Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating actual x-ray flux measurement

  1. Nov 1, 2009 #1
    Hi,

    Hopefully I'm posting this in the correct forum.

    I'm using Nuclear Associates 07-451 flux detector with a gigaohm input impedance such that the measurement read is logarithmically compressed. The x-ray tube is outputting a constant number of rads. The flux sensor is mounted to the left side of tube (not directly below the output window). The x-ray beam is deflected such that it starts farthest away from the sensor and is progressively brought closer to the sensor. The beam is passing through different thicknesses of metal along the path that it travels and a reading is taken at known distances from the sensor (the metal is between the tube and the sensor).

    So I know that my measurement is logarithmically compressed and x-ray falloff is 1/(distance^2). Maybe I'm just having a bad day but my Math is not so great to understand how to undo the loss in the signal to compensate for the distance between the beam on the output side of the metal and the sensor, the distances are known by Pythagorean theorem (for the moment I just need to correct this part of the measurement). This part should be easy. Also I'm not clear how to undo the logarithmic compression of the measurement due to the input impedance (mathematically).

    Once I correct the measurement then I'm hoping to compensate for the 4.5 million measurements that I have to take before plotting them.

    Hopefully this makes sense and I got the right forum.

    Appreciate any assistance.

    Thanks,

    Jeremy
     
  2. jcsd
  3. Nov 1, 2009 #2
    Hi fuzzychaos-
    You have brought up several important questions.
    1) A high gigohm input impedande does not make a log response. I had to use the base emitter junction of a transistor to get an ideal diode response for a log response ion chamber:
    I = expt(-eV/kT) + I0
    to get
    V= (kT/e)log(I/I0)

    2) I don't understand the geometry and in particular why the distance between x-ray tube and sensor is changing.

    3) How do you deflect an uncharged x-ray beam?

    4) A metal filter hardens the x-ray flux and changes the sensor response. Do you want to know the flux or the rads per gram?

    Bob S
     
  4. Nov 2, 2009 #3
    Hi Bob,

    1) I found the log response in the manual, it says that low input impedance is linear but 10MOhm input impedance is logarithmic (mine is 1GOhm). I didn't look up in any electronics manual why that would be so. Section 1.2 pg 5 of http://assets.fluke.com/manuals/07_451__omeng0000.pdf states "...with very low input impedance to the measurement device, the output is linear with the input x-ray intensity...operated into a very high impedance ... 10 megohms, the output will be logarithmic."
    - Should I expect it to still be linear?
    - The formula that you mention is the diode equation? If = Io exp(-eV/kT)

    2) The distance is changing because the beam is deflected to a point starting 7" (7 inches) from the detector, the beam is slowly deflected to 2" from the detector.

    3) We are deflecting the beam with a current deflection coil before it hits the Beryllium window. We can deflect the beam 5" this way (+/- 2.5" from the center) on our large window. I'm not a Physicist to know much of the details about this.

    4) I need to make an adjustment to the sensor response, in my logging software, based on the drop in x-rays having to travel 7" vs 2". I have the same titanium thickness at 7" and 2". I understand that x-ray drops off at 1/(distance^2). For the life of me I can't figure out this simple equation.

    Maybe I should try to write it down as best I can.

    7", flux output = X1
    2", flux output = X2

    X-ray falls off at a rate of 1/(distance^2). Find the adjusted output for X1 and X2. That is at least my very simple understanding of x-ray.

    I'm not sure if I need to undo the log response (if there is one). I thought that it might be easier to comprehend and set a low limit if the measurement was linear. Occasionally the target/window has something the equivalent of holes in it which don't produce x-rays/low output and I need to properly capture this by adjusting for the distance so that I know when the tube is not properly producing x-rays I can flag the measurements. Since each tube is slightly different I get a flux reading at x-rays on/off when the beam is centered (~4.5" from the detector) so that I'll know roughly what the maximum/minimum output is (this also requires adjustment for the x-ray fall off distance between the spot and the detector).

