You will generate carriers within the depletion region, but it will be drowned out by the forward bias current.

Drar Arpit

Actual effect will depend on the type of radiation, energy of radiation and also on the semiconductor material used to make diode. High energy irradiation in general  create defects in the semiconducting material and depletion region may be affected which may change the barrier height. Once barrier height is changed current - voltage characteristics will change.

How high the energy of the radiation should be to change the voltage characteristics? If the radiation is a laser light with energy greater than bandgap of semiconductor, will it change the voltage characteristics?

It depends if the PN junction has a direct band gap. In this case if the energy of incoming photons is near or slightly higher than the band gap, you will be able to create free carriers (optical pumping) and possible cause a population inversion. If the energy is much higher then you will cause defects in the material. If the energy is smaller than the band gap the majority of the photons will be absorbed by the material causing heating, whereas a small portion will allow free carrier excitation through multi-photon processes like two photon absorption. 

On the other hand if the material used has an indirect band gap (Silicon) then the injected photons will be absorbed and allow refractive index changes, or you will be able to produce phonon emission. 

I hope I helped somehow 

Thanks Charis,

But when I have already forward biased the PN junction, will this optical population inversion create ant measureable difference ?

It depends on the energy of the optical input. If the forward bias is strong enough and has caused a population inversion then the optical carriers will further deplete the free carriers and you will monitor a twofold event. 

1) Optical output of the PN junction will increase

2) Free carriers will be reduced due to depletion  

It means that Quasi Fermi Level difference will decrease and there will be a decrease in the voltage across the terminals as well.

Dear Arpit,

I cam late to your question. But i hope you find my answer useful to you. However your question is conceptual and the answer may be useful for the others. Your question can be fully answered within the the response of p-n or pin photo diode.

It is so that the diode current will be the superposition of two currents; the photo current and the dark current. The photocurrent Iph is depending on the intensity of the incident light and its photon energy. It is so that one photon generates one electron hole pair.

These photogenerated electron hole pairs will be be collected of the diode if they succeeds to arrive the depletion region other wise they will be lost by recombination. The external quantum efficeincy determines the generated photocurrent in response to the incident power of the photon flux.

As the diode is forward biased it will have its own current due its forward bias. This current is called dark current Id

So, the total current will be I= -Iph +Id,

So, the diode current will be negative so long as Iph is greater than Id up to the socalled open circuit voltage Voc, the net diode current will be positive.

For any positive diode current the diode voltage after illumination will be greater than without illumination provided the temperature remains constant.

Best wishes