[Halld-pid] ADC

[Alexander Ostrovidov]

On Tuesday, October 04, 2011, Beni Zihlmann wrote:
> Hi Sasha,
> I have some questions about this.
> to point
> 1) the Landau peak position represents the value with the
>      highest probability. What we need however is the mean
>     of the distribution which is above the peak. Only for a Gaussian
>     is the mean and the peak position the same. So what is the mean?

The mean is at 1036 counts. This results in (1036-350)/(890-350)=1.27,
i.e., 27% more photoelectrons

> 2) I do not quite understand why a 50/50 splitter
> introduces a factor of 4. I would have expected a factor of 2 when
> a 50 Ohm termination is put to the unused output and a factor
> of 1 when no termination is present.

The terminator is not on the unused output. It's at the same
output as a signal cable. Without it, there are clear reflections
at twice the cable length (~150 ns) which are seen both with
a scope and in TDC values (and yes, the other end of the cable
is terminated as well). So, the first factor of 2 signal split
happens at the anode between terminator and cable. Then
a cable is split in two with a 50-50 splitter between ADC and TDC.
I'll show some plots at the meeting.

> 
> 4) 2138 photo electrons is huge. that would represent a relative
>      error 1/sqrt(2138) of 2%. That is too good to be true. Are you sure
>      you run the PMT at 1750V and not at 2000?
> 

Our LeCroy HV power supply tells me it is at 1750V, and Hamamatsu
spec sheet for PMT with this serial number tells me the gain is 1.576e6
at 1750V.


> cheers,
> Beni
> 
> 
> 
> > Beni,
> >
> > OK, here is an estimate for the number of
> > photoelectrons from a minimum ionizing peak
> > for a hit in the middle of a TOF paddle:
> >
> > 1) Peak of Landau distribution is at 890 ADC counts,
> > pedestal is at 350 counts. With 0.25 pC per ADC count,
> > the integrated charge is (890-350)*0.25 pC = 135 pC
> >
> > 2) There is exactly factor of 4 loss due to a 50-50 splitter
> > and 50 Ohm terminator at the PMT. This was verified
> > with a pulser. This results in 135 pC * 4=540 pC at
> > the anode.
> >
> > 3) Therefore, the number of electrons at the anode is
> > 540e-12 / 1.602176565e-19 = 3370415045
> >
> > 4) PMT has a gain of 1.576e6 at 1750V. So, the number
> > of photoelectrons from a single hit is
> >   3370415045 / 1.576e6 = 2138
> >
> > Sasha
> >
> > On Friday, September 30, 2011, Beni Zihlmann wrote:
> >> Hi Sasha,
> >> thanks for the info.
> >> the cable length should not matter if the ADC gate is
> >> sufficiently long.
> >>
> >> cheers,
> >> Beni
> >>> On Friday, September 30, 2011, Beni Zihlmann wrote:
> >>>> Hi Sasha,
> >>>> in the link
> >>>> https://halldweb1.jlab.org/talks/2011-3Q/ADC%20spectra%20and%20Time-walk%20corrections_files/
> >>>> you show an ADC spectrum of a paddle PMT: adc_l_par_nottrig.gif.
> >>>> what is the HV of the PMT set at and what type of ADC are you using?
> >>>>
> >>>> cheers,
> >>>> Beni
> >>>> _______________________________________________
> >>>> Halld-pid mailing list
> >>>> Halld-pid at jlab.org
> >>>> https://mailman.jlab.org/mailman/listinfo/halld-pid
> >>>>
> >>>>
> >>> Beni,
> >>>
> >>> This particular plot is done with:
> >>>
> >>> 1) CAEN ADC v265 in 12-bit mode. This  corresponds to
> >>> 0.25 pC per count
> >>>
> >>> 2) 10-stage PMT with a nominal gain of 1.5*10^6 (at 1750 V).
> >>> However, it was used at 2000 V in this particular measurement.
> >>> So, its actual gain is unknown in this case.
> >>>
> >>> 3) What makes things more complications is a signal attenuation
> >>> on its way from PMT anode to the ADC due to a 50-50 splitter,
> >>> a 50 Ohm terminator right at the PMT,  and 150ft of cable. Right now,
> >>> I'm trying to measure what is the correction coefficient I need to apply
> >>> to compensate for this attenuation. I should have an estimate of the
> >>> number of photoelectrons later today.
> >>>
> >>> Sasha
> >>>
> >>>
> >>>
> >>>
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> >>
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