Needing more than a spark test?

[graham-xrf]

It's likely that noise coming out of the PIN detector is larger than the amplifier's voltage and current noise so there's not much to be gained there.

I know that reverse biased diodes can be noisy, but are we sure of this?
It makes my choice of low noise front end look unnecessary.

On a slightly different need, do I recall you mentioning some trick circuit you had to keep the ADC save from overvoltage (2.5V)?

 

[homebrewed]

I know that reverse biased diodes can be noisy, but are we sure of this?
It makes my choice of low noise front end look unnecessary.

On a slightly different need, do I recall you mentioning some trick circuit you had to keep the ADC save from overvoltage (2.5V)?

No, I'm not certain that the noise is coming from the diode. Something I was going to do was play with the diode bias voltage to see if that affects the noise, but now I'm distracted by my S/W hang problem. I may be homing in on that though.

On preventing overdriving the ADC, I just have used a clamp diode. At least, that's what my current signal conditioning circuit does -- but it can use the Teensy's 3.3V. The AD7667's Vref is available on a package pin but switched-capacitor ADCs really want a low impedance drive. Putting a series-connected clamping resistor in there may cause problems, so I'd put the clamping network on the input side of the analog buffer IC.

 

[graham-xrf]

Perhaps for @RJSakowski : ?
You have provided the answer that an incoming photon energizing an atom of some stuff, if it has energy "left over" can then continue to potentially energize some other electron shell in the same atom, or go on to hit some other atom.

I feel foolish to ask about some "what if" scenarios.

1. Suppose an incoming 60KeV photon hits (say) an atom of gold. There is not enough energy in it to excite either of the two electrons in the K-shell. Therefore one supposes, since the photon hit the atom already, we might still hope to get some 9.7KeV and a 11.07KeV out of the L-shell, and have some 38.4KeV continue onward to get some responses from other atoms it runs into. Gold has six shells, filled as 2, 8, 18, 32, 18,1. I would have thought the lone electron in the outer shell (what is that one called?) would be the weakest bonded, and the most prone to getting "increased" by a photon. So then, it is so that the K and L shells get preferentially excited in to new energy states?

2. If we consider (say) tungsten. Might an incoming photon cause 8.4KeV and 9.7KeV responses from the L-shell, even though there is a 59.3KeV shell right there? The incoming is at 59.54 KeV.If there is enough energy, does the inner K-shell get to glow first? Is it that the photon might energize electrons shell without any preferred order, just randomly?

Yeah - I know.. truly ignorant! I do have an actual cause to ask the question, related to figuring out whether we can expect returns from pretty near all the heavy end elemets, relying on their L-shells alone. We would be able to guess at the alloys, and tell whether there is lead in the free machining steel, or thorium in the TIG rod.

Data sheet regardless, who knows what happens if you show it some uranium ore from the back yard. I somehow doubt that all it would show is a 13.6KeV and a 17.2KeV count. That would be a bit like visiting Chernobyl to get a sun tan!

 

[homebrewed]

Another fool? bought the Pocket Geiger. @homebrewed what did you end up using on the Pocket Geiger board? I think you replaced the TIA, but I don't recall your final bits and pieces. Do you have a current marked up "working" schematic? I have a spare ELS PCB with a Teensy 4.1 and display pre-wired that I can hack. Might help me get up and running, I hope. Have some old smoke detectors that haven't been discarded (I think).

See attached, in combination with the signal conditioning circuit I posted a little while ago.

 

[WobblyHand]

See attached, in combination with the signal conditioning circuit I posted a little while ago.

Thanks. Think I can cobble something together. That 0-27V supply must be pretty quiet. I don't have any low noise supplies, will have to make up a battery pack for the bias. Going to have to get some more 18650's.

 

[homebrewed]

Thanks. Think I can cobble something together. That 0-27V supply must be pretty quiet. I don't have any low noise supplies, will have to make up a battery pack for the bias. Going to have to get some more 18650's.

The bias voltage is pretty heavily filtered by that 100K resistor & 1uF capacitor. The capacitor is a ceramic so there also could be some microphonics that would contribute some noise. I need to look into that since the laptop is sitting on the bench, along with the detector board.

I'm not seeing any evidence of 60Hz (or 120Hz) noise on the output of my signal conditioning board.

 

[graham-xrf]

See attached, in combination with the signal conditioning circuit I posted a little while ago.

Hi Mark
I know you are getting great pulses, and working on the processing, and you probably don't want to be soldering stuff there right now, but just looking at it, I see a circuit aiming at a transimpedance gain of 6.6e+7 followed by a another 40dB (x100) voltage gain. Then in the signal conditioner, there are more opamps, with another non-inverting 20dB (1+90) voltage gain, and I guess offset adjustment.

The TI opamp LMC662 has a Gain-Bandwidth Product of 1.4MHz, and a voltage gain of 126dB (19.95 million)

With input bias current of 2fA, and an input resistance >1Tera-Ohm, it would make a fantastic electrometer, pH meter, or ECG amplifier, but here I have to admit that this circuit goes beyond my experience. Even currents across circuit boards insulation would be quite tricky to get lower than that! The data sheet has a major section on guard rings explanation and layout examples for the PCB if using LMC662.

