Difference between revisions of "Some specifications on PMTs"
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# 5 mm thick scintillators | # 5 mm thick scintillators | ||
# 10000 initial photons per tagged electron | # 10000 initial photons per tagged electron | ||
| − | #* assume that 2000 photons are collected (20 %) | + | #* assume that 2000 photons are collected (20 % light collection efficiency) |
#* 400 p.e. (assuming 20% quantum efficiency) | #* 400 p.e. (assuming 20% quantum efficiency) | ||
| + | |||
| + | |||
| + | ''' Counter rates (according to the latest Dan's design) ''' | ||
| + | * 1.1 μ A electron current, <math>10^{-4} X_{0}</math> radiator | ||
| + | : (corresponds to the photon flux of <math>5x10^{7}</math> phot/sec in the cohrent peak region) | ||
| + | * Counter rate around E_gamma = 9 GeV: '''1 - 1.5 MHz''' | ||
| + | : (counter with ~20 MeV) | ||
| + | * The rate is about 4 times larger for the last counter (at 3 GeV) | ||
| + | |||
| + | |||
| + | ''' Operating gain ''' | ||
| + | * Assume '''4 MHz counter rate, 400 p.e., 50 μA anode current''' | ||
| + | * The maximum anode current is typically 100 μ A. We have to operate tubes at at least 2 times smaller current. | ||
| + | * Gain = A / (Rate x Np.e. x 1.6 10^-19) = <math>2x10^{5}</math> | ||
| + | |||
| + | |||
| + | ''' Pulse amplitude''' | ||
| + | |||
| + | * Assume a triangular pulse 20 ns wide | ||
| + | * Umax = ( 2 / 20 ns ) x Np.e. x Gain x 1.6 10^-19 x 50 Om = '''64 mV''' | ||
| + | * 128 mV for operating at two times smaller rate of 2 MHz (i.e., two times higher gain) | ||
| + | * The signal pulse has to be splitted (FADC250 and LE/F1TDC) | ||
| + | * We should consider to use amplifiers | ||
Latest revision as of 09:54, 21 September 2012
Light yield estimates
- 5 mm thick scintillators
- 10000 initial photons per tagged electron
- assume that 2000 photons are collected (20 % light collection efficiency)
- 400 p.e. (assuming 20% quantum efficiency)
Counter rates (according to the latest Dan's design)
- 1.1 μ A electron current,
radiator
- (corresponds to the photon flux of
phot/sec in the cohrent peak region)
- Counter rate around E_gamma = 9 GeV: 1 - 1.5 MHz
- (counter with ~20 MeV)
- The rate is about 4 times larger for the last counter (at 3 GeV)
Operating gain
- Assume 4 MHz counter rate, 400 p.e., 50 μA anode current
- The maximum anode current is typically 100 μ A. We have to operate tubes at at least 2 times smaller current.
- Gain = A / (Rate x Np.e. x 1.6 10^-19) =
Pulse amplitude
- Assume a triangular pulse 20 ns wide
- Umax = ( 2 / 20 ns ) x Np.e. x Gain x 1.6 10^-19 x 50 Om = 64 mV
- 128 mV for operating at two times smaller rate of 2 MHz (i.e., two times higher gain)
- The signal pulse has to be splitted (FADC250 and LE/F1TDC)
- We should consider to use amplifiers
