[BDXlist] Calorimeter saturation effect

Wed Mar 30 04:43:24 EDT 2016

Hi Misha,
thanks, you're right, there is a 10^3 label lost under the color scale!

With Luca we looked further at this effect: the attached plot is the 
amplitude (mV) vs charge (phe) measured for MPPC 25 um.
The amplifier saturation at 750 mV is clearly visible.
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).
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.
For the 50 um MPPC, saturation happens at ~ 1000 phe (50 MeV).

* 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.

* 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

--> Q1 < 900 AND Q2 < 1000: both amplifiers here are in a linear regime 
(E1<92 MeV AND E2<50 MeV)
--> 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
--> Q1>900 AND Q2>1000 : both amplifiers are in the saturation regime.

* To summarize:
--> 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.
--> 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.
(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)

Bests,
Andrea

On 03/30/2016 09:37 AM, osipenko wrote:
> Hi Andrea,
>
> in the plot on X-axis you have probably lost 10^3 (phe).
>
> 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.
>
> Cheers,
> Misha.
>
>
>
>
> On 03/29/2016 11:39 PM, Andrea Celentano wrote:
>> Dear all,
>> I investigated the saturation effect as seen in the calorimeter, and 
>> I'd like to share with you what I found.
>>
>> 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.
>>
>> 1B) The single-phe amplitude measured by DAQ (i.e. after splitting) is:
>> - For the 25 um MPPC: 12 mV
>> - For the 50 um MPPC: 9.7 mV
>> 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)
>>
>> 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:
>>
>> - For the 25 um MPPC: 62.5 phe
>> - For the 50 um MPPC: 77 phe
>>
>> *Important:* 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.
>>
>> 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.
>>
>> 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:
>>
>> Nphe =  Ncells * (1-exp(-N0*PDE / Ncells))
>>
>> 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.
>>
>> 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)
>>
>> - For the 25 um MPPC: 9.73 phe/MeV
>> - For the 50 um MPPC: 19.67 phe/MeV
>>
>> Note that these two numbers were derived without correcting for the 
>> intrinsic MPPC saturation.
>>
>> 4) I took run 1338 and plotted, for all the events, the two MPPC 
>> charges, one against the other, in phe.
>> Attached is the result.
>> 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:
>>
>> - Completely ignoring saturation, Q(50) = Q(25) * 19.67 / 9.73
>> - 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))
>>
>> 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.
>> 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.
>> At higher charge neither curve reproduces data - but here also the 
>> amplifier saturation is important too..
>>
>> Bottom-line messages:
>> - 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
>> - I'd suggest to use the two MPPCs to measure different energy regions:
>> --Low energy measured by the 50 um MPPC (higher PDE)
>> --High energy measured by the 25 um MPPC (higher number of cells)
>>
>> 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?
>>
>> Please, let me know what you think about.
>>
>> Bests,
>> Andrea
>>
>>
>>
>>
>>
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>

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