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Measuring Solar Radiation based on Photovoltaic Cells
(27-08-2017, 00:03)jgveill Wrote: Somewhere in the thread, someone (alan I think) came with the idea to use a panel with more cells. 

Hi,

Welcome to the forum.  No it wasn't me, it was Werk who has proposed using the multiple diodes of a single PV panel right from the start.  I was originally working with a single BPW34 (which can be made to give excellent results even interfacing directly with a microcontroller) but I'm now a "convert" that the PV panel is a good solution for this particular application.   

If you look at the cell (on the right hand side of the picture) that Werk supplies with his kit of parts, it has at least 3 and probably 6 separate sections (diodes) in series.  Although the Op-Amp does have some voltage gain, I don't believe that is its main purpose. For example, a method to calibrate (change the voltage gain) of the interface is needed and a "1 Ohm Potentiometer" is NOT the way to do it !

IMHO, the advantages of the PV panel are much more "practical" than theoretical.  One important requirement of a Solar sensor is that it must be absolutely horizontal, which is much easier to do with a small PV panel than a packaged photo-diode.  Also, the sensitivity range (or conversion efficiency)  of a photo-diode may be quite large (+/- many tens of percent) whilst a PV panel will be "expected" to be optimised for a specified/maximum output from the solar radiation.

Bear in mind that the purpose of the PV panel is to convert solar energy ("heat") into electrical energy, but from then on, "heat is your enemy", because nearly all, electrical components "dislike" getting hot (becoming inaccurate and/or unreliable).  So large currents (i.e. large panels)  are to be avoided, particularly noting that if the PV panel is "Short-Circuited" then the electrical energy, which would normally have been fed out to do useful work, remains in the panel and heats it up. 

PV panels (and silicon photo-diodes in general) have a very significant temperature coefficient.  Davis (whose equipment is often considered as "reference quality") quote the temperature coefficient of their (silicon photo-diode) pyrometer, but I'm quite surprised that they don't appear to make any provision to actually measure that temperature.

Cheers,  Alan.
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Hi jgveill,

When I read your first post, I did a interpretation mistake, which I only realized when I read your second post ("10 series cells giving an open circuit of 5 volts").
I understood that you are talking about grouping solar panels (which have many individual solar cells), instead of adding individual solar cells.

As I said before, the main problem I see, in trying to get a higher short circuit current, is the power dissipation on the shunt resistor and on the diodes itself. Ok, 0,23A on a 3 Ohm resistor causes a dissipated power of around 0.16 Watt, not a big deal for a 1/4 Watt resistor.

However the use of and OP, as referred by AllyCat in previous post, has some advantages, mainly for calibration, as its easier to deal with a potenciometer to change the OP gain, than making changes in the shunt resistor.
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(27-08-2017, 11:01)AllyCat Wrote: If you look at the cell (on the right hand side of the picture) that Werk supplies with his kit of parts, it has at least 3 and probably 6 separate sections (diodes) in series. 

You are right. The solar panel I'm using is composed by 8 individual solar cells, arranged in a parallel of two series of four cells, thus its a 2V solar panel. Short circuit current is between 110 and 120mA (not all panels are equal, one more reason to have a way to allow some calibration).
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Hello, 

For those of you less familiar with electricity and maths, I bring your attention to some specific sections of Chuck Wright document and give my understanding.  Then, it will be easier to have a commun understanding on what's good and what's not.  

1- Basic Electricity : " A Current Source is an element that delivers a constant current, regardless of the voltage across it." 

The voltage across the PV comes from the shunt resistance we add. As we expect constant current, voltage will be higher if we use a large shunt resistance and lower with smaller resistance.

2 - Appendix A, pv cells : "Over part of its operating region, PV cells act like a current source." 

That's where we want to operate : when PV is acting like a current source which gives constant current. Constant current gives reading proportional to sun irradiance, required for a pyranometer.   

3- "Note that when voltage is very low, the cell acts like a current source. That is, it delivers a rather constant current, independent of the voltage." 

To get proportional reading, we have to have the output voltage low. How low ? 

