- Joined
- Jul 28, 2017
- Messages
- 2,625
You have some good comments here. I'm thinking that the incident photons do experience a sequential drop in energy as they excite fluorescence in atoms they hit and also pump carriers into the conduction band that eventually contribute to the current pulse height. The fluorescence x-rays also do their bit to add to the current pulse, but since they can't excite any more fluorescence x-rays all they can do is move carriers up into the conduction band. So in a fast cascade the incident photon is converted into back-radiated fluorescence photons, hole-electron pairs and heat (phonons). For detection efficiency we mostly want hole-electron pairs.This is a point on which I looked hard for what is known. I know when the photon arrives, it might miss. Most of what is inside of what we think is solid hard stuff is mostly empty space. There is a probability of collision involved. If it misses, there is more to hit beyond. If the stuff is thin enough, it can go right through. It's higher energy X-rays stuff (Gamma), after all!
If it hits, it excites the element electrons into a higher energy state. They don't stay that way. They drop back into their "normal" state. The amount they took in to get to the excited state is released as a fluorescence photon, having a new wavelength determined by the energy change involved, and Planck's constant. It comes out as X-Rays.
So - what happens to the excess?
A 60KeV photon hitting iron (Fe) uses up only 6.4KeV and 7.6KeV to excite the K-shell, and presumably, at the same time, uses up another 705eV and 718eV getting the L-shell electrons into a higher state. That total is 15.4KeV, leaving another 44.57KeV yet to do anything.
Does the leftover energy simply excite the same atom electrons another couple of times until it can't quite manage a last K-shell event?
Does the remainder go on to keep working the same atom L-shell up and down again and again until even that runs out?
The wavelengths are getting pretty long for the small energies. Would that be into infra-red?
Does it end up shaking the atom about somewhat, as in it "warms the stuff up" a bit?
Does it not happen that way at all? Does the remainder 44.57KeV keep going to strike some other atom instead?
------------------------
All of the above is about the fluorescence scintillation. The next bit is about what happens in the photodiode detector.
Although I have trawled many videos about what happens in photodiodes, particularly from the advanced set of lectures from a Indian university, I have not seen a clear explanation of exactly how a arriving photon turns into a current amid the conduction band carriers in a material. All the equations are about a energy flux of lots of photons. There is an efficiency involved. We do not get a current of energy equal to 100% of the incoming. Some of it, I think, ends up as heat.
In our design, we have a transimpedance amplifier capable of seeing a current started by only one photon, which is then amplified.
A whole bunch of other noise currents will be amplified along with it. Some of it is thermal generated, but I don't propose cryogenic amplifier design. Some is induced from outside fields, but we can shield it from such interference by design. We even can deny magnetic interference, and we can get up to circuit design tricks that can cancel differential noise, and avoid common mode noise. In the end, this will be about signal to noise ratio, and the only way to preserve that, locked in, is to start with extreme low noise amplification with very high gain, so that any noise from later stages is dwarfed by comparison to our (amplified) original noise.
I am pretty sure that we will run into pulse and real signal situations we did not anticipate, meaning the difference between theory and practice. We just have to give it our best shot. You are doing all the right things to anticipate most everything. The aluminium plates I had first thought was overkill, but I revise my opinion.
The energy needed to get a carrier into the conduction band of doped silicon is pretty low, down in the single electron-volt range so it should be possible to suck most of the energy out of any spare x-ray photons flying around the crystal lattice.