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Message ID: 1992     Entry time: Fri Jul 31 15:21:19 2015
Author: Alessandra 
Type: Misc 
Category: Seismometer 
Subject: QPD calibration 

QPD calibration

 

In a previous entry I described the setup used to build the QPD calibration curve and the method used to take measurements.

Here I show the results of our measurements and calculate the Volts/displacement ratio and the laser's beam radius.

 

X axis calibration

 

  • In the following plot the X axis QPD calibration curve is shown and a fit in the linear region is made:

 

The errorbars on the plot where estimated by looking at the fluctuations of the voltage output.

 

The fit in the linear region with the function

 

f(x)=ax+b

 

returns:

 

a=(2.34\pm{0.02})\hspace{0.2cm}V/mm

b=(-10.942\pm{0.004})\hspace{0.2cm}V

 

where a gives us the Volts/displacement ratio.

 

  • The following plot shows the fit of the QPD calibration curve with the error function

 

 

 

The fit with the function:

 

f(x)=a_1+b_1\cdot{erf(c_1x+d_1)}

 

returns:

 

a_1=(0.252\pm 0.007)\hspace{0.2cm}V\\ b_1 = (1.961\pm 0.009)\hspace{0.2cm} V\\ c_1 =(1.236\pm0.007) \hspace{0.2cm}\tfrac{1}{mm}\\ d_1 =-5.94\pm 0.04

​a_1=0.2517 \hspace{0.2cm}V\\ b_1 = -1.961\hspace{0.2cm} V\\ c_1 =-1.236 \hspace{0.2cm}\tfrac{1}{mm}\\ d_1 =5.939​a_1=0.2517 \hspace{0.2cm}V\\ b_1 = -1.961\hspace{0.2cm} V\\ c_1 =-1.236 \hspace{0.2cm}\tfrac{1}{mm}\\ d_1 =5.939

The QPD's X output is a voltage given by:

 

\Delta{V}=\alpha(P_{right}-P_{left})

 

where \alpha is a constant, while P_{right}-P_{left} is the difference between the power on the right and left side of the QPD. Thus:

 

\Delta{V}=\alpha P_{0}\cdot{erf\left(\frac{\sqrt{2}(x-x_{0})}{w_x} \right )}

where P_0 is the total power transmitted by the beam, x is the distance from the beam's center to the QPD center, x_{0} is an offset and w_x is the beam radius.

 

From the expression above and the fit's results we obtain:


w_x=1.14 \hspace{0.2 cm} mm\\ x_0=4.81 \hspace{0.2cm} mm

 

Y axis calibration

 

  • Linear region fit

 

The fit with the function:

 

f(x)=px+q

 

returns

 

p=(2.89 \pm 0.03)\hspace{0.2cm}V/mm

q=(-28.647\pm0.003)\hspace{0.2 cm} V

 

where p gives us the Volts/displacement ratio.

 

  • Error function fit

The fit with the function:

 

f(x)=a_2+b_2\cdot{efr(c_2x+d_2)}

 

returns:

 

a_2=(0.041\pm 0.007)\hspace{0.2cm}V\\ b_2 = (2.005\pm 0.008)\hspace{0.2cm} V\\ c_2 =(1.311\pm 0.008)\hspace{0.2cm}\tfrac{1}{mm}\\ d_2 =-13.02\pm 0.08

 

The expression for the QPD's Y output is analogous to the one for the X axis:

 

\Delta{V}=\alpha P_{0}\cdot{erf\left(\frac{\sqrt{2}(y-y_{0})}{w_y} \right )}

 

From the fit results we obtain:

 

w_y=1.08 \hspace{0.2 cm} mm\\ y_0=9.93 \hspace{0.2cm} mm

 

  • Observation: Both w_x and w_y are compatible with the beam spot size we observed in laboratory.

 

The Volts/displacement ratios obtained here will be used to measure the resonant frequencies of the rhomboid motion.

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