What is the Effective Quantum Efficiency of my Sensor?

Now that we know how to determine how many photons impinge on a sensor we can estimate its Effective Quantum Efficiency, that is the efficiency with which it turns such a photon flux into photoelectrons (e^- ), which will then be converted to raw data to be stored in the capture’s raw file:

(1)   \begin{equation*} EQE = \frac{N_{e^-} \text{ produced by average pixel}}{N_{ph} \text{ incident on average pixel}} \end{equation*}

I call it ‘effective’ because it represents the probability that a photon arriving on the sensing plane from the scene will be converted to a photoelectron by a typical digital camera sensor.  It therefore includes the effect of microlenses, fill factor, CFA and other filters on top of silicon in the pixel.  It is usually expressed as a percentage.  For instance if  an average of 100 photons per pixel within the sensor’s passband were incident on a uniformly lit spot of the sensor and on average each pixel produced a signal of 20 photoelectrons we would say that the Effective Quantum Efficiency of the sensor is 20%.  Clearly the higher the eQE the better for Image Quality parameters such as SNR and DR.

Photons arriving onto the sensor directly from a light source or after reflection at the scene need to go through the lens, various filters on top of the sensor (AA, IR, UV), the microlenses and the Color Filter Array (if Bayer) before hitting a certain amount of silicon (determined by Fill Factor) and finally being converted to e^- according to a charge collection efficiency determined by the type and quality of doping used during manufacturing.  We will ignore the lens and – since to calculate eQE we only need to know the number of incoming photons and outgoing  e^- independently of what goes on in between – the fact that most of these processes are wavelength dependent.

If we assume a studio lamp source of ‘daylight’ 5300K temperature shone directly onto the sensor we can estimate the number of photons hitting the average pixel as discussed earlier.  Below is the Spectral Photon Distribution of such a source depicted in yellow.Visible Photons Spectral Distribution

The other curves are the photopic eye response function (Sharpe et al 2005) and the Nikon D610’s CFA Spectral Power Distribution I measured recently in somewhat similar conditions (in that case there was a lens involved though).  The relative scales of these other curves are arbitrary.  The ‘visible’ light passband is arbitrarily taken here to be 395-718nm because those are the wavelengths at which the photopic function is about 10 stops below its peak and they represent a reasonable estimate of the usable overall passband for the Color FIlter Array of a typical digital camera above the sensing material, silicon.

If we sum the number of photons in the ‘visible’ spectral range within each 1nm bandwith at every wavelength in the picture above we obtain about 11,260 photons/um^2/lx-s as calculated in the earlier post.  So we have the denominator to calculate eQE per equation (1).  What about the number of photoelectrons generated by the sensor as a consequence of that many photons hitting it, for the numerator?

To determine that figure all we need is some Signal to Noise data in a uniform shot-noise-only area of a couple of raw captures (typically around 1-5% of full scale at low ISO for Exmor sensors)*.  One of the benefits of photons arriving and photoelectrons being generated according to Poisson statistics is that if we square the SNR (defined as the ratio in raw units of the mean and standard deviation of a uniformly lit neutral patch in the raw file) as measured by, say, RawDigger we obtain the signal in photoelectrons – the very N_{e^-} that we are after for the numerator of eQE.

That’s right: SNR^2 in, N_{e^-} out.  This little bit of magic is called Photon Transfer. Should you be interested in learning  more about it I would recommend the similarly named excellent little book by J. R. Janesick.  If we do that for the D610 from DxO Full SNR curves we obtain an average channel* N_{e^-} = 74,971 at full scale at base ISO.  How many photons incident on the sensor produced that photoelectron signal?

We know that DxO accurately measures the Exposure at which modern digital cameras saturate (or clip) under a Daylight illuminant.  I assume they look at the green channel which in such conditions should clip first.  They publish data in terms of sensor sensitivity S_{sat}, what they call Measured ISO.   We can recover the saturation Exposure that they actually measured by using the formula

(2)   \begin{equation*} H_{sat}= \frac{78}{S_{sat}} \text{ lx-s} \end{equation*}

For the D610 at base Measured ISO is 74, which means that DxOmark actually measured saturation at an exposure of 1.054 lx-s.  With this information we can calculate the D610’s 5.9 micron pitch average channel EQE using equation (1)

    \[ \begin{split} Average \: EQE &= \frac{74,971}{11,260 \times 1.054 \times 5.9^2} \\ \\ &= 18.1\% \end{split} \]

18.1% is the probability that a visible photon from a blackbody radiating source at 5300K impinging on the D610 sensor at base ISO will make it through its IR filter, UV filter, AA low pass filter, microlenses, average Color Filter – and produce a photoelectron upon hitting some silicon according to Fill Factor.  It’s a decent figure, with most modern sensors’ average eQEs in the 15-23% range.

The procedure can be repeated at every camera ISO.  Best average Effective Quantum Efficiency estimates are normally obtained around ISO 2-400 for Exmor sensors and around ISO 800 for most others.  Here is the series for the D610 from DxO Full SNR curves:

D610 Sensor Data

Some folks measure peak QE instead of effective QE.  With some assumptions the two are proportional to each other.  Here is an article that explains the difference.


* If you don’t have  a camera with an Exmor sensor or access to such captures you can always turn to DxOMark‘s Full SNR curve database to determine an average channel Effective QE.  Continuing with the D610 as an example, we read an SNR of 34db at 3.476% of full scale off the base ISO curve.  34dB corresponds to a linear SNR ratio of 50.1, which when squared results in a signal of 2512 e^-.  At full scale this would be about 72,264 e^- (=2512/3.476%).

For a better estimate one can perform a fit to the full SNR curve data as described here, which results in a full scale signal of 74,971 e^- for the D610 at base ISO.

Since we also know that the number of photoelectrons generated is proportional to the raw value (ADU/DN) written to the raw file and the D610 clips around raw level 15,778 at base ISO we can calculate the camera’s ‘gain’: at ISO 100 the camera produces a raw output of 1 ADU for every 4.75 photoelectrons rolling off the sensor.

Keep in mind that these figures are obtained from averages of data from the four raw channels.  DxO says that they use a ‘daylight’ source – so far so good.  However they then average the SNR from the four RGBG raw channels to produce Full SNR curves, which may mean that the saturation signals derived from them are also the average of the channels’ signal.  The Green raw data typically being higher than Red and Blue with such an illuminant (as shown in the figure above) would therefore be the result of a larger signal in photoelectrons than the average, with R and B a bit lower than the average.  This of course is not going to be reflected in the average Effective QE produced by DxO’s averaged channel data.