I'd like to simulate photodiode current for a silicon photodiode, one example device the project may use has the following specifications:

Photodiode Specifications

Based on the above plus a guess at Rs for the moment from a similar package / die size device where it's specified I've come up with the following so far to use as a model. I'm planning to use the device in photovoltaic mode so I believe Id may be treated as zero.

Photodiode equavlent circuit

However I'm unsure how to convert lux into a ballpark for the photodiode current Iph. For example Wikipedia gives 100 lux as a very dark overcast day, assuming .40 A/W which I realise is amps per watt I'd like to derive an estimate of the current. From the physical package diagram I can see the active area is 3mm2 so I believe that would give 3.0 × 10-6 lumens of light on that area.

But presumably now to convert to watts I need to know the efficiency of the photodiode and I don't seem to have been able to find any information for this or for other photodiodes. Am I missing something or is their a standard 'near enough' value for all silicon photodiodes?

The reason I'd prefer to simulate rather than just measure is that some devices under consideration are selective photodiodes with fairly long lead times and subject to an MOQ and I need to cover a few possible options for the future.


2 Answers 2


The responsively number given in the table is specific to your device and is exactly what you need (but only partly - see below). There is no "single" parameter for all sensors as it changes from manufacturer to manufacturer. This is primarily determined by QE (Quantum Efficiency) both internal and external QE that is all bundled up in the one number of responsivity.

What you need is a mapping from Lux to Watts, and then the responsivity maps from watts to current.

All detectors will need a passivation layer on top of them to protect the underlying detector material (Here it's Si) so you'll have layers of SiO2 and other material on top. This is important as the External -QE is concerned with getting the light into the Si. This is explained using fresnel equations, but is best understood by the need to match the index of refraction in air (~ 1.0) to that of Si (~ 3.8), the use of AR (Anti-reflection) coatings, and the interaction of light with the passivation layers greatly affects the external QE of the sensor. Once the light gets into the sensor, internal QE is now the concerning factor. As the light penetrates the Si, it leaves a trail of E/H pairs (electron/hole) which are then swept up in E-fileds in the Si substrate. While the E/H generation is understood the E-fields are what determine which electrons/holes get collected. If you generate a E/H pair but it doesn't get collected then you lose internal QE. The electric fields are in turn created through the distribution of dopants and the applied voltages to the device.

In short, even though the Si absorption characteristics are well understood, individual diodes can vary wildly with design. The good news is that this can determined with the appropriate experimental setup. For example the QE of image sensors (say in the green) can vary between manufacturers from as low as 20% up to 98%. In teh NIR (say around 850 nm) these values diverge even more from 1% to 40%.

Radiometry is the measurement of light in quantitative units, Lux is the same curves with the human photopic response laid over top. Consider that mapping as a dimensionless attenuation factor that is dependant upon wavelength.

Ideally what you have is the illumination vs wavelength spectra, the photopic curve again vs. wavelength (which is easily found on-line) and the sensor response vs. wavelength and from those you'd calculate the amount of current flowing.

You have two deficiencies though. One is that you have not identified your illumination spectra and two, the sensor is only defined at 3 points.

A short hand way of calculating is to use the simple estimate (and it will be only an estimate) of 1 lux =\$\frac{1}{683} \frac{watts}{m^2}\$ @ 556 nm (green). Basically this is saying that if you have a green laser at \$ 1 \frac{w}{m^2} \$ then it will appear as 683 Lumens to the human eye.

You will need to understand the difference between luminance and illuminance. So this means you will need to also say what the imaging/collections system is and in particular it's F/#.

Knowing the relationship between wavelength and energy for light \$ E = \frac{hc}{\lambda}\$ where h = planck's constant, C = speed of light. Will allow you to determine the photon flux. And from that you can come up with the shot noise of the system.

Once you can provide the illuminant wavelength dependance, the collection optics f/# and various other parts I'll come back and fill in the details. Or if you want to use the pointers here to answer the question I can check out the answer for you.

  • 1
    \$\begingroup\$ "What you need is a mapping from Lux to Watts, and then the responsivity maps from watts to current." This is the key thing I think PeterJ is missing --- you said it much more clearly than I did, I think. \$\endgroup\$
    – The Photon
    Commented Feb 4, 2013 at 6:27

I see two areas where you seem to have some misconceptions about how to approach this problem:

  • You give the output of the source you're measuring in lumens. Lumens is a unit related to the human perception of brightness, scaled by a function approximating the human eye's response to different wavelengths of light.

    A silicon photodiode, however, will respond to a different spectrum than the eye. In particular, a silicon photodiode will respond to infrared wavelengths out to about 1 um, while the longest wavelength the eye can see is somewhere in the neighbor hood of 780 nm.

    This means that if you have a source characterized in lumens, you will have to find out its actual spectrum to be able to find out its output in the wavelengths seen by your photodiode.

  • You say you don't know the "efficiency" of your photodiode. But you showed that you do know the responsivity, given in A/W. The watts in this characteristic are the watts of the light applied to the detector. So the responsivity already contains within it the efficiency (in terms of light that isn't converted to photodiode current) of the device. Since you know the responsivity, you don't need to know the efficiency.

To solve your problem, you need to know your source's spectrum and intensity (W/m2) when it strikes the photodiode. The response curve of silicon vs wavelength is well known, and your device will almost surely follow this curve closely if it hasn't been deliberately filtered. You can spot check this by the three different responsivity values given for different wavelengths. You need to integrate the product of the light spectrum with the photodiode response to get the total response.

If your light signal is striking the photodiode from all angles (like if you're measuring ambient light) instead of in a narrow beam (like if you place the detector in front of a collimated laser source), you will also need to take into account the varying reflectivity of the diode surface as a function of input angle.

  • 1
    \$\begingroup\$ PeterJ, fixed now I think. \$\endgroup\$
    – The Photon
    Commented Feb 4, 2013 at 6:21
  • 1
    \$\begingroup\$ The Ioffe Institute page doesn't necessarily use the most convenient units if you're not a physicist. For another reference on the response spectrum of silicon, you can look at the response for the 1601-AC detector from New Focus shown here (the blue curves): newport.com/125-MHz-1-GHz-Photoreceivers/917956/1033/info.aspx \$\endgroup\$
    – The Photon
    Commented Feb 4, 2013 at 6:26

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