Thus if the open loop gain is 20 million, then it's hard see how an attempt at 66 million through the feedback resistor in that one stage can work. Yet, it somehow does, because you have pulses!

The 66MegaOhms is going to make noise, although in a transimpedance amplifier, it does not add in in quite the same way as a voltage amplifier. The input referred current noise is only 0.0002pA/√Hz.
--------------------------------------
Even so, I calculate
Noise V = √(4**k*T*B*R)
Plugging in 298K, and Boltzmann's Constant, and 10kHz bandwidth and 66 Mega Ohms, we get the result 104uV at the input.
That is more than I expect to see even as power rail ripple.
-------------------------------------
Then, with GBW at 1.4MHz, as I mentioned way back, the value is less than the gain. Thus I think the bandwidth is fractional Hertz, and I end up on unsteady ground - because you have pulses.
-----------------------------------
The input bias current is so low that you can use high value resistors, but even so, I think you would get a huge performance increase if you redistributed the gain. If you make Rf (say) 1MegOhm, and made R6 (say) 1K and R6 (say) 220K, and then add gain in the signal conditioning amplifier if you need to. Although this total is much less gain than shown in your circuit, it is still huge.
Even with the "low" gain, it may still be way too much!
1e6 x 220, x 10 = 2.2 Billion.

If you had a full 2V at the ADC to count, then it would have started out as 0.9 pA
So I revise the Rf to 470K. At least get a bandwidth about 3Hz. Make Rf yet lower if you like.
-----------------------------------
The input bias current 2fA may be spectacular low , but the input offset voltage is a whole 6mV (if using LMC662C).
The capacitor isolated U1B turns this into 600mV DC, which needs another C3 capacitor in the signal conditioner to lose it, and I am guessing that R11 is used to adjust the offset out.
--------------------------------
Even with using a slightly reworked Pocket Geiger tracking, and some different opamps, I am thinking you could have a circuit that works just like mine simulated. What I have no answer for is that somehow, you get pulses, and I cannot figure how that happens.
The thing is, you have it powered, and it does stuff.

 

[graham-xrf]

Hi Mark
Recall your integrator from the past?
On the last page of the TI datasheet for LMC662, I see this trick in use to drive the offset voltage out of the op-amp's output.
Is that something like what you used to do?

 

[rwm]

Perhaps for @RJSakowski : ?
You have provided the answer that an incoming photon energizing an atom of some stuff, if it has energy "left over" can then continue to potentially energize some other electron shell in the same atom, or go on to hit some other atom.

I feel foolish to ask about some "what if" scenarios.

1. Suppose an incoming 60KeV photon hits (say) a atom of gold. There is not enough energy in it to excite either of the two electrons in the K-shell. Therefore one supposes, since the photon hit the atom already, we might still hope to get some 9.7KeV and a 11.07KeV out of the L-shell, and have some 38.4KeV continue onward to get some responses from other atoms it runs into. Gold has six shells, filled as 2, 8, 18, 32, 18,1. I would have thought the lone electron in the outer shell (what is that one called?) would be the weakest bonded, and the most prone to getting "increased" by a photon. So then, it is so that the K and L shells get preferentially excited in to new energy states?

Graham- This is an interesting question. I will take a stab at it but I reserve the right to be wrong!
In short, I think the answer lies in the probability of photoelectric interactions.
"In X-ray absorption spectroscopy, the K-edge is a sudden increase in x-ray absorption occurring when the energy of the X-rays is just above the binding energy of the innermost electron shell of the atoms interacting with the photons. The term is based on X-ray notation, where the innermost electron shell is known as the K-shell. Physically, this sudden increase in attenuation is caused by the photoelectric absorption of the photons. For this interaction to occur, the photons must have more energy than the binding energy of the K-shell electrons (K-edge). A photon having an energy just above the binding energy of the electron is therefore more likely to be absorbed than a photon having an energy just below this binding energy or significantly above it." (Christensen's Physics of Diagnostic Radiology)
Also, note the sample below. Yes there is a low level of background interaction, but there are much stronger absorption lines at the K alph and K beta energies.

I also think this is why you prefer a monochromatic or narrow energy beam rather than a wide spectrum of incoming xrays to excite the sample.

 

[graham-xrf]

A 3 Op-Amps PCB - $29.42 from Mouser (for the op-amps we use)

If anyone wants to have a ready-made transimpedance amplifier PCB that may have the potential to let one play, there is this..


I have included the circuit diagram, and all information I could grab. (DC2414A.zip)
Given that it comes unpopulated, it may not be a bargain if you are going to have an uphill task soldering the bits on. Most parts are 0603, but there are two components that are 0402.

I have not checked it out in detail, but it looks like it only uses those tiny TSOT-23-6 packages with only one opamp in them. Even so, one of these may be all anyone needs. I fished the circuit PDF out of the pack, and posted here.