    Jeremy
     
  5. Nov 2, 2009 #4
    Hi jeremy-
    Here is a description of the typical transdiode log amplifier using the diode equation:
    http://www.analog.com/library/analogDialogue/bestof/pdf/19_1.pdf
    Are you using a beryllium or titanium window as a x-ray souce? What is the energy of your electron beam? If the thickness of your window varies, it will change (filter out the low energy x-rays) the x-ray spectrum.
    Sorry; I miswrote the diode voltage-current equation.
    Bob S

    [Edit} On reading your rad sensor instruction manual, I see that a silicon diode is being used as the radiation detector. Radiation creates conduction electrons in the diode, leading to a current through the diode. The instruction manual states "However, if the detector is operated into a very high impedance such as an oscilloscope, which is commonly 10 megohms, the voltage output will be logarithmic" (I inserted the word voltage). The silicon diode is being used as an approximate current-to-voltage logarithmic converter of the input current. It is not a precise log converter like the transdiode connection used by Analog Devices and other mfgrs.
    Bob S

    Bob S
     
    Last edited: Nov 2, 2009
  6. Nov 2, 2009 #5
    Hi Bob,

    Thanks for checking it out and following up.

    We are using a Beryllium window that has one side coated with Tungsten. The thickness changes only because I have a Tungsten ring attached (of a certain thickness and width) covering the rim of the window. Yes there is filtering and that is intended by adding in the Tungsten ring. I'm using 160KV @ 100uA.

    Although the x-ray energy is filtered I'm only worried about the measurement and if it is compensated for correctly by the different distances travelled. Unless the results are different for different energies? For instance I'll know that I'm over the Tungsten ring and the measured output looks like X, while it will be different again when it is not over the Tungsten ring, but that it should be the same X again when it is back over the Tungsten ring (but now closer to the sensor requiring some compensation). I understand that I might have to compensate a little for the angle of beam pointing away from the sensor when it is at 7" and pointing a little toward the sensor when it is at 2". I'll eventually be making measurements everywhere within the circular window.

    Thanks for the clarification on the voltage. Yeah, it probably isn't the best detector but it is cheap. Maybe you have a recommendation that I can check out? Unfortunately I don't have much of a seemingly useful electronics background any more (all lost to the more than decade of inactivity) but it is slowly crawling back to me :D

    I didn't quite understand why the low impedance vs. high impedance made it linear vs. logarithmic. V = IR no? Ugh, seems like I need to be schooled in basic electronics again... Or maybe there is a way to understand the megaohm series resistance causing the log voltage measurement?

    Jeremy
     
  7. Nov 3, 2009 #6
    The silcon detector produces an amount of current depending on the amplitude of the radiation field. If you short the output through a low impedance, such as a transimpedance amplifier, then you can measure the current directly. If you terminate the signal with a very high impedance, then you will measure a voltage across the silicon detector that follows the diode equation: V = (kT/e)Ln(I/I0). The voltage is not an accurate measure of the radiation. Build your own low-impedance linear amplifier, and measure the current accurately.
    Bob S
     
  8. Nov 4, 2009 #7
    Hi Bob,

    Thanks for the help and for the application assistance. I appreciate your patience and I just want to check my understanding of what you just said.

    I see the transimpedance amplifiers are used for current to voltage applications right, so when you say measure the current directly you mean that I'll be measuring the voltage output from the transimpedance amplifier which is using the 'current' output from the sensor. And this output will be linear or follow a different equation?

    I'm not certain if you are saying that the transimpedance amplifier will be an inaccurate measure or the voltage directly from the sensor is an inaccurate measure?

    So with the high impedance that means that my present set up with the 1GOhm input impedance will follow the diode equation, correct?

    You are suggesting that it is better to use a low-impedance linear amplifier and measure the current directly? It sounds like these two set ups are doing almost the same thing but with the low-impedance linear amplifier will the current output still be linear from the sensor? I tried to search a bit for what you might be talking about but I really wasn't sure. Do you have any links for what I might be looking for?

    I see that there are some challenges to building a good transimpedance amplifier. http://www.mobilehandsetdesignline.com/howto/201400084" [Broken]

    Do you know if there is something that I can purchase rather than making this myself that I can plug into BNC connectors?

    The previous engineer and PhD guys who were working on this (who are gone now so I can't ask why) plugged a 15 foot cable (actual distance is more like 6 feet) into the sensor and connected it to a differential voltage input on a National Instruments DAQ board (so it starts as a single-ended connection on the sensor and ends up as 2 BNC connectors on the differential inputs of NIDAQ, each of the 2 inputs are tied to ground with a 20MOhm resistor). It does measure signals this way. This seems quite incorrect to me but maybe they tried to get rid of power line noise by using the differential input. Based on your feedback it sounds like it shouldn't be that simple that someone can just plug wires into it that way.

