Question related to a gamma spectrometer

[patric44]

Homework Statement

how does the single channel analyzer work in the gamma spectrometer ?

Relevant Equations

no equations

hi guys
i am kinda have a vogue understanding of the working principle of the gamma spectrometer and have some questions

1 - i need some one to verify my interpretation of the working principle : the sodium iodide crystal emits photons of visible light the intensity of these photons is related to the energy of gamma radiation → these photos is turned into an electrical signal with a certain pulse height by the Photo Multiplier tube → by using a single channel analyzer with a specific window width we are able to count exactly the number of pulses within an interval $ΔV$ apart , a histogram of the number of pulses with their height is plotted indicating the different peaks .
is that right ?

2 - our professor stated that increasing the gain of the linear amplifier stage to high value will reduce the resolution of the produced peaks in the sense that it will make the pulse width wider ? i don't understand why is that ?!

 

[gleem]

1.) Ok. The light is almost all in the blue to the violet range for NaI.

2.) The resolution of a photopeak is usually taken as the 1/2 width of the peak Δn₀ where n₀ is the channel number of the peak. If you increase the gain by a factor of m you increase n₀ and all the channels allocated to the peak by the same factor so the 1/2 width becomes mΔE₀.

Is that significant? Since the amplifier changes everything by the same factor all peaks maintain their relative position. So although the peaks are wider (more channels) their separations are also wider.

The true resolution of a scintillation spectrometer is determined by the processes associated with scintillation crystal and the photomultiplier tube and is typically about 10% of the energy associated with the peak. So a Cs137 gamma of 662 Mev will have a width of the number of channels associated with 66 KeV.

 

[patric44]

so stating that " increasing the gain will reduce the resolution " is not correct at all , though it seems that way because the peaks gets wider only . but in the fact the separation distance of the peaks increases as well , and they remain just as distinguishable as before , right .
i am not sure i understood the part with the channel number , i am assuming the channel number is associated with pulses that have the same height , like pulses with height 20mv will be collected at a specific channel number

 

[gleem]

Your ability to distinguish two peaks does not depend on the amplifier gain to the extent that they are distinguishable at a lower gain.

Sorry about my use of channel number. in the real world, the pulse output of the PM tube is digitized by an analog to digital converter. The digitized pulse amplitudes are sorted into bins by computer circuitry. Each bin or channel contains pulses whose heights fall into a small range of amplitudes. Typically you want your highest energy gamma ray to be put into the highest amplitude bin. If the Cs137 gamma is your highest and you had 512 bins then if your resolution is 10% you would try to make sure you got all the counts from that gamma in your bins. This might mean putting the peak channel in a bin about one full width or more from the highest channel. If you did this then each channel width would contain pulse amplitudes corresponding to an energy range of 1.42 KeV and your 662 KeV (oops I see I said MeV in my first post, it is KeV) gamma would have a width of about 46 channels.

 

[patric44]

thank yo so much its clear now

 

[Steve4Physics]

so stating that " increasing the gain will reduce the resolution " is not correct at all , though it seems that way because the peaks gets wider only . but in the fact the separation distance of the peaks increases as well , and they remain just as distinguishable as before , right .

Hi. I'll throw my two-pennyworth in. Maybe your professor meant this...

If the amplifier gain is *too* high, saturation may occur. By this I mean the output from a large pulse entering the amplifier would exceed the maximum amplifier output (saturation).

This image shows the effect for an over-amplified sine wave: https://qph.fs.quoracdn.net/main-qimg-1b10e586fd0b3b0025d27fa421e313a5

The output has the top of the pulse 'clipped', leaving the broad lower part of the pulse and a flat top. The overall effect is that the efflective pulse-width is increased.

You can imagine a situation in which 2 close large pulses would then appear as a single pulse because their lower parts overlap - and the amplifier output remains saturated from the start of the first pulse to the end of the second pulse. (A bit like resolution of diffracted point sources in optics.)

 

[patric44]

Hi. I'll throw my two-pennyworth in. Maybe your professor meant this...

If the amplifier gain is *too* high, saturation may occur. By this I mean the output from a large pulse entering the amplifier would exceed the maximum amplifier output (saturation).

This image shows the effect for an over-amplified sine wave: https://qph.fs.quoracdn.net/main-qimg-1b10e586fd0b3b0025d27fa421e313a5

The output has the top of the pulse 'clipped', leaving the broad lower part of the pulse and a flat top. The overall effect is that the efflective pulse-width is increased.

You can imagine a situation in which 2 close large pulses would then appear as a single pulse because their lower parts overlap - and the amplifier output remains saturated from the start of the first pulse to the end of the second pulse. (A bit like resolution of diffracted point sources in optics.)

you mean something like this :

now i got what he meant !
thank you so much

 

[patric44]

i am sorry i still have another question , while i was searching more on the working principle of the MCA i came across this statement " Since the amplification necessary to increase the level of signals to that required
by the amplitude analysis system or an MCA ..."

(1) does that mean that when the ADC of the MCA divide the energy range into different number of channels let's say 512 which may be associated with a voltage range from 0 to 10V , if i introduced a pulse directly from the pre amplifier which might have a pulse height of few microvolts it will not be enough to tiger a count as it doesn't even lie between the 0→1 bit , am i interpreting this right .

