconductor physics.

Edit: It's actually possible to choose a metal and a doped semiconductor combination that make the bands bend the other way, but this is the usual situation.
 
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1. What is a Schottky Diode?

A Schottky diode is a type of semiconductor diode that is composed of a metal contact and a semiconductor material, typically silicon. It is named after its inventor, German physicist Walter Schottky. Unlike conventional diodes, a Schottky diode has a metal-semiconductor junction, which gives it unique properties and applications.

2. How is a Schottky Diode formed?

A Schottky diode is formed by creating a metal-semiconductor junction between a metal contact and a semiconductor material, such as silicon. The process involves depositing a thin layer of metal, such as gold or platinum, onto a semiconductor material using techniques like evaporation or sputtering. This creates a barrier that allows current to flow in only one direction, similar to a regular diode.

3. What are the advantages of Schottky Diodes?

Schottky diodes have several advantages over conventional diodes. They have a lower forward voltage drop, which means they require less energy to operate. They also have a faster switching speed and lower reverse recovery time, making them ideal for high-frequency applications. Additionally, Schottky diodes can handle higher temperatures and have a higher power density.

4. What are the applications of Schottky Diodes?

Schottky diodes have a wide range of applications in electronics and electrical engineering. They are commonly used as rectifiers in power supplies, voltage clamps in electronic circuits, and in radio frequency (RF) applications. Schottky diodes are also used in solar cells, switching regulators, and in high-speed digital circuits.

5. What are the potential drawbacks of Schottky Diodes?

One of the main drawbacks of Schottky diodes is their higher leakage current compared to conventional diodes. This can lead to increased power consumption and potential reliability issues in some applications. Additionally, Schottky diodes have a lower breakdown voltage, which limits their use in high voltage applications. They also have a higher cost compared to regular diodes due to the use of precious metals in their construction.

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