Author Topic: Duane-Hunt limit  (Read 6155 times)

Nicholas Ritchie

  • Professor
  • ****
  • Posts: 141
    • NIST DTSA-II
Re: Duane-Hunt limit
« Reply #30 on: February 12, 2024, 06:08:23 PM »
We know from electron energy loss spectroscopy (EELS) that the electron source is essentially monochromatic from the perspective of X-ray analysis.  Today, monochromated cold field emission beams are used which can produce beams with milli-eV energy spreads.  However, early EELS was performed with thermal sources which had resolutions on the order of 1 eV.  This is the number we should compare to.

Do you have any literature references for the thermal sources that you can share with us?

K. Kimoto, G. Kothleitner, W. Grogger, Y. Matsui, F. Hofer Micron, 36 (2005), p. 185

Quote
The bandgap is one of the principal properties of electronic materials. Electron energy-loss spectroscopy (EELS) in transmission electron microscopy (TEM) gives information on the direct and indirect transitions with high spatial resolution (Egerton, 1996). There are several reports on bandgap measurements of insulators and semiconductors (Batson et al., 1986, Bangert et al., 1997, Rafferty and Brown, 1998, Batson, 1999). Since a bandgap measurement requires high energy resolution, the previous authors employed a cold field-emission gun (CFEG), whose energy spread (>0.25 eV) is smaller than that of a Schottky emitter (>0.5 eV) and those of other thermionic electron sources (>1 eV). Recently, several types of monochromator for a TEM have been developed to reduce the energy spread (Terauchi et al., 1999, Tiemeijer, 1999a, Kahl and Rose, 2000, Mook and Kruit, 2000, Tanaka et al., 2002), and some of them are now commercially available. Several authors have demonstrated the efficiency of monochromators (Terauchi et al., 1998, Terauchi and Tanaka, 1999, Mitterbauer et al., 2003). There is a pioneering study for bandgap measurements (Lazar et al., 2003), but the previous papers mainly focused on core-loss spectroscopy or valence-band spectroscopy.
"Do what you can, with what you have, where you are"
  - Teddy Roosevelt

Probeman

  • Emeritus
  • *****
  • Posts: 2823
  • Never sleeps...
    • John Donovan
Re: Duane-Hunt limit
« Reply #31 on: March 06, 2024, 09:33:32 AM »
I just thought I would follow up with some WDS and EDS measurements I did last weekend on Si, Ge and W at 10 keV, looking specifically at the Duane-Hunt limit.

Each of these scans (simultaneous WDS and EDS) took ~7 hours (300 points at 80 sec) for each sample. The EDS had a dead time of ~28% at a 2000 nsec pulse processing time.

Here's the EDS data:



And here's the WDS data:



Due to the convolution by the detector spectral resolution (as mentioned previously by SEM-geologist) the extrapolated Duane-Hunt limit appears to be higher than the expected value of 10 keV.

Of course it could be that our high voltage power supply is also not calibrated accurately.

We need a better method to determine the actual value of of electron beam energy, perhaps using the method suggested by Nicholas Ritchie where we measure a characteristic emission line at various over voltages from 1.5 and down close to the edge energy and see where that trend approaches zero as shown here (from Edax):



For example, for calibrating our beam energy at say 5 keV we could use Ti K edge at 4.967 keV and measure the Ti Ka emission intensities at say 8 keV, 7 keV, 6 keV and 5 keV (the last one being problematic to say the least!).

For calibrating at 10 keV we could utilize Zn Ka (edge energy 9.66 keV) and at 15 keV we'd have to use EDS and perhaps the Rb Ka (in RbTiOPO4 synthetic) with an edge energy of 15.2 keV.
The only stupid question is the one not asked!

Probeman

  • Emeritus
  • *****
  • Posts: 2823
  • Never sleeps...
    • John Donovan
Re: Duane-Hunt limit
« Reply #32 on: March 13, 2024, 09:12:18 AM »
I think we can explain what we are seeing in the above Duane-Hunt limit measurements.

