Changes

Jump to navigation Jump to search
11 bytes added ,  00:17, 14 July 2008
m
no edit summary
Line 3: Line 3:  
[[Image:S20photocathode_QE.jpg|frame|HPD Photo-cathode efficiency as a function of wavelength.]]
 
[[Image:S20photocathode_QE.jpg|frame|HPD Photo-cathode efficiency as a function of wavelength.]]
   −
The quantum efficiency of this device is shown in the adjacent figure. The DEP-supplied device has an inherent capacitance (C<sub>HPD</sub>) of 200&nbsp;F and provides a gain factor of 2700 at the HV of 12&nbsp;kV and additionally requires 40&nbsp;-&nbsp;80&nbsp;V bias to collect the electrons across the diode junction. Built into the package is a transimpedance amplifier with 50&nbsp;k&Omega; || 1.5&nbsp;pF feedback, requiring ±5&nbsp;V supplies. The amplifier circuit also contains a 51.1&nbsp;&Omega; resistor in series with the output. Reading out the HPD signal from the amplifier with 50&nbsp;&Omega; termination creates a factor of two (50&nbsp;&Omega;/(50&nbsp;&Omega; + 51.1&nbsp;&Omega;)) on top of the 50&nbsp;k&Omega; current to voltage conversion.  
+
The quantum efficiency of this device is shown in the adjacent figure. The DEP-supplied device has an inherent capacitance (''C<sub>HPD</sub>'') of 200&nbsp;F and provides a gain factor of 2700 at the HV of 12&nbsp;kV and additionally requires 40&nbsp;-&nbsp;80&nbsp;V bias to collect the electrons across the diode junction. Built into the package is a transimpedance amplifier with 50&nbsp;k&Omega; || 1.5&nbsp;pF feedback, requiring ±5&nbsp;V supplies. The amplifier circuit also contains a 51.1&nbsp;&Omega; resistor in series with the output. Reading out the HPD signal from the amplifier with 50&nbsp;&Omega; termination creates a factor of two (50&nbsp;&Omega;/(50&nbsp;&Omega; + 51.1&nbsp;&Omega;)) on top of the 50&nbsp;k&Omega; current to voltage conversion.  
      Line 28: Line 28:       −
Thus, in the complex space, this integral is equivalent to the residues of two poles. Depending on the constants provided, the poles are situated either on the positive imaginary axis or along the -ib/2a line. For the physical constant of our circuit, the former set of poles is relevant. The resulting response function is:
+
Thus, in the complex space, this integral is equivalent to the residues of two poles. Depending on the constants provided, the poles are situated either on the positive imaginary axis or along the ''-ib/2a'' line. For the physical constant of our circuit, the former set of poles is relevant. The resulting response function is:
    
<math>g(t) = \frac{2\pi\alpha}{\omega_2} e^{i\omega_1 t} \sinh \omega_2 t </math>
 
<math>g(t) = \frac{2\pi\alpha}{\omega_2} e^{i\omega_1 t} \sinh \omega_2 t </math>
   −
Since the response seen on the HPD circuit output is the convolution of the light pulse shape (f<sub>L</sub>) with the response function of the HPD, the pulse shape can be recovered by deconvolving the measured signal (f<sub>M</sub>) with the g(&omega;). By the convolution theorem:
+
Since the response seen on the HPD circuit output is the convolution of the light pulse shape (''f<sub>L</sub>'') with the response function of the HPD, the pulse shape can be recovered by deconvolving the measured signal (''f<sub>M</sub>'') with the ''g''. By the convolution theorem:
    
<math>f_L(t) = \mathcal{F}^{-1} \left\{ \frac{f_M(\omega)}{g(\omega)} \right\} </math>
 
<math>f_L(t) = \mathcal{F}^{-1} \left\{ \frac{f_M(\omega)}{g(\omega)} \right\} </math>
1,004

edits

Navigation menu