photovoltaic current

[Pooja-S]

Hi all, i need to design a software tool that can be used model the Photovoltaic cells. The software should be able to measure the amount of power generated by a PV device which considers:
The type and area of the material
The intensity of the incident light

This is purely software simulation. No hardware is being used.

I = IL - ID = IL - IO (e^(qV/kT)-1)

For now i just need to know how to draw the block diagram that can plot current I vs voltage

[Pooja-S]

Sorry ignore the 1st question....pls consider this one. I am new to labview. i need to design a software tool that can be used model the Photovoltaic cells.

This is purely software simulation. No hardware is being used.

I = IL - ID = IL - Is (e^(qV/kT)-1)

For now i just need to know how to draw the block diagram that can plot current I vs voltage.

I have drawn the vi for diode current [ Is (e^(qV/kT)-1) ]....see attachements. Now i need to draw the vi for total current [ IL - Is (e^(qV/kT)-1) ] which is equal to the current I_L generated by the photoelectric effect minus the diode current I_D.

PLS HELP!!

[Pooja-S]

Sorry ignore the 1st question....pls consider this one. I am new to labview. i need to design a software tool that can be used model the Photovoltaic cells.

This is purely software simulation. No hardware is being used.

I = IL - ID = IL - Is (e^(qV/kT)-1)

For now i just need to know how to draw the block diagram that can plot current I vs voltage.

I have drawn the vi for diode current [ Is (e^(qV/kT)-1) ]....see attachements. Now i need to draw the vi for total current [ IL - Is (e^(qV/kT)-1) ] which is equal to the current I_L generated by the photoelectric effect minus the diode current I_D.

PLS HELP!!

[Zach-H]

You want I graphed instead of ID correct?

If ID = Is(e^(qV/kT)-1) the you need to subtract ID from IL to get I. You can do that by adding a subtract block right before the ID value leaves the while loop. 

I've attached an example with IL set as a control you can enter on the front panel. Is that what you're looking for? You can replace the control with how ever you want IL to be generated.  

[Pooja-S]

Hi Zach-H, thank u so much. I am really grateful to u. Theres a problem, i use labview 2010, & u hv done this in labview 2011. Can u pls convert it for me to labview 10. Thanks a lot.

[Zach-H]

Sure Can. Sorry about that.

[Pooja-S]

Thanks Zach....i need to draw the block diagram so that the plot look like this, [see the image] . It should go the the negative axis also. There should be controls to control the light intensity ,
the area of cell. And also it should display the performance parameters such as fill factor. I really have no idea how to do this. 😞 The block diagram should be simple. 

[Zach-H]

I'm sorry I realize I didn't initialize that shift register to 0. If you right click the left side of the shift register and click create constant, it should make it so your IL line will show up properly every time you run it the VI with different IL values.

Is IL related to light intensity? It appears to be. Right now you can change the intensity by increasing the IL. You could change the IL control to light intensity then modify the value input to the control based on the relationship between light intensity and IL. Then run it into the same subtract that we currently use once you've converted it to IL. 

I'm not sure how you want to calculate the fill factor, but you should be able to use numeric blocks to compute it off of the wires we already have and output them to a numeric indicator, which will display on the front panel. 

Now that no light case, is that simply a base line that never changes? (achieved by setting IL to 0?) if so you could hard code a second set of xy data and add that to your graph to plot both functions. Do this by adding a second subtract function and attaching a 0 constant instead of the IL control. Bring down the same ID wire to the other input and output it to a new auto index out of the while loop like you're other two data points. If you look at the help it will tell you how to display multiple plots on the XY Graph. Just search "Displaying Multiple Plots on XY Graphs" in the help bar. There is an easy to follow example linked there too. 

You can change the scales of the graph by editing the graph properties. (right click on the graph) Go to scale, turn off auto scale and set the minimum and maximum values.

Try working these changes in and see if you get the functionality you want. 

[Pooja-S]

Hello, Zach, I have used 0.182 ....[see image] means if it exceeds 0.812, it will stop, i need to put different variables like x1 = 0.812, x2 = 0.850, x3 = 0.900
so that it will choose one between these three variables and it will stop the loop. I guess it will use case structure to do that. How can this be done?
Also, how can we calculate the power from the vi.
Power = V x I
 Can u do this for me?

Ill be grateful. Thanks lots

[Zach-H]

Based on what you've told me I have an updated example for you. You'll still need to play around with the graph settings to get the data to look better, and any light intensity formulas as well.   

[Pooja-S]

Hi Zach. Sorry, we shouldnt calculate the power from the block diagram, instead we should plot another graph: power versus voltage. The graph would be less confusing if it is plotted separately, i mean on the front panel i should have two xy graphs, one for current vs voltage & one for power vs voltage. power = V x I . How can we do this? The graph should look like this [see image] and the vi u sent me, i just deleted the power [see image], else everything works fine.