N-doped region tend to flow to the p-doped region

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In diode manufacturing, electrons from the n-doped region flow to the p-doped region, creating a depletion region until an equilibrium is established. Despite this equilibrium, a minimum forward voltage of 0.3-0.5 V is necessary to overcome the built-in potential that opposes current flow. This built-in potential must be surpassed for the diode to conduct, as it counteracts the diffusion effect. The discussion highlights the importance of understanding the relationship between the depletion region and the required forward bias voltage. Therefore, a voltage is essential to turn on the diode and allow current to flow.
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When a diode is first manufactured, the electrons in the n-doped region tend to flow to the p-doped region due to diffusion effect. In the process, a depletion region is formed until the electric field in the depletion region cancel out the diffusion effect. Hence an equilibrium is reached. (or so I thought)

However, since this two effect cancel out each other exactly, why would we need a minimum voltage of about 0.3-0.5 V to forward bias it? Wouldn't any voltage cause current to flow, since the diffusion+external voltage source is now greater than the electric field in the depletion zone alone?
 
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You're correct in that a depletion region is formed, and that a "built-in" potential across the depletion region results, but incorrect in the ramifications of this. The built-in potential opposes the direction of current flow as it normally occurs when the diode is acting in forward-bias mode.

So you need to overcome the built-in potential (the forward diode voltage) before the diode turns on and current starts flowing:
http://en.wikipedia.org/wiki/Diode#Current.E2.80.93voltage_characteristic
 
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