# How to get the maximum power point from I-V curve for Organic Photovoltaics (OPV)?

I am taking organic solar cell course on Coursera, and I have a question about I-V curve for a certain solar cell. The question asks for the solar cell characteristics such $V_{oc}$, $P_{max}$ and $FF$. I get the $V_{oc}$ correctly but I couldn't calculate the other two correctly as they provide me with an excel sheet of the current and voltage.

I plotted the voltage on the x-axis and current,power on y-axis below. I need to get the correct $I_{max}$ and $V_{max}$ from the graph with (I-V) in blue and (P-V) in orange.

Here is the question from Coursera:

Download this dataset DataSheet (, decimal mark: dataset) and calculate the maximum power point (Pmax). Give answer in either W or mW, a valid form of answer would be '94' or '0.094'.

Hint: You may need to do an interpolation to get the correct answer, if the value you are looking for does not exist.

• Can you post a link to the original question? I suspect that something is wrong with your data because the curves look upside down compared to typical ones, and I don't know what it means to have negative power. – Chris Mueller Sep 2 '15 at 12:23

As you have the power-voltage curve, finding the maximum power point is easy: it's the turning point on the power-voltage curve.

The problem is that the absolute values in your graph look completely wrong. If the graph is flipped upside down, and based at zero power, zero current and zero voltage, then the blue curve does indeed look like an I-V curve for a photovoltaic cell, and the orange curve looks like the corresponding power curve.

So it looks like the shape of your curve is right (albeit upside down from the usual presentation), but the absolute values are wrong.

You need to work out what the right absolute numbers are.

You know that the lowest the actual voltage is going to be is zero, and the lowest observed value is currently marked as -10V. The curve is the right shape going left to right (it's upside down, but does not also need reflecting in a vertical axis), so the correction coefficient for voltage is going to be positive. You know that the lowest the actual current and the actual power can be is also zero, and their lowest observed values are marked as 0.1A and about 6.9W respectively - I say lowest, because we need to flip the curve upside down. And because we do, you know the coefficient to correct them will be negative. And you know that $P_{real}=I_{real}V_{real}$.

Given the shape of the curves, it's safe to assume that current and voltage each need a linear transformation (i.e. $V_{real} = aV_{obs}+b$, where you need to find $a$ and $b$)

This gives you a set of simultaneous equations you can solve, to transform observed voltage (current/power) [$V_{obs}, I_{obs}, P_{obs}$] into actual voltage (current/power) [$V_{real}, I_{real}, P_{real}$].

Write out those simultaneous equations, and solve them.

That's the recipe on how to solve it. You'll need to follow the recipe yourself, though: given that this is coursework, there's no use in me doing it for you. Be sure to go through the recipe yourself, and make sure you understand why each step exists - try to work out how I derived it.

• What do you recommend to do as I need to find it correctly ? – Mahmoud Zidan Sep 11 '15 at 7:28
• @MahmoudZidan I've elaborated the answer to give you the recipe on how to solve it. You'll need to follow the recipe yourself, though: given that this is coursework, there's no use in me doing it for you. Be sure to go through the recipe yourself, and make sure you understand why each step exists - try to work out how I derived it. – 410 gone Sep 11 '15 at 8:13
• I tried using the minimum value in the Power-Voltage curve and I got the answer correctly. The problem with me was I take a value greater than the minimum. The question is designed to be as it is. I will try the above also. Thanks for your help. – Mahmoud Zidan Sep 11 '15 at 8:26
• @EnergyNumbers I am following the same course and have also been given a set of data featuring negative values which I can't conceive. The course itself features seemingly reversed curves. The linked kind of IV curves with positive only values makes more sense (stephenstuff.wordpress.com/2013/04/20/…). – BzzBzz Jan 5 '17 at 14:06