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Hi Misha,<br>
thanks, you're right, there is a 10^3 label lost under the color
scale!<br>
<br>
With Luca we looked further at this effect: the attached plot is the
amplitude (mV) vs charge (phe) measured for MPPC 25 um. <br>
The amplifier saturation at 750 mV is clearly visible. <br>
The charge when this happens is ~ 900 phe (92 MeV): this is much
larger than 750 mV / 12 mV = 62 phe (12 mV is the single phe
amplitude for this MPPC). <br>
The reason is the CsI(Tl) long decay time: a signal with 750 mV
amplitude has a large contribution to the charge given by photons
hitting the MPPC after the signal head, in the scintillation tail.
This effect is a factor ~ 900 / 62 ~ 15.<br>
For the 50 um MPPC, saturation happens at ~ 1000 phe (50 MeV).<br>
<br>
* By "amplifier saturation" I mean the effect shown in the
"saturatedSignal.png" plot: the head of the signal has an amplitude
> 750 mV, hence gets "truncated" due to the amplifier limited
output range. Clearly, for these signals, there is still a relation
between the deposited energy in CsI(Tl) and the measured charge, but
it is non linear.<br>
<br>
* The "amplifierSaturation2.png" plot is, again, the Q1 vs Q2 plot
for the crystal, run 1138. I reported with red lines the points
where MPPC 25 um and MPPC 50 um starts to saturate due to the
amplifier. We can identify 3 regions<br>
<br>
--> Q1 < 900 AND Q2 < 1000: both amplifiers here are in a
linear regime (E1<92 MeV AND E2<50 MeV)<br>
--> Q1<900 AND Q2> 1000: MPPC 50 um output is > 750 mV,
hence the amplifier saturates. However, 25 um MPPC is in a linear
regime<br>
--> Q1>900 AND Q2>1000 : both amplifiers are in the
saturation regime.<br>
<br>
* To summarize: <br>
--> Saturation effects we see are due to the limited amplifier
output range. Intrinsic MPPC saturation here, in particular for the
25um MPPC, is not expected to play a significant role. <br>
--> In order to measure properly the high-energy part of the
spectrum, 25 um MPPC can be used, provided we lower the amplifier
gain. A factor of 6 reduction is feasible (more than this is
complicate): this results in a saturation happening a ~ 5400 phe,
i.e. 540 MeV. To do so: R10=51 Ohm should be removed in the
amplifier, R11=270 Ohm should be kept as it is, R12=51 Ohm should be
replaced with a 0 Ohm resistor.<br>
(Note that in my previous e-mail I proposed a factor x3 reduction in
gain. Probably, this is not enough if we really want to reach the
500 MeV region)<br>
<br>
Bests,<br>
Andrea<br>
<br>
<br>
<br>
<br>
<div class="moz-cite-prefix">On 03/30/2016 09:37 AM, osipenko wrote:<br>
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Hi Andrea,<br>
<br>
in the plot on X-axis you have probably lost 10^3 (phe).<br>
<br>
Unlikely the difference has to do with amplifier bandwidth, it
shall be a minor effect. Perhaps the real saturation would need
"an effective" number of cells to account for cross talk. Looking
on your plot I would say that for 50 um MPPC the effective number
of cells id 2,000, not 3,600.<br>
<br>
Cheers,<br>
Misha.<br>
<br>
<br>
<br>
<br>
<div class="moz-cite-prefix">On 03/29/2016 11:39 PM, Andrea
Celentano wrote:<br>
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Dear all,<br>
I investigated the saturation effect as seen in the calorimeter,
and I'd like to share with you what I found.<br>
<br>
1A) The single-phe charge, for both MPPC mounted on the crystal,
is ~ 712 pVs. This is consistent with the measurement that was
performed in Genova, where we found 1.4 nVs for both: in Catania
there is a x2 splitter between each MPPC and the DAQ.<br>
<br>
1B) The single-phe amplitude measured by DAQ (i.e. after
splitting) is:<br>
- For the 25 um MPPC: 12 mV<br>
- For the 50 um MPPC: 9.7 mV<br>
The two are different, but the area is the same, since the
single phe-signal has a slightly different shape between the
two, the 50 um is longer (it is reasonable: for the same total
MPPC capacitance, the 50 um capacitance per cell is larger,
hence the single phe signal is longer)<br>
<br>
1C) Our amplifiers are really NOT optimized for output dynamic
range. According to the schematic (attached), the last stage is
OPA694-based. This has an output dynamics of 3V. However,
there's also a 50 Ohm-50 Ohm x2 voltage divider, between the
output resistor (R12) AND the 50 Ohm impedance of the splitter.
