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Based on the transfer length method (TLM), one can accurately calculate the contact resistivity for an ohmic contact, by evaluating the absolute resistance measured through the test structure and plotting it as a function of the gap spacing between the two ohmic contacts. By extrapolation, the contact resistance and transfer length (and thus, the contact resistivity) can be calculated.

However, what if a measurement of the contact resistivity of a Schottky contact was desired? In this case, the forward biased current is non-linear (does not follow Ohm's law), and thus the absolute resistance measured is a function of voltage. Is there another way to measure the contact resistivity in this case?

On the flip side of the coin, I have only seen the Schottky barrier height calculated for Schottky contacts. However, some ohmic contacts (e.g. tunneling ohmic contacts) still have a positive Schottky barrier height. How is the height measured in this case?

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If you put a Schottky device into very-forward-bias, the IV curve becomes a straight line whose inverse slope is the resistance as usual. So you can still use TLM. Of course, people don't normally bother to figure out contact resistance because the Schottky aspect of the contact has a much much bigger effect on the device than the resistive aspect.

For an ohmic contact that arises from tunneling through a Schottky barrier, again, people don't normally bother to figure out the "barrier" height because the "barrier" is irrelevant for the electrical behavior of the device. But if you had to figure it out for some reason, the only method I know of is internal photoemission.

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  • $\begingroup$ Thank you! This is exactly what I was looking for. Is the reason why the IV curve linearizes due to the fact that after a certain forward bias, a very large proportion of the carriers will be at an energy greater or equal to the SBH, and any additional bias doesn't largely change the ability for carriers to emit over the barrier, but rather just changes the velocity? Also, when you say very-forward-bias are you referring to greater or equal bias to achieve flat band conditions in the semiconductor? $\endgroup$
    – Caedar
    Commented Mar 20, 2012 at 13:26
  • $\begingroup$ I more-or-less agree with all your comments. Very-forward-bias is probably best determined in practice by actually looking at the IV curve. If the SBH is 0.5V, and you apply 5V, then at least 90% of the voltage drop will be due to resistance. I say "at least" because I don't know whether you ever quite get all the way to flat-band. But either way, the barrier eats up a more and more negligible amount of the applied voltage, or in other words the barrier plays a more and more negligible role in limiting the current. $\endgroup$
    – Steve Byrnes
    Commented Mar 20, 2012 at 18:36

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