Doubt on Depletion Layer Width in Biased pn-Jnction

In summary, the depletion layer in a diode is a region consisting of immobile ions of p-type and n-type dopants that do not move during biasing. This makes it a good insulator and allows for semiconductor switching at much lower potentials compared to metals. The width of the depletion layer changes due to the movement of these immobile ions when the diode is biased.
  • #1
Himabindu
1
0
" Depletion Layer is the region in diode which consists of ions( formed due to formation of octet configuration )of p-type and n-type dopants"
What happens to the ions present in the depletion layer when a diode is biased?
why the depletion layer width changes??
 
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  • #2
Himabindu said:
" Depletion Layer is the region in diode which consists of ions( formed due to formation of octet configuration )of p-type and n-type dopants"
What happens to the ions present in the depletion layer when a diode is biased?
why the depletion layer width changes??

Ions are part of the semiconductor crystal lattice. They were placed there during the doping process. They're supposed to not move during biasing. (If the ions move around, it means that your doping profile is diffusing, and the characteristics of your diode/transistor/etc. are being altered.)

Remember that 'insulator' basically means 'a material lacking mobile charge carriers.' The depletion zone is a fair insulator because those ions aren't mobile.

Analogous situation: Metals are composed of a crystalline grid of positive ions immersed in a "sea" of mobile electrons. If it was possible to sweep all the electrons out of a piece of copper, then the copper would become insulating, and we would have all-metal transistors. But now go and estimate the range of potential needed to remove all electrons from even a very thin film of copper. But while copper contains ~1 mobile electron per atom, doped semiconductor contains ~10^-9 mobile charge carrier per atom (or even less for lighter doping.) Thus the required potentials are many, many orders smaller for semiconductor switching, when compared to those required for metals. That's why it's possible to sweep the carriers out of a region of doped silicon, but it's not really feasible to do the same with a metal.
 

Related to Doubt on Depletion Layer Width in Biased pn-Jnction

1. What is a depletion layer in a pn-junction?

The depletion layer in a pn-junction is a region where there are no free charge carriers due to the diffusion of majority carriers across the junction. This creates a region with a fixed electric field that acts as a barrier to the flow of current.

2. How does bias voltage affect the width of the depletion layer?

Bias voltage can either increase or decrease the width of the depletion layer, depending on the polarity. A positive bias voltage will widen the depletion layer in the p-side and narrow it in the n-side, while a negative bias voltage will have the opposite effect.

3. What factors influence the width of the depletion layer?

The width of the depletion layer is primarily influenced by the doping concentrations of the p and n regions, the applied bias voltage, and the physical properties of the semiconductor material such as its dielectric constant and bandgap energy.

4. How does the depletion layer width affect the performance of a pn-junction?

The width of the depletion layer plays a crucial role in determining the characteristics of a pn-junction. A wider depletion layer results in a higher breakdown voltage and lower leakage current, while a narrower depletion layer allows for a higher current flow and lower forward voltage drop.

5. Can the depletion layer width be controlled or adjusted?

Yes, the width of the depletion layer can be controlled by changing the doping concentrations of the p and n regions or by applying a bias voltage. In some cases, the width can also be adjusted by varying the temperature or using specialized techniques such as ion implantation.

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