Junction diode - unequally doped sides

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In a p-n junction diode with a lightly doped p-side and a heavily doped n-side under reverse bias, the depletion layer penetrates the p-side due to the movement of electrons from the n-side to the p-side, filling holes and creating a region devoid of charge carriers. This occurs because the electric field established by the reverse bias causes electrons to migrate, leading to a wider depletion region on the less doped side. The balance of charge carriers remains constant, but the density of holes in the p-side is lower, resulting in a more significant penetration of the depletion layer there. The assumption that the heavily doped side would be more penetrated by the depletion layer is incorrect, as the electric field and carrier dynamics dictate the behavior. Understanding this mechanism is crucial for grasping the operational principles of diodes in reverse bias.
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Why does the depletion layer penetrate a lightly doped side of a p-n junction diode?? Consider that there is a lightly doped p-side and a heavily doped n-side in a p-n junction diode , which is connected to a dc power supply in reversed biased. Now, I've read that depletion layer penetrates the p-side in this case. But i could not figure out the reason behind it. So, Can u please explain me the mechanism involved in terms of the movement of electrons in the circuit.
Im in a faulty assumption that the side in which the diode is more heavily doped is bound to be penetrated more by the depletion layer..
(-I've completed my intermediate)
 
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The sum of electrons and holes stays constant (the whole thing does not get a significant net charge). If the hole density is larger, the volume with holes is smaller, and vice versa.
 
ok,, but i dint get the answer to my question!
 
In this case, I don't understand the question. You have depletion at both sides as electrons from one side move into the holes at the other side, unless your diode is in a conducting mode.
 
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