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

[Harmony]

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?

 

[MATLABdude]

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

 

[oswaler]

Your understanding is partially correct. When a diode is first manufactured, the electrons in the n-doped region do tend to flow to the p-doped region due to the diffusion effect. This creates a depletion region, where there is a lack of free charge carriers. However, as you mentioned, once the electric field in the depletion region reaches an equilibrium with the diffusion effect, current flow is halted.

In order for current to flow in a diode, there needs to be a net flow of charge carriers. This can only occur when the electric field in the depletion region is overcome by an external voltage source. When a forward bias voltage is applied to the diode, it creates an electric field that is in the same direction as the built-in electric field in the depletion region. This results in a decrease in the width of the depletion region and allows for current flow to occur.

The minimum voltage of 0.3-0.5 V is needed to overcome the built-in electric field in the depletion region. Any voltage lower than this will not be enough to decrease the width of the depletion region and allow for current flow. Additionally, the external voltage source must be greater than the diffusion effect in order to create a net flow of charge carriers. This is why a minimum voltage is necessary for current flow in a diode.

In summary, the combination of the built-in electric field in the depletion region and the external voltage source is what allows for current flow in a diode. The minimum voltage required is necessary to overcome the built-in electric field and create a net flow of charge carriers.