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I Why does LED shunt resistance decrease with increasing T?

  1. Nov 21, 2016 #1
    I'm not entirely clear what shunt resistance is, it's fairly hard to find things on google which refer to it in the context of LEDs. So we did an experiment in which we measured the IV characteristics of InAs at different temperatures, and observed the gradients of these graphs (apparently those give the shunt resistance?) to be decreasing with increasing temperature.

    As far as I understand, this shunt resistance is something caused by manufacturing defects. So why would temperature have an effect? I have not got an extensive mathematical knowledge of the subject so qualitative answers would be really helpful, although if the maths is accessible I'd definitely be interested in seeing it! Thanks for any help! :)
  2. jcsd
  3. Nov 21, 2016 #2
    Diodes is general are not perfect that is why with a reverse bias there is a small measurable current flowing. The shunt resistance is the ratio of the voltage change to the current change in the VI curve at
    v ≅ 0

    Temperature sensitivity in the characteristics of semiconductor device is to be expected. Temperature increases the energy of electrons in the semiconductor raising more into the conduction band from the valence band thus reducing resistance.
  4. Nov 22, 2016 #3
    I would like to discourage the notion that shunt resistance of a diode is "something caused by manufacturing defects". When glee says "diodes are not perfect" he means real diodes do not behave as ideal diodes. However I want to make clear that for the most part that is not because of some manufacturing defect that can be made better. It is an intrinsic characteristic of a semiconductor. The band gap is finite. The temperature is finite, so there will always be a thermal population of charge carriers in the conduction band as given by the Fermi distribution. As glee points out this is also the reason for the variation with temperature. These thermal charge carriers mean a semiconductor will always act at least partially like a conductor. That's the reason for the name after all.
  5. Nov 22, 2016 #4
    In an ordinary resistor though, increasing temperature increases resistance. What makes this type of resistance different? In an circuit with an ordinary resistor would adding more electrons result in lower resistance too?
  6. Nov 22, 2016 #5
    Yes! Exactly. That's one of those tell tale signs that something is different. In a resistor the higher temperature means there are more lattice vibrations, and so more phonon states for an unfortunate charge carrier to interact with and so higher resistance. This happens in a semiconductor too. However the very rapid increase in the number of charge carriers with temperature completely trumps the fact that each of them has a little tougher time getting around. So the resistance goes down.
  7. Nov 22, 2016 #6
    Great, thanks for your help!
  8. Nov 22, 2016 #7
    You are welcome.
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