What exactly is your setup? At 5keV, it's probably more a problem of absorption and not of detector material efficiency.

Dear Andrey,

Erik is right.

It could be a matter of x-ray attenuation.

Do you have got an x-ray window in front of your detector and what is the distance in air which the x-rays have to pass?

X-ray source and filtration?

How do you have measured the energy dependence of the detector sensitivity?

Details of the experimental set up are wellcome ....

25kV lower band edge, is the specification of Gd2O2S:Tb scintillator we use.

And the steepness of image brightness variation does not allow us to believe in absorption only. We are putting the sensor at a distance of 10-15 cm; it is not like medical fluorography with its 1-2 meter distances.

I am not sure why this happens. Maybe there is an absorption pike around 25 keV, followed by a low absorption valley down to several keV ? Common sense was telling me that at lower energies scintillation should be more intensive, but the manufacturer and our experiments tell opposite.

I am interested in better material, but still less expensive than CsI:Tl crystals.

Dear Andrey,

K-edge of Gd is around 50keV and that of Tb around 52keV.

For low energy the attenuation strongly increases due to quite high atomic number. There is no absorption peak at 35keV. Gd L-edges are around 7,2 to 8,3 keV.

For Cs the situation is more relaxed due to smaller atomic number (K-edge at ~36keV).

Some time ago we have dealt with CsI:Tl investigating EXAFS around the Ca K-edge (it is around 4keV ) and this material works pretty well down to that photon energy.

But what is about your x-ray source (tube? , gamma source(s)? ...). You might have low photon flux at low energies. For example air is attenuating at lot at 5keV. Transmission is around 50 to 60% at 5keV (15cm). So any bulk filter or window will further decrease your low energy transmission.

By the way:

a) what is the term '35kV lower band edge' ?

b) What does lead you to think having perhaps a 35 keV absorption peak? (Where ; in the detector ? or in the sample? or ...)

c) how do you 'see' that your detector does not perform at lower photon energies ( down to 5keV) ?

>But what is about your x-ray source

It is tube with 0.1 Be window. Unfortunately, we do not have metrologic equipment to measure the exact radiated energy. But when you say the absorption can be around half of the energy we see it as the overexposed sensor at 25 keV becomes very dark below 10keV. Even taking into account the narrowed radiated bandwidth, by readouts it should be a hundred or a pair of hundred of times. And I thought that 15keV is not that big a difference.

>what is the term '35kV lower band edge' ?

Sorry, I was wrong 25keV, not 35. 25keV-160keV is in the specifications by the manufacturer.

>What does lead you to think having perhaps aт absorption peak?

I mean, the scintillator. Shouldn't the response be proportional to absorption? So I just guess, the response degradation can be a valley below some absorption peak.

But it is just a wild guess.

>how do you 'see' that your detector does not perform at lower photon energies

by very low detected intensity (read-out). And the manufacturer confirms, that it can work specs, down to 15 keV, but we should not try to pull anything out of it below.

In air, the mass absorption coefficient is around 100 cm2/g at 5 keV. The areal density of 15cm air is around 0.02 g/cm2. So you lose around 80% of your radiation in that layer, as already said. Between 25 and 10 keV the mass absorption coefficient of light materials increases by the factor of about 10, so absorption increases almost a factor of 10'000.

Dear Andrey,

thanks for sharing more details.

But I am still confused yet.

a) X-ray source is a tube having a Be window of 0,1mm.

Even for 5keV photons the transmission of your window is quite ok (quite close to 100%).

b) In your question and in your answers above you talk about energies; 5keV, 10 keV, 15keV, 25keV, 160keV. But on the other hand you state, that you are not able to determine the energy. Even the manufacturer's specification is 25keV to 160 keV, as your have written above.

Here something must be wrong or a misunderstanding present.

c) The only parameter which corresponds to the energy (distribution) of your incoming x-ray photons, and which you are able to change, is the tube voltage.

d) Resolving the problem

You know that the distribution of the photon energies is inbetween 0keV and a maximum value in keV, which is equal to the electron charge times the applied tube voltage (mainly in kV) .

In order to avoid confusion between the kV value (kilo -Volt) of the tube from the keV energy values (kilo electronvolt) of the x-ray photon, the voltage value often is named as peak voltage value kVp. So the maximum photon energy Emax = e*kVp.

I think there is just a misunterstanding in these numbers.

The manfacturers specification should read: 25kV to 160kV (tube voltages), a voltage region which is used for Mammography up to whole body scans and industrial applications. These tube voltage ranges can be found in the literature for your scintillator material.

However, when you go to 15kV or 10kV or even 5 kV, you should bear in mind, that the maximum photon energy is only 15keV, 10keV or 5keV respectively (having almost no photons at that energy !!!) .

The light output of the scintillator is proportional to the total x-ray photon energy deposited in your scintillator. However the amount of photons or (better named ) the photon flux f is roughly proportional to the square of the tube voltage and the tube current I (mA): f~ I[mA]*kVp² .

Now you can imagine that when going from 25kVp to 10kVp or even down to 5kVp in tube voltage, the photon flux leaving the tube window goes down a factor of about 6 when ending up at 10kVp, and by a factor of about 25 when ending up at 5kVp.