Appendice A, Pitfalls : "The method described above limits PV output voltage to about 0.4 volts, which keeps the measurement reasonably linear with radiation variation."

As an example, for the blue line (1 sun), current is really constant from 0 to 0.285v.  So I would not go up to 0,4 volts as indicated in the document to limit heat. I would stay close to 0,1 - 0,15 volts / cell.  So linearity does not require to be exactly on 0,1 v per cell at full sun. No need to pay for 1 % resistance.  All we need is to keep the shunt resistance to have a voltage lower than 0,285 volts (at full sun) and we are sure current will be proportional to irradiance, But target around 0,1 volt per cell to limit heat.  Later I will give you an easy way to avoid the pitfalls in measuring short current.   

Next, I'll look at one cell panel vs solar panel with multiple cells.  

J Guy 

I just hope image can be seen

   
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Hello, 

Now let me explain why using a solar panel with multuple cells has advantages over a single cell. 

Fisrt, as soon as your panel is rated for more than 0,5 volts when you buy it, you buy multiple cells connected in series. Voltage specified divided by 0,5 gives you the number of cells in parrallel panel has. 

I rapidly did some drawings to explain more easily how it doesn't change the theory and why we benefit from multiple cells. Theory is : we MUST have cell operating to low voltage (0,1v) to have constant current and then current proportional to sun irradiance. 

Fig. 1 shows 3 separate copies of a unique cell. I used the values from C. Wright. Each circuit has a short current 1.6 Amp, a 0,0625 resistance and the cell voltage is 0,1 volt. Voltage is low and require an opamp to higher the output level to have a good precision. These values being for 1 sun irradiance. 
   

Now join the 3 circuits together like in fig 2. As you can see in the leg DE - CF and in the leg HI - JG we have 2 times 1,6 Amp in each leg, but one being the reverse of the other. So we have + 1,6 A and -1,6 A. In fact we can remove this leg as no current will go through it. They cancell each other. Same thing for HI and JG. Each cell has a voltage of 0,1 volt and current going through each cell is still 1,6 amp. 
   
Now look at fig 3. We have 3 cells in parrallel, each one providing 0,1 volt and 1.6 A current going through. Now if you look voltage between B - C we have 0,3 volts. Like me, if you have a 5 volt panel, it has 10 cells in parraller (10 X 0,5 volts). To have each cell at 0,1 volts ( to keep the cell operating in constant current mode), we must have 10 X 0,1 volts = 1 volt.

As each cell in the panel should be really similar, we can expect good results with multiple cells. Having 1 volt swing voltage we do not need the opamp any more as it's job was to increase the 100 mv voltage.

J Guy


Attached File(s) Image(s)
   
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last one ... 

Knowing this, you should feel comfortable using multiple cells panel for better precision and less components. Also, we do not really need to calculate the short circuit which requires an expensive multimeter having low resistance in current measurement. We can just use a voltmeter and measure the voltage across a trial value as shunt resistor and adjust it to be close or right on the target : 0,1 v / cell.   

Start with higher value (like 10 ohms for a 5 volts panel) and go down till you get 1 volt across the shunt (for 5 volts panel or 0,1 volt/cell your panel has) at full sun. Not possible to have full sun at this time, like for me in Québec. How will you know how much sun you have while testing ? Apogee instruments has a Clear sky calculator http://clearskycalculator.com/pyranometer.htm  that gives you the theorical value of irradiance according to you location, temp, humidity, etc ... It's used by pyranometer's owner to validate their pyranometer calibration to see if it requires calibration. So that's perfect to calibrate ours. We are going to pick the shunt and do our calibration in 2 steps. 

Calibration : 
Lets suppose your yearly maximum sun should be 1300 w/m2.  Suppose Clear Sky calculator gives you 720 W/m2 when you do your testing. Current being proportional to irradiance, at this sun level, target voltage should be :  0,1 volt X ()720 /1300) = 0,0554 volts per cell.  If you have a 10 cells panel like me, target would be 10 times X 0,0554= 0,554 volt. You probably won't be have a shunt giving exactly that voltage. Don't go too much over the target voltage, a bit lower is safer, closer gives more precision but generates more heat. Solder your shunt and do another test with clear sky value. 

