Dear Mouleeswaran G.S ,

If you connect the positive electrode to the n-side and the negative electrode to the p-side, then the circuit is closed. It is said that the diode is reverse biased by a voltage equal to the battery voltage.

It is so that all the reverse voltage will be dropped on the transition region in the p-n junction. This voltage will be absorbed by widening the space charge region.

The space charge region itself is composed of an electric dipole.

Which means that the its positive charges = its negative charges.

I would like that follow the charge distribution in the p-n diode at the different biasing conditions in the book at the link:Book Electronic Devices

Best wishes

Hi Mouleeswaran G.S ,

the condition "not in a closed circuit" would mean that the positive terminal of the 2 V supply is unconnected. But then the pn junction is not reversed biased. Both terminals of the diode would be on ground potential, and the positive side of the supply on a potential +2 V above ground.

So, regarding the diode there is no difference between this setup and being completely unconnected. Generally, "biased" and "not in a closed circuit" are mutually exclusive.

Dear Mouleeswaran G.S ,

If you connect the positive electrode to the n-side and the negative electrode to the p-side, then the circuit is closed. It is said that the diode is reverse biased by a voltage equal to the battery voltage.

It is so that all the reverse voltage will be dropped on the transition region in the p-n junction. This voltage will be absorbed by widening the space charge region.

The space charge region itself is composed of an electric dipole.

Which means that the its positive charges = its negative charges.

I would like that follow the charge distribution in the p-n diode at the different biasing conditions in the book at the link:Book Electronic Devices

Best wishes