# A pyranometer, a device that measures sunlight intensity

1. Apr 5, 2006

### TSN79

I am to use a pyranometer, a device that measures sunlight intensity. I am confused about the output unit it produces. It says on it's side $4.55 \cdot V/\left( {W \cdot m^{ - 2} } \right)$ but I'm not sure how to interpret this unit. All help will be appreciated...

2. Apr 5, 2006

### Integral

Staff Emeritus
It appears to output Volts, if it reads 1 V out you have 4.55 $\frac W {m^2}$

If you could post the name and model number perhaps we could give better information.

3. Apr 5, 2006

### FredGarvin

That's exactly what it is doing. That's your scale factor for calibration. You should have a calibration curve with the instrument. If it's relatively linear, then it's easy to display what you are reading. If it's not linear the manufacturer would supply a curve fit for the cal interval.

4. Apr 6, 2006

### TSN79

Thanks. The number wasn't quite correct though. It is 4.55 µV instead of V. My logging software measures mV and when I tested it on a semi-overcast day it read about 0.02 mV which results in a value of 0.02 mV/4.55 µV = 4.39 W/m2, which doesn't make sense. The number should be much higher, despite the fact that the sky is overcast. One uses 7-800 W/m2 as an estimated maximum for direct sunlight, so 4.39 is ridiculously low. Any idea why? Or am I doing something completely wrong?

Last edited: Apr 6, 2006
5. Apr 6, 2006

### FredGarvin

Perhaps on a voltage that low, your acquisition might not be capable to measure that low of a voltage. Perhaps you need some signal conditioning prior to the readout.

6. Apr 6, 2006

### Staff: Mentor

That doesn't seem rediculously low to me. Our perception of light intensity is not linear, so it doesn't seem at all unreasonable for the intensity of a cloudy day to be 1/200th of direct sunlight.