## donor and acceptor levels in band diagrams

1.in band diagrams, books usually show parabolic bands for the conduction band and the valence band. They will also make note of the acceptor and donor leves in the band gap.
However, they always draw straight horizontal lines for these levels.
Is that correct? The impurity levels are independent of k?

2. Another thing that really bothers me about band diagrams is when people draw them
filled. If I understand things, electrons can only exist in states on the actual curve of the band. For instance the conduction band at k=0 is parabolic and at energy 0. There cannot be an electron at k=0 at energy E above E=0 (unless it's on another band). I'm correct, right?
 PhysOrg.com physics news on PhysOrg.com >> Atomic-scale investigations solve key puzzle of LED efficiency>> Study provides better understanding of water's freezing behavior at nanoscale>> Iron-platinum alloys could be new-generation hard drives

Mentor
Blog Entries: 27
 Quote by BeauGeste 1.in band diagrams, books usually show parabolic bands for the conduction band and the valence band. They will also make note of the acceptor and donor leves in the band gap. However, they always draw straight horizontal lines for these levels. Is that correct? The impurity levels are independent of k?
Not always. This is only a simplified description. But remember that even your "parabolic" bands are highly simplified. If you look at the actual band structure of many materials, you'll see spaghetti-like lines in various crystallographic directions. So in essence, if you already are using the simplified parabolic band, then drawing the donor/acceptor levels as being k-independent is acceptable.

 2. Another thing that really bothers me about band diagrams is when people draw them filled. If I understand things, electrons can only exist in states on the actual curve of the band. For instance the conduction band at k=0 is parabolic and at energy 0. There cannot be an electron at k=0 at energy E above E=0 (unless it's on another band). I'm correct, right?
You are right. The shading of those "occupied" part also annoys me because it causes confusion. What would be better is simply to draw the Fermi energy level and indicate that the band is occupied below that level.

Zz.
 Yeah, I was just thinking about the simplified description. Why, though, are the impurity levels independent of k?

## donor and acceptor levels in band diagrams

Wait! So what are we assuming when we simplify that parabolic conduction and valance bands to a straight line?

edit: Really bad grammar. Let me rephrase, " Wait! So why is it OK to approximate these parabolic conduction and valance bands by straight lines?"

Mentor
Blog Entries: 27
 Quote by BeauGeste Yeah, I was just thinking about the simplified description. Why, though, are the impurity levels independent of k?
Because we approximate the dopant to be uniformly distributed and having a simple Rydberg-type energy state. That's why you get a distinct energy state.

 Quote by Swapnil Wait! So what are we assuming when we simplify that parabolic conduction and valance bands to a straight line?
We did not. It is the DOPANT acceptor or donor level that is approximated to be that.

Zz.

 Quote by ZapperZ We did not. It is the DOPANT acceptor or donor level that is approximated to be that.
I'm sorry. I meant to ask you why was it OK to approximate the conduction and valence bands with straight lines to begin with (i.e when we are drawing energy-band diagrams)?

Mentor
Blog Entries: 27
 Quote by Swapnil I'm sorry. I meant to ask you why was it OK to approximate the conduction and valence bands with straight lines to begin with (i.e when we are drawing energy-band diagrams)?
This is a different question. The typical energy-band diagram has integrated out the k-dependence of the band dispersion. Thus, all you have is the energy bands, and the "straight line" doesn't mean anything other than to indicate where the band energy starts and stops.

Zz.

 Quote by ZapperZ This is a different question. The typical energy-band diagram has integrated out the k-dependence of the band dispersion. Thus, all you have is the energy bands, and the "straight line" doesn't mean anything other than to indicate where the band energy starts and stops.
You you are saying that the energy E_c that we denote as a straight line in the Energy-band diagrams is actually the total energy which we got from summing up all the energies at all the different k values?

Mentor
Blog Entries: 27
 Quote by Swapnil You you are saying that the energy E_c that we denote as a straight line in the Energy-band diagrams is actually the total energy which we got from summing up all the energies at all the different k values?
Well, what do you think happened to the k-dependence, then?

Those "block" diagrams are more of an energy "density of states" sketch. This means that it is a "momentum averaged" band diagram. The FULL band diagram must include the band dispersion, which, by definition, includes E vs. k dependence.

Zz.
 Mentor My solid state prof distinguished between two types of band diagrams: 1. Dispersion relation diagrams. These show the dispersion curves E(k). These are often depicted as parabolic as you described. So here we're working in k-space. 2. Generic band diagrams, where straight horizontal lines denote certain energy boundaries (e.g. top of the valence band, bottom of the conduction band), and the vertical axis is still energy. Here, it is VAGUE what the x-axis means. All we know for sure is that this is not a k-space diagram. One interpretion you could choose is that this is a "real space" band diagram...(the energies mentioned are constant as a function of position in the specimen...and correspond to the maximum and minimum respectively of the two parabolas in the dispersion diagram). He asked us to do so on a test (in fact, he just said, "draw the real-space band diagram", and we were expected to know what he meant). Or...you could interpret the x-axis as being totally meaningless (i.e. this is not a plot, just a diagram, a pictorial representation of a band gap). Or you could interpret it as...uh...what ZapperZ said. I'm not sure if I've contributed anything over and above what he said, but whatever...