Here are some continuum models using PENEPMA which show two different artifacts which occur together in our empirical measurements. First a "binning" effect where the last several bins (just under the electron beam energy) will always have positive intensities (with a long enough integration time), so the trend never extrapolates exactly to zero at the beam energy as seen here:



And second a convolution effect due to detector spectral resolution (which spreads out the photon energies) as seen here using the same model data as above:

The only stupid question is the one not asked!

Probeman

  • Emeritus
  • *****
  • Posts: 2823
  • Never sleeps...
    • John Donovan
Re: Duane-Hunt limit
« Reply #33 on: March 24, 2024, 01:34:13 PM »
So here's the basic problem with trying to determine the accuracy of one's electron beam energy using the Duane-Hunt limit:

First, even with WDS, the spectral resolution using an LiF crystal (using the Ge Ka emission line of ~9.87 keV), is still around 100 eV (FWHM) as seen here measured at 15 keV in Ge metal:



Running the same region at 10 keV (below the Ge K edge energy of 11.1 keV), we see both the binning artifact and the spectra resolution artifact imposed on the continuum (100 nA, 120 sec per wavescan point):



Compare this measurement to the continuum intensities modeled using PENEPMA in the previous post above.

This is why we need a better method to estimate our electron beam energy accuracy.  Here are some preliminary measurements I performed on Zn metal using a range of accelerating voltages from 10 to 15 keV this weekend (three points per keV):



Extrapolating the curve it appears the net intensity does *not* intercept zero exactly at the Zn Ka edge energy... is this evidence that our electron beam energy is slightly higher than the nominal value reported by the instrument?.
The only stupid question is the one not asked!

sem-geologist

  • Professor
  • ****
  • Posts: 300
Re: Duane-Hunt limit
« Reply #34 on: March 25, 2024, 03:46:57 PM »

Running the same region at 10 keV (below the Ge K edge energy of 11.1 keV), we see both the binning artifact and the spectra resolution artifact imposed on the continuum (100 nA, 120 sec per wavescan point):



Compare this measurement to the continuum intensities modeled using PENEPMA in the previous post above.

What intrigues me is that on WDS we see some background much above 10.1kV (ignoring the questionable region where Duane-Hunt limit could be). Then I see that Your data (which You had sent in mail thread) and mine WDS scans shows that this faint background over Duane-Hunt limit seems clearly dependant on average atom number. The higher atom number - the higher this fain background intensity is. Differently to EDS there (on WDS) is no pulse pile-up to blame. I see no physical possibility to create such artifact, and maybe we should consider it is not artifact, but a tiny window showing what some rare electrons (from massive stream of electrons in the beam) goes through.  Another clue is PENEPMA simulations. Abrupt cutoff at set beam energy is probably artificial implemented limitation in the modeling to not produce any X-rays above Duane-Hunt limit. But intensities at last channel below Duane-Hunt limit in those simulations, and its clear dependency to material average atom number, hints that this cut off at Duane-Hunt is rather artificially forced in the modeling.

So are there some electrons accelerated more than set acceleration voltage - I doubt it. But I think there is important concept missed here:
1) Bremstrahlung is not just from breaking (Stopping) of electron - it is inertia breaking radiation - be it acceleration, de-acceleration or deflection. Observed bremstrahlung radiation does not directly map to the kinetic electron energy
2) if deflection of electron is more than 90 degree (thus a bit backwards than initial trajectory) - the difference in ripple of updated electric field radiating from the charged particle will have shorter wavelength than it would be if halted to 0 velocity (in such complete halt case giving bremstrahlung radiation at Duane-Hunt limit). In case the electron would be ricochet in perfect 180 degrees - the wavelength of ripple would be double to that of halting it to 0 velocity. What Do I mean with ripple in changed electric field? Such ripple would be virtual photons as I had shared some interesting concept found in internet in some previous post:
Here, I find this material quite complementing my understanding on Bremstrahlung:
https://physics.stackexchange.com/questions/186361/why-does-accelerating-electron-emits-photons
and especially this animated particle acceleration looks like a picture worth many thousand words:

The red dot represents a charged particle, where lines represent the field, the ripple in field is seen by us as some electromagnetic radiation.
Source and further explanations: http://www.tapir.caltech.edu/~teviet/Waves/empulse.html
In most of our cases electron has many small deflections (As we know from MC simulations) before loosing completely its velocity and thus many much lower (than set beam energy) continuum radiation will be observed from those numerous deflections. But if we send millions of electrons in short time-span to relatively same spot, some will have opportunity to get deflected backwards at the first interaction with matter which (following the pseudo-photon/e-field ripple concept) will produce higher energy pseudo-photon than energy of the beam electron. The denser the matter - the greater the chance to get such first-interaction backward deflected electrons. Additionally - lets remember that other electrons from the beam do not instantaneously magically dissapear after hitting the matter, but after loosing its velocity (energy) they need to be drained from the hit sample surface back to the HV generator, but as there is resistance on the surface - it takes some time (impedance all the way from the spot-coated conductive surface of the sample, stage contacts, wires, picoamp, HV generator...)... and so some "unlucky" electrons from the beam with full velocity has higher chances to hit such "free" drifting electrons which are traveling outward to ground and in some rare circumstanced such beam electron can be deflected backwards.
 
One more detail to not overlook - Electron has nearly no mass and reacts to electric field near instant (that is why waveguide - aka conductor wire and its dialectric space allows sending signals nearly at speed of light, where electrons travel only some cm/s on conductor surface). Such stunts are much harder to make with positive ions which have mass (i.e. FIB) (PIXE- needs few MeV acceleration).
....
I just got one big extremely crazy idea, which probably wont work miserably...
« Last Edit: March 25, 2024, 03:57:14 PM by sem-geologist »

Probeman

  • Emeritus
  • *****
  • Posts: 2823
  • Never sleeps...
    • John Donovan
Re: Duane-Hunt limit
« Reply #35 on: March 30, 2024, 03:22:15 PM »
Last weekend I again did some overvoltage measurements on some pure metal standards, but this time using 0.2 keV increments above the edge energies as proposed by Nicholas Ritchie (using the net intensities avoids the detector convolution issues of the Duane Hunt limit determinations we have seen in this topic). Here are Ti K edge net intensity measurements:



Can we say something about the accuracy of the electron beam energy at 5 keV on this instrument?   :-\

I think next I need to try 0.1 keV increments and this time I will polish the metal standards to remove the carbon coating which at very low overvoltages could affect this measurement.

By the way, Donovan mentioned to me that there is now a new right click menu item in Probe for EPMA for easy export of these over voltage measurements which do not require primary standards:

The only stupid question is the one not asked!

Sander

  • Post Doc
  • ***
  • Posts: 11
Re: Duane-Hunt limit
« Reply #36 on: April 06, 2024, 01:50:09 AM »
I wanted to weigh in on this discussion, if only to sharpen my own understanding of things.

First, the Wehnelt voltage.  I don't think this should really be subtracted from the initial HV.  I think the Wehnelt determines how many electrons emerge from the gun and from which part of the crystal/tungsten filament, but not their final energy.  Typically, the Wehnelt voltage is relative to the emitter, so whereas the electrons emerging from the filament see a decelerating field on their way towards the Wehnelt, they are accelerated again by the same amount after leaving the Wehnelt aperture.  I think of it this way: When you consider the whole gun module as a "closed system" ensemble which is at voltage E0, then electrons coming out of it will have that potential, regardless of what's happening inside that ensemble.

I tried to do some bookkeeping of all the energy deltas.  Here's my list, feel free to comment.

