Wrong statement about a p-n junction

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Homework Help Overview

The discussion revolves around the characteristics and behavior of a p-n junction formed by bringing a p-type semiconductor into contact with an n-type semiconductor. Participants are examining various statements regarding the behavior of the junction under different biasing conditions, specifically focusing on identifying which statement is false.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the validity of multiple statements regarding the behavior of p-n junctions under forward and reverse bias conditions. There is a focus on logical reasoning to determine the truth of statements C and D, which are seen as opposites. Questions arise about the implications of biasing on the charge carriers and the depletion region.

Discussion Status

There is active engagement with participants questioning the assumptions behind the statements and discussing the implications of reverse bias on the p-type and n-type materials. Some participants suggest that both C and D could be incorrect, while others emphasize the need for clarity on the definitions and implications of the terms used in the statements.

Contextual Notes

Participants express uncertainty about the definitions of "less positive" and "less negative," and how these relate to the behavior of the p-n junction under bias. There are also mentions of potential breakdown conditions and the behavior of minority charge carriers in the depletion region.

songoku
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Homework Statement


A p type semiconductor is brought into contact with an n type semiconductor to form p - n junction. Which statement is false?

A. During forward bias condition, if applied p.d. overcomes the junction potential, electron will cross steadily from n - type side to p - type side

B. During reverse bias condition, the width of the depletion region becomes larger as the externally applied p.d. adds to the junction potential

C. During reverse bias condition, the p type semiconductor becomes less negative

D. During reverse bias condition, the p type semiconductor becomes less positive

E. Under increasingly high reverse bias p.d., current can increase sharply through the p-n junction

Homework Equations


Not sure

The Attempt at a Solution


I think option A is correct but I am not sure about "if the applied p.d. overcomes thr junction potential". Let say I have a diode connected to dc source having a small emf (maybe 1 mV) in forward bias condition, is it possible that current not flowing because 1 mV can not overcome junction potential?

Option B is correct because holes in p type will be attracted to negative terminal of the power source and electrons in n type will be attracted to positive terminal of power source so potential barrier increases.

Option C and D, not sure. Can n type or p type of p-n junction become less positive / negative in forward / reverse bias condition? If it is possible, it means that in reverse bias condition p type becomes less positive because electrons from the negative terminal of power source fill the holes in p type?

I think option E correct, breakdown of p-n junction happens but how to explain it with regards to depletion layer? In reverse bias condition, the width of depletion layer increses and if the p.d. of power source is increased again and again there will be a point where the depletion layer decreases until zero instead of keep increasing?

Thanks
 
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"C" is incorrect. Reverse bias apply negative potential to P area, therefore it become more negative, not less
 
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Interesting this can be (almost) answered by knowing very little about semiconductor electronics and using laws of logic only.

We can notice (after some careful thinking) that options C and D are the opposite of each other so one is false and the other is true (given the law of logic that for a statement P only one of P, or (not P) can be true, and the other is false). So the false statement is one of C or D.
 
trurle said:
"C" is incorrect. Reverse bias apply negative potential to P area, therefore it become more negative, not less

By applying negative potential to P area, is it correct to say that electrons from the negative terminal of power source fill the holes in p type? And electrons from n type move to positive terminal of the power source?

Delta2 said:
Interesting this can be (almost) answered by knowing very little about semiconductor electronics and using laws of logic only.

We can notice (after some careful thinking) that options C and D are the opposite of each other so one is false and the other is true (given the law of logic that for a statement P only one of P, or (not P) can be true, and the other is false). So the false statement is one of C or D.

How about the reasoning to each option? Is it correct? And can you please address to my questions regarding to each of the option?

Thanks
 
No sorry i can't comment much on the reasoning for each option as i am afraid i don't know much about semiconductor electronics.

I can invoke @gneill and @CWatters who i believe they know a lot more than me on this specific area.
 
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Delta2 said:
Interesting this can be (almost) answered by knowing very little about semiconductor electronics and using laws of logic only.

We can notice (after some careful thinking) that options C and D are the opposite of each other so one is false and the other is true (given the law of logic that for a statement P only one of P, or (not P) can be true, and the other is false). So the false statement is one of C or D.
Or it could be that neither C nor D is correct. Not suggesting that that's the case; just critiquing the reasoning ...
 
rude man said:
Or it could be that neither C nor D is correct. Not suggesting that that's the case; just critiquing the reasoning ...

That could be the case indeed, but then we would have two possible answers for this multiple choice test, and this test implies that the answer is only one (it asks which (i assume which one ) of the following is wrong). But yeah deep down we can't exclude the possibility that this test has two (or more) wrong statements.
 