Clear sky irradiance : ex. 1050 w/m2
Your maximum sun irradiange : 1300 W/m2
Full sun target : 1 volt
The measured voltage with you shunt in place : 0,791 volt

How many w/m2 is you pyranometer ?  

791 mv -> 1050 w/m2
1 mv = 1050/791 = 1,327 W / m2

Suppose now, you want to calculate irradiance from a voltage : 
791 mv X 1,327 W/m2/mv = 1050 W / m2 ( as expected it was ou calibration value.) 

As basic Arduino gives 4.88mv / step. Our precision would be : 4.88 X 1,327 = 6,48 w / m2 if we take the 5 volts as reference. 

To have better precision, someone could buy a 16 bits I2C ADC board with 4 inputs to be used elsewher in your Weather station or use 2 shunt resistances to split the total shunt voltage and be able to use the 1,1 v reference.  An easier way is to use this 1,1 v reference, is to ensure you are below 1 volt at full scale and even higher.   As in the example, you could consider using 1500 W / m2 even if your maximum is 1300. This way your shunt voltage should never be over the reference.  

I hope these long posts could help some people. 

J guy
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Hi,

Chuck is quite "honest" that his "pyrometer" is a "fun" project, with his conclusion that "In spite of the limitations, the simple device shown here is useful as a learning tool.".  The current from his PV panel is ridiculously large, necessitating the use of copper wire as the "load resistor"; a really bad idea because of the large temperature coefficient of copper*.

To make a "scientific" measurement you don't normally use a "load" resistance at all, but measure directly in the current domain; typically using an Op-Amp configured with a "Virtual Earth" input that gives zero voltage drop (<~1mV) across the panel/diode.  Alternatively, configure the amplifier as a Current-to-Frequency converter (aka an oscillator), which can also perform the ADC function with high accuracy and dynamic range.  By remaining in the current-domain, it's possible to use a single photodiode and a $1 microcontroller to achieve a dynamic range from peak sunlight down to moonlight, or almost 1,000,000 : 1 (20 bit resolution), but that's not the purpose of this thread.

As said above, the main aim of this project / thread is to make practical and reliable long-term outdoor measurements. For these, there are much more important practical design issues, such as the Enclosure / Window, with its impact on the Cosine response of the sensor, and the reliability (and low resistance) of the electrical connections, etc..

*EDIT: For reference: Davis quote the Temperature Coefficient of their 6450 Solar Radiation Sensor as +0.12% per degree C, and the TempCo. of copper is +0.4% per degree C (a load resistor is also likely to get hotter than a solar panel, of course). The TempCo. of a low-resistance Carbon Film resistor is around 0.02% (200 ppm) and Metal Film around 0.001% (10 ppm) per degree C.

Cheers,  Alan.
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Again, just for fun, I found a white paper studying thermopile pyranometer uncertainties compared to PV uncertainties.  Conclusion may not go the way most people would think : here's an extract from conclusion : 

" The PVRD achieves an expanded uncertainty that is a factor of ~2 lower than the thermopile pyranometer. During the majority of the day uncertainties are on the order of +/-5% for a pyranometer, and +/-2.4% for a PVRD, both stated with 95% confidence intervals. We therefore conclude that 6 PVRDs provide superior irradiance measurements for PV power plant monitoring applications. 

The white paper is not for those allergic to maths, but you could learn few things having an eye on it. 

 http://imtsolar.com/wp-content/uploads/2...-Paper.pdf 

J Guy
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Hi,

But isn't the title rather a clue:  "Comparison of Pyranometers vs. PV Reference Cells for Evaluation of PV Array Performance".

Basically, they are saying that if you want to predict the output from an array of PV panels, then it's best to measure the output from a PV panel.     Big Grin

Cheers,  Alan.
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Hi,

I was looking for a waterproof box and saw garden led light.  This model seems to have a better solar panel (crystalline silicon), more efficient compared to others, it's small, waterproof and really not expensive (less than 4 $ can) as there's a special for 4 days. 

https://www.aliexpress.com/item/LED-sola...st=ae803_3

J Guy
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