[Shiraz (Ray)]

Hi.
The question often arises - how sensitive is this imaging system compared to that one. You may wish to compare different setups when deciding what to buy or how to configure what you have. The following provides a way to do this. I am considering asking Mike to put this in the how-to area, but thought it would be useful to run it by the forum beforehand - so would be grateful for comment on the content, the way it's expressed, whether you think it has value and if you have seen this sort of approach published elsewhere. thanks for looking. Regards ray


The question often arises as to which combination of imaging scope and sensor will be the most sensitive. Relative system sensitivity can be found with reasonable accuracy by comparing the S values for systems, where S is:

S = QE * OE * p^2 / FNo^2

Where QE is detector quantum efficiency, OE is optics efficiency, p is pixel size and FNo is F number.
* is multiply, / is divide and ^2 means squared.

To work through an example; say you want to compare:
a 300mm f3.8 Newtonian + 16803 camera (system 1)
with a 200mm f8 RC + 8300 camera (system 2).

The explanatory notes (later on) show how to estimate the parameter values, but for now appropriate numbers are:
For system 1; quantum efficiency QE=0.53, optics efficiency OE=0.65, pixel size p=9, FNo=3.8
For system 2; quantum efficiency QE=0.47, optics efficiency OE=0.75, pixel size p=5.4, FNo=8

from which,

S(system 1) = 0.53 * 0.65 * 9 * 9 / 3.8 / 3.8 = 1.9

S(system 2) =0 .47 * 0.75 * 5.4 * 5.4 / 8 / 8 = 0.16

Since the Newtonian (system 1) S value is ~12 larger than that of the RC (system 2), then the Newtonian system is about 12 times as sensitive. Similar calculations for a range of systems are included in the attached image.

Of course overall sensitivity is not the whole story. Some other equally important issues are resolution, image scale, field of view and required sub length - for the chosen examples, the RC has finer sampling, so it would resolve much finer detail in very good seeing and it would produce much larger scale images of small objects. However, the fact remains that, with the example RC system, you would need to stay out under the stars for ~12 times longer than you would with the Newtonian in order to get to the same SNR (for sky limited imaging). ie, if you got a nice smooth image in a total of 3 hours with the example Newtonian system, you would need ~36 hours to get the same image smoothness with the example RC system.

The other thing that needs to be said is that this should not be read as implying that either large pixel size or low FNo is of prime importance in isolation – neither is inherently better for sensitivity. To show that this is the case, look at the equation – if you increase the FNo by a factor of 2 and also increase the pixel size by a factor of 2, the two increases cancel out and there is no change of sensitivity – ie if you must use a high FNo because that is the characteristic of your chosen scope, just match it with large pixels to retain sensitivity. However, once you have decided on your pixel size and focal length, a faster scope of the same focal length will give you a sensitivity advantage (if such a scope is available with the desired specs). Alternatively, if you decide on a particular focal length and aperture, larger pixels will give you higher sensitivity – but at the price of reduced resolution.

*****************
Explanatory remarks:

The calculated S value is not a standard quantity, but you can use it to say that “system A is 1.4 times as sensitive as system B” by comparing the calculated S values. It was derived from first principles, but could also be extracted from the absolute spectral sensitivity formulae used in professional astronomy.

The S value comparison is only valid for broadband imaging, where shot noise from the sky is the limiting factor. However, it still provides some idea of relative performance with narrow-band imaging. It is based on Signal to Noise Ratios (SNRs), not ADU values – the ADU values you get from different cameras will vary depending on camera gain and they do not tell you much about image quality.

The only parameter with units is p. Use microns for consistency.

QE (Quantum Efficiency) is the average over the system bandwidth, not the peak. Typical average broadband QE values (400-700nm) for a few popular chips are:
16803 = 0.53
11002 = 0.4
8300 = 0.47
694 = 0.7
OSC/DSLR ~ 0.2 (probably) EDIT: it could be as low as 0.1 - see post 16

The optics efficiency is the nett effect of all optical losses such as reflection loss, transmission loss and central obstruction. For example, my Newtonian telescope has 2 reflective surfaces, each with estimated reflectivity of about 0.87, a central obstruction that reduces the total light to about 0.9 and a coma corrector that has 6 surfaces and 3 bits of glass and probably has something like 0.95 transmission. The total optics efficiency is the product of the individual efficiency factors: 0.87 x 0.87 x 0.9 x 0.95 ~= 0.65. A refractor with a field flattener could possibly be around 0.9. An RC or SCT with large obstruction, high reflectivity coatings and a field flattener may be around 0.75.
In general, geometrical effects are more important than optics efficiency and you probably won’t be too far out if you use:
Refractor = 0.9,
SCT/RC/CDK/premium Newtonian = 0.75,
corrected standard Newtonian = 0.65.

have fun