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Measuring Solar Radiation based on Photovoltaic Cells

Hi,

The optimum resistance depends on the rated Current and Voltage from the PV Panel. A higher voltage and corresponding lower current (which would occur with a panel of comparable total area) both require a higher resistance value.

The PV panels are made from phototiodes connected in series, but the graph in post #2 shows just one diode. The curved lines show the I/V characteristic at various light levels. The diagonal "load lines" correspond to different resistance values and the requirement is that the I/V lines should cross them at proportional distances from the origin (0). That occurs with a vertical load line (short-circuit), but the corresponding voltages are all zero, so that's not a useful measurement.

The top I/V curve begins to fall at about 300 mV (R = 0.3 volts / 1.3 amps = 0.25 ohm), so a load with a resistance less than that is essential. Note how "bad" is the lineararity (proportionality) where the curves cross the horizontal axis (open circuit load).

A "5 volt" panel, for example, may have about 12 "diodes" in series and for the same current as in the graph would be rated at perhaps 6 watts. For such a panel, a resistance of around 12 * 0.25 (=3) ohms might be acceptable, but a lower value would be preferred.

Cheers, Alan.
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Hi,
I have built a solar sensor (not calibrated very well because i'm waiting for a clear summer day) and i can see some (low) spikes during night! (see attached trend)
I'm using a LM358 instead of the AD822 for the moment.
Does someone have an explanation from what those spikes could come from ?
Thanks
Laurent
   
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(22-01-2016, 14:11)laulau Wrote:  I'm using a LM358 instead of the AD822 for the moment.
Does someone have an explanation from what those spikes could

I never tested the circuit with an LM358, so I don't know if it makes any difference, but most likely the spikes you describe are due to some noise induced on the cable, between the solar / UV interface and the input of the ADC in the TX unit.
(Cable between de solar cell and the OP should always be very short, 20cm máx).

To eliminate the possibility of low spikes, the software on the RX rejects any reading below 4 Wm2, but if you are getting spikes above this threshold, better than raise this threshold is check the effect of the cable length in this behaviour.

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Cable between solar cell and the OP is very short, about 15cm.
The cable between the solar / UV interface and the TX unit is about 2.5m for the moment.
I just put a "torus" on the cable to see if it helps to reduce noise. If not i will reduce the cable to 1.8m but i think this is the shorter i can have for the final settlement.
I also order a AD822 OP to see if it makes difference.
I will share my investigations.
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Laurent, this may be a crazy idea, but is it possible the night-time spikes are caused by some kind of bright light hitting the sensor??

Allan.
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I covered the sensor for a while last night, but still had spikes.
The "torus" on the cable didn't help.
I'll shorten the cable to see if it helps.
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Hi,

Do the spikes definitely occur only at night? It looks like they might be there also during the day.

Personally, I'd be looking for a source of radio interference (any frequency), perhaps a nearby mobile phone, or maybe "spikes" on the (mains) power supply.

Cheers, Alan.
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Hi,
I replaced the LM358 by the AD822 and the little 'night' spikes are gone.

Laurent
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Hi, Laurent

The AD822AN doesn't play in the same league than the LM358, something should explain the price that is almost ten times more. The AD822AN it's a high quality OP, according to data sheet mostly used for precision instrumentation.
Thank you for reporting.

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Question 

(07-09-2014, 11:38)uncle_bob Wrote:  Step 4 - Determining the value of Rsh

As in the example presented in the study referred to in the first topic, I will also choose to calculate the value of Rsh, in order to obtain a voltage drop in Rsh with the ratio of 1 mV for each 10W / m2. So for a Solar Radiation 1300W / m2 intend to get a voltage drop of 130mV Rsh. Using Ohm's law as R = V / I, we Rsh = 0,130V / 0,075A ie Rsh = 1.7333333 Ohm.


Hi for all

Please I want illustration about the ratio ( 1 mV each 10 W/m2 ) how to use or supposed these ratio ???

thank a lot for brilliant job.

I get it ....

when solar radiation is 1300 w/m2 , Isc was 75 mA
so voltage drop is : Isc = 75mA * Rsh = 1.74 ohm = 130 mV
therefore
1300 w/m2 --------> 130 mV drop
10 w/m2 --------> X mV drop
so X= 1 mV
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