First, the E0 itself (as set by the microscope) will likely have some (systematic) "setpoint error", for example because the voltage entering the HV cascade is determined by some DAC with a finite number of steps.  The "problem" here is that this will then be a fixed percentage of the "full range", so if your HV subsystem can go up to 30kV and uses a 10-bit DAC to set the value, you're looking at a discretization error of ~30V just from this.  The HV supply should also specify a certain maximum "ripple", which is typically an order of magnitude better than this (because it would otherwise give bad ("fluctuating") imaging results).

There may be a small voltage that needs to be subtracted from the supplied HV because that's applied to the filament itself to heat it up, but that would be just a few volts.

Then, there will be a spread in the energy with which the electrons escape from the filament or crystal.  This is worst for tungsten (because you typically heat it up the most) and best for a cold FEG.  Regardless, it will also be only a few eV, so I don't think this should worry us.  This is one of the causes of chromatic aberration which is why microscope builders try to keep it as small as possible, or even insert monochromators for ultimate imaging performance (at the cost of intensity, so I suppose nobody here would use them unless you're also doing EELS).

Then, electrons can also undergo further ("perpendicular") acceleration inside the scanning column, so that increases their energy as a function of their final distance from the optical center.  In case you're scanning a big field of view, this may be significant.  Speaking for the microscope my team is producing, we're scanning electrostatically with a few hundred volts on the scanning poles, so my guess is that electrons aimed for the edges of the maximum field of view have their energy increased by the same amount (and worse: that means the electron energy varies over the image - interesting when doing mapping!)

Finally, I think in many cases there will be some charging happening at the sample surface.  This is the hardest to quantify.  But unless you're analyzing some piece of perfectly grounded metal, I'm sure there will be some equilibrium between the electrons you're blasting onto a single tiny point on the sample and them flowing away to earth again.

And when the X-rays are finally being created in the sample, there is (as already mentioned here) the gaussian spread in the detector itself (and the resulting "convolution" of the measurement) and I think that should not be underestimated.  After all, even for a perfect SDD with a MnKa FWHM of 119 eV, the FWHM of a 15kV peak would be something like 190eV.  So, I think that even if "perfect" 15000.0eV electrons are hitting your specimen, you should not be surprised to find some signal in your spectra a few hundred eV higher than that.

By the way, this is all not to say that I don't find the whole bremsstrahlung phenomenon super weird.  Since we are seeing photons with energies up to E0 in our spectra, that means that these were emitted by electrons losing all their energy during deceleration (so far, I'm nodding) but since each photon corresponds to a single quantum mechanical event, that must mean that some electrons come to an abrupt stop from full speed in one infinitesimal event.  How?! 

Wasn't it Bohr himself who said "Anyone who is not shocked by quantum theory has not understood it" - I'm definitely still in the "shocked" phase.

Probeman

  • Emeritus
  • *****
  • Posts: 2823
  • Never sleeps...
    • John Donovan
Re: Duane-Hunt limit
« Reply #37 on: April 06, 2024, 08:34:49 AM »
I wanted to weigh in on this discussion, if only to sharpen my own understanding of things.

I'm with you here.  I still consider myself a student because I try and learn something new every day!

And when the X-rays are finally being created in the sample, there is (as already mentioned here) the gaussian spread in the detector itself (and the resulting "convolution" of the measurement) and I think that should not be underestimated.  After all, even for a perfect SDD with a MnKa FWHM of 119 eV, the FWHM of a 15kV peak would be something like 190eV.  So, I think that even if "perfect" 15000.0eV electrons are hitting your specimen, you should not be surprised to find some signal in your spectra a few hundred eV higher than that.

Yes, this convolution effect is the main reason we don't see the so-called photon "shelf" or "cliff" artifact that we've been seeing in PENEPMA I think. Here is a simulation from PENEPMA that is not convolved to the detector resolution:



And here convolved:



The photon cliff is completely obscured.  Yes, yet one more reason not to trust the Duane-Hunt limit to check the accuracy of one's electron beam energy!

