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Negative Resistance in Resonant Tunneling

  1. Jun 27, 2011 #1

    I am a student who's learning about quantum mechanics. I was reading about the RTD (Resonant tunneling diode) and I got the general idea. However, I was looking at a I vs V graph and came up with a question. Since Resistance = V/I, the inverse slope of the I vs V graph has to be the resistance. (for a simple model of RTD, it had a giant peak). Now, the problem is, for the rising part of the peak, I understand, but what is the physical interpretation of the falling part of the peak? (the negative resistance) Does that simply mean the flow of the electron is greater?

  2. jcsd
  3. Jun 27, 2011 #2
    Negative Resistance simply applies in a situation where an increased current causes a decreased voltage. Remember there is STATIC resistance, where R=V/I and there is a very similar Differential resistance, which is R=dV/dI, and if you look at your graphs, while dV/dI is clearly negative,V/I remains positive, as both values are positive.

    Negative resistance causes and increasing amplitude over time, however, it doesn't violate energy conservation, because we can't get this amplification to go on forever, and the energy that goes into it must obviously some from something. I believe it does mean that the electrons in the current are speeding up.
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