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Entry  Wed Aug 4 16:52:59 2010, Razib, Aidan, Update, Phase Camera, Sideband power measurement (updated) Setup_08_04_2010.jpgpower_40MHz.pngpower_25MHz.png
    Reply  Thu Aug 12 16:52:02 2010, Razib, Update, Phase Camera, Sideband power measurement (updated) SCR.jpgDC_sig_sideband_profile.jpgsideband_profile.jpgsine_trig.jpgsine_trig.jpg
       Reply  Thu Aug 12 17:10:07 2010, Koji, Update, Phase Camera, Sideband power measurement (updated) 
          Reply  Thu Aug 12 17:28:28 2010, Razib, Update, Phase Camera, Sideband power measurement (updated) 
Message ID: 3360     Entry time: Wed Aug 4 16:52:59 2010     Reply to this: 3411
Author: Razib, Aidan 
Type: Update 
Category: Phase Camera 
Subject: Sideband power measurement (updated) 

Aidan and I made some attempt to measure the power of the sidebands so that we can calculate our expected signal strength.

Our setup looks like the following:

Setup_08_04_2010.jpg

 

As light from the laser is split into two at BS1, the transmitted beam has higher power as our BS1 is only coated for 1064nm. We get two reflected beams from BS1, one reflected of the front surface and the other from the back surface. We took the stronger back reflected beam to the EOM driven at 40 MHz (also at 25 MHz at  a later time). The AOM produced a reference beam with 40 .000 005 MHz offset which we recombined with the sidebands obtained from the EOM. The beat produced is sent off to PDA 10CF connected to 4395A spectrum analyzer.

The plots for 40MHz sidebands and 25 MHz sidebands looks like this:

power_40MHz.png

From the above spectra, at 40 MHz sideband regime:

Power of the carrier @ 40 MHz = -39.72 dBm

Power of the sideband @ 80 MHz = -60.39 dBm

 

 

power_25MHz.png

At 25 MHz sideband regime,

Power of the carrier @ 40 MHz = -40.22 dBm

Power of the upper sideband @ 65 MHz = -61.72 dBm

Power of the lower sideband @ 15 MHz = -60.99 dBm

 

Power Measurement:

We made some necessary power measurement using a PD connected to a voltmeter after the EOM and the AOM when the EOM is driven at 40 MHz:

___________________________________________________________

Dark :  0.025 V

AOM on: 4.10 V    (EOM blocked)

EOM : 2.425 V      (AOM blocked)

___________________________________________________________

 From the earlier calculation (ref: Elog entry July 28) the power that we expect to see at the PD is,

P= A_c ^2 + A_r^2 + A_(-sb)^2+ A_sb ^2 +2* A_r* A_sb * cos ( w_(r,sb) t ) ,                         where A_c= carrier;   A_r= reference beam;     A_sb=Upper sideband;    A_(-sb)= Lower sideband,     w_(r,sb) = w_r - w_sb

P = A_c ^2 + A_r^2 + A_(-sb)^2+ A_sb ^2 +2* A_r* A_sb  , letting cos (w_(r,sb) go to 1) is order to approximate the maximum signal

So the signal that we expect to see relative to the DC ( i.e    A_c ^2 + A_r^2 + A_(-sb)^2+ A_sb ^2,    the first four terms of the power equation) is,

Sig = 2* A_r* A_sb    / { A_c ^2 + A_r^2 + A_(-sb)^2+ A_sb ^2 },

Since the modulation index is small, the power in the sideband is very small compared to carrier and the reference beam. So we can ignore the sideband power for the signal expression.

So,

Sig = 2* A_r* A_sb  /  ( A_c ^2 + A_r^2 )

So if we want to maximize this signal w.r.t the reference then,

d (sig)/ d(A_r) = 2 { ( A_c ^2  - A_r^2) *A_sb } / {( A_c^2 + A_r^2)} ^2

Thus, the signal is maximized when,

A_r^2 = A_c^2

 

We adjusted the AOM to be driven at + 7.7 dBM so that the new power at the AOM matched the EOM power, which is 2.397 in the voltmeter.

So the power at both the AOM and the EOM are:

P_AOM = ( V_AOM - V_dark) / (PD responsitivity * Transimpedance gain)

               = ( 2.397 - 0.025 ) / ( 0.45  * 1.5 x 10 ^5 )

               = 3.51 x 10 ^ - 5  W

P_EOM = (V_EOM - V _dark) / (PD responsitivity * Transimpedance gain)

               = ( 2. 425 - .0.025) / ( 0.45 * 1.5 x 10 ^5 )

               = 3.55 x 10^ - 5  W

 

From the spectra of the 40 MHz sideband above, the ratio of the carrier and the sideband amplitude is:  A_c / A_sb = 10.8 .

P_EOM = A_c ^2 + 2 A_sb ^2

Therefore, A_sb = sqrt ( P_EOM / 118.64) = 5.47 x 10^ - 4   V/m

Thus,     A_c = 5.908 x 10^ -3   V/m

and    A_r = sqrt ( P_AOM) = 5.92 x 10 -3    V/m.

 

This measurement can be used to calculate the signal to contrast ratio (SCR) that we expect to see:

SCR = 2 A_r * A_sb  / ( A_c^2  + A_r^2 )  = 0.09

 

Our next step is to measure the actual signal to constrast ratio as seen by the camera. Details of that will be posted soon.

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