[semiconduction] Does reverse bias decrease depletion region?

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SUMMARY

The discussion centers on the behavior of the depletion region in a reverse-biased pn junction. When a positive voltage is applied to the n-pole, the depletion region increases due to the movement of electrons from the p-region to the n-region, leaving behind positively charged ions on the n side and negatively charged ions on the p side. This process results in a wider depletion region as free carriers are removed from their parent atoms. The conclusion is that the depletion region indeed increases with reverse bias, contrary to some initial assumptions.

PREREQUISITES
  • Understanding of pn junctions in semiconductor physics
  • Knowledge of charge carrier dynamics (electrons and holes)
  • Familiarity with Poisson's equation in electrostatics
  • Basic concepts of biasing in semiconductor devices
NEXT STEPS
  • Study the effects of reverse bias on pn junctions in semiconductor theory
  • Learn about Poisson's equation and its application to semiconductor junctions
  • Explore the differences between forward and reverse bias in semiconductor devices
  • Investigate the role of charge carriers in the formation of depletion regions
USEFUL FOR

Electrical engineers, semiconductor physicists, and students studying semiconductor devices will benefit from this discussion, particularly those interested in the behavior of pn junctions under different biasing conditions.

nonequilibrium
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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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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 different 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).
 
thank you very much
 

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