    Even after I do all this to get the correct measurement I still have to adjust the measurement for the x-ray dropping after travelling different distances through the air. Any idea about this one? I found a simple illustration which I am trying to work from here http://www.e-radiography.net/radtech/i/inverse_square.htm"

    Jeremy
     
    Last edited by a moderator: May 4, 2017
  9. Nov 4, 2009 #8
    Hi Jeremy-
    You ask a lot of good questions. I will answer all of these in sequence.
    A good transimpedance amplifies has a low virtual impedance to ground, uses a good hi-meg resistor of known resistance that is linear and has a low temperature coefficient, and uses a low-leakage op-amp with low input bias current and voltage offset. The voltage measured directly across the silicon diode sensor is inaccurate because it does not follow the ideal diode equation. You will need two transdiode-connected transistors that are in a balanced circuit, and a temperature-correcting output circuit, because kT/e = 26 mV is proportional to temperature.

    Your present circuit voltage output is inaccurate because your sensor is not an ideal diode and is not temperature corrected.

    To build a good transimpedance amplifier, you need to use the neg (inverting) input, and surround the input on the pc board with a ground ring to intercept leakage currents.

    Your best choice may be to go to Keithley Instruments and purchase an ammeter.

    You can use differential inputs as long as the neg input is a virtual ground and the input bias current leakage resistance paths to ground are equalized, and the only ground is at the amplifier input to eliminate ground loops. Minimize the length of input cable to minimize the shunt capacitance.

    As far as 1/r-squared is concerned, I believe it, as long as you correct for finite source size, etc. Also, if the x-rays pass through a variable thickness of material, correct for the variable radiation hardening. This is minimized by using a beryllium window. I still have not seen a drawing of your setup.

    Bob S
     
  10. Nov 5, 2009 #9
    Hi Bob,

    Thanks, this is helping. I'm a bit unsure about manufacturing my own transimpedance amplifier, let alone 2 that are balanced, it doesn't sound easy to do this properly. However let me keep at this and see what I can find. I'll get a shorter recommended type of cable, the RG-58u.

    I found some information from what you've mentioned http://books.google.com.sg/books?id...e&q=transdiode-connected transistors&f=false" in a practical circuits book on pg 164. Pg 180 log/anti-log circuit sounds a little like the two transdiode-connected transistors in a balanced circuit that you referring to. Is that correct?

    Surrounding the input with a ground ring means I add a ferrite core around the wire or I need to have the cable shield that is connected to ground?

    When you say get an ammeter, I can use it without the extra transimpedance amplifier circuitry since I'll be measuring the current directly or I will still need the extra circuitry? I need to use the NIDAQ because I have to take 4.5 million measurements for automated analysis and displaying a picture of the results, so I can't do this manually.

    I've attached a rough layout of the system and some of the relevant dimensions, however I can't go deeper than this. If it is appropriate I'll wire the sensor up correctly so that I don't need the differential input (the NIDAQ BNC-2090/PCI-MIO-16XE-50 can be switched to single-ended mode). I forget what the thickness of the Tungsten (W) ring is, at least 0.125". In the present set up I usually measure about 30mV (x-ray off) to 100mV from the sensor (in this less than ideal configuration). Each time I deflect the beam I take a measurement, I scan the entire target this way in small steps.

    Jeremy
     

    Attached Files:

    Last edited by a moderator: Apr 24, 2017
  11. Nov 5, 2009 #10
    First, let’s address the physical setup. As you may not know, at non-relativistic energies, the bremsstrahlung + x-ray flux peaks at an angle to the direction of the beam, and classically, the intensity is distributed around the beam direction with a pattern that looks something like a doughnut. At intermediate energies, the “rabbit ears” fold forward, such that the opening angle between the radiation maximum and the electron beam direction is about θ =[STRIKE] E/mc2[/STRIKE] mc2/E, where E is the electron beam energy and mc2 is the electron mass (510 KeV). As the beam is sweeping, this angle, as well as the distance from the radiation source to the sensor, is changing. So modeling the radiation at the sensor as a function of beam position will be difficult. If you want to research this more, see the GEANT4 equation for bremsstrahlung angular distribution in
    http://geant4.cern.ch/G4UsersDocuments/UsersGuides/PhysicsReferenceManual/html/node60.html
    based on the Koch and Motz formulas.

