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How does a work function difference cause band bending in equilibrium in Metal-Oxide-Semiconductor (MOS)? I don't understand the physical meaning behind this sentence from Sze:

to accommodate the work function difference, the semiconductor bands bend

I understand that the work functions are the the energies required to move an electron from the fermilevel to the vacuum. I do not understand why this implies band bending when a metal and a semiconductor of different work functions are brought together.

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    $\begingroup$ Because the work functions must align in equilibrium, or else you could move an electron from one material to the vacuum, back to the other material, and then in to the first and gain energy. The only thing that can allow that to happen is band bending and rearrangement of charge to equalize the work functions between the materials. $\endgroup$ – Jon Custer Oct 14 '16 at 19:45
  • $\begingroup$ Could you please specify what you mean by 'align'? I have added a figure above. I naively would have thought that the vacuum level should be at a constant value, but it seems that is wrong. I just don't quite see what exactly is being aligned by th bending in (b) in comparison to (a). $\endgroup$ – ari Oct 14 '16 at 20:11
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    $\begingroup$ The electrons in the solid, to be in equilibrium, need to be at the same chemical potential, or else they will move in response to deltas in the chemical potential. So, the Fermi energies must come in to alignment or there will be net carrier movement. $\endgroup$ – Jon Custer Oct 14 '16 at 20:20
  • $\begingroup$ To me that just explains the alignment of the fermi energies. Why does that imply band bending? Thank you for your patience! $\endgroup$ – ari Oct 14 '16 at 20:23
  • $\begingroup$ How else are the Fermi energies going to come in to alignment? $\endgroup$ – Jon Custer Oct 14 '16 at 20:25
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The work function is the energy necessary to move an electron from the Fermi level to the vacuum level. When you bring two materials in contact, here the metal and the semiconductor, thermodynamic equilibrium will require that the two Fermi levels (electrochemical potentials) equalize. This can be thought of to happen by the transition of electrons from the material with lower work function to the material with higher work function which increases the electrostatic potential of the low work function material with respect to high work function material until the Fermi levels are equal. Now, far from the interface the vacuum levels differ by justnthe work function difference and the vacuum potential level has to make a continuous transition when going from one material to the other. For energy conservation (movement of an electron along a circular path including the vacuum level and the internal metal/semiconductor interface), the same potential difference has to build up at the internal junction of the materials. The potential difference is also called contact potential. This potential difference is produced by space charge regions (e.g., electron depletion and accumulation zones at the interface) which lead to the mentioned band bending at the metal-semiconductor interface. In the MOS system the potential difference corresponding to the work function differences drops over the depletion/inversion space charge region at the surface of the semiconductor and the linear potential drop over the gate insulator $SiO_2$.

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  • $\begingroup$ Sadly I am still confused. As I understood it, in the MOS, electrons cannot travel through the oxide - and how does a work function difference imply an electrical field physically? How can I imagine that in terms of actual field-inducing charges? $\endgroup$ – ari Oct 14 '16 at 21:28
  • $\begingroup$ @ari - In the MOS, equilibrium between the metal gate and the semiconductor is established by an electrical connection. This happens, e.g., when you apply zero voltage between the two. The diagram (b) shows the case that gate and semiconductor are in equilibrium, i.e. have the same Fermi level. Then an internal potential difference has to build-up which corresponds to the difference of work functions as depicted. This internal potential difference drops over the depletion/inversion zone+insulator, is produced by the space charge there and causes the band bending at the surface. $\endgroup$ – freecharly Oct 14 '16 at 22:10
  • $\begingroup$ Disappointing thing about this site is good answers like this get one up vote. $\endgroup$ – boyfarrell Oct 14 '16 at 23:38

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