# Bench Testing a Solar Panel - math please?

Discussion in 'General Electronics Chat' started by allan.w.macdonald, Nov 3, 2010.

1. ### allan.w.macdonald Thread Starter New Member

Nov 3, 2010
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Hi,

This is my first post!

As a learning exercise, I am trying out a few small solar panels I have kicking around and, for convenience, I want to test them using an ordinary 60W light bulb and then extrapolate for sunlight conditions. I have tried to analyze the problem as follows but I am not sure I am doing this right. Can anyone check my work/assumptions? I am not all that good at math so bear with me.

Problem description: A 0.1m x 0.1m solar panel of unknown properties is placed 0.1m away from the filament of a 60W incandescent bulb and its output is measured. How can this output be used to determine its outputs under full sun?

Assumptions:

Assumption 1: 60W bulb throws away 90% of its energy as heat. Therefore, only 10% of the emitted energy is in the form of useable light so the actual usable light power output is roughly 6W.

Assumption 2: The remaining non-thermal energy given off by the bulb is roughly white light and, for the purposes of this exercise, is assumed to represent an approximation of a proportion of full sunlight. (I know this is a terrible assumption - just bear with this for the remains of the exercise).

Assumption 3. For the purposes of simplifying this exercise we assume the bulb is a point source isotropic radiator (not perfect but close enough for my purposes).

Calculations:

1. The bulb's intensity is: 6W÷(4*pi) = 0.477 W/sr

2. The panel's active area is 0.01 m².

3. The angle of illumination for the panel, in radians, is artcan(dist/panellength) = artcan(0.1m/0.1m) = 0.785 rad.

4. This works out to 0.785÷( 4*pi) = 0.0625sr.

5. The amount of light power incident on the panel is 0.0625 * 0.477 = 0.0298W

6. The panel's irradiance works out to 0.0298÷0.01m² = 2.98W/m²

7. One standard sun is 1000W/m². So this light source is roughly equivalent to 0.3% of one standard sun. (Seems awfully small!)

Cheers,
Allan

2. ### debe Well-Known Member

Sep 21, 2010
947
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Hi Allan 100mm x 100mm solar cell is pretty small, wouldnt expect much op from a globe. I put my solar panels in the sun & check no load voltage, then put varying loads on & measure the current it can supply @ rated op voltage. Daryl

3. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,795
951
Don't know about the math you used to get that figure but the scale of the answer seems a very good fit for reality. If you are sure of the math you used I would say that you hit the nail on the head. The assumptions you made previously with the light bulb, are another matter. There is DATA available on a wide variety of light sources. For a directional source that gives good light output data - See any LED datasheet. High output white LED's might be a better light source to use. The facts needed to figure output power levels are documented, and would be a better starting reference to figure from. Also many other types of lighting give light output data, that you don't get for a 60 watt, Edison style, incandescent

If you went with several different color temps of the same LED's even better because then you have response variation
vs kelvin color temp. (I think the sun comes in at 2700-2900K) I think

4. ### allan.w.macdonald Thread Starter New Member

Nov 3, 2010
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Thanks. However I am merely trying to learn what is happening... I know real sun is better but I am trying to understand the arithmetic. I have more powerful lamps available as well.

Nevertheless, with the setup I described, but with the bulb very close (the glass envelope of the bulb a mere 2cm away), the panel I tested gave me: ISC = 90mA, VOC = 10V.

5. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,795
951
So that output converted to watts and using a standard conversion factor of say 12% for the panel ---- how does the answer compare to the input you figured out?

6. ### DonQ Active Member

May 6, 2009
320
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Light to electrical converters (like solar panels) are wavelength dependent, so it will be difficult to make a direct comparison with an incandescent. Like comparing apples and bananas. (bananas are always better!)

Since power is what you really want to get from the panel, and P=IE, you might be tempted to just go P=IOC*VOC. But this isn't right. It is easy to see that ISC always gives you zero power (since E=0), and VOC also gives you zero power (I=0). And besides, you can never do both at the same time.

If you were to graph the current vs voltage output of the panel (an experiment I recommend), you would find that the I curve slopes slightly downward until the Voutput approaches VOC. Then it curves sharply and intersects I=0 at VOC. Different panels have vastly different curves. Even for the same panel, these curves also change with age, temperature, light source, etc.

To get the max power... If you imagine a rectangle within the curve, with opposite corners at the origin and on the curve, that place where the area of the rectangle is the greatest is the point of max power (this is just a visual version of I*E). It will occur somewhere in the vicinity of the sharp corner of the graph. Since this is unlikely to be the exact voltage you want, or the exact current you are needing, some form of electronic interface (involving inductors and switching transistors) can allow you to operate at that point on the curve regardless of your V or I requirements at the time.

