Solar Sensor

Finaly got to doing the experiment. Did not use the 1 wire solar sensor. Put the CLD240E in series with both the 390 ohm and 4.7K ohm resistors accross a 5vdc supply. Measured the voltage drop accross the sense resistor with a digital multimeter. And, recorded the multimeter volts and my SP1065 solar energy meter reading. I have attached an excel spread sheet with the data and calculations. The chart shows the two readings follow very closely. Probly some errors in trying to read both meters at the same time.

Observation, with the 390 ohm resistor one is likely to only be able to tell when the sun comes up and goes down. The readings are in millivolts. Maximum of about 150 mv. Since the DS2438 resolution is 10 mv, this does not give much range. There would only be 15 steps between daylight and night. Using the 4.7K resistor gave more range. Since the highest voltage was 1.75Vdc, there would be 175 steps daylight to dark. One could likely go to a higher resistance value, but the day I measured was a bit overcast and would not recomend going to a larger resitor than 4.7K untill I have a very clear day.

The meter I have reads in BTU/HrFt^2, so I found a convertion factor to W/M^2. Watts seems to be a defacto standard for solar flux.

I did some trial calculations on the spread sheet and came up with a multiplier to use to get W/M^2. Remember that the reading returned from the DS2438 is the voltage drop across the sense resistor, and one needs the current. So, I = E/R to get the current. One should use a multimeter to read the real value of the resistor, espesialy if it is not 1% tolerance.

The formula:

DS2438 volts / sense resistor ohms = sensor current

sensor current * 1157598 = W/M^2 (solar energy)

Note: this number is calculated with the data from the 4.7K resistor.

The chart shows that this comes very close to the solar meter.

Look at the chart and maybe things will be more clear ;)

I still use excel 2000, hope it comes up in a newer version.

Richard
 

Attachments

  • CLD240E_mv_watts.xls
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I forgot one thing in the last post. Dividing the voltage in the current A/D register by 4096. The resolution of the current A/D converter is .2441 mV. The resolution of the voltage A/D converter is 10 mV.

The statement that one would only be able to see night and day would be wrong when the current A/D converter is used as is the case on the Hobby Boards solar sensor board.

The maximum reading I got with the 390 ohm resistor was about .150 Volts. Dividing that by .0002441 Volts read by the current A/D converter results in about 1000 steps between total dark and full sun.

The formula to get mVolts from the current register is:
(Current register counts)/4096 = (current register volts)
Note: 1/4096 = .0002441

The formula to get current thru the sense resistor
(current register volts)/390 ohms = (sensor current)
Note: asumes a standard Hobby Boards solar sensor

The formula to get solar energy is:
(sensor current) * 1157598 = W/M^2 (solar energy)
Note: the readings for solar radiation taken with a SP1065 meter from edtm

http://www.solarstop.net/edtm/1_solar_measurement.htm

thanks for tolerating all the reiterations and mistakes in coming up with this.
am also reposting this as a new thread
 
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