    Next, what is the peak radiation level (rads/sec) and what radiation intensity variation do you expect during the beam sweeps. You need to start thinking about

    a) two sensors, one on each side,
    b) Is the radiation field high enough to use gas ionization chambers?
    c) a single range, high dynamic range current transimpedance amplifier-digitizer (how fast do you need to digitize?). If your sensor goes directly into a NIDAQ with a 10-volt full scale, you need a preamp. if your max sensor current is 100 nanoamps for example, the transimpedance amplifier gain should be ~ 100 megohms.

    Before we go any further, what is the maximum rad dose rate, what is the max radiation intensity variation, and what is the max current from your sensor? I am guessing from your 30 mV to 100 mV numbers and the diode equation that the sensor current range might be from 1 to 100 nanoamps, with a dynamic range of ~e3.3 = 30:1. Is that correct? What bandwidth do you need?

    Bob S
     
    Last edited: Nov 5, 2009
  12. Nov 5, 2009 #11
    Jeremy-
    Here is a good resource on low current measurement.
    http://www.keithley.com/support/data?asset=6169 [Broken]
    Here are others
    http://www.keithley.com/support/keidoc_searchresult?keyword=6514&item_type=All [Broken]
    Bob S
     
    Last edited by a moderator: May 4, 2017
  13. Nov 6, 2009 #12
    Hi Bob,

    Thanks for the useful pages on the current measurements.

    The previous engineer left me a similar picture (attached) to the rabbit ear one that you describe. With the appropriate cross-section that is exactly what you are describing. Thanks for the GEANT4 equation reference, I'll take a more thorough read through that.

    The energy from the tube is ~58KeV. This number was given to me but I understand, without knowing why exactly, that it is about 1/3 of [tex]E_{max}[/tex] which is 160KV/3.

    a) I'm not sure that I understand why I would need 2 sensors because I have a circular target. Would I need 4 or 2 mounted perpendicular to each other? I'm not sure that I can afford 2 sensors and their complete set up, so I'll have to try to work with just the one. How about if it was dead center under the tube target?

    b) Yes, I think that it should be high enough. I presently use a hand held one for surveys outside the lead box.

    c) I'm not sure of the exact digitizing speed. There are 2600 indexes to make it across the entire horizontal and vertical of the target and it only uses every 3rd one. So [tex]\pi*(2600/3/2)^2[/tex] ~= 589,921 points. I found an error in the stepping formula that's being used so it is x2 or 1,179,842 points. And it takes roughly 45 mins to complete that. So it should be taking a measurement about every 2.3 mSecs (that seems a lot faster than I originally thought). The response from the detector states a 1 micro second rise time.

    I don't really know the maximum dose rate. You mean something like the rads/min? If so it is about 506 rads/min at 0.66" from the source (dead centre) when the beam has been deflected 0.3" from the centre and rotated in a complete circle around the centre (measurement was taken with a TLD).

    I don't have a current measurement from the sensor, only the voltage which peaks at about 100mV and reads about 3mV in the x-ray off state (sorry not the 30 previously mentioned). This is all measured at the PC software through all the previously mentioned hardware.

    I'm sorry but I'm not sure what you mean by the dynamic range/bandwidth that I need. Is there a reference that I can look at to help answer that?

    Jeremy
     

    Attached Files:

  14. Nov 6, 2009 #13
    Hi Bob,

    I've been reading up a bit to try and better understand what you are saying.

    Can I use the average Bremsstrahlung calculated for a certain beam angle and then decay it by the distance and angle travelled to the detector to get my actual output?

    Most of the Math that I've come across so far talks about calculating the Bremsstrahlung angles and doesn't say much about the detector to spot source angles except in the experiment apparatus (basically made perfect for the best measurement). My simple brain seems to think that at some distance from the source everything is fuzzy, with no definite shape, and so the radiation falls off at 1/(distance^2). Is that correct?

