For a project, I have to model a solar cell just with a voltage source and a resistor (Thevenin equivalent model.)

I searched on the internet and all of the models I found are with a current source, diode and Rs & Rp.

The characteristics of my solar cell are:

  • Voc = 4.6V
  • Isc = 160mA
  • Vmp = 3.3V
  • Imp = 150mA
  • Pmp = 0.5W
  • Dimension : 36cm²

This is a first try at modeling what I meant:

Solar cell connected with a variable load

Pin and Pout

I went directly for a power adaptation with the same load at the input and output (Pout=Pin? or Pout=0.5*Pin ?) to maximise the power adaptation. Is my model correct for what I want to do?

  • \$\begingroup\$ You should draw the exponential characteristic of the cell, intersect it with the load line of your resistor and then find the value of the slope DeltaV/Delta I of the characteristic at such intersection. That is R. The value of the voltage is the intersection of the tangent line in that point to the V axis. But why do that? \$\endgroup\$ Mar 18 at 16:01
  • \$\begingroup\$ To be clear, the first resistor I alluded to is the load resistor, which you should know in order to linearize the characteristic of the cell. \$\endgroup\$ Mar 18 at 16:15
  • \$\begingroup\$ Yes ! exactly I think they want me to linearize the characteristic of the cell. And then to change the load resistor to have the mppt (the cycle duty is fixed) Can you see the picture at the bottom ? \$\endgroup\$
    – ktnl527
    Mar 18 at 16:24
  • \$\begingroup\$ Diagram So this is the curves i have, the load R is just Vmpp/Impp ? and then the linear model of my solar cell will be for this case a voltage source of 6.6V, with a internal resistor of 22ohms ? is that correct ? \$\endgroup\$
    – ktnl527
    Mar 18 at 16:28
  • \$\begingroup\$ Well, the value you chose for RL is Vmp/Imp, and then you assumed the characteristic intersected the load line orthogonally to find VTh. My guts say this could not always be the case, so VTh could be different depending on the cell. But le me do a few calculations... \$\endgroup\$ Mar 18 at 16:29

1 Answer 1


You cannot accurately model a solar cell with these two components.

So you must decide what is important, and your model must model that accurately.

Your own answer is (was) correct as far as it goes, and will accurately model the cell at a single point, the MPP point (and sufficiently close to it) but will be inaccurate at both extremes of open circuit and short circuit.

As you are modelling a simple curve with a linear approximation, you can intersect the curve at 2 points, providing 2 accurate answers. Three obvious choices are:

  • correct at open and short circuit points, which will woefully underestimate the cell's performance at MPP
  • correct at short circuit and MPP, which will woefully overestimate the open circuit voltage
  • correct at open circuit and MPP, which will woefully overestimate the short circuit current.

The last of these is probably most useful for most purposes; the Thevenin calculations should be easy. (However it should be used with caution for simulating MPP algorithms because the gradient is wrong; for that purpose, your original answer is better)

Open circuit and MPP are important conditions for e.g. solar inverter design, performance at MPP, and overvoltage protection, hence my choice of the third option.

However, for testing short circuit protection, I would switch to the first option, so I would calculate both of these. (The second option would give you false results on overvoltage protection circuits, I can't see a good reason to ever use it)

  • \$\begingroup\$ Thanks a lot for your response, I understand better now the thing. And just, do you know why on my first picture, the power at the input is 1W, I did the calculation to have 500mW at the input. And also is is normal to have 50% percent of the power transmitted right ? \$\endgroup\$
    – ktnl527
    Mar 18 at 17:05

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