Measuring Irradiance with Photodiode

[mechguy]

Hello,

I have a project to build a photodiode circuit with microcontroller to measure solar irradiance. The function of OpAmp is to limit the analog voltage to A/D converter, as shown in attached picture below. Vout will be connected to A/D converter (Vout = Ip * R1).

My problem is, how to measure irradiance? (power per diode area). For my understanding in measuring power, I will use a voltmeter to measure voltage across photodiode, and multiply it with the photocurrent (Vp * Ip). Is that correct?

Kindly appreciate any help and suggestion. Thanks in advance! ;-)

 

[tpetar]

Insolation is measured with Pyranometer in mV, later this can be racalculated into W per m2.




You can see this great projects of great prof. Wichit Sirichote :

Solar Recorder
**broken link removed**

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Solar Recorder V2.0
**broken link removed**

**broken link removed**

 

[enjunear]

Diodes are current-based devices, so you need a circuit to convert the small photodiode current into a voltage.

Check out this presentation for a simple op-amp circuit implementation, slide 9.
ecee.colorado.edu/~ecen4827/hw/hw1/PhotodiodeAmplifers.pdf

As for calculating irradiance, I found some useful equations on this site: **broken link removed**

Eqn 4 states: Irms = R*Ee*Ad
Where Irms is the average current out of the diode,
R is the Responsivity of the diode in Amps/Watt,
Ee is the irradiance in Watts/cm^2,
and Ad is the illuminated capture area of the diode.

The op-amp will give you a conversion from current to voltage. If you measure that voltage and scale it according to the op-amp's conversion gain, with a little algebra (and the photodiode characteristics from the spec sheet), you can calculate Ee (irradiance in W/cm^2).

Be sure to watch out for the units and make sure they cancel out correctly, so you know if you are in W/cm^2 or W/m^2 (depends on what's given on your diode's spec sheet).

 

[tpetar]

If you dont whant to spent some wallet green unit you can make your :

Measuring Sunlight at Earth's Surface: Build Your Own Pyranometer
https://www.instesre.org/construction/pyranometer/pyranometer.htm

 

[mechguy]

Eqn 4 states: Irms = R*Ee*Ad
Where Irms is the average current out of the diode,
R is the Responsivity of the diode in Amps/Watt,
Ee is the irradiance in Watts/cm^2,
and Ad is the illuminated capture area of the diode.

Hi enjunear,

First, thanks for you suggestion.

I have also thought of the equation which you suggested to me. I can get the value Ee (irradiance) after measuring Irms (photocurrent) with an ammeter, or using the equation Vout / R1. However, for Responsitivity value (or Sensitivity as shown in datasheet of most photodiodes), it only gives sensitivity at specific wavelength, while the wavelength of sunlight is ranging differently from time to time.

Correct me if I'm wrong.

 

[enjunear]

Hi enjunear,

First, thanks for you suggestion.

I have also thought of the equation which you suggested to me. I can get the value Ee (irradiance) after measuring Irms (photocurrent) with an ammeter, or using the equation Vout / R1. However, for Responsitivity value (or Sensitivity as shown in datasheet of most photodiodes), it only gives sensitivity at specific wavelength, while the wavelength of sunlight is ranging differently from time to time.

Correct me if I'm wrong.

You are correct. The physical design and composition of the diode will dictate it's peak spectral sensitivity. You are also correct that sunlight will have different spectral composition, however, this change is only really dramatic at sunrise/sunset, when the sun is heavily filtered by the increased thickness of the atmosphere (why the sun looks red/orange at sunrise/sunset).

Looking at the red line (which is what we'll see near the ground), you can see that the two peak wavelengths for the diodes you selected are at two of the peaks in the spectrum, so they should remain relatively unaffected by atmospheric changes (other than their intensity throughout the day, which is what you want to measure).

 

[mechguy]

You are correct. The physical design and composition of the diode will dictate it's peak spectral sensitivity. You are also correct that sunlight will have different spectral composition, however, this change is only really dramatic at sunrise/sunset, when the sun is heavily filtered by the increased thickness of the atmosphere (why the sun looks red/orange at sunrise/sunset).

To enjunear:

I have been thinking of this assumption before.

For your info, instead of just yellow sunlight emit to the photodiode's surface, we have to consider the surrounding light, especially from the blue sky in which the sunlight is scattered by earth's atmosphere, blue light is absorbed and some reaches to our eyes. So, I consider the photocurrent Iph is actually affected by both background light and direct sunlight. Unless we want to ignore the background light (in which the effect on photodiode is very small?), I doubt whether the equation Irms = R*Ee*Ad is applicable.

Anyway, thanks for your explanation and feel free to comment again here.

 

[FvM]

You didn't yet mention your irradiance specification. The solar spectrum is wider than the spectral range of a Si photodiode. Technical irradiation measurements are either "total" irradiation measurements, that need to use thermopiles, or measurements with a specified spectral window.

 

[deetarris]

Hello,

I have a project to build a photodiode circuit with microcontroller to measure solar irradiance. The function of OpAmp is to limit the analog voltage to A/D converter, as shown in attached picture below. Vout will be connected to A/D converter (Vout = Ip * R1).

My problem is, how to measure irradiance? (power per diode area). For my understanding in measuring power, I will use a voltmeter to measure voltage across photodiode, and multiply it with the photocurrent (Vp * Ip). Is that correct?

Kindly appreciate any help and suggestion. Thanks in advance! ;-)

please, i have an exact same project at hand. i really want to know how far you went with yours and to know if you can be of any assistance to me.
thanks