By the way, this is all not to say that I don't find the whole bremsstrahlung phenomenon super weird.  Since we are seeing photons with energies up to E0 in our spectra, that means that these were emitted by electrons losing all their energy during deceleration (so far, I'm nodding) but since each photon corresponds to a single quantum mechanical event, that must mean that some electrons come to an abrupt stop from full speed in one infinitesimal event.  How?! 

Wasn't it Bohr himself who said "Anyone who is not shocked by quantum theory has not understood it" - I'm definitely still in the "shocked" phase.

Yeah, I'm always shocked!   And it is weird that an electron could come to a complete halt. Here is a figure that a colleague of mine in the Physics Department at the University of Oregon , Andrew Ducharme, wrote me recently in discussing the above photon "cliff":

Quote
The established physics is that bremsstrahlung cross-sections are nonzero for complete energy losses, or for production of photons with the beam energy E (see figure from Penelope manual), while the probability of producing a photon with energy E + 1 eV is 0. Wait long enough (assuming ideal detector, electron source) and you will count photons at the Duane-Hunt limit while never getting counts in the energy bin of E + 1 eV. The longer you wait, the more DH-energy photons, and the higher the cliff.



I think of it in the sense that the coulombic field of an atom is actually a pretty big target, so a few electrons are going to hit some atoms directly right at the surface...

Here's a question I have: I was recently chatting with someone from the industry and they said that they only use the DH limit as a crude measure of the electron beam energy, say to check for sample charging (as you mentioned). And that in fact, on their high end FEG electron beam instruments they utilize a special electron detector situated inside the electron column (gun?) so that when the objective lens is adjusted to reflect electrons back up the column, they use this detector to measure the electron energy with extreme precision and accuracy. He mentioned that it was based on a "Wine" detector (pronounced "vine" in German?) but as it's a trade secret he couldn't say more.

Does anyone here know anything more about this electron beam energy measurement system?
« Last Edit: April 06, 2024, 04:21:50 PM by Probeman »
The only stupid question is the one not asked!

yuji

  • Student
  • *
  • Posts: 1
Re: Duane-Hunt limit
« Reply #38 on: April 07, 2024, 06:33:24 AM »
Wien filter?
It is used as monochromater in TEM.

Probeman

  • Emeritus
  • *****
  • Posts: 2823
  • Never sleeps...
    • John Donovan
Re: Duane-Hunt limit
« Reply #39 on: April 07, 2024, 07:41:50 AM »
Wien filter?
It is used as monochromater in TEM.

Wien filter pronounced like "vine" in German?  That sounds about right.

This person mentioned that this Wien (he called it a detector) could be used to calibrate the electron beam energy of an SEM (and possibly a microprobe). Does that sound right?

Looking through the literature I can't find anything specific to using Wien device for electron beam energy calibration except maybe for very low beam energies...
« Last Edit: April 07, 2024, 12:05:14 PM by Probeman »
The only stupid question is the one not asked!

John Donovan

  • Administrator
  • Emeritus
  • *****
  • Posts: 3264
  • Other duties as assigned...
    • Probe Software
Re: Duane-Hunt limit
« Reply #40 on: April 08, 2024, 07:32:52 AM »
Then, there will be a spread in the energy with which the electrons escape from the filament or crystal.  This is worst for tungsten (because you typically heat it up the most) and best for a cold FEG.  Regardless, it will also be only a few eV, so I don't think this should worry us.  This is one of the causes of chromatic aberration which is why microscope builders try to keep it as small as possible, or even insert monochromators for ultimate imaging performance (at the cost of intensity, so I suppose nobody here would use them unless you're also doing EELS).

