# Solar Sensor

Started by
bjoern174
, Apr 04 2006 05:11 AM

16 replies to this topic

### #1

Posted 04 April 2006 - 05:11 AM

Hi,

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?

Thanks,

Bjoern

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?

Thanks,

Bjoern

### #2

Posted 04 April 2006 - 01:05 PM

Bjoern,

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.

Eric

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.

Eric

### #3

Posted 05 April 2006 - 05:51 AM

Eric,

thank you for you fast reply. Do you have any schedule for your testing?

/Björn

thank you for you fast reply. Do you have any schedule for your testing?

/Björn

### #4

Posted 06 April 2006 - 12:51 AM

Björn,

We don't have a schedule yet. It will have to fit in around the edges.

Eric

We don't have a schedule yet. It will have to fit in around the edges.

Eric

### #5

Posted 21 December 2006 - 04:00 PM

While looking for info on the CLD140 I found that it is obsolete. Replaced by the CLD240E apparently from the physical appearance. The CLD240E has a current of 13 uA at a radiation of 1mW/sqcm. The effective area of the CLD240E is .0036 sq inches, .06 inches on each side. From that you could calculate the W/sqM.

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.

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.

### #6

Posted 21 December 2006 - 04:41 PM

Actually we do use the CLD240E, I just haven't fully updated the site yet.

Eric

Eric

### #7

Posted 21 December 2006 - 10:25 PM

Good to know that it is the CLD240E.

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?

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?

### #8

Posted 21 December 2006 - 11:42 PM

My last post can not be correct. Using the max current in the other posts, .250A, would calculate out to only a fraction of a watt/sqM. I know that in full noon sun, at summer solstice there is about 1000 W/sqM. Will just have to wait and compare the numbers to my meter.

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.

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.

### #9

Posted 22 December 2006 - 12:23 AM

After reading the data sheet for the DS2438 a bit closer. The number returned from the DS2438 is the voltage drop across the sense resitor, and you have to calculate the current with ohms law E = IR.

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?

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?

### #10

Posted 22 December 2006 - 11:08 AM

I set up an experiment. I origionly intended to connect the CLD240E to the analog voltage input of the DS2438, not the current inputs, and played with just the sensor and a resistor size untill I got a result in the 0-5V range. Found that a 4.7K resistor produced about 4volts with the sensor in full sun.

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?

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?

### #11

Posted 22 December 2006 - 02:35 PM

I'm not familiar with these sensors, but curious where you are getting the following from:

Like I said, not familiar with the sensor or Dallas one-wire so I'm probably missing something very basic.

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.**

### #12

Posted 22 December 2006 - 02:45 PM

The Hobby boards solar detector uses a CLD240E photodiode in series with a 390 ohm sense resistor. The sense resistor is accros the current input A/D of a DS2438. The DS2438 records the voltage drop accross the sense resistor. The current thru photo diodes is proportional to the solar flux. So we need to convert the 12 bit number read from the DS2438 to base 10 voltage by dividing by 4096 then divide by the sense resistor ohms to get current. The formula for getting current is from the Dallas Semiconductor data sheet, back in the section that describes the current measuring.

### #13

Posted 22 December 2006 - 02:57 PM

But where does that 4.7 K resistor fit in? I understand the A-D bit span but not understanding exactly what is being fed into the A-D input itself.

### #14

Posted 22 December 2006 - 06:24 PM

Keep working on this till you get it nailed down.

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.

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.

### #15

Posted 25 December 2006 - 09:01 PM

The 4.7K resistor is just an experiment I did on a bread boad. To get some idea as to what the characterisitics of the CLD240 are. The data sheet for it does not have a graph of the current/solar flux relation. Most other photodiods and photo xistors have a graph of that relationship, and it is usualy linear. I am waiting for a couple of things lost in the xmass mail to arrive. Then I can do some better analisis.

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