What Happens When We Apply a 10V Source to a Diode in Reverse Bias?

In summary: Again, I don't understand your question. The depletion region at zero bias has a certain width, depending on the doping densities of the two sides.
  • #1
CaptainMarvel1899
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Assume we have a diode . We connenct the negative terminal of the battery(anode) to the n type region and we connect the positive terminal of the battery to the p type region of the battery.The net force forces electrons to start flowing from the n type region to the p type region.As long as we increase voltage,we increase the propability of a bond to be broken(Si-B-) so the depletion region gets narrower.Once we reach 0.7V the depletion region gets so tiny that we assume it doesn't exist.Now if we connect the diode to a different way (reverse bias) electron flow until the depletion region will get wide enough so that it would oppose the external voltage(Battery).Since now the electric field is very big why don't the electrons flow from the p type to thw n type to reduce the electric potential?I know that during reverse bias condition there is a reduce in chemical potential , but in zero bias condition the chemical and the electric potential reach equillibrium at 0.7V so breakdown voltages 10V can't be explained . Help appreciated.
 
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  • #2
P type semiconductors conduct by moving electron holes. But there are no holes available from the depletion region. Similarly, the n type semiconductor moves electrons - but there will be no electrons available on the n side of the depletion region.
 
  • #3
Moving electron holes are basicly electrons moving at the opposite direction and what this has to do with my issue?1
 
  • #4
CaptainMarvel1899 said:
...Since now the electric field is very big why don't the electrons flow from the p type to thw n type to reduce the electric potential?..

The reason is that there are essentially no electrons in the p type region. The few electrons that are generated by thermal processes do flow into the n-type region. This is why diodes have a small reverse current.
 
  • #5
phyzguy said:
The reason is that there are essentially no electrons in the p type region. The few electrons that are generated by thermal processes do flow into the n-type region. This is why diodes have a small reverse current.
If there are no electrons in the p type region how is it negatively charged?
 
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  • #6
CaptainMarvel1899 said:
If there are no electrons in the p type region how is it negatively charged?

To be clearer, I should have said there are no free electrons in the p region. There are of course bound electrons at each of the lattice atoms, but these are not free to move. To answer your question, the p region is not negatively charged, it is charge neutral. If you mean in the depletion region on the p-side, it is negatively charged because holes have moved out of the region, exposing the negative lattice charge. You can also look at this as extra electrons have moved into the depletion region to fill the hole sites, but again, these extra electrons are bound and are not free to move. The depletion regions on both sides contain no free charge carriers (they are depleted of free carriers).
 
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  • #7
If we offer more energy than the bond energy , the bonds will be broken and electrons will start moving . That is my question.The net potential must be more than the bond energy(because the diode reaches equillibrium in reverse bias at 0.7V).What am I missing?
 
  • #8
CaptainMarvel1899 said:
If we offer more energy than the bond energy , the bonds will be broken and electrons will start moving . That is my question.The net potential must be more than the bond energy(because the diode reaches equillibrium in reverse bias at 0.7V).What am I missing?
I'm not sure what your question is. The free electrons will move with a small applied bias. But there are no free electrons in the charge neutral p-type region, and no free carriers of either type in the depletion regions.

The bound electrons will not move until the applied voltage is very large. At this point the semiconductor breaks down and is typically destroyed.
 
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  • #9
phyzguy said:
I'm not sure what your question is. The free electrons will move with a small applied bias. But there are no free electrons in the charge neutral p-type region, and no free carriers of either type in the depletion regions.

The bound electrons will not move until the applied voltage is very large. At this point the semiconductor breaks down and is typically destroyed.
I am asking how the depletion region in a forward bias mode is tiny when thw breakdown voltage is 10V for example?
 
  • #10
CaptainMarvel1899 said:
I am asking how the depletion region in a forward bias mode is tiny when thw breakdown voltage is 10V for example?

Again, I don't understand your question. The depletion region at zero bias has a certain width, depending on the doping densities of the two sides. In forward bias, the depletion region shrinks. What does this have to do with the breakdown voltage? The breakdown voltage refers to reverse bias.

Note there is also a certain built-in field across the depletion region at zero bias. Forward biasing the diode reduces this field, it doesn't increase it. Is this what you are missing?
 
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  • #11
If the electrons have a small propability of not being in the conduction band at 0.7V , then imagine what would be at 10V
 
  • #12
CaptainMarvel1899 said:
If the electrons have a small propability of not being in the conduction band at 0.7V , then imagine what would be at 10V
I have no idea what you're talking about.
 
  • #13
phyzguy said:
I have no idea what you're talking about.
When we apply a forward voltage of 0.7V the depletion region becomes very narrow so the electrons have a big propability to be in the conduction band . Imagine what would happen with 10 volts.
 
  • #14
CaptainMarvel1899 said:
When we apply a forward voltage of 0.7V the depletion region becomes very narrow so the electrons have a big propability to be in the conduction band . Imagine what would happen with 10 volts.

Maybe I see what you are asking. It is impossible to apply 10 volts of forward bias to a diode. Since the diode current increases as [itex] I=I_0 e^{\frac{q V}{kT}}[/itex], you cannot apply more than ~0.7 volts of forward bias before the current becomes very large. Any remaining applied voltage gets dropped in resistances leading up to the diode. Is this your question? Maybe you need to read more about semiconductor physics. Do you have a textbook you are studying?
 
  • #15
It is kind of the opposite.I know the voltage drop of the depletion regiin can be max 0.7V.If the energy whixh breaks the B- Si bonds is given by a 0.7V source then what would happen if we replaced it with a 10V source?
 

Related to What Happens When We Apply a 10V Source to a Diode in Reverse Bias?

1. What are semiconductors and how do they work?

Semiconductors are materials that have properties in between those of conductors and insulators. They have the ability to conduct electricity, but not as well as conductors. They work by having a small number of free electrons, which can move around and conduct electricity under certain conditions.

2. What causes confusion about semiconductors?

Semiconductors can be confusing because they have properties that are different from both conductors and insulators. Additionally, their behavior can be affected by factors such as temperature, impurities, and electric fields.

3. How are semiconductors used in technology?

Semiconductors are used in a wide range of technologies, including computer chips, solar cells, and LEDs. They are also used in transistors, which are essential components in electronic devices.

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

Intrinsic semiconductors are pure materials, such as silicon or germanium, that have a small number of free electrons. Extrinsic semiconductors have impurities added to them, which can increase the number of free electrons and make them more conductive.

5. How do semiconductors contribute to the advancement of technology?

Semiconductors have played a crucial role in the development of modern technology. They have allowed for the miniaturization of electronic devices, making them smaller, faster, and more efficient. They also enable the creation of complex integrated circuits, which have revolutionized the fields of computing and communication.

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