[semiconduction] Does reverse bias decrease depletion region?

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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