Understanding the width of the depletion region in a p-n junction

In summary, the width of the depletion region in a p-n junction is determined by the concentration of dopants on either side of the junction and the strength of the electric field. This region is essentially a barrier that prevents the flow of majority charge carriers, allowing the p-n junction to act as a diode. The width of the depletion region plays a crucial role in the functionality and efficiency of various electronic devices, as well as in the understanding of semiconductor physics. Understanding and controlling the width of the depletion region is key in the design and optimization of these devices.
  • #1
skweiler
7
0
I am studying to become and electrical engineer and am currently taking Electronics I. I am having trouble understanding the change in the depletion region that results from a bias. I understand that at the time of manufacture the free electrons from the n-type semiconductor are attracted to the vacancies in the p-type semiconductor and move until the electric field produced by the newly created ions prohibits further majority carriers from crossing the junction. What I don't understand is exactly which mechanism causes the depletion region to shrink when a diode is forward biased (+ on the p-side, - on the n-side) or vice versa when reversed biased? Is the depletion region defined as the width of the region of ions surrounding the p-n junction or the gap created when a potential "pushes" or "pulls" the majority carriers towards or away from the junction? If it is defined by the ions how does the applied potential fill (or empty) more holes and thus create more (or less) ions? How do the ions and their electrons surrounding the junction react to the applied potential?
 
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  • #2
This is how I understand it, with limited knowledge of any quantum physics.

Forward bias: Negative voltage on the N-region, positive on the P-region.

Negative voltage pushes more free electrons into the N-region toward the PN junction, and the positive voltage pulls them out of the P region, in effect pushing holes toward the junction. This makes the depletion region smaller. At about 0.7V, the free electrons and the holes meet in the middle and the diode starts to conduct.

Reverse bias: Negative voltage in the P, positive voltage on the N.

Negative voltage pushed more electrons into the P region, which fill up the holes, restricting how much current can flow through there. Positive voltage pulls the free electrons out of the N region, which reduces its ability to conduct. Increasing the voltages strip out more and more holes and electrons, which widens the depletion region.

With high enough reverse bias though, breakdown occurs and the diode starts to conduct again. That part I have to read about a couple more times to understand.
 
  • #3
Accidental double post
 
  • #4
Jiggy-Ninja said:
This makes the depletion region smaller.

This is what I don't fully understand. Kindly look at this http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html#c3" in the slide called "Depletion Region." In order for the depletion region to become larger more holes in the P-type need to be filled and more extra electrons in the N-type region need to be gotten rid of. Looking at the diagram the region of ions needs to expand.

Could it be that when the diode is reversed biased the negative potential pushes the minority carriers (electrons) in the P-type towards the junction, to recombine with the holes there in an attempt to enter the valence band and counterbalance, the forces exerted from the negative potential on the one side, and the line of negative ions on the P-side of the P-N junction?

When the diode is forward biased the opposite must occur.
 
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  • #5
skweiler said:
This is what I don't fully understand. Kindly look at this http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html#c3" in the slide called "Depletion Region." In order for the depletion region to become larger more holes in the P-type need to be filled and more extra electrons in the N-type region need to be gotten rid of. Looking at the diagram the region of ions needs to expand.

Could it be that when the diode is reversed biased the negative potential pushes the minority carriers (electrons) in the P-type towards the junction, to recombine with the holes there in an attempt to enter the valence band and counterbalance, the forces exerted from the negative potential on the one side, and the line of negative ions on the P-side of the P-N junction?

When the diode is forward biased the opposite must occur.
That's exactly what I was saying, just in different words. In reverse bias, the majority carriers (holes in P type and free electrons in N type) are all pulled away from the depletion region, making it wider. In forward bias, the majority carriers are pushed toward it. At about 0.7V, the depletion region closes completely.
 
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  • #6
Thank you for your help. These different ways of explaining the phenomena are, I believe, practically the same; any majority carriers will react in the opposite manner from the minority carriers because they have opposite charges.
 
  • #7
Welcome to PF skweiler.

It looks like Jiggy-Ninja has answered your question and I see you have found http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html" website (I love it).

Now for transistors. Here are a couple of images that I feel are good references.

tran10.gif


tran12.gif


http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html"

Enjoy
 

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1. What is a p-n junction and what is its role in electronics?

A p-n junction is a boundary between a p-type semiconductor (with positively charged holes as majority carriers) and an n-type semiconductor (with negatively charged electrons as majority carriers). It is a fundamental component in electronic devices such as diodes and transistors, allowing for the control and flow of electric current.

2. How does the depletion region form in a p-n junction?

The depletion region forms due to the diffusion of majority carriers from one side of the junction to the other. When the p-type and n-type semiconductors are brought into contact, electrons from the n-side diffuse to the p-side, and holes from the p-side diffuse to the n-side. This creates a region near the junction where there are no majority carriers, known as the depletion region.

3. What factors affect the width of the depletion region in a p-n junction?

The width of the depletion region is primarily influenced by the doping levels of the p- and n-type semiconductors, as well as the magnitude of the applied voltage across the junction. Higher doping levels result in a narrower depletion region, while a larger applied voltage leads to a wider depletion region.

4. How does the width of the depletion region affect the behavior of a p-n junction?

The width of the depletion region determines the barrier potential that must be overcome for current to flow across the junction. A wider depletion region means a higher barrier potential, which translates to a lower current flow. This is why p-n junctions are often used as diodes, as they only allow current to flow in one direction.

5. How can the width of the depletion region be measured or calculated?

The width of the depletion region can be measured using specialized equipment such as a capacitance-voltage (CV) meter or a scanning capacitance microscope. It can also be calculated using the built-in potential and the doping levels of the p- and n-type semiconductors. Alternatively, the depletion width can be estimated using the depletion approximation, which assumes a linear variation of carrier concentration across the depletion region.

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