Band diagram of intrinsic semiconductors

In summary: Because that is the definition of the slope. If the potential is linear, then the electric field is constant. If the electric field is constant, then the potential is linear. They are related to each other by a constant, in this case q. This is just a simple straight line equation.
  • #1
shaikss
33
0
How to sketch the band diagram of intrinsic semiconductors including the fermi level with the electric field present verses distance? Its not a homework question.
 
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  • #2
Er.. your question is very vague. What exactly do you not know? A search on the energy band diagram of intrinsic semiconductor would have given you plenty of results. Did you try it? If you did, where exactly are you having difficulties?

And what "electric field"?

Zz.
 
  • #3
ZapperZ said:
Er.. your question is very vague. What exactly do you not know? A search on the energy band diagram of intrinsic semiconductor would have given you plenty of results. Did you try it? If you did, where exactly are you having difficulties?

And what "electric field"?

Zz.

Zapper,

I know the energy band diagram of intrinsic semiconductor where Fermi energy level lies in the middle of conduction band and valence band. If we consider that characterstics, its the graph which is plotted w.r.t energy.

What I have asked is to sketch the energy band diagram of intrinsic semiconductor which includes fermi level with uniform electric field verses distance.

I googled but I didn't found the relevant sketch.

Thanks!
 
  • #4
shaikss said:
Zapper,

I know the energy band diagram of intrinsic semiconductor where Fermi energy level lies in the middle of conduction band and valence band. If we consider that characterstics, its the graph which is plotted w.r.t energy.

What I have asked is to sketch the energy band diagram of intrinsic semiconductor which includes fermi level with uniform electric field verses distance.

I googled but I didn't found the relevant sketch.

Thanks!

I can only guess at what you are asking, because you are still not explaining it clearly.

Are you asking for the situation where a perpendicular electric field is applied to the surface of the semiconductor, and you want the effect on the semiconductor bands due to this external field from the surface and into the bulk of the material?

If it is, then you should be searching for "bend bending" diagram.

Zz.
 
  • #5
ZapperZ said:
I can only guess at what you are asking, because you are still not explaining it clearly.

Are you asking for the situation where a perpendicular electric field is applied to the surface of the semiconductor, and you want the effect on the semiconductor bands due to this external field from the surface and into the bulk of the material?

If it is, then you should be searching for "bend bending" diagram.

Zz.

please clarify me with respect to your second question.
The question i have posed was asked in an iit interview.
But i still do not find the answer
 
  • #6
shaikss said:
please clarify me with respect to your second question.
The question i have posed was asked in an iit interview.
But i still do not find the answer

This is getting sillier. Now it is *I* who have to clarify what I thought you were asking?

1. Uniform E-field. I assume you know what that means.

2. Surface of semiconductor is perpendicular to this uniform E-field. Again, I assume you know what this is.

3. E-field affects the bands in the semiconductor. OK so far?

4. Is this what you are asking?

I think if I don't get a definite answer to what you are asking after this, I'm done.

Zz.
 
  • #7
The energy bands are related to the electric potential through E=q*V where E is the energy, q the electron charge. If your Electric field [itex]E_{field}[/itex] is constant through the device, then your potential, and thus your energy Bands will be linear in position. That is [itex]qV=-q\int^x_0 E_{field} dx'=-qE_{field}x=E_c + Const.[/itex] where [itex]E_c[/itex] is your conduction band, and the valence band is just [itex]E_v=E_c-E_g[/itex] where [itex]E_g[/itex] is the bandgap. So, just draw a straight line with a slope [itex]-qE_{field}[/itex] to get the shape of your bands with respect to position in your semiconductor.
 
  • #8
Let me pose the Question in this way:

Sketch the energy band diagram (E versus x) including Fermi level of an intrinsic semiconductor under uniform electric field in x-direction.
 
