Needing more than a spark test?

I have a bit more information plus a thought regarding that ever-present low-energy peak in the spectrums I'm getting so far. The new information: I cut a small piece of 1/16" thick lead sheet and taped it over the aperture hole in my x-ray source assembly. The higher-energy bump disappeared. Right now the overall count rate is well below 1 count/second. Closer to 1 every 10 seconds! That is, if my code is doing what I think it is :). It's probably time to upload the latest version(s) of my pulse-processor code and MCA library so folks can look it over for misteaks <joke>.

Anyway, the thought I had is that at least some of the low-energy peak most likely is due to wideband noise coming from the detector. The TIA and low-pass filter implemented by the second opamp in the pocket geiger circuit will convert any relatively fast current pulse coming out of the PIN diode -- regardless of how it was generated -- into a waveform that looks pretty much like a pulse generated by an xray photon.

This isn't necessarily a problem. If our incoming XRF photons are significantly more energetic than those phantom noise pulses, the MCA should be able to distinguish them.
 
I think the purpose is to avoid overdriving the ADC. If you look at my signal conditioning board you will see I have a resistor + diode hung on its output. The diode is connected to the Teensy's +3.3 so it's a clamp, designed to accomplish something similar.
I can understand a clamp on the ADC. Or an offset shift. Does the first stage, the TIA need this offset supply? +2.5, -0.8?
I'm asking about the output of the diode and the TIA. It's the most sensitive stage and mostly sets the noise figure. Or at least it should.
 
It is a circuit I have used before, and I completely understand how it can confuse. Imagine temporarily that the output is not connected to 0V, but instead is called Vout. Then imagine what is called 0V is indeed connected to the "0V". We then have a simple circuit where "0V" is the negative supply for the opamp, 3.3V is the positive supply. and the voltage divider delivers +0.8V, which is buffered by the opamp unity gain follower.

Now if we "disconnect" the 0V from the bottom of the 3.3V supply, and connect the output of the opamp to that 0V, the circuit still operates in exactly the same way, except instead of asserting that the opamp output is +0.8V, it has to assert that V+ is 2.50 volts, and that V- is -0.8V, up to the limit of the current the opamp output is capable of giving. It becomes a regulator for the TIA low current circuits.

The key thing is that the 3.3V supply negative end, if it had a 0V from anywhere else, is not the same 0VA as we are using for the TIA.
The scheme works well, but has it's inconveniences. You get the same result using two separate regulators, but the need is to contrive a no-noise negative supply. The really important thing is that no signal drives or powers the ADC above +2.5V. It's absolute maximum is 2.7V, and not recommended ever.

You may be right that the good way is just to use batteries.
Your circuit reminds me of an old audio amplifier design that used the opamp in a sort of current-output mode. The output of the amplifier was connected to ground through a relatively low-value resistor, and current variations in the opamp's Vcc/Vee pins became the drive current to a pair of complementary transistors. It worked well because the current bandwidth of inexpensive opamps often is much higher than their voltage bandwdith.

A blast from the past. A fun variation I made was to use a dual opamp and I used the second one as an integrator to drive the overall amplifier offset to zero. The reason it worked: the supply currents are common to both amplifiers.
 
I can understand a clamp on the ADC. Or an offset shift. Does the first stage, the TIA need this offset supply? +2.5, -0.8?
I'm asking about the output of the diode and the TIA. It's the most sensitive stage and mostly sets the noise figure. Or at least it should.
I don't think so. The TIA is like other opamp circuits, it just acts to force the differential voltage between its + and - inputs to zero.

Some circuits might be sensitive to the fact that the feedback _also_ enforces a specific voltage on the inverting input, which could change the effective bias on the detector diode. That's not an issue for our PIN diode but it could be a factor when using an avalanche detector or a SiPM.
 
I still have concerns about the AC-coupled nature of the PIN diode in my circuit.

All my simulations show that the pulse does not pass through into the TIA in the way a voltage amplifer would have it, losing it's base line, and settling to an area average. It acts like a DC couples amplifier. Also, one opamp is comparing the the input to zero, and driving to make the offset be zero.

I have trawled all the TIA circuits I could find, and there are a lot! First we take out all those operating in zero bias photovoltaic mode. Of the remainder, where I find the question of biasing the diode, or persuading the TIA amplifier itself to provide the bias, they always get into complications, or don't address the question properly.

