How was this graph taken? The classical way with an 'external' voltage (or current) source?

Looking at the difference between "dark" and "light", the builtup seems to develop its own voltage when illuminated. Regarding the 'size' of voltage, I'd suspect a thermal offset voltage, not a PV one. Then the resulting voltage would be zero at zero current - effectively shifting your red curve down until it crosses at (0,0).

Does this help?

Since the Iphoto drops below the dark current right after crossing, it shows the generation of photocurrent. Below this voltage, it seems that the IV curve is showing normal characteristics of a solar cell. However, above this voltage, Iphoto is shown in 2nd quadrant. This can be caused due to many reasons:

1- Measurement fault: The first thing you want to check is to make sure you have acquired the data correctly. I recommend you to measure the I-V characteristic of this cell two, three times more (with a time difference between each measurement), and then see if this behavior still exists. Comparing these results with your previous results unveils many things. 

2- Effect of temperature: increasing the temperature increases the energy of electrons be freed from Valence to Conduction band and move along the depletion zone. Again, coming back to point 1 of making sure the measurement is done in the right environment.

3- Effect of dark current: Since the current difference between Iphoto and Idark is in the range of nano-amperes, I would also consider the effect of dark current, which exists in a solar cell. It seems that this solar cell is not capable of producing much photocurrent, and only current circulating in the circuit is the dark current which has a opposite direction to photogenerated current. This current can be caused due to poor contacts, band alignment and etc. 

I recommend you to check this two possible causes in sequence (first repeating measurement to see if anything changes, and if nothing changes, checking the effect of temperature and dark current). 

Good Luck!

@Dreher : The I-V data was taken at room temperature(20 C) using Keithley Current source meter from -1 to +1 volt. I would like to understand why there is no photovoltaic effect in the discussed device structure?

@Ghods : When I-V data was taken with same structure but over FTO substrate instead of AZO, the data gives photovoltaic nature.Please suggest the possible causes

Though it is not impossible, about 0.01 V (if you got the x-axis correct) is very little for PV effects while reasonable for thermoelectricity. THUS the assumption. . . that is not necessarily correct. Just reasoning :)

Can somebody helps me to get light current graph from dark current. I have simulated a simple pin solar cell in comsol and obtained its generation,recombination and current density graphs. i am confused with dark current plot and trying to figure out what equation or method i should integrate more to get short circuit current and open circuit voltage graph. 

 @Dreher: Both the graphs with AZO and FTO as substrate is attached in this reply.I know 0.01 V is so low but the I-V curve should never go through the second quadrant as for PV cells.

Oh. The AZO graph is about the graph of the whole assembly...

AZO as in AlZnO ?? A short check with wikipedia gave me the information that this 

• is semi-conducting (II-VI)
• has luminescence. When considering luminescence as some 'inverse' of PV - could this explain the pos. offset?

I honestly don't know why this does not happen in the dark. Eventually some kind of 'activation energy' is required.

@Dreher: Yes AZO is Aluminium doped ZnO acts like a transperant conducting oxide alike FTO,ITO etc.

Ok, I will check your suggestions.Thank you

@Maria: I can't understand your requirement.sorry

You may find the following discussion very interesting and useful regarding the increase in dark current under illumination. It may answer some of your questions.

https://www.researchgate.net/post/why_is_solar_cell_illuminated_current_above_VoC_finally_matches_the_dark_current_What_does_it_indicates_if_it_doesnt_match

 @Maria Jabeen:

As a simple answer: recombination current determines the reverse saturation current (denoted in equations with "I0"). But if you are specifically talking about "dark current", you may want to have a I-V curve which extends to negative voltages and measure the current at a voltage like -0.5V 

Reverse saturation current has a formula that you can easily find it by searching on internet, and therefore you can plot it :)

1.) How large is the active area of your your devices?

2.) What is the unit of measurement of the current in the FTO graph?

3.) How does the AZO graph change when varying the illumination intensity? How does the AZO graph change when varying the device temperature?

@Wagner : My device area is 0.04 cm2.My device is annealed so I don't think that temperature variation will may cause it. Thank you for your co operation

@ Ghods : Thank you for this link.It will definately help me.

Dear Ayon Das Mahapatra,

is it, possible, to

1) include, please, the measurement (exper.) points in your (line) graph ?

2) say, also, the (total) time of the measurement (in seconds) ?

What is the unit of measurement of the current in the FTO graph? Is the FTO sample area also just 0.04 cm2?

How does the AZO graph change when varying the illumination intensity (at constant sample temperature)?

It depends on the kind of characterization you're doing and materials you're working on? 

Dear Ayon,

I came very late to your question. But as you question is very fundamental and other colleagues my benefit for the answer i decided to give an answer.

The simple explanation of your curves is that you exchanged your anode and cathode and consequently the current direction so that your third quadrant is the first quadrant and your first quadrant is the third quadrant and your second quadrant is the fourth quadrant.

Verification:

Exchange the electrodes designation and measure the current in the direction out of the anode.

One important note that your device works as a solar cell with low shunting resistance causing the triangular shape of the I-V curve in the fourth quadrant which is here the second quadrant.

Best wishes

The standard characterization technique of a PV device consists in the determination of the DC[1] Current-Voltage curves[2]. Additional information can be obtained using time domain and frequency domain measurements. Electrochemical impedance spectroscopy (EIS)[3] offers the possibility to investigate the behavior of the device in the frequency domain under some (Vdc polarization and other) operating conditions[2,3,4].

So, dear Ayon Das Mahapatra, try, if possible, to measure some EIS[3] (10mHz < f < 1MHz), using values of the Vdc polarization parameter in the Vdc-range, at least, of your special interest: -0.02V to 0.00V.

1. Photovoltaics Autolab:Application Note PV01 http://www.metrohm-autolab.com/download/Applicationnotes/Autolab_Application_Note_PV01.pdf

2. Under different incident light intensities.

3. Part 2 –... impedance measurements http://www.metrohm-autolab.com/download/Applicationnotes/Autolab_Application_Note_PV02.pdf

4. Autolab Application Note PV03 Photovoltaics Part3 – Dye sensitized solar cells, IMVS and IMPS measurements http://www.metrohm-autolab.com/download/Applicationnotes/Autolab_Application_Note_PV02.pdf

Thank you sir for @Samaras & @Zekry valuable informations, it will definately help me