Semiconductor Doping: Impurity Effects on Energy States

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Homework Help Overview

The discussion revolves around the effects of doping in semiconductors, specifically examining how impurities influence energy states and charge carriers in n-type and p-type materials.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants explore the implications of different doping options, questioning the validity of statements regarding energy states, electron-hole pairs, and the behavior of holes in valence and conduction bands.

Discussion Status

There is an ongoing exploration of the definitions and characteristics of holes in both n-type and p-type semiconductors. Some participants express uncertainty about the interpretations of options presented, while others suggest that certain statements may not accurately reflect semiconductor behavior.

Contextual Notes

Participants note potential misunderstandings regarding the roles of valence and conduction bands, particularly in relation to the presence of holes and the effects of doping. There is also mention of language barriers affecting clarity in the discussion.

songoku
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Homework Statement


Which one of the following describes the result of doping in semiconductors?
a. The impurity provudes additional energy states within the conduction band
b. More electron - hole pairs are introduced as a result of doping
c. For n - type semiconductor, the number of "holes" in the valence band remains small
d. For p - type semiconductor, there are more "holes" in the conduction band
e. None of the above

Homework Equations


None

The Attempt at a Solution


Option (a) is wrong because energy states stay the same

Option (b) is wrong because doping only adds only electron or hole, not electron - hole pair

Option (c) I think is correct because for n - type the number of free electron in conduction band increases and number of hole in valence band stays the same

Option (d) I am not sure

Is my reasoning for (a) - (c) correct? And what about (d)?

Thanks
 
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For C: the number of holes remains small - that is, zero. But I don't think the author is taking that to be an answer.
Actually, the number of holes in the valence band isn't really small. The valence band of every atom has holes that share electrons with neighboring atoms. What is small (zero) for P-type is the number of holes in the conduction band.
 
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.Scott said:
For C: the number of holes remains small - that is, zero. But I don't think the author is taking that to be an answer.
Actually, the number of holes in the valence band isn't really small. The valence band of every atom has holes that share electrons with neighboring atoms.
I learn that valence band is the highest populated band, filled completely with electrons. How can valence the number of holes in the valence band not really small?

What is small (zero) for P-type is the number of holes in the conduction band.
So, option D should be: For p - type semiconductor, there are more "holes" in the VALENCE band. Am I correct?

And why for p - type the number of holes in conduction band zero? Conduction band is the first band that is not fully occupied by electrons but why the holes are zero?

The answer of this question should be (e)?

Thanks
 
songoku said:
I learn that valence band is the highest populated band, filled completely with electrons. How can valence the number of holes in the valence band not really small?So, option D should be: For p - type semiconductor, there are more "holes" in the VALENCE band. Am I correct?

And why for p - type the number of holes in conduction band zero? Conduction band is the first band that is not fully occupied by electrons but why the holes are zero?

The answer of this question should be (e)?

Thanks
There are more holes in the CONDUCTION band for p-type semiconductors.
I am taking "valence band" to be an attribute of the atoms. Although I am not sure how the author intends it to be interpreted in this context.
I believe the answer is D.
 
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.Scott said:
What is small (zero) for P-type is the number of holes in the conduction band.
.Scott said:
There are more holes in the CONDUCTION band for p-type semiconductors.

English is not my strong point but I feel that the two sentences above contradict each other. Maybe I misinterpret your intention?

Is it not correct if I say in p - type semiconductor, the addition of holes make the electrons in valence band be able to move to fill the holes so there are more holes in valence band?

Also, I think in n - type semiconductor, the addition of electrons will result in more electrons in conduction band. Am I correct?

Thanks
 
In post number 2, I mistyped. I was referring to option C so I should have said: "What is small (zero) for N-type is the number of holes in the conduction band.".
songoku said:
Is it not correct if I say in p - type semiconductor, the addition of holes make the electrons in valence band be able to move to fill the holes so there are more holes in valence band?
Since the author seems to be differentiating between the valence band and the conduction band, I am not sure that is correct.[/QUOTE]
songoku said:
Also, I think in n - type semiconductor, the addition of electrons will result in more electrons in conduction band. Am I correct?
Yes
 
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.Scott said:
Since the author seems to be differentiating between the valence band and the conduction band, I am not sure that is correct.

You mean valence band and conduction band should be the same?
 
semiconductor physics not my strong point but I THINK the only option that seems reasonable here is C, for the reason you stated in your first post.

Doping increases electrons in the conduction band (for n type) or holes in the valence band (for p-type).
 
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I think Delta is right.
I just read up on bands.
In n-type semiconductors, charge moves as electrons through the conduction band.
In p-type semiconductors, charge moves as holes through the valence band.

So D cannot be right because there are never holes in the conduction band.
C is right because doping an n-type affects the conduction band, not the valence band.
 
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thank you very much for the help
 

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