The capacitor has to be about 1.5nF. Any bigger starts losing the gain, and waveshape. Less than about 800pF also spoils things. It seems robust, and non-critical. Also, we dont get any excursions below the base line.

I know now it is time to build some, and get it working. I have to have enough options on the board to be able to change tack if I run into trouble. I think I have wrung all I can out of simulations.
 
A blast from the past. A fun variation I made was to use a dual opamp and I used the second one as an integrator to drive the overall amplifier offset to zero. The reason it worked: the supply currents are common to both amplifiers.
Your blast from the past is at work here. If you look at one of my earlier TIA circuits, you can see there is the integrator U1 at the bottom of the circuit, driving U3 to remove the offset, so that the J1 FET gate will stay at 0V. Sorry about the image size. You have to hit "zoom" to see it.

JFET TIA- 45pA input.png
 
@WobblyHand
Re: your question about "does it need -0.8V" ?
The 0.8V is what just what is left over when you regulate to take 2.5V out of a 3.3V total.
The idea was to make a 2.50V supply, so that signals would be limited to that.
The leftover -0.8V does have the handy serendipity of allowing the inverting opamps to never need to hit the negative rail.
 
I still have concerns about the AC-coupled nature of the PIN diode in my circuit.

All my simulations show that the pulse does not pass through into the TIA in the way a voltage amplifer would have it, losing it's base line, and settling to an area average. It acts like a DC couples amplifier. Also, one opamp is comparing the the input to zero, and driving to make the offset be zero.

I have trawled all the TIA circuits I could find, and there are a lot! First we take out all those operating in zero bias photovoltaic mode. Of the remainder, where I find the question of biasing the diode, or persuading the TIA amplifier itself to provide the bias, they always get into complications, or don't address the question properly.

The capacitor has to be about 1.5nF. Any bigger starts losing the gain, and waveshape. Less than about 800pF also spoils things. It seems robust, and non-critical. Also, we dont get any excursions below the base line.

I know now it is time to build some, and get it working. I have to have enough options on the board to be able to change tack if I run into trouble. I think I have wrung all I can out of simulations.
I looked at this particular issue soon after reading through the Theremino stuff. They claim to use a pole-zero cancellation network to address it.

The undershoot is a consequence of the coupling capacitor charging up a little during the pulse, so when the voltage going into the capacitor declines to zero the built-up charge on the capacitor requires that the output side has to go below the quiescent set point. That, actually, is the main reason I used a really long time constant for the C-R input network on my signal conditioning board -- if the RC time constant is very long compared to the pulse duration the undershoot should be close to zero. However, the pocket geiger's lowpass filter network takes no such precaution. My simulations show that increasing the value of some of the low-pass elements on the pocket geiger board reduce the undershoot, but at the cost of changing the filter cutoff frequency. Options are few here.

I'm wondering if the undershoot may be a non-issue if we're integrating the pulse rather than performing a peak-find using interpolation as described by the Theremino folks. One potential "fly in the ointment" is my current pulse-qualification code, which sets pulsewidth based on excursion above the quiescent baseline -- so voltages that go BELOW the baseline aren't integrated. This whole situation probably requires some simulation data plus a standalone program to see how it all works out. Yes, a full analytical approach is possible but I think it could be pretty messy. If there are any takers, go for it!

An alternative might be to use LTspice to do it. Use a switch to route pulse voltage into an integrator. The switch would be gated using various threshold voltages as chosen by your favorite theory :).

Finally, I'm wondering if errors due to that pesky undershoot might not just become a simple offset that could be compensated for.
 
I looked at this particular issue soon after reading through the Theremino stuff. They claim to use a pole-zero cancellation network to address it.

The undershoot is a consequence of the coupling capacitor charging up a little during the pulse, so when the voltage going into the capacitor declines to zero the built-up charge on the capacitor requires that the output side has to go below the quiescent set point. That, actually, is the main reason I used a really long time constant for the C-R input network on my signal conditioning board -- if the RC time constant is very long compared to the pulse duration the undershoot should be close to zero. However, the pocket geiger's lowpass filter network takes no such precaution. My simulations show that increasing the value of some of the low-pass elements on the pocket geiger board reduce the undershoot, but at the cost of changing the filter cutoff frequency. Options are few here.
Yes - I agree entirely, and I am puzzled why I can't see "undershoot". I have very carefully simulated, down to pico-aps, all that a PIN diode can deliver from "pulses", into an entirely artificial load. It does require re-working a whole lot of LTSpice default settings.

