Diode operation from reverse bias to zero bias

[CaptainMarvel1899]

Assume we have a closed diode circuit .We connect the n type region of the diode to the positive terminal of the battery.We connect the p type region of the diode to the negative terminal of the battery.The depletion layer is increased.Now we open the circuit.Why the diode returns to its zero bias mode ?The lattice of silicon is more uniform and electrons have created bonds with unpaired electrons from Si atoms so they could not "return". What am I missing?

 

[dlgoff]

The depletion region is where charge is concentrated and it can increase or decrease in width depending on the applied voltage. The crystalline material atoms don't get displaced; just the charge gets displaced.

 

[CaptainMarvel1899]

The depletion region is where charge is concentrated and it can increase or decrease in width depending on the applied voltage. The crystalline material atoms don't get displaced; just the charge gets displaced.

Hmm I know that .I was just wondering how the covalent bonds between Si and B- can be broken so the diode would return to its unbiased state.

 

[dlgoff]

Isn't a covalent bond one where there's a sharing of an electrons? Same answer as before.

 

[CaptainMarvel1899]

Yes it is.

 

[CaptainMarvel1899]

Once they form the covalent bond , how can they be free charge carriers again?

 

[Rive]

What am I missing?

This part
Semiconductor crystals has something special between the 'pure' covalent bonds and metalic bonding. That's what makes them special. So the 'trapped' electrons/holes in the enlarged depletion layer are still mobile, their state is forcibly maintained just by the appropriate external electric field.

 

[essenmein]

Assume we have a closed diode circuit .We connect the n type region of the diode to the positive terminal of the battery.We connect the p type region of the diode to the negative terminal of the battery.The depletion layer is increased.Now we open the circuit.Why the diode returns to its zero bias mode ?The lattice of silicon is more uniform and electrons have created bonds with unpaired electrons from Si atoms so they could not "return". What am I missing?

Leakage current does the discharging? Technically this is charge carrier drift due to the now comparatively large electric field.