Here is a reference Nicholas Ritchie posted above for the energy spread of an electron beam:

K. Kimoto, G. Kothleitner, W. Grogger, Y. Matsui, F. Hofer Micron, 36 (2005), p. 185

By the way, here is one (quite old) paper for measuring electron beam energies:

Schulson, E. M. "Method for Measuring the Incident‐Beam Energy in Scanning Electron Microscopy Using Electron Channelling Patterns." Journal of Applied Physics 42.10 (1971): 3894-3899.
John J. Donovan, Pres. 
(541) 343-3400

"Not Absolutely Certain, Yet Reliable"

Probeman

  • Emeritus
  • *****
  • Posts: 2823
  • Never sleeps...
    • John Donovan
Re: Duane-Hunt limit
« Reply #41 on: April 08, 2024, 09:10:47 AM »
I wanted to try and push the envelope this last weekend and see if we can evaluate this overvoltage method for attempting to check our electron beam energy accuracy by utilizing 0.1 keV increment intervals for the Ti K edge (~5 keV) and the Ge K edge (~11 keV).

I finally realized that this overvoltage method is essentially a trace element problem because the net intensity approaches zero asymtotically.  And zero intensity is presumably where our overvoltage is 1.0 for the edge energy.  Therefore I tuned up at 100 nA using 120 seconds on-peak and 60 seconds on each off-peak, carefully selected my backgrounds.

I also carefully re-polished (removed the carbon coat) on our Ti and Ge standards to avoid any electron energy loss from a coating or oxide layer, and immediately put them in the instrument (yeah, this old man ran from the polishing lab to the microprobe lab!).

Here is the first attempt using Ge Ka starting at 11.2 keV and increasing 0.1 keV to 12.2 keV using a 2nd order polynomial extrapolation:



The red arrow is approximately where we would like to see the intercept cross zero on the Y axis (11.103 keV), but maybe we are underfitting, so let's try a 3rd order polynomial:



Hmmm, well now we're clearly overfitting, as we never cross zero intensity!  OK let's try again with a second dataset, once more with a 2nd order polynomial:



OK, and now with a 3rd order polynomial:



Well so what can we conclude from this?  The high voltage at 11.2 keV is not too far from the nominal voltage I would guess...

It sure would be nice to test at around 15 keV (where most users tend to run), but then we would need an LiF220 Bragg crystal and use the Rb K edge or Sr K edge, but those aren't great as pure metals (I think we need to use pure metals without a conductive coating or oxide layer for best accuracy...).

Maybe we need to use 0.05 keV increments?  Does anyone know what the minimum high voltage step sizes on a Cameca SX100 are?
« Last Edit: April 08, 2024, 11:46:16 AM by Probeman »
The only stupid question is the one not asked!

sem-geologist

  • Professor
  • ****
  • Posts: 300
Re: Duane-Hunt limit
« Reply #42 on: April 09, 2024, 12:56:20 AM »
Maybe we need to use 0.05 keV increments?  Does anyone know what the minimum high voltage step sizes on a Cameca SX100 are?

I think I know. SX100 HV control board is controlled by analog DC signal of 0-10V. As for EHT that correspond to range of acceleration voltages 0-50kV. i.e. for 30kV it will be 6V DC, for 15kV it will be 3V. Analog DC signal is created at Column Control board with 12bit DAC, thus one bit is 50kV/2^12 = 12.20703125V, so theoretically that would be minimal step...

...which brings me to point that 15kV would require 1228.8 bit, where closest round bit would give 15002.7V - essentially close enough to the 15kV.

First, the E0 itself (as set by the microscope) will likely have some (systematic) "setpoint error", for example because the voltage entering the HV cascade is determined by some DAC with a finite number of steps.  The "problem" here is that this will then be a fixed percentage of the "full range", so if your HV subsystem can go up to 30kV and uses a 10-bit DAC to set the value, you're looking at a discretization error of ~30V just from this.

