Simple answer: the cell absorbs more photons so it's likely to generate more charge carriers and hence improve the short circuit current. 

It also depends on how higher you mean. In fact, solar spectrum exhibits significant photon flux in infrared region.

http://en.wikipedia.org/wiki/Sunlight

@ Hung

What is the relation between number of photons and its wavelength? Could you please give the physics behind this.

Dear shamjid,

I agree with Hung, as the cell absorbs more photons and generate more electron hole pairs, their photo current will increase and consequently their conversion efficiency provided that the generated e-h pairs will be collected and not recombine again.

Concerning your inquiry to Hung, the incident solar radiation has certain spectral distribution. It covers a large range of wavelengths from far infra red to ultraviolet.

The photon energy is related only to the wavelength of the radiation, where Ep= hC/ Lambda, with C the speed of light and Lambda is the wavelength of the radiation.

The incident solar radiation is measured in watts per m^2. Let us assume that  this power is S, then S= Phi x Ep , where Phi is the photon fux per cm^2.

The material absorbs the radiation having Ep => Eg where Eg is the gap energy.

Moreover, the material is characterized by absorption coefficient  curve depending on it energy band structure.This means that the material has certain spectral response which determines it ability to absorb the incident radiation.

For more information please see the Link:

https://www.researchgate.net/publication/236002879_A_course_on_photo-voltaic_array_systems

wish you success.

Book A course on photo-voltaic array systems

How number of photons and wavelength of light is related? In best of my knowledge as wavelength increases number of photos increases. I'm expecting answers about the basic physics principle behind this trend.

This is if you assume that you have  constant incident solar power/  cm^2 for all wave lengths.

Then to get the number of photons corresponding to certain incident solar power S as assumed before,  the number of photons Np =  S/Ep = S x wavelength/C h photons/second/ cm^2. Which is the same as you mentioned in your lat comment.

Shamjid,

Photons with energy less than energy needed to move an electron against output voltage cannot participate in current generation (for a usual simple mechanism). Energy of a photon in electron-volts is E=1240/lambda, where wavelength is in nm. This numerically corresponds to the voltage. This actually gives a theoretical optimum for red side limit of the cell absorption, around 700-800nm. Tandem cells with different OC voltages can improve total efficiency by adding power collected from longer wavelength photons.

It's not difficult to find an AM1.5 spectrum (or a similar standard) as a dependence  of photon flux from photon energy, or recalculate this from more common dependence of power per nanometer per sq. meter from wavelength. For _very_ rough estimations you can use Planck formula for black body emission.

I have to clarify one more point,

Does a red shift in the active layer absorption peak of solar cell lead to any enhancement in photovoltaic characteristics of solar cells?

It depends on the amount of the red shift and the type of device you are considering.

Sun light spectrum is peaked in the green range (peak at 530 nm), this means that a PV cell receives more power density in that range. So you should first try to absorb light in this peak range. Then, since there is  a meaningful portion of the sun light spectrum in the red and near infrared range, it is convenient trying to extend the absorption spectrum in those light ranges, too.

So, to answer to your new question, a red shift may enhance the performances if the absorption in the sun light peak range is kept, i.e. if the absorption spectrum of the cell and mainly its external quantum efficiency spectrum are extended towards the higher wavelenght region more than just being red shifted.

If you simply red shift the absorption of the active layer, let's say red shifting the peak from 530 nm (sun light peak) to 700 nm, this means that in order to keep the same energy production from your device (that is the actual performance you aimed to) you need a device with a higher photoconversion efficiency as the sun light in that range has a lower power density.

As Vladimir wrote above, tandem cells are a solutions to achieve this spectral extension, including active layers of materials with red shifted absorption peak with respect to the green absorbing underlayer. Other solutions are the use of dye blends in Dye Solar Cells, the synthesis of wide range absorbing active materials or the use of wavelength converters (down converters and up converters). Tandem cells in my opinion are the best feasible and effective solution up to now.

.

Assuming that only the absorption peak is shifted towards larger wavelength without any change in the absorption curve which is in form of band pass filter, then we have two answers:

If the peak is initially lying at a wavelength smaller than that of the incident solar radiation peak at the green wavelength of .55 um then the cell will absorb more solar radiation and consequently its photocurrent increases.

In the opposite of that if the peak is lying initially at higher wavelength of the solar radiation peak and it is shifted to higher wavelength, then the material will absorb a smaller fraction of the solar radiation and the photocurrent decreases.

I just want to make one additional point to add to what previous people have said.  So far, most of the comments have focused on how much current is generated under different conditions based on the photon flux at a particular wavelength and on the absorption of the material at that wavelength.  However, it is important to remember that a high current alone will not produce an efficient solar cell.  You also need a high open-circuit voltage.  

The open-circuit voltage is approximately proportional to the band gap (or "effective band gap" in donor-acceptor blends).  As a result, a red shift in the absorption peak (corresponding to a reduced bandgap), in addition to any changes in current output, will also cause a reduction of the open-circuit voltage.  To obtain the optimal power conversion efficiency, you have to find the point where the product of the current and the voltage is maximum.  As Stefano mentioned, this optimal point should occur near where the power density of the incoming solar radiation is the highest.  

Dear Abdelhalim Zekry ,

Can you give some references for your explanation as concern to my question