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Electric field and built-in potential in metal-insulator-metal

  1. Jun 4, 2013 #1
    Hi guys

    I'm currently trying to get in depth about how the energy levels/bands in organic semiconductors behave when sandwiched between different metal electrodes. More precisely, it's about organic photovoltaics.
    I'm a trained chemist, so to me there seems to be a lot of terms and theory that is not explained in detail - like 'it's like this and everybody knows that'.

    So, to get to my question: People usually model OPVs and similar like metal-insulator-metal assemblies. But I have a hard time to figure out how the built-in potential and the electric field developes when going from seperate metal electrode and insulator to contact (with and without short-circuiting the whole thing).
    To me there is no charge carriers in insulators, and therefore I can't see that electrons should move one way or another and create an electric field.
    Moreover, people usually say (or, I understand it so) that the highest open-circuit voltage obtainable is equal to the difference in the work functions of the electrodes - but how is that if I don't short-circuit the terminals and there is no charge-transfer to build up an electric field?

    Also, it seems like the vacuum energy level changes throughout M-I-M assemblies, how is that?

    I hope you have some explanations and that I've not completely misunderstood the whole thing. In return, I might help in the chemistry section :)
  2. jcsd
  3. Jun 15, 2013 #2
    I don't know about OPVs specifically, but MIM structures develop a built-in bias with asymmetric electrodes because the difference in chemical potential (due to different work functions) causes charge flow until the chemical potential is flat everywhere.

    If there is no circuit connecting the two electrodes, then it will still happen due to tunneling. At finite temperature, all particles have some x, y, z momentum components due to thermal energy. So some will be randomly directed towards the opposite electrode. This happens both ways, but the electrode at the higher chemical potential will on average inject more electrons than it receives. Eventually, an equilibrium is reached (chemical potential is flat, net current is 0). Now, if the insulator is a very high and/or thick barrier, this equilibrium may take a long, long time to achieve without the assistance of an alternate current path. It may well take forever compared to human time scales.

    Vacuum energy changes because usually these diagrams are drawn at equilibrium. The charge transfers that establish equilibrium cause electric fields (linear drops in vacuum energy level) or dipoles (nearly discontinuous drops). Energies are relative and you could just as well draw the vacuum level flat, but then you need to distort everything else to have the proper distance from the vacuum level, which is silly because it's the device that we're actually trying to visualize.
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