  • #9
Just draw to lines )one for conduction band, one with the valence band)with a slope -qEfield, separated by a distance Eg. Then, since for an intrinsic case, the fermi level Ef is almost nearly right at the midgap, just draw a dotted line in between your bands with the same slope. You have to be careful though, since this is a quasi fermi level. By definition, when you apply a voltage (and hence create an Efield) you are taking the system out of equilibrium and putting it into a steady state. So, to get an exact answer, we need to know what is on either side of your device (n+p junction, metal-semiconductor interface, oxide and semiconductor interface, etc.). Then you would draw a constant fermi level at you boundries, and the difference in the two sides of your device wll give you the applied voltage time q (or -q*d*Efield, where d is the total depth or length of your device).
 
  • #10
cbetanco said:
Just draw to lines )one for conduction band, one with the valence band)with a slope -qEfield, separated by a distance Eg. Then, since for an intrinsic case, the fermi level Ef is almost nearly right at the midgap, just draw a dotted line in between your bands with the same slope. You have to be careful though, since this is a quasi fermi level. By definition, when you apply a voltage (and hence create an Efield) you are taking the system out of equilibrium and putting it into a steady state. So, to get an exact answer, we need to know what is on either side of your device (n+p junction, metal-semiconductor interface, oxide and semiconductor interface, etc.). Then you would draw a constant fermi level at you boundries, and the difference in the two sides of your device wll give you the applied voltage time q (or -q*d*Efield, where d is the total depth or length of your device).

Why the slope of -qE is required?
 
  • #11
shaikss said:
Why the slope of -qE is required?

Because the Energy of the bands is equal to [itex]qV+Const.[/itex]. So, the conduction band for example is [itex]E_c=qV+Const.=-q\int_0^x E_{field} dx'+Const.=-qE_{field}x+Const.[/itex] for a constant electric field, which I have denoted by [itex]E_{field}[/itex]. The Valence band is just taken by [itex]E_v=E_c-E_g=-qE_{field}x -E_g+Const.[/itex] where [itex]E_g[/itex] is just the bandgap energy. So the slope of the bands is [itex]-qE_{field}[/itex]
 

1. What is a band diagram of intrinsic semiconductors?

A band diagram of intrinsic semiconductors is a graphical representation of the energy levels of electrons in a pure, undoped semiconductor material. It shows the energy bands, or ranges of allowed energy levels, for both the valence band (where electrons reside) and the conduction band (where electrons can move freely). It also includes the energy gap, which is the energy difference between the two bands.

2. What is the significance of the band diagram in semiconductors?

The band diagram is significant because it helps to explain the electrical behavior of semiconductors. The energy gap between the valence and conduction bands determines whether a material is a conductor, semiconductor, or insulator. Intrinsic semiconductors have a small energy gap, allowing some electrons to move into the conduction band under certain conditions, making them useful for electronic devices.

3. How does temperature affect the band diagram of intrinsic semiconductors?

The band diagram of intrinsic semiconductors is affected by temperature in two ways. First, as temperature increases, more electrons are excited into the conduction band, causing the material to become more conductive. Second, the energy gap decreases with increasing temperature, making it easier for electrons to move into the conduction band. This is known as bandgap narrowing.

4. What is the difference between intrinsic and extrinsic semiconductors?

Intrinsic semiconductors are pure, undoped materials with a balanced number of positively charged holes and negatively charged electrons. Extrinsic semiconductors, on the other hand, are doped with impurities to increase their conductivity. This can be done by adding elements with more or fewer electrons than the semiconductor material, creating either p-type or n-type extrinsic semiconductors.

5. How does the band diagram of intrinsic semiconductors change with the introduction of impurities?

When impurities are added to intrinsic semiconductors, the band diagram changes. In p-type semiconductors, the impurities create acceptor states, which are energy levels in the energy gap that can accept electrons. This results in a shift of the Fermi level towards the valence band. In n-type semiconductors, the impurities create donor states, which can donate electrons, causing the Fermi level to shift towards the conduction band. This results in a narrower energy gap and increased conductivity.

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