You can see diagrams here: http://www2.warwick.ac.uk/fac/sci/physics/current/postgraduate/regs/mpags/ex5/devices/hetrojunction/ohmic/
the answer is easy: these band diagrams plot the potential energy versus the distance d from the interface. Applying a voltage just adds a linear slope to the total potential energy of charge carriers (the band structure) since voltage is nothing else but a difference in potential energy (U2 - U1) for a charge sitting at different positions 1 and 2 in an electric field. For a flat interface the electric field E is homogeneous and the potential energy can be expressed as U = E*d. This leads to a linear slope in this type of band diagram. Obviously, the steeper the slope, the larger the electric field corresponding to a larger applied voltage.
I did not understood the issue in this diagram..may be my limited understanding but still i notes few point
1) if you have n semiconductor under positive bias then electrons will flow form semiconductor to metal without barrier because the contacts are pure conductors
2) same semi-conductor under negative bias will force electron from metal to semi-conductor the barrier comming in picture is simply due to the the barrier space charge region created by cross flow of drift and diffusion current under thermal equlibrium
3) bending of the band slope away from junction simply shows that the net flow of charge is higher from metal to semi-conductor since it is negative bias you could take it as (ohmic contact at higher energy than semi n region)
4) if question was about tunneling behaviour then it is quite apparent that more less will be forbidden gap higher will be the chance of carrier transporting from from valence band to conduction band a phenomenon uncontrolled seen in zener breakdown
i hope i was able to put some light on it
thanks a lot , but my question was about the reason of band slope , not band bending.
I clarify my question: why do we see band slope in ohmic contact under bias but see horizontal band in schottky contact under bias?
the answer is easy: these band diagrams plot the potential energy versus the distance d from the interface. Applying a voltage just adds a linear slope to the total potential energy of charge carriers (the band structure) since voltage is nothing else but a difference in potential energy (U2 - U1) for a charge sitting at different positions 1 and 2 in an electric field. For a flat interface the electric field E is homogeneous and the potential energy can be expressed as U = E*d. This leads to a linear slope in this type of band diagram. Obviously, the steeper the slope, the larger the electric field corresponding to a larger applied voltage.
The answer by Bert Hecht is convincing with the Band diagrams. But I have a question in continuation with the explanation above.
@ Bert Hecht: Do you mean the energy levels of the semiconductor are pinned at the interface in Ohmic type contact?
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