    Jeremy
     
  15. Nov 6, 2009 #14
    Hi Jeremy-
    First, I would like you to look into exactly what is inside the Nuclear Associates 07-451 x-ray output detector. Please find out what the expected accuracy, linearity, and calibration is. Does it have the necessary linearity and calibration for your application? I attach a sketch of what I would expect it to look like. The radiation sensing diode should be back-biased sufficiently to have a large depletion area, where the radiation-induced charge carriers are created by the radiation flux. The forward biased log response diode is there to provide a log response if the output is not connected to a low input-impedance trans-impedance amplifier. A single diode like the one shown is not an ideal log-response diode. Does the Nuclear Associates unit have an internal battery? What voltage? What is the radiation sensing diode? What is the log response diode? What is the recommended impedance (resistance) termination to get a linear current response?

    Do you also monitor and record the 160 kV x-ray beam current? The current is a very good monitor for average radiation dose. I agree that the average photon energy is probably ~1/3 of the beam voltage.

    506 rads/min (8.4 rads per sec) could yield ~6 nanoamps in a 10-cc argon ion chamber. See
    http://beamdocs.fnal.gov/DocDB/0010/001068/001/A%20tutorial%20on%20beam%20loss%20monitoring.pdf [Broken]
    This is probably similar to the current from your sensor. Look at the PIN diodes mentioned in the same paper.

    I mentioned 2 (or maybe 4) sensors so that the current sum would be less dependent on the variation of sensitivity to beam scanning.

    Bob S
     

    Attached Files:

    Last edited by a moderator: May 4, 2017
  16. Nov 7, 2009 #15
    Hi Bob,

    I read through the whole beam monitoring paper, thanks for the link.

    Unfortunately the manual for the detector is very basic and doesn't show or explain any of the technical details that you are asking about. I'll check with Fluke again, last time they didn't answer me.

    The device does not run on any power source. The ionizing energy causes the diodes to emit the negative current/voltage. The device has a single BNC connector on the back of it to hook to whatever you intend to measure it with. From the operator manual it just says the following, "The detector utilizes silicon diodes that have very high sensitive volume and very low mass encapsulation. The characteristics of diode detectors, when used with x-rays, are similar to that of a silicon diode detector when used for light. When such a detector is used as a current generator with very low input impedance to the measurement device, the output is linear with the input x-ray intensity." So it sounds like there is more than 1 diode inside it.

    Yes, I do record the beam current, which doesn't usually ever change. This is because there is a feedback circuit to keep the current at exactly 100uA by adjusting the voltage of a grid in the gun assembly. I'm pretty sure that when there is a small hole burned through the Tungsten plating of the target the x-ray flux drops off and my digital image goes dark at that location on the target but the current still reads 100uA. The flux detector will show a sudden drop as well. In practise, if I was to position the beam carefully then I can find the size of the whole and the degree of flux production loss. If I plot readings from the detector for the hole and the surrounding area I can get a something like a heat map of flux production in that area. I can do this for the whole target to find where the holes are and how bad they are (but presently I don't make any adjustments to the measurements which I need to do).

    To summarize our current discussion. Now I know that I need to compensate for the distance, as well as the angle of the beam to the sensor due the Bremsstrahlung characteristic. I also need to fix the sensing circuit for the flux detector so that I get better readings. I would buy more sensors but I don't have any money for that. The manual also doesn't provide any way to calibrate the sensors so that I can do what you suggest with multiple sensors, although I can manually apply a voltage offset.

    Jeremy
     
  17. Nov 7, 2009 #16
    Right now, neither you nor I understand how the Nuclear Associates radiation detector works. I personally would not trust any detector that I did not understand for any kind of quantitative linear response.
    I still vote for one that uses some kind of dc bias voltage. Among other things, a reverse bias voltage increases the size of the depletion region, and therefore also the sensitivity.
    Bob S
     
  18. Nov 13, 2009 #17
    Hi Bob,

    Looks like I need to get a different sensor in the future. This is reply that I got from Fluke.

    The device includes a photodiode (anode and a resistor connected to center pin) and a shunt resistance of about a megohm. This device is intensity dependant and reasonably accurate for time measurement. We do not offer any intensity vs . voltage or current data on this operation though. The device is simply for diagnosing malfunctions in a xray generator. This includes timer calibration ,loading,rectifier malfunction,cable or connector arcing. Any single-ended voltage oscilloscope with 1 or 10meg impedance shall work fine. A classical photodiode resistive feedback inverting low bias current opamp circuit can work also with the understanding that possible amplifier response time is then added to the waveform.

    Jeremy
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculating actual x-ray flux measurement
  1. X Rays (Replies: 8)

  2. X-ray diffraction (Replies: 1)

Loading...