- #1
paul_harris77
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Dear All
I am having problems with how to draw band diagrams with band bending for Schottky Barriers and PN junctions. My issue is that I don't know how you determine which Fermi level remains fixed at its equilibrium level and which Fermi level moves to align with it when the two materials are brought into contact.
My textbook gives the following diagram for an N-type semiconductor in contact with a metal with a higher work function than the semiconductor:
[URL]http://homepage.ntlworld.com/beehive77/images/ntypeschottky.jpg[/URL]
It looks to me here that the metal fermi level remains constant at its equilibrium value, and the n-type semiconductor moves downwards to meet it. I understand the direction in which the bending must take place and the theory behind it, but I don't understand why it is the metal fermi level that remains constant, and not the n-type fermi level.
Could it not just as easily be the n-type fermi level remaining constant and the metal fermi level moving upwards to meet it?
My textbook then gives me the following diagram for a p-type semiconductor in contact with a metal with a lower work function than the semiconductor:
[URL]http://homepage.ntlworld.com/beehive77/images/ptypeschottky.jpg[/URL]
This time, the semi conductor fermi level remains constant and the metal fermi level moves up to align with it. Why is it the opposite way round?
I originally assumed that both fermi levels would move to meet each other and meet at some "middle ground" energy level, but the textbook seems to suggest otherwise!
Please could someone shed some light onto what happens and why?
Any answers would be greatly appreciated!
Many thanks
Paul
I am having problems with how to draw band diagrams with band bending for Schottky Barriers and PN junctions. My issue is that I don't know how you determine which Fermi level remains fixed at its equilibrium level and which Fermi level moves to align with it when the two materials are brought into contact.
My textbook gives the following diagram for an N-type semiconductor in contact with a metal with a higher work function than the semiconductor:
[URL]http://homepage.ntlworld.com/beehive77/images/ntypeschottky.jpg[/URL]
It looks to me here that the metal fermi level remains constant at its equilibrium value, and the n-type semiconductor moves downwards to meet it. I understand the direction in which the bending must take place and the theory behind it, but I don't understand why it is the metal fermi level that remains constant, and not the n-type fermi level.
Could it not just as easily be the n-type fermi level remaining constant and the metal fermi level moving upwards to meet it?
My textbook then gives me the following diagram for a p-type semiconductor in contact with a metal with a lower work function than the semiconductor:
[URL]http://homepage.ntlworld.com/beehive77/images/ptypeschottky.jpg[/URL]
This time, the semi conductor fermi level remains constant and the metal fermi level moves up to align with it. Why is it the opposite way round?
I originally assumed that both fermi levels would move to meet each other and meet at some "middle ground" energy level, but the textbook seems to suggest otherwise!
Please could someone shed some light onto what happens and why?
Any answers would be greatly appreciated!
Many thanks
Paul
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