Delta2 said:
We can notice (after some careful thinking) that options C and D are the opposite of each other so one is false and the other is true (given the law of logic that for a statement P only one of P, or (not P) can be true, and the other is false). So the false statement is one of C or D.

If "becoming less positive" means: "It's positive without any bias, and less positive with reverse bias", then that is not the logical negation of "becoming less negative".
 
  • #10
willem2 said:
If "becoming less positive" means: "It's positive without any bias, and less positive with reverse bias", then that is not the logical negation of "becoming less negative".
Can you expand on this because according to my understanding "becoming less positive" is logically equivalent to "becoming more negative"

I mean it is like saying "becoming less tall" which obviously is equivalent to "becoming more short"
 
  • #11
Delta2 said:
That could be the case indeed, but then we would have two possible answers for this multiple choice test, and this test implies that the answer is only one (it asks which (i assume which one ) of the following is wrong). But yeah deep down we can't exclude the possibility that this test has two (or more) wrong statements.
You're right. It is implied that there is only one wrong answer. I missed that little subtlety. o:)
 
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  • #12
C and D are both wrong if you consider the that the depletion region (positive in the n material and negative in the p material) increases in size in the reverse bias. i.e. The p material becomes more negative and the n material becomes more positive due to the increase in size of the depletion region.
...+++-+---
p side +++-+--- n side
...+++-+---
 
  • #13
CWatters said:

I have read the link and I have several questions:
1. How can the minority charge carriers pass through the depletion layer? Is it something to do with barrier tunneling?

2. In reverse - biased condition, the number of electrons that move from n - type to positive terminal of power source will be equal to number of electrons that move from negative terminal of power source to p - type so that the net charge in the p - n junction is zero (it is still neutral)?

3. What happens to the depletion layer when breakdown occurs? As external p.d. increases, the number of minority charge carriers increases and large reverse current will be produced hence breakdown occurs. So it means that the the depletion layer will increase at first and then it vanishes because minority charge carriers have enough energy to penetrate the layer?

4. Does holes can move around or actually only electrons move? I mean holes just like the conventional current (flow of positive charge), it actually does not flow and the one flowing is electron.

Thanks
 
  • #14
WesMisenheimer said:
C and D are both wrong if you consider the that the depletion region (positive in the n material and negative in the p material) increases in size in the reverse bias. i.e. The p material becomes more negative and the n material becomes more positive due to the increase in size of the depletion region.
...+++-+---
p side +++-+--- n side
...+++-+---

Maybe I misinterpret your post but I think you are saying that p material becomes more negative which suits option D?
 
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  • #15
songoku said:
I have read the link and I have several questions:
1. How can the minority charge carriers pass through the depletion layer? Is it something to do with barrier tunneling?

The applied voltage pushes minority carriers towards the depletion layer. Then, if I remember correctly, there are thermally created electron hole pairs in the depletion layer that allow a very small current to pass. This is called the reverse bias saturation current. Its typically a constant current (eg independent of the applied voltage) but is dependent on temperature.

2. In reverse - biased condition, the number of electrons that move from n - type to positive terminal of power source will be equal to number of electrons that move from negative terminal of power source to p - type so that the net charge in the p - n junction is zero (it is still neutral)?

I think that's correct.

3. What happens to the depletion layer when breakdown occurs? As external p.d. increases, the number of minority charge carriers increases and large reverse current will be produced hence breakdown occurs. So it means that the the depletion layer will increase at first and then it vanishes because minority charge carriers have enough energy to penetrate the layer?

Explanation off breakdown here..

https://ecee.colorado.edu/~bart/book/book/chapter4/ch4_5.htm

4. Does holes can move around or actually only electrons move? I mean holes just like the conventional current (flow of positive charge), it actually does not flow and the one flowing is electron.

A hole is an atom missing an electron. The atoms don't move.
 
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  • #16
CWatters said:
A hole is an atom missing an electron. The atoms don't move.
So the phrase "holes are attracted towards..." or "holes diffuses towards..." or "holes moves to..." all have the same meaning that actually holes do not move and electrons are the one moving filling the holes?

Thanks
 
  • #17
songoku said:
So the phrase "holes are attracted towards..." or "holes diffuses towards..." or "holes moves to..." all have the same meaning that actually holes do not move and electrons are the one moving filling the holes?

Thanks
A hole is the absence of an electron. Holes do have characteristibcs like mobility which differ from that of electrons (electron mobility is roughly twice that of holes).
This is all a matter of quantum mechanics, and as I'm sure you've heard before, it's inappropriate to use "coomon sense" in this area.
 
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  • #18
A Hole is a bit like an IOU note. It's not really money but can move from person to person like money. Although not money it does carry wealth with it.

OK yes it a bad analogy. Sorry I mentioned it.
 
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  • #19
I am so sorry for late reply.

Thank you very much for all the help
 

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