At least 12 bit is a standard. What kind of a**h**** would make SEM/EPMA EHT with 10bit DAC? EHT is not (can't be) fast changing as i.e. scanning coils - there is absolutely no excuse using lower resolution. Even my grandma would had used 12bit for that...

Some HV PSU use even 14bit for EHT.

  The HV supply should also specify a certain maximum "ripple", which is typically an order of magnitude better than this (because it would otherwise give bad ("fluctuating") imaging results).
It is kind of ironic that HV tank is powered with high frequency AC which we don't see (on images) thanks to smoothing properties of Cockcroft–Walton voltage multiplier on its own (often on SEM/Probe there is up to 10 multiplication stages) and additional RC filtering stages just takes any residual ripple out of equation. The HV ripple is order of magnitude smaller than accuracy of HV, because it is few orders of magnitude easier technically to achieve (basically - no effort, and accuracy as we see here is enormously challenging to keep in check).

There may be a small voltage that needs to be subtracted from the supplied HV because that's applied to the filament itself to heat it up, but that would be just a few volts.
Subtracted or added? If it is DC used to heat the filament, mostly those are up to 5V. As example,  lets say filament is supplied with 5V - the HV is applied through symmetric matched resistors to plus and minus rail of that filament circuit, lest say 10kV (this is biasing the filament). The negative rail of filament then has -10002.5V and positive rail of filament then have -9997.5V. So as current in ampere range flows through filament, what voltage (relative to anode) will be at the middle  (the electron emitting) point of the filament? At center point the filament will be 10kV - there is nothing to be subtracted or added. Well... unless some SEM manufacturer would bias filament not symmetrically, but directly to the single filament supply rail... Some of HV PSU supplies filament with AC, but again at middle point of it the added/subtracted filament voltage is going to be near 0V. (The advantage of AC is the equal filament current from both sides - the tungsten evaporates more evenly - no thinning from single side as on DC). 

Then, electrons can also undergo further ("perpendicular") acceleration inside the scanning column, so that increases their energy as a function of their final distance from the optical center.  In case you're scanning a big field of view, this may be significant.  Speaking for the microscope my team is producing, we're scanning electrostatically with a few hundred volts on the scanning poles, so my guess is that electrons aimed for the edges of the maximum field of view have their energy increased by the same amount (and worse: that means the electron energy varies over the image - interesting when doing mapping!)
this! This is very interesting concept. Would this be able to recognize with Duane-Hunt limit on EDS? The limit should be shifted I guess?

By the way, this is all not to say that I don't find the whole bremsstrahlung phenomenon super weird.  Since we are seeing photons with energies up to E0 in our spectra, that means that these were emitted by electrons losing all their energy during deceleration (so far, I'm nodding) but since each photon corresponds to a single quantum mechanical event, that must mean that some electrons come to an abrupt stop from full speed in one infinitesimal event.  How?!
That's the whole point of D-H limit, to catch these electrons which get to an abrupt stop from full speed in one infinitesimal event. And one of challenging part of D-H is that these are very very rare as most of them are stopped in few or many many steps (thus we get 0 to E0 continuum). If electrons can be instantly accelerated (from cathode) why they could not be instantly deaccelerated? Electrons can be instantly deflected why they could not be made to stop? electrons can instantly jump the "band-gap" and can do weird tunneling and other weird weird stuff... From all of weird workings of electrons, indeed, bremstrahlung effect looks the least weird in my opinion.
 
Wasn't it Bohr himself who said "Anyone who is not shocked by quantum theory has not understood it" - I'm definitely still in the "shocked" phase.
But do we need quantum theory at all in this case? (probably I am the one who do not fully understand it (or completely don't understand) as I don't feel shocked :D)
« Last Edit: April 09, 2024, 04:03:12 AM by sem-geologist »

Probeman

  • Emeritus
  • *****
  • Posts: 2823
  • Never sleeps...
    • John Donovan
Re: Duane-Hunt limit
« Reply #43 on: April 09, 2024, 07:31:24 AM »
Maybe we need to use 0.05 keV increments?  Does anyone know what the minimum high voltage step sizes on a Cameca SX100 are?

