If greater potential energy is toward pos charge, then why PN energy band diagram show lower energy toward pos charge across depletion region?

Question

If in an electric field, greater potential energy and voltage are toward the positive charge and away from negative charge, then why do PN energy band diagrams indicate lower energies toward the positive charge across the depletion region?

Answer 1

Think of it like this, if you let go of a ball, it will fall until the next stable point it finds right? because of gravity and the gravitational field, and one obvious thing to note is, that the ball is going from a higher energy level to a lower one.

If you look at a normal battery, the electrons are moving from the negative end to the postive end, and this is the same story as it was with the ball.

So in the depletion region, after stability has been reached, you have an electric field from your N-side to your P-side (as shown in your pictures). Now think about the battery again, the electric field is from the positive end to the negative end, but the voltage (as its scientific term) is in the opposite direction.

So this was an intuitive understanding of fields and energies, but if you are interested enough in maths like me, you would see that in physics, when you are talking about work being done on sth or by sth, there is usually a negative sign invovled.

To expand, lets imagine you have a fixed charge that doesn't move (Q). The field generated by this electrostatic charge is: $$E= Q/(4πεr^2 )$$ So to calculate potential, you will need to integrate the field, but then a problem pops up! Integration is done over a range (definite): $$∫_O E(r).dr =∫_O Q/(4πεr^2 ).dr =[-Q/(4πεr )]_O$$

The point O is our reference, and you need a reference to measure your potential against. The way physicists go around this is by putting infinity the reference point, so the answer to the above integration would be: $$potential = (-Q/(4πεr )) - (-Q/(4πεR ) $$ $$=$$ $$\lim_{R\rightarrow ∞}(-Q/(4πεr )) - (-Q/(4πεR )$$

where

$$R = distance\ from \reference$$

So, 1/∞ is really close to 0! and we will only be left with: $$potential = (-Q/(4πεr ))$$

And thus, the negative sign! That is why in your depletion region, where you have an electric field from your N-side to P-side, the energy (In the way that I explained above) and thus the voltage, is in the opposite direction. it is worth noting that energy is a scalar value and does not have a direction, but I am trying to give an intuitive answer.

I hope this will help undestand the relationship better. For your understanding of P-N junctions, start looking into fermi level, energy band diagrams, band gap enegy and Recombination/generation in a semiconductor.

Answer 2

in an electric field, greater potential energy and voltage are toward the positive charge and away from negative charge

This is correct from the view point of very definition of Electric field.

then why do PN energy band diagrams indicate lower energies toward the positive charge across the depletion region

this is because, in energy band diagram, on y-axis energy is not the energy of a positive test charge (remember, to define electric field, we take positive unit test charge) but the energy of the electron in particular state in the semiconductor.

Answer 3

E=-qv in the band diagram The energy is defined minius the potential times the charge (in elcectron volte units)