The energy deposited in detector is proportional to the photon flux times the mean photon energy Emean of the tube spectrum.

For a rough estimation here Emean is assumed to be proportional to the kVp value: Emean ~ kVp.

So the mean photon energy flux is roughly :

~ f*kVp = I[mA)*kVp²*kVp = I[mA]*kVp³.

Now when going from 25kVp to 10kVp or even 5kVp the photon energy flux reduces by a factor of ~ 2,5³~16 when going to 10kVp and by a factor of 5³ =125 when going to 5kVp respectively.

When additionally taking into account the attenuation in air, you see that, your are in severe troubles with respect to your detector signal.

You see that a loss factor of over 100 is easily achieved when going to low kVp.

Please recognise, that up to now we do not have had a look at your scintillator material. So when going to any other material you will have the same behaviour.

Thus going to CsI:Tl or any other material will not really help you.

Thanks, my confusion has gone now.

Erik, thank you for the data! I was googling it around, but could not find anything below 100keV; neither I expected the function to be that steep at soft energies.

Gerhard, thank you for the answer

> In your question and in your answers above you talk about energies; 5keV, 10 keV, 15keV, 25keV, 160keV. But on the other hand you state, that you are not able to determine

And it is exactly as I said. 25-160keV is the specification issued by the manufacturer. I do not know why they really do not rate it below. They said, it is incapable of detecting there) . I can say only for 5-25 kV, which we feed to the source controller. Where practically we work with 10-20kV.

But we can not determine the radiated power (not energy). Still, we have an image at different energies, generated by the scintillator, and digitized by the sensor.

> The only parameter which corresponds to the energy (distribution) of your incoming x-ray photons, and which you are able to change, is the tube voltage.

Well, for incoming photons that is 1)Target material and 2)voltage profile (we work with DC) 3)Be filtering 4)Air attenuation. 5)probably some technological issues at scintillator side, like absorption at fiber substrate, or protective coating...And I do not know anything about this.

For outgoing, yes, we stop at (3). So pardon me for misspelling keV for kV sometimes.

> (having almost no photons at that energy !!!)

...emm, yes you are right; integrating the Kramer's law, we get a power of two. Probably that is the case. The problem was, why is there so much difference between 10kV and 25 kV. And seems, Erik has already cleared this question up.

So, thank you for explaining, Now I am sure that we are dealing with attenuation.

Moreover, I found a large comparative datasheet on scintillators (though, for 511 keV), and Gs2O2S:Tb seems the best material; probably even better than CsI:Tl for video-scanning. (Carel, Eijk, Inorganic scintillators in medical imaging)

And by the way, I found two controversial characterizations of Gd2O2S:Re; one showing maximum at 22kV, and drop to 0.5 efficiency at 10kV; which does not explain total reduction in visible brightness.

Andrey,

I mostly refer to

https://public.wsu.edu/~collins/Phys415/writeups/massabscoeff.gif

60 years old, but a really nice and handy compilation.

As others have mentioned losing the signal could be due to absorption in some material between the source and scintillator. Not being able to see the signal could also be caused by noise in your system. It would be interesting to see what your expected count rate in the ROI if interest is and how that compares with what you are observing. We have built NaI(Tl) based detectors that have a noise threshold of less than 1.0 keV. Send an email to [email protected] and I can provide a data sheet, if you would like more information.

Erik, thank you for sharing this invaluable scan!

I just want to confirm my understanding. When estimating air absorption, N is almost close to O, and can be approximated by the same curve, and H from water vapor, which reduces absorption, has mass part around 2%(saturation humidity)*60%(vapor humidity) *2/3(atomic ratio)

Frank, thank you so much for offering the data. I guess it is very important in spectrometry, for Na:Tl is definitely famous for short decay time, and no afterglow. I just can not be sure that it useful in imaging. The decay of scintillators(phosphors) in imaging is 3 orders longer, and I guess, for a reason. The scintillation of Gd2O2S:Tb, CsI:Tl appears to be 30% stronger, if there are no dips and spikes at 10kV of which I am not informed. But the worst thing of all, is that we do not have any energy measuring tool for X-ray, only the sensor, and programmable tube-based source. So I will rely on your word. And I guess there is a annoying noise floor, as my coworker says. But it looks, the problem is the difference in attenuation at 10kV and 20kV of several orders, which was so confusing.

Eric, sorry, I am trying to built a calc. table, but I do not see how did you get 10000 factor. If I am correct, the absorption increase is exp( (mu2-mu1)rho*x), which gives just 15% difference by your table for oxygen between 10kV and 25 kV exp(0.02*(8.5-1.7)), or using NIST data and air density from google just 8% exp(0.01776*(5.12-0.57)). When roughly integrating the spectrum by the energy between 5kV and 20kV I get decrease of detectable brightness around 14 times only when reducing voltage from 20 to 10 kV. Did I miss something in calculation?

(I finally found the data here: https://physics.nist.gov/PhysRefData/XrayMassCoef/ComTab/air.html)