Silicon photodiodes respond to a wavelength range of around 190-1100nm (source: wikipedia). I understand that photodiodes function by having a photon of sufficient energy create an electron-hole pair in a semiconductor via the photoelectric effect, which in turn increases the number of charge carriers, which in turn increases the photocurrent.

I know that the energy of a photon is proportional to its frequency as per the Planck-Einstein relation: $E=hf$. The minimum energy required to stimulate the creation of an electron-hole pair in the semiconductor material should therefore be a product of its bandgap. The bandgap for silicon at 302 Kelvin is approximately 1.14eV. The associated wavelength of the minimum energy should therefore be:

$$f = \frac E h = \frac {1.14eV} {6.62607015 \times 10^{−34}} = 2.75650774 \times 10^{14} Hz$$

$$\lambda = \frac c f = \frac {3 \times 10^8} {2.75650774 \times 10^{14}} = 1088 nm$$

This tallies with the 1100nm figure quoted on the wikipedia page.

The maximum energy is where I get a little confused. My intuition says that the ionisation energy would be an upper limit; I found 8.15168eV quoted for silicon. Repeating the equation finds the associated wavelength:

$$f = \frac E h = \frac {8.15168eV} {6.62607015 \times 10^{−34}} = 1.97106745 \times 10^{15} Hz$$

$$\lambda = \frac c f = \frac {3 \times 10^8} {1.97106745 \times 10^{15}} = 152nm$$

This is somewhat smaller than the 190nm figure claimed above.

My initial guess was that the maximum energy level is lowered as to avoid exceeding the free exciton binding energy, but the figure I found listed for that is 14.7eV (84nm) so that doesn't line up.

Where does the 190nm lower wavelength bound (upper energy bound) figure come from? Am I also right in thinking that not all wavelengths in the 190-1100nm range would elicit photocurrents, due to the requirement that photon energies match the quantum energy state transition levels?

  • 1
    $\begingroup$ (+1) If the photodiode has a fused silica window, as mine do to protect the die surface, then the 190 nm may just be the typical fused silica short wavelength cutoff. $\endgroup$
    – Ed V
    Oct 31, 2022 at 2:11

2 Answers 2


It's somebody's practical limit. There's usually some sort of thin "dead layer" on the diode surface. It's close to perfectly transparent at long wavelengths, but as you move into the UV, it becomes opaque. At x-ray wavelengths it becomes transparent again.

With careful, fussy surface treatment it is possible to reduce the dead layer for better UV and soft x-ray response.

  • $\begingroup$ Which energy value for the semiconductor material sets the maximum photon energy in combination with the optical attenuation from the surface layer? First ionisation energy? Free exciton binding energy? Or is it solely down to the optical attenuation with no limit on the quantum side? $\endgroup$
    – Polynomial
    Oct 31, 2022 at 0:58
  • 1
    $\begingroup$ @Polynomial It's complicated, not just a simple material property. Depends on the surface. $SiO_2$ is opaque below ~200 nm unless it's very thin. The penetration depth for UV in semiconductors is very short, so the photoionization may occur in undepleted semiconductor near the surface. The depleted bulk material starts at some depth below the surface, depending on doping. If the minority carriers from a shallow event can't make their way through the undepleted material, the diode won't collect them. $\endgroup$
    – John Doty
    Oct 31, 2022 at 13:26
  • $\begingroup$ At the least there is the top doped layer that is one side of the junction. Electron-hole pairs generated there are less likely to result in current flow. And the passivation layer and metallization (may be a small fraction of active area, but the thinner the top layer the more you need some. $\endgroup$
    – Jon Custer
    Oct 31, 2022 at 16:21

I do believe the doping has an effect on the wavelength response considerably given the band gap can be widely tuned with doping, and cmos detectors are doped. However, I have yet to find a concrete upper limit of a silicon cmos spectral response. Through my own testing I have unequivocally detected light at 1153nm no problem with a Sony A7s cmos, which is above the theoretical wavelength one would derive from pure Si’s bandgap alone, it seems that based off my own spectroscopy measurements, it is sensitive up past 1180nm as well, but I’m hesitant to give that figure until further testing is done. Signal gets buried into noise there doesn’t seem to be a clear upper wavelength limit, longer integration times yield yet deeper IR wavelengths thus far so I’ve yet to determine the limit, though I’m sure one exists.

  • $\begingroup$ In the IR, it's temperature dependent. $\endgroup$
    – John Doty
    Dec 26, 2022 at 1:53

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