I think that if the pulse charge is small enough, and the coupling capacitor is orders of magnitude larger than the PIN diode reverse capacitance, it cannot "charge" the coupling capacitor. The coupling capacitor looks like a short-circuit straight into the TIA input. It only takes about 6fA into that 500MHz GBP to provoke it into cancelling the little shot of current
I'm wondering if the undershoot may be a non-issue if we're integrating the pulse rather than performing a peak-find using interpolation as described by the Theremino folks. One potential "fly in the ointment" is my current pulse-qualification code, which sets pulsewidth based on excursion above the quiescent baseline -- so voltages that go BELOW the baseline aren't integrated. This whole situation probably requires some simulation data plus a standalone program to see how it all works out. Yes, a full analytical approach is possible but I think it could be pretty messy. If there are any takers, go for it!
I think I may have already been a taker, and gone for it hard, and it was messy!

Preserving the resolution
At the small signal stage, well before we are attempting to measure it's energy analogue by any kind of final integration, this attempt is to faithfully preserve that wave shape, with it's precious information. I don't yet know if it is unrealistic to expect that we could count up (say 20) places across the waveform, and add them up, and hope to see a bucket filling with 5.5KeV as separate to a bucket filling with 5.9KeV for (say) chromium, but it is what I hope for. I do feel we won't have much chance unless that pulse is preserved.

Low Pass Filtered Versions

I accept that low-pass filtered versions might possibly retain the energy analogue, or enough of it to be useful, so long as the poles of the filtering do not get on top of each other. I fully understand that the prime attraction for this is that it provides more time to process the slower pulse through slower ADC samplings, at the expense of a pipeline delay that may let other pulses that might have happened go by without getting looked at. That is not really a big deal. We can just sample for longer. I simply liked it better that the pulse we have takes a time which depends on the PIN diode capacitance, and so I chose to sample fast enough to measure that.

Arguably, if we allowed a small amount of Butterworth type filter, we could let the 13uS pulse become (say) 50uS, and get a finer resolution measure of the energy under it. That was a passing thought, but I also thought that if I got 20 samples of the original, then it left room to play with software higher resolution sampling later, if slowed down versions turned out to be a raging success!
An alternative might be to use LTspice to do it. Use a switch to route pulse voltage into an integrator. The switch would be gated using various threshold voltages as chosen by your favorite theory :).

Finally, I'm wondering if errors due to that pesky undershoot might not just become a simple offset that could be compensated for.

When I reduce the coupling capacitor, in simulation, to anything less than about 800pF, then I do start to see the unwanted effects on the waveform. We will never go to that 100pF or so as seen on the Pocket Geiger. The Thermino "signal conditioning circuit" using two BCW60B transistors, working on a 200mV pulse, is also, in my view, pretty useless.

Working with electrons
Simply having a pile of electrons that resulted from a X-ray photon arriving into the PIN diode is not enough to get a grip on it. Those electrons are not yet a current. One needs to introduce the per second time over which they made a capacitance arrive at a voltage from that charge.

Every equation and physics analysis I found, (or watched Indian University lectures on YT) always involved a rate of arrival, meaning a radiation flux. Photons per second. There was nothing on how to figure how much pulse current would happen from a single photon.

The answer was to calculate via the energy known to be in there. That is how I came up with the currents pulse estimates.

It should be OK

Regardless of what the actual pulse might or might not do using my low noise TIA, I expect I will have a way. Even if I have to use the DC coupled TIA, and find a way to lose the inevitable offset. For now, I am happy that my most careful simulations do not show a problem, provided that first TIA current comes in via a capacitor so big that it cannot start storing significant charges of it's own.
 
I need a noise free negative supply
Deciding that the pristine +2.5V supply for the critical ADC voltage AVDD is the ideal safe one to be using for the positive supply to any op-amps driving it, we have the need for a similar, pure-as driven snow, negative supply.

Just enough to allow that a true zero volts can be had at an output without hitting the negative rail. The op-amps can take 5.25V total.
Some kind of charge pump thing?

OK then, maybe a simple 1.5V AAA battery :grin:
 
Last edited:
  • Like
Reactions: rwm
Back
Top