Trying to replicate what a pulsed wide-beam source from the gallary does to camera with lab-bench equipment

Ideas

  • Z_Substrate_pulse_proposal.pdf
  • Substrate Pulse Cals
    # ePix10kA
    # 20230210
    # Scott T. Block
    # TRYING TO FIGURE OUT IF WE CAN SIMULATE DUMPING ENOUGH CHARGE
    import numpy as np
    
    # From Bojan Markovic
    # Here is the conversion from photon energy to generated sensor charge:
    # Q[C]= PhotonEnergy[eV] * (electron/3.6eV)*(1.6*10^(-19)[C/electron]) 
    # basically each 3.6eV generates an electron and the charge of an electron is 
    # 1.6*10^(-19) coulombs
    
    t_plz = 50e-15 # [sec] pulse time of beam
    PixTot = 4 * 176 * 192 # number of pixels
    C_pix = 150e-15 # [F] estiamted cap/pixel at bias???
    C_tot = PixTot * C_pix
    print('C_tot = {}[nF]'.format(C_tot*1e9))
    
    # how do we go from eV to number of electrons, 2400 keV --> 6.6e5 electrons,
    # assuming per sensor
    E_trp = 2400e3 # [eV] energy <-- what beam produces to trip sensor in AML
    E_max = 10e6 # [eV] max energy
    rto_E = E_trp / E_max # ratio of trip point to max
    print('trp/max = {}'.format(rto_E))
    # The pair creation energy in Si is ~3.6[eV]
    e_N = E_trp / 3.6
    # oh ok, we know we are getting energy electron volts... 
    print('e_N = {}[#-of-electrons]'.format(e_N))
    # number of elementry charges to current
    e = 1.602176634e-19 # coulombs (charge)
    Q = e_N * e # coulombs (charge)
    print('\nThe amount of charge the accelerator puts on to a single pixel')
    print('Q = {}[fempto-Coulombs]'.format(Q*1e15))
    Iplz = Q / t_plz # this is accelerator pulse (not used)
    
    # voltage to get the same amount of charge, Q = CV
    Vplz = Q / C_pix
    print('\nWith our current C/pix = {}fF, we need a voltage of...'.format(C_pix*1e15))
    #print('[Vplz,Iplz] = [{},{}]'.format(Vplz,Iplz))
    print('Vplz = {}[mV]'.format(1000*Vplz))
    print('To generate the same amount of charge in less than 100usec')
    
    # WE are generating a postiave pulse of +5V for 100nsec, how much attenuation do 
    # we need to get down to Vplz, desired 10MeV
    Vmax = Vplz / rto_E
    Attn = -20*np.log10(Vplz / Vmax)
    print('\nWe can generate a pulse from 0 to 5V, attenuate?')
    print('Attn = {}[dB]'.format(Attn))

Testing

  • NSUB Capacitance vs. Bias-Voltage???

Hardware

Q&A

  • What is the amplitude of the voltage pulse required to emulate beam-line?
    • Clue: Not sure, but output voltage capabilities of DAC might give a clue also need to make sure not over-driving till break down
    • How much charge do we expect tripping to happen at?
      • "medium -> low is 2400 keV/pixel so 6.6*10**5 electrons/pixel. High->low is probably 1/3 that, but we don't use AHL currently, so I'd focus on AML if possible." – Philp Hart
    • Beamline Pulse duration?
      • "The pulse is a delta function - 50 femtoseconds or less, typically.  The pair creation energy in Si is about 3.6 eV." – Philp Hart
  • What is typical bias voltage applied on Detector Bias line to reverse bias the substrate?
  • At this bias or some known bias voltage what is the expected substrate capacitance for a entire sensor (there are asic 4 in parallel)? 
    • "The detector capacitance per pixel should be around 150fF" – Bojan
    • An ASIC is ~2x2[cm^2] at 176x192 pixels
    • Total Number of pixels = 2x (176x192) x 2 = 352x384 = 135168 → 20.3nF
  • When the beam hits the substrate, which way does the current flow (in/out of pixel front-end) or (assuming substrate is negatively biased) that the current being drawn from the pixel front-end is less/more negative when a pulse hits (positive/negative voltage pulse)?
    • "It's for sure positive (not sure the actual voltage value). The baseline of the preamplifier is around 2V and it swings down when there is a signal."–Bojan 
  • What is the capacitance vs. bias curve of substrate?  
    • See testing section above
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