These sort of experiments/calculations will get you much more out of the panel than all that 'isotropic' stuff.

7. ### allan.w.macdonald Thread Starter New Member

Nov 3, 2010
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Thanks to all who wrote.

I know the bulb spectrum is different from sunlight (apples, bananas) but what if you only have an apple - you still have to eat!

Anyway, I am just trying to get familiar with the panel and its behaviour and set up my apparatus well before I actually take it outside - the weather here in Nova Scotia is very unpredictable.

I haven't actually tried applying a load to find the MPP yet. That's the next thing I'll try (I have to get a few things together such as some sort of variable load). I'll keep working with the bulb as a light source for now until I have everything in order and then then I'll go outside (when the sun is out) and do the real measurements.

Also, if I can get my setup working predictably, and gain more confidence in the area, I might consider buying some panels online to try out.

I'm not powering a house here. My panel will be charging a battery of some sort to store the energy to be used by the load and I'm hoping to come up with a daily average average power of around 10-20 milliwatts or so. Therefore, I am guessing I might need a panel in the order of something like a 0.25W - the panel I'm testing now is really just a learning experiment.

Can anyone suggest any good simple MPP circuit/algorithm/tutorial resources? (Googling seems to cast a very wide net and sifting through all the google spam is very time consuming).

Cheers,

Allan

8. ### allan.w.macdonald Thread Starter New Member

Nov 3, 2010
5
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Late breaking developments:

I changed my apparatus a bit. For my lamp, I am now using a gooseneck desk lamp with a white parabolic reflector and am resting the reflector right on top of the panel. The reflector completely surrounds the panel edges and suspends the glass of the bulb by about 1.7cm away from the panel surface. This seems to provide much more light than a bare bulb. So, here are my findings (I've simplified my analysis somewhat):

1. Bulb is 60W.
2. Light power is 10% or 6W.
3. About 70% of that light in the lamp seems to be falling on the panel so panel is getting about 4.2W of light.
4. The panel is still 0.1m x 0.1m (area 0.01m²)
5. With this setup, I get about 420W/m².
6. Using a variable load, I found the MPP to be about 0.59W.
7. The panel efficiency is 14.11%.

Not bad, I guess. Of course, this efficiency figure only shows how well it converts incandescent light into power, not sunlight. At least I now have a repeatable experiment and my numbers seem to be reasonable. Now I should try this panel out in the sun.

9. ### timrobbins Active Member

Aug 29, 2009
318
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First things first - you need to ensure the light intensity is uniform across the total area of the solar panel/cell. If the intensity is not uniform then all your subsequent equations become very suspect due to non-linear conversion within the panel. It's a bit like the problem of 'shading' parts of a panel.

Practically, this means moving the source much further away, and hence having a lower intensity at the panel/cell.

You also don't mention temperature - a very significant parameter when doing comparisons, especially with those that are made in the sun (if the panel is allowed time to heat up).

Ciao, Tim

10. ### allan.w.macdonald Thread Starter New Member

Nov 3, 2010
5
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Thanks Tim.

Yes, you are right - those things must be considered for further analysis but for the time being (as the storm rages outside):

1. In my setup, the frosted light bulb inside the white-painted, roughly-parabolic lamp reflector seemed, superficially, at least, to give fairly consistent light across this small panel. The rim of the reflector is roughly 165mm in diameter while my panel itself is 100mm x 100mm. Some parts of the panel were undoubtedly brighter than others but I am only approximating - I am not trying to be precise at this time.

Nevertheless, if I want to make absolutely sure the illumination is consistent by moving away the light source, I'll need to come up with a much brighter source for testing indoors. I don't have that at the moment. I don't have any sun, either.

2. I will factor in temperature later. Although, I did notice that the bulb was effective in warming the panel up quite nicely, over time, which seemed to have a negative effect on the power output, which is to be expected (I didn't quantify the effect, I just observed it). For more detailed measurements, I have a thermocouple wire temperature sensor which connects to my DMM that I can tape to the surfce of the panel so I will use that to determine any temperature effects.

Nevertheless, my tests do appear to suggest that I am approaching the analysis of the panel well enough to go on to more rigorous experiments. The solar cells used in these panels, which were made somewhere between 12 and 15 years ago, probably yielded in the vicinity of 15% efficiency anyway so the resulting number I got is close enough in the ballpark.

Thanks for all your help, everyone.

Cheers,

Allan

11. ### timrobbins Active Member

Aug 29, 2009
318
16
Ummm, not quite sure what you are trying to achieve. Part a) seems to be just a light intensity meter - something you can probably pick up off eBay for nuts. Part b) seems to be to check/confirm the IV-PP-FF parameters of the panels/cells. I think you are making 'a rod for your back' by merging the two.