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- May 27, 2016
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- 3,477
I think I found the way it actually works!
The X-ray photon hits the silicon PIN diode.
For silicon, every little 3.66 ± 0.03eV pulls an electron away from a silicon atom, creating a free electron that can move under the influence of electric fields. It leaves a "vacancy" in the silicon atom - a classic "hole" acting like a positive charge. It's an electron-hole pair. Left to itself, in something even slightly conducting, it will very rapidly re-combine, though possibly promiscuously. (It finds a different -oh whatever..)
In regular diodes, just having P+ type holes material up against N- type electrons-to-spare material results in a brief current, and a depletion region where the electrons and holes found each other. Reverse bias can widen the region, and forward bias can narrow it, until it disappears, and diode conducts. In a PIN diode, there is a built-in intrinsic pure silicon layer between the P and the N. Any electron-hole pairs generated in this layer will extremely rapidly cross to the P and N layers. Except for some electron-hole pairs made by being shook up by heat, or encouraged to get pulled free by bias current, (ie. noise), every 3.66eV gets you an electron's worth.
Current is not the flow of electrons
Oh yes it is! Oh no it isn't!
Actually it's the latter. Electrons can take weeks to get around a circuit, and many may never do if the current is a reversing AC. Think if the power generator is pushing electrons from a windmill in the North Sea, and your light bulb is on the grid 100 miles inland. It is the effect of their fields is what transfers the energy. The analogy is a bunch of snooker balls in a line. Hit the the one on the end, and the one on the other end goes on it's way. So it is with our PIN diode.
So - let's say we whack our alloy with some 59.5keV from the smoke detector thingy, and it hits a molybdenum atom.
The L-shell is outermost, with electrons that are the easiest to get excited. It lets go some 2.29316keV.
That one is going to deliver a pulse of 2293.16 ÷ 3.66 = 626.54 electron-hole pairs. Others too - but lets do this one.
Lets gloss over what is 0.54 of an electron If that 626 electons moves in the circuit, it's a pathetic tiny current, and I would think somewhat too small to ever see against the other racket. Nor will all of them make it. There is an efficiency involved. Yet our X-100 specification sheet shows it has a 20% probability of making it.
OK then, just because it got lucky, and was among the 20%, that does not mean our circuit amplifier can see enough of it to give it gain, and offer at an ADC to be reliably have it's count put in a "2.29keV" bucket - does it?
So how much actual amps is 626 electrons? Each one is 1.602176634E-19 coulombs.
626 of them is 1.602176634E-19 x 626 = 1.0029625728E-16 coulombs.
That load of charge makes a funny, roughly triangular pulse about 13uS long, the integrated area under being proportional to the electron-volts.
If current is Coulombs/Sec, and we use 1/2 base x height approximation, then it's 1/2 x 1.00296257288e-16 ÷ 13E-6
The answer is 3.86 pico-amps !
That's disheartening! I know it was a rough-and-ready calculation, but now, hopefully, someone can tell me where I messed it up.
The X-ray photon hits the silicon PIN diode.
For silicon, every little 3.66 ± 0.03eV pulls an electron away from a silicon atom, creating a free electron that can move under the influence of electric fields. It leaves a "vacancy" in the silicon atom - a classic "hole" acting like a positive charge. It's an electron-hole pair. Left to itself, in something even slightly conducting, it will very rapidly re-combine, though possibly promiscuously. (It finds a different -oh whatever..)
In regular diodes, just having P+ type holes material up against N- type electrons-to-spare material results in a brief current, and a depletion region where the electrons and holes found each other. Reverse bias can widen the region, and forward bias can narrow it, until it disappears, and diode conducts. In a PIN diode, there is a built-in intrinsic pure silicon layer between the P and the N. Any electron-hole pairs generated in this layer will extremely rapidly cross to the P and N layers. Except for some electron-hole pairs made by being shook up by heat, or encouraged to get pulled free by bias current, (ie. noise), every 3.66eV gets you an electron's worth.
Current is not the flow of electrons
Oh yes it is! Oh no it isn't!
Actually it's the latter. Electrons can take weeks to get around a circuit, and many may never do if the current is a reversing AC. Think if the power generator is pushing electrons from a windmill in the North Sea, and your light bulb is on the grid 100 miles inland. It is the effect of their fields is what transfers the energy. The analogy is a bunch of snooker balls in a line. Hit the the one on the end, and the one on the other end goes on it's way. So it is with our PIN diode.
So - let's say we whack our alloy with some 59.5keV from the smoke detector thingy, and it hits a molybdenum atom.
The L-shell is outermost, with electrons that are the easiest to get excited. It lets go some 2.29316keV.
That one is going to deliver a pulse of 2293.16 ÷ 3.66 = 626.54 electron-hole pairs. Others too - but lets do this one.
Lets gloss over what is 0.54 of an electron If that 626 electons moves in the circuit, it's a pathetic tiny current, and I would think somewhat too small to ever see against the other racket. Nor will all of them make it. There is an efficiency involved. Yet our X-100 specification sheet shows it has a 20% probability of making it.
OK then, just because it got lucky, and was among the 20%, that does not mean our circuit amplifier can see enough of it to give it gain, and offer at an ADC to be reliably have it's count put in a "2.29keV" bucket - does it?
So how much actual amps is 626 electrons? Each one is 1.602176634E-19 coulombs.
626 of them is 1.602176634E-19 x 626 = 1.0029625728E-16 coulombs.
That load of charge makes a funny, roughly triangular pulse about 13uS long, the integrated area under being proportional to the electron-volts.
If current is Coulombs/Sec, and we use 1/2 base x height approximation, then it's 1/2 x 1.00296257288e-16 ÷ 13E-6
The answer is 3.86 pico-amps !
That's disheartening! I know it was a rough-and-ready calculation, but now, hopefully, someone can tell me where I messed it up.
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