[semiconduction] Does reverse bias decrease depletion region?

In summary, in a pn junction, the depletion region widens as electrons move from the p-region to the n-region, reversing the polarity of the junction.
  • #1
nonequilibrium
1,439
2
Hello,

For simplicity's sake, let's take a np junction.

Say we connect the left side (n-pole) to a POSITIVE voltage; this is called a reverse bias, because it will result in a small current from n to p. In other words: electrons will travel from p to n across the junction (and holes will go from n to p, but let's just talk about the electron movement to keep it clear). It is clear that the voltage across the junction increases.

Now, what happens to (the width of) the depletion region? I would say, it decreases, because the electrons move from the p-region to the n-region, and the depletion region was originally formed by electrons going from n to p ("annihilating" holes in the p-section), which is now partially being undone. Yet I think sources are saying the region increases. What is the truth? And if I'm wrong, could anybody tell me why it increases (using the motions of the charge carriers).
 
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  • #2
The depletion region increases in a reverse biased pn junction. As you apply a voltage, the electrons and holes from the n and p layers respectively will be shed from their parent atoms, leaving behind positively charged ions on the n side and negitively charged ions on the p side. The region where you have removed the clothes (removed the free carriers, electrons or holes) from the atom to make them ionized is referred to the depletion region. In both forward and reversed biased junctions you are removing the clothes from the atom and creating a depletion layer, but the current will be differnt in both cases since in one case you are injecting free carriers into the junction (forward biased) and the other case you are moving them away from the junction (reversed biased).

You could just solve poisson's equation with the appropriate boundary conditions for a pn junction and find the depletion region increases with increasing bias voltage (could be negative or positive depending if you want forward or reversed biased).
 
  • #3
thank you very much
 

Related to [semiconduction] Does reverse bias decrease depletion region?

1. Does reverse bias affect the depletion region in a semiconductor?

Yes, reverse bias has an effect on the depletion region in a semiconductor. It causes the depletion region to widen, reducing the concentration of charge carriers in the region.

2. How does reverse bias decrease the depletion region?

Reverse bias decreases the depletion region by creating a potential barrier that opposes the flow of charge carriers. This barrier causes the depletion region to widen as the concentration of charge carriers decreases.

3. Can reverse bias completely eliminate the depletion region?

No, reverse bias cannot completely eliminate the depletion region. The depletion region will always exist in a semiconductor, but the width of the region can be reduced by applying a reverse bias.

4. Is reverse bias the only factor that affects the depletion region?

No, there are other factors that can affect the depletion region in a semiconductor, such as temperature, doping concentration, and applied voltage. Reverse bias is just one of the factors that can alter the width of the depletion region.

5. What is the significance of the depletion region in semiconductors?

The depletion region plays a crucial role in the functioning of semiconductors. It acts as a barrier that controls the flow of charge carriers and helps in the formation of a p-n junction, which is essential for the operation of many electronic devices.

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