Solid State Circuit Structure Question

A Schottky diode has - as you well said - a semiconductor region (normally it is an N-type material that contain high energy electrons) and a metallic region (normally gold, silver or platinum - almost never copper). This means that in the metallic region you have many conduction electrons (this is the nature of metals) and in the semiconductor region (if it is an N-type material) you have free electrons and valence electrons as well.

When the Schottky diode is created and no bias is applied to it (like when you put it on a table) no conduction takes place, becase, as you mentioned, there are electrons in both sides of the junction and they repel each other, so they are pushed away from the junction. This is not the case with standard P-N diodes. When these diodes are manufactured (when you put together the N and P materials) during the fabrication process, there is a small amount of current flowing acroos the junction for a very short period of time, and then the current stops. This movement of carriers across the junction creates the potential barrier in both sides of the junction (this is what makes a diode so special).

Now, if you want to have current (for any type of diode) you must provide a potential difference across the diode. This voltage that you must apply in order to produce current (across the junction!) is called the bias voltage. The amount of current produced depends on the polarity of the bias. If the bias is such that the electrons from the N material are pushed toward the junction and the voltage is big enough so that they will be pushed across the junction, then you will have a big amount of current. This condition is called forward bias condition. In a Schottky diode you achieve forward bias when you apply a negative (lower) potential to the N region and a positive (higher) potential to the metal region.

Suppose you use a battery of 5 volts and you connect the negative terminal to the N region and the positive terminal to the metallic region of the Schottky diode, then this is what happens: The negative terminal will repel the negative electrons. This means that the electrons will be pushed toward the junction, and if the voltage is big enough (5 volts is more than enough, because the minimum voltage to produce current in Schottky diodes is around 0.3 volts), then the electrons will be pushed (injected) into the metallic region (they may like it or not! Who cares!). When the electrons cross to the metallic region, then the positive terminal of the battery that is connected to the metal will attract the electrons toward the end of the metal (away from the junction) and into (if you allow me to be simplistic) the wire of the battery. So an electron that was nice and quite in the N region it is now injected into the metal by mean of a battery connected in forward bias. If the battery is connected in reverse bias (negative terminal to the metal positive to the N region), then no current across the junction takes place.

In summary: you must provide a bias to inject the electrons across the junction. If the voltage is big enough then the electrons from the N material will be forced to cross the junction with energies bigger than the energy of repulsion of the electrons in the metallic region.

I hope this answer your question.

Abe Michelen

Dear Abe Michelen,

You have taken pains to write in lot of details. Nice work.

I will like to point out that in metals, so-called free electrons are not all that free and this is so in semiconductors. They form their own energy levels of the bulk due to waveform extending the entire bulk up to boundaries.

This makes the electrons to fall into several energy zones and these zones are too many in surface electrons. Deeper electrons have somewhat different discrete levels.

One more parameter that plays the role is the mobility of electrons which is very high in Schottky diodes and reverse voltage recovery much lower. They are very very fast.

Rest of the story is just fine.

Dear Shyam, You are right. I did not wanted to mention mobility or energy levels just because i did not want to make the response too lenghty. I teach Electronic Devices in local college and when I cover diodes, of course I explain with much more details their behavior. Thanks you very much for your nice comments. Abe

Dear Abe Michelen,

You make a good Professor and that will help students for sure.

This from Microsemi may be of some interest

www.microsemi.com/micnotes/401.pdf

In this design, Schottky diodes use both n+ and n- layers and also use special metal that act as Schottky barrier metal. Selection of metal can raise the potential limit and use of n+ and n- layers in highly dopped state have high reverse leakage current.

This research group is working on THz Schottky diodes and Heavily-Doped Cryogenic Schottky Diode.

www.cfa.harvard.edu/srlab/THz-2002/abstracts/Schle cht.pdf

I use them for ns high voltage pulsers of 1kV to 10kV range. Schottky diodes are excellent for fast killing of reverse voltage Ground level bounter pulses. I get these from http://www.ixys.com/ or
http://www.directedenergy.com/

I think there is lot more difference in design of Schottky diode for practical use then just theoretuical understanding of its basics.