(2) from these previews statements its some how obvious to me that the value of the gain i choose depend on the value of the gamma energies that i will introduce to the MCA (for small gamma energies i will need larger gain to put the signal into proper range for the MCA to count ) , why my experimental nuclear physics book has an experiment for choosing the best "gain" for the linear amplifier ? i guess its not in the context of this "lower energy gamma " but rater for the clipping effect .

 

[gleem]

(1) does that mean that when the ADC of the MCA divide the energy range into different number of channels let's say 512 which may be associated with a voltage range from 0 to 10V , if i introduced a pulse directly from the pre amplifier which might have a pulse height of few microvolts it will not be enough to tiger a count as it doesn't even lie between the 0→1 bit , am i interpreting this right .

Small voltages are not of interest. In fact, one may introduce a low level discriminator to eliminated these low amplitude peaks so as they do not affect the dead time of the MCA with unnecessary digitalization.

why my experimental nuclear physics book has an experiment for choosing the best "gain" for the linear amplifier ? i guess its not in the context of this "lower energy gamma " but rater for the clipping effect .

Doesn't your book give the reason for optimizing the gain? You want the highest energy gamma peak of interest to be fully analyzed so it must be position so that all counts in it are displayed.

 

[Steve4Physics]

i am sorry i still have another question , while i was searching more on the working principle of the MCA i came across this statement " Since the amplification necessary to increase the level of signals to that required
by the amplitude analysis system or an MCA ..."

(1) does that mean that when the ADC of the MCA divide the energy range into different number of channels let's say 512 which may be associated with a voltage range from 0 to 10V , if i introduced a pulse directly from the pre amplifier which might have a pulse height of few microvolts it will not be enough to tiger a count as it doesn't even lie between the 0→1 bit , am i interpreting this right .

(2) from these previews statements its some how obvious to me that the value of the gain i choose depend on the value of the gamma energies that i will introduce to the MCA (for small gamma energies i will need larger gain to put the signal into proper range for the MCA to count ) , why my experimental nuclear physics book has an experiment for choosing the best "gain" for the linear amplifier ? i guess its not in the context of this "lower energy gamma " but rater for the clipping effect .

I’m no expert but as far as I know, in reply to your questions...

(1) The minimum voltage to set bit-1 of the ADC will depend on the number of bits in the ADC and the voltage-range of the ADC. In addition I believe systems typicaly have an offset, so voltages below some threshold are ignored. Very small voltage pulses will be ignored

(2) You don’t say what the experiment for setting the “best gain” is. I would imagine it means “best” for the particular energy-range under investigation. So the experiment would have to account for this in order to avoid clipping within the energy-range under investigation.

 

[patric44]

its just the first experiment in the book where we try to draw the characteristic curve of the scintillation detector
and from that plot we should determine the best working voltage for the PMT and the best gain , i am assuming the term best here is related to the radioactive source and thus the gamma emitted that we use in the experiment ? , but our professor was talking about the best gain as its the best in general ! not for a specific case .
in which he said " the best gain will be determined by the flattest characteristic curve ??! "

 

[Steve4Physics]

its just the first experiment in the book where we try to draw the characteristic curve of the scintillation detector
and from that plot we should determine the best working voltage for the PMT and the best gain , i am assuming the term best here is related to the radioactive source and thus the gamma emitted that we use in the experiment ? , but our professor was talking about the best gain as its the best in general ! not for a specific case .
in which he said " the best gain will be determined by the flattest characteristic curve ??! "
View attachment 272566

I wonder if there is some confusion between the photomultiplier's (PM's) gain and the amplifier's gain. I think changing the PM's gain (by changing the applied voltage) will change the PM's linearity. So there will be a 'best gain' for the particular PM being used. This will be a 'constant'. The amplifier gain is then separately adjusted for the gamma energy range.

Is there no way to simply ask your professor what he meant?

 

[patric44]

I wonder if there is some confusion between the photomultiplier's (PM's) gain and the amplifier's gain. I think changing the PM's gain (by changing the applied voltage) will change the PM's linearity. So there will be a 'best gain' for the particular PM being used. This will be a 'constant'. The amplifier gain is then separately adjusted for the gamma energy range.

Is there no way to simply ask your professor what he meant?

from what was written in the book i guess its specifically was pointing to the amplifier gain ?! :

and the next page

our professor will not be avaliable for the next couple of weeks so i have no way of asking him .

 

[Steve4Physics]

from what was written in the book i guess its specifically was pointing to the amplifier gain ?! :
View attachment 272568
and the next page
View attachment 272569
our professor will not be avaliable for the next couple of weeks so i have no way of asking him .

Looking at step 12, the procedure seems to be about selecting the best combination of voltage (effectively PM gain) and amplifier gain, not simply amplifier gain alone.
I'm not clear why settings giving maximum length and minimum slope (of region IV) should be the 'best'. Could relate to optimising linearity but that's just a guess.

I would just put it 'on hold'. It can be cleared up when your professor gets back (unless someone else can help). It shouldn't be a barrier to further related work/study.

Sorry I can't help more.

 

[gleem]

The reason that one wants the PMT voltage to be where the signal size changes slowly with tube voltage is if the PMT voltage changes during an experiment it will produce an additional variation in the signal associated with a peak resulting in a wider peak and thus lower resolution.

Δsignal = slope x ΔV

In addition, the count rate and therefore efficiency of a PMT also depends on the voltage. Stable PMT voltage supplies are essential.