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Transmission Line Model (TLM) End Resistance (Re)

  1. Dec 15, 2015 #1
    Hi guys.

    I'm looking into the modelling of a transmission line model (TLM) and feel that I'm understanding it fairly well. Although, one parameter keeps popping up with very little explanation as to what it actually is.

    It seems that to characterize contact resistance for a e.g. GaAs TLM with AuGe ohmic contacts, that the contact resistance cannot just be characterized by the sheet resistance beneath the contact (Rsk) and the semiconductor resistance (Rsh) when these two are not equal. Many resources direct me towards Re in this case.

    Now, the measurement of Re appears simple, with current flowing between two contacts the voltage is measured from the end contact and a neighbouring unloaded contact, contacts 2 and 3, and this value is divided by the current between contacts 1 and 2 (fig. 6.10).

    N4qfjDp.png )

    It can be characterized as:

    FLPx6m5.png

    where
    Rsk = Sheet resistance beneath ohmic contact,
    W = width of the contact,
    d = length of the contact,
    Lt = transfer length (the length from the front of the contact to where the current is 1/e*io beneath the contact).

    My question is:
    -Is Re the resistance from Lt to the end of the contact?
    -Is Re the resistance blocking current from dispersing from the contact to the next (unloaded) contact?

    P.S. the reason I believe this is because if this is the case then and increase in Lt would reduce the area from Lt to end of the contact (x=d) whereas the beginning of the contact (x=0) will have a larger area, contributing to a decrease in contact resistance (Rc) but the "leftover" current after x=Lt will be more dense due to the decrease in area at the contact end. Of course, with greater with this will reduce the resistance of both Rc and Re and finally 1/sinh(d/Lt) or sech(d/Lt) shows that there's an exponential decrease in resistance due to the ratio that current density is shared (Rc/Re = cosh(d/Lt)).
    I hope this helps with explanations.

    Thanks, in advance.
     
  2. jcsd
  3. Dec 20, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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