Shift in energy levels of transistors

Question

As I was reading basic electronics, PN-junction diodes and two terminal MOSFETs, I understood the position of the fermi energy level and position of valence and conduction band in those devices. But I would like an explanation on how the bending of the valence or conduction bands occurs with the application of proper forward bias for either an N- or P-type PN-junction diode or a two-terminal MOSFET.

Answer 1

AFAIK the statement that energy levels actually bend is wrong. We are just extrapolating diagrammatically to show how the two conduction bands and valence bands are linked with a space charge region in between.

Lets just consider the case of a pn-junction. First, when the two parts, p and n are not forming a junction, then, you will agree that the fermi level shifts upwards and downwards in the p and n type materials from the middle position of energy band gap? How? Take the case of an intrinsic material where number of electrons in the conduction band will be equal to the number of holes in the valence band. This means that the overall energy balances somewhere in the middle of the forbidden band gap. Thus, for an intrinsic semiconductor, the fermi level lies approximately at (Ec+Ev)/2. Now, when we dope it with a pentavalent impurity, thus creating an n-type semiconductor, we are effectively adding more electrons in the conduction band than there are holes in the valence band (owing to the extra discrete energy level inserted by the dopant near the conduction band) Thus, the balance of energies tilts in favor of the conduction band and the fermi level now lies somewhere closer to the valence band. Similarly, the fermi level for a p-type semiconductor lies near the valence band.

Now, let us create a junction with these two types of semiconductors. By creating a junction, we create a concentration gradient and that results in diffusion. How long would this continue? As long as there is graident, or there is some force that can equally oppose this process of diffusion across the junction. But what does that equilibrium condition mean in terms of energy? When no more diffusion can take place, it basically means that the fermi levels of both p-type and n-type materials are now equal (since unequal energy levels creates a gradient?) Thus, the Fermi levels of the two materials are equal. But what we did there was to reduce the fermi level from the center of band gap in case of p-type and raise it above in case of n-type. Thus, for an electron to jump from the valence band of the p-type material to the conduction band of the n-type material, it now needs energy that is equal to the total displacement of fermi levels, i.e. downward shift on p-type fermi level plus upward shift in n-type fermi level. This is the minimum energy that must be provided to the electron through external bias (forward bias) for the junction to conduct and would be called the barrier potential. So, I think, it is not the forward bias that causes the bend, forward bias just helps cross this small barrier or hill. The bend is due to the shift of fermi levels and as a consequence, the entire energy profile of the n-type material shifts downwards, relative to the p-type material (because both will have the same fermi level)

I hope this helps, it is a little difficult to visualize things in text.