I think I know. SX100 HV control board is controlled by analog DC signal of 0-10V. As for EHT that correspond to range of acceleration voltages 0-50kV. i.e. for 30kV it will be 6V DC, for 15kV it will be 3V. Analog DC signal is created at Column Control board with 12bit DAC, thus one bit is 50kV/2^12 = 12.20703125V, so theoretically that would be minimal step...

Very good!  OK so 12.2 eV (0.012 keV) would be the smallest step size?  I'll give 50 eV increments a try next weekend.  So far the overvoltage plots (above) seem very smooth using 100 eV steps:

https://probesoftware.com/smf/index.php?topic=1063.msg12536#msg12536

Then, electrons can also undergo further ("perpendicular") acceleration inside the scanning column, so that increases their energy as a function of their final distance from the optical center.  In case you're scanning a big field of view, this may be significant.  Speaking for the microscope my team is producing, we're scanning electrostatically with a few hundred volts on the scanning poles, so my guess is that electrons aimed for the edges of the maximum field of view have their energy increased by the same amount (and worse: that means the electron energy varies over the image - interesting when doing mapping!)
this! This is very interesting concept. Would this be able to recognize with Duane-Hunt limit on EDS? The limit should be shifted I guess?

Based on the D-H testing we've performed above:

https://probesoftware.com/smf/index.php?topic=1063.msg12442#msg12442

due to convolution effects of the EDS detector on the photon "cliff" binning artifact, I don't think we can do better than +/- 100 or 150 eV using the Duane-Hunt limit test:

https://probesoftware.com/smf/index.php?topic=1063.msg12472#msg12472

Therefore I think the overvoltage method on the characteristic net intensity, suggested by Nicholas Ritchie, is the only way forward for absolute determinations of electron beam energies at high voltages:

https://probesoftware.com/smf/index.php?topic=1063.msg12536#msg12536

A blind test I've thought about trying someday is to acquire some EDS spectra at say 15.2 keV or 14.9 keV and see if anyone can actually back out the correct accelerating voltage-  I think if we didn't already know what the high voltage was we couldn't do it...
« Last Edit: April 09, 2024, 11:51:54 AM by Probeman »
The only stupid question is the one not asked!

Nicholas Ritchie

  • Professor
  • ****
  • Posts: 141
    • NIST DTSA-II
Re: Duane-Hunt limit
« Reply #44 on: April 10, 2024, 11:51:26 AM »
Here is a little data and analysis to inform the conversation.
I measured some deep (12,000 s) spectra from Pt at very low probe current (170 pA) to minimize pulse-pileup.
These spectra were then processed to extract an estimate of the Duane-Hunt limit.  By sub-sampling the deep spectra, I was able to construct 450 more realistic dose spectra.  These too were fit.

The short story...  The D-H was consistently about 70 eV higher than I expected (based on the assumption that my instrument's beam energy is calibrated correctly) and that for 40 nA.s spectra, the one-sigma uncertainty was about 24 eV.




When I compared the measured spectra with simulations from PENEPMA and DTSA-II, both were capable of modeling the general shape of the continuum quite well.  Red is measured, blue is penepma, and green is DTSA-II.  All are scaled such that the continuum intensity near 5 keV is the same.



However, at the highest energies (within a couple hundred eV of the D-H) PENEPMA did a more realistic job.  It has been suggested that the continuous slowing down model underestimates the intensity at energies just below the D-H. This seems credible.

Red is measured, blue is penepma, and green is DTSA-II.   All are scaled such that the continuum intensity near 5 keV is the same.


If you are curious about the details, I've attached a Jupyter notebook as a self-contained HTML file. (I had to ZIP it to fit within the site's file size constraints.)
"Do what you can, with what you have, where you are"
  - Teddy Roosevelt