I am interested in measuring the relative power output of a solar cell when it is placed behind glass and other materials. For example, one might guess that the cell's power output is reduced by 70% when placed behind frosted glass.

One thought I had to measure this was to use a very small resistor. Going by this diagram for the cell (Maxeon C60), we can see that we don't want voltage going over ~0.5 V, or the current drops. A small resistor ensures we stay in the constant-current region of the cell:

enter image description here

Assuming an absolute maximum current of 10 A, that means we need a 0.5 / 10 = 0.05 Ohm resistor.

Since $$P = \frac{V^2}{R}$$ we can see that e.g. if we measure half the original voltage, our power output is 25% of the original.

Does that make sense? I am wondering how accurate this will be. Are there any cases, like very low light, where this can break down? Thanks.


That will describe the power produced by the cell in your arrangement, but not the power the cell is capable of producing.

To do that most accurately would require MPPT tracking of some sort (which could be as simple as turning a wirewound pot while continuously logging V and I, and finding the peak in a spreadsheet or Python script).

But a simpler way would be to observe from the graph above, that independent of the illumination (within the ranges on the graph) the current remains relatively constant below 0.5V. So, measure current and multiply by 0.5 to get approximate power.

If you're interested in illumination ranges below the graph, you would have to establish this relationship still held, by separate measurements.


That is a reasonable way of doing it.

You could also just measure the current with an ammeter - they have a small resistor and measure the voltage across it, just like you are proposing.

The voltage across the meter is designed to be a maximum of a few hundred millivolts so it should keep it below the point where the current starts dropping off.

It will still work at low light levels but you can change the scale of the meter.

A couple of warnings:

Make sure you keep the leads between the cell and the meter short and of suitable wire thickness or the voltage drop couple exceed the 500mV limit.

This is one of the rare cases where you would connect a current measurement meter directly to a source rather than in series. Make sure of your connections, with a maximum of 600mV you are unlikely to damage anything though.

Why are you using such a large cell just for measurement? You could use a much smaller, cheaper cell.

You could also use other light sensing measurements but unless they match the spectral sensitivity of the cell they might not be accurate.


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