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Intensity and measurement

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Jo

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Jul 18, 2001, 7:41:17 AM7/18/01
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Hello,
I would like to verify some results of my own experiments.

When measuring total irradiation(W/m^2) on a Si reference cell, these
results have to be corrected because of temperature-dependency. Can one
consider this amount as significant? Example:
12:45h; june 21 :
1224 W/m^2, (isn't that really high??)
and 1214 W/m^2 after correction
21,1°C ambient temperature, estimated 57,8°C cell temperature

Does any of this makes sense to someone? I mean, an amount of 32,8°C above
STC would only decrease the available intensity with a lousy 10W/m^2?

As I understand, available power should decrease with ca. 0,5% per °C,
i.c. 16,4%.
Wrong?

--
-----The above are my opinions only. I am not consciously
trying to attack anyone's views. I am only trying to
add my own.----- Jo


Bill Kaszeta / Photovoltaic Resources

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Jul 19, 2001, 12:02:43 AM7/19/01
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The proper way of using a silicon solar cell to measure irradiance is to
operate the cell at short circuit (zero volts).
The main reason for doing so is that the temperature coefficient of short
circuit current is very low, about 0.05% per deg C. It is a positive
coefficient, current increases with temperature. If you purchase a calibrated
silicon pyranometer, the calibration will be in terms of current at close
to zero voltage (or of voltage across a resistor that will keep voltage
below 50 mv, essentially short circuit current).

The opposite is true of power outout and voltage, they both decrease
with increasing temperature. The voltage at the maximum power point
decreases by about 0.4% per deg C, and power decreases at about
0.5% per deg C. Most PV manufacturers publish these coefficients and
they vary with the technology used to produce the solar cell.

For this (and other) reasons, the irradiance does not directly relate to
photovoltaic (PV) array outputs. The PV output can be calculated from
irradiance, temperature, and operating voltage if the coefficients are
used correctly.

How did you measure the 1224 W/m^2?

Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
bi...@kaszeta.org

kammar...@gmail.com

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May 2, 2013, 2:05:56 PM5/2/13
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Hello, I have some questions regarding pyranometer calibration!

After doing a linear regression analysis in excel plotting the pyranometer response (y-axis) vs the reference cell (x-axis) and inserting a "line of best-fit" we get the R-squared value and the line-equation. If the R-squared is 0.95 it indicated that the correlation is good, but how do I calculate and allow for the offset in future experimental values obtained by the pyranometer? Do I simply substitute the x in the equation by the experimental value or can I calculate some correction factor to multiply by?

Please help, I'm really confused about this!

Cheers

Lu Wei

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May 5, 2013, 10:09:09 PM5/5/13
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If you are calibrating the pyranometer using reference cell, what you
should do is to substitute y and solve x in the equation.

--
Regards,
Lu Wei
PGP key ID: 0x92CCE1EA
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