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Posted 04 April 2006 - 05:11 AM
how do I convert the AtoD Voltage of the for the Solar Sensor that is delivered by the Humidity / Temperature / Solar Board to irradiance, heat flux density?
For humidity, I found a fromula for calculating the true relative humidity in an article from the Sensors Journal. Is there anything similar for the CLD 140 photodiode?
Posted 04 April 2006 - 01:05 PM
Currently there is no defined way to translate the output of the CLD140 into any standard reading like w/m2. We are going to be doing some testing to see if we can come up with a formula but none exists at this time.
Posted 05 April 2006 - 05:51 AM
thank you for you fast reply. Do you have any schedule for your testing?
Posted 06 April 2006 - 12:51 AM
We don't have a schedule yet. It will have to fit in around the edges.
Posted 21 December 2006 - 04:00 PM
Have you considered replacing the CLD140 with the CLD240E?
I have a solar radiatin meter that reads in W/sqft. Have played with other photodiods and phototransistor and have found none of them are linear with my meter. Also atmospheric conditions will greatly affect the result.
I am putting together a one wire data colection system to calulate the btu and efficiency of hot air solar collectors. When I get it working I will post a conversion for the CLD140 on the solar radiation detector current reading vs the reading on my meter. This will be a reading on as clear a day as I can get. The conversion will not be linear but should be close in full sun.
Posted 21 December 2006 - 04:41 PM
Posted 21 December 2006 - 10:25 PM
This is a formula for getting Watts/square meter, from the current reading obtained on the Solar detector board.
Assuming that the current reading from the DS2438 on the solar detector is in Amps:
Solar detector Amps/13ua/sqcm = milliwatts/square centimeter
The effective area of the CLD240E is .0036 square inches.
converted to square cm:
1 sqare cm = .155 square inches approx.
.0036sqinch/.155 square inches per sqcm = .02323 sqcm effective area
since we only have .02323 sqcm:
mW output per 13 uA = milliwats/sqcm * .02323 sqcm
Converting to Watts/square Meter
W/sqM=(Detector Amps / .000013)*10,000 sqcm/Meter*1W/1000mW*.02323sqcm
This works out to :
W/sqM = SolarDetector Amps * 178.66 (a bit of rounding off here and there)
Multiplying the W/sqM times the area of a solar collector gives you the total watts of sun falling on it. This can be converted to btu's. By calculating the flow thru the collector and the delta of the temperature one can get the actual btu output of the collector. Then using the btu's of sun that falls on the collector one can judge the efficiency of the design.
Would you verify my math?
Posted 21 December 2006 - 11:42 PM
From reading the specs on the CLD240, it should be a better sensor for solar flux than some others. It is sensitive to a broader range of light than some of the infrared sensitive detectors I have looked at. It also has a wider angle of detection. Some sensors have a very narrow angle and are better for recieving data in that incedent light does not affect them. However the light on a collector or area of land does not all come directly from the sun. Some is reflected light from other objects and enters at other angles. The underlying reason we can see other objects.
Posted 22 December 2006 - 12:23 AM
The formula they give is :
Current = current register base 10 / (4096 * sense resitor ohms)
the current is proportional to the watts/meter squared, so the current is what we want not the voltage drop.
getting a bit complex for me.
any other suggestions?
Posted 22 December 2006 - 11:08 AM
First since the spec sheet says that a solar flux of 1 mW/cm^2 will cause a current of 13uA to flow, one can divide 13uA into the current from the sensor and know the solar flux in mW/cm^2. Simply multiplying by 10 will give W/M^2.
In my experiment:
I = E/R so 4 Volts/ 4700 ohms = .00085 Amps
.00085 Amps/ .000013Amps = 65.5 mW/cm^2
65.5 mW/cm^2 * 1W/1000mW * 10000 cm^2/M^2 = 655 W/M^2
the 655 W/M^2 sounds reasonable.
so a formula for getting W/M^2 from the current register on a DS2438 is:
assuming the sense resistor is the 390 ohms in the kit.
DallasRegister Volts/(4096 * 390 ohms) = Amps
Amps/.000013 = mW/cm^2
mW/cm^2 * 10 = W/M^2
does this sound reasonable to any one else?
Posted 22 December 2006 - 02:35 PM
I'm guessing the "4096" is a 12 bit A-D converter, but I'm not sure how the internal 390 ohms is fitting in.
DallasRegister Volts/(4096 * 390 ohms) = Amps
Like I said, not familiar with the sensor or Dallas one-wire so I'm probably missing something very basic.
Edited by BraveSirRobbin, 22 December 2006 - 02:35 PM.
Posted 22 December 2006 - 02:45 PM
Posted 22 December 2006 - 02:57 PM
Posted 22 December 2006 - 06:24 PM
I am interested in your progress and appreciate your efforts.
If it's not linear, I guess you'll be dialing in a formula???
I suppose there will be a +- margin compromise but close enough will do.
Posted 25 December 2006 - 09:01 PM
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