It means the maximum output voltage MEASURED by the DAQ is 750
mV for both channels (3 V of OPA 694 -> 1.5 V after voltage
divider-> 750 mV out of splitter). This means that the
maximum number of phe is:<br>
<br>
- For the 25 um MPPC: 62.5 phe<br>
- For the 50 um MPPC: 77 phe<br>
<br>
<b>Important:</b> this limit holds for a signal where all the
phe are syncronous, i.e. detected almost at the same time, such
as a fast plastic scintillator. For a CsI detector, that has a
very long decay time compared to the amplifier response time,
the actual MAXIMUM number of measurable phe is much larger
(since they arrive at different times). If I consider the
single-phe signal time ~ 100 ns, and the CsI(Tl) decay time ~ 1
us, the ratio x10 suggests that AT LEAST we can measure 10x phe
than the two numbers above.<br>
<br>
2) Other than the amplifiers saturation, there's the intrinsic
saturation of the MPPC, that can't fire more than Ncells, where
Ncells=3600 for the 50 um MPPC, and Ncells=14400 for the 25 um
MPPC. <br>
<br>
IF all the incoming photons were hitting the two MPPCs in a time
interval shorter than the MPPC-cell recovery time (~10 ns?),
then, for a input signal of N0 photons, the response of the
MPPCs would be:<br>
<br>
Nphe = Ncells * (1-exp(-N0*PDE / Ncells))<br>
<br>
Here, the situation is again more complicated, since photons ARE
NOT hitting the MPPCs at the same time, given the long CsI(Tl)
decay time.<br>
<br>
3) From cosmics-ray calibrations, the two MPPCs have a different
overall gain, i.e. the number of phe seen per MeV is different,
probably due to a different optical coupling / PDE (Hamamatsu
quotes 25% for 25um and 40% for 50um)<br>
<br>
- For the 25 um MPPC: 9.73 phe/MeV<br>
- For the 50 um MPPC: 19.67 phe/MeV<br>
<br>
Note that these two numbers were derived without correcting for
the intrinsic MPPC saturation.<br>
<br>
4) I took run 1338 and plotted, for all the events, the two MPPC
charges, one against the other, in phe. <br>
Attached is the result.<br>
Using the two cal. constants before *assuming cosmics are in a
low-charge area, where saturation can be neglected*, one can
derive the expected charge of MPPC 50 um as a function of the
measured charge of MPPC 25 um:<br>
<br>
- Completely ignoring saturation, Q(50) = Q(25) * 19.67 / 9.73 <br>
- Ignoring saturation for 25 um (since Ncells is "large"), but
considering for 50 um saturation-for the case of all photons
hitting the MPPC together: Q(50) = Ncells(50) * (1 - exp(-(Q(25)
* 19.67 / 9.73)/Ncells(50))<br>
<br>
The two super-imposed curves refer to the two above scenarios.
One can see that, although the two curves reproduce the order of
magnitude of the data, neither agree with it well.<br>
It seems that the "no-saturation" curve is better at low charge
- because photons hitting the MPPC are distributed in time,
hence for a single event the SAME MPPC cell can fire twice.<br>
At higher charge neither curve reproduces data - but here also
the amplifier saturation is important too.. <br>
<br>
Bottom-line messages:<br>
- Saturation and non-linearity effects in the crystal are very
complicate - analytical formula we are used to do not apply so
easily, given the fact CsI(Tl) photons hit the MPPCs in a longer
time than the MPPC intrinsic one<br>
- I'd suggest to use the two MPPCs to measure different energy
regions: <br>
--Low energy measured by the 50 um MPPC (higher PDE)<br>
--High energy measured by the 25 um MPPC (higher number of
cells)<br>
<br>
To do so, it would be good to increase the dynamic range of the
25 um amplifier: this can be done by changing, in amplifier
n.10, resistor R11 from 270 Ohm to ~ 100 Ohm, thus decreasing
the amplifier gain by a factor of ~3. Marzio, Mariangela, can
you do so?<br>
<br>
Please, let me know what you think about.<br>
<br>
Bests,<br>
Andrea <br>
<br>
<br>
<br>
<br>
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