R&D (FilterCavity)
NaokiAritomi - 19:27, Monday 27 May 2019 (1369)
Alignment method of homodyne
 
[Aritomi, Chien-Ming, Yuhang]
 
Our BS for homodyne (BSW41-1064) is R = 49.14513016% (p-pol) and T= 50.49834442% (p-pol) when the injection angle is 45 deg. That means we have 1.35% unbalance when the injection angle is 45 deg. So far we made the injection angle 45 deg and actually we had 74 mV homodyne DC offset even if the homodyne is aligned. Each PD has 5V so the unbalance is 1.48%. To reduce this unbalance, the injection angle should be below 45 deg.
 
The alignment method of homodyne will be as follows. The rough sketch is also attached.
 
1. inject BAB to homodyne PD along the hole line of optical bench
2. adjust the angle of BS so that homodyne DC of BAB is 0
3. align BAB to AMC with two  mirrors after homodyne BS
4. align LO to AMC with two mirrors before homodyne BS
 
Our BS for homodyne (BSW41-1064) is R = 49.14513016% (ppol) and T= 50.49834442% (ppol).
That means we have 1.35% unbalance when the injection angle is 45 deg. So far we made the injection angle 45 deg and actually we had 70 mV homodyne DC offset even if homodyne is aligned. Each PD has 5V so the unbalance is 1.4%. We have to make the injection angle below 45 deg to get 1:1 beam splitting ratio.
Our BS for homodyne (BSW41-1064) is R = 49.14513016% (ppol) and T= 50.49834442% (ppol).
That means we have 1.35% unbalance when the injection angle is 45 deg. So far we made the injection angle 45 deg and actually we had 70 mV homodyne DC offset even if homodyne is aligned. Each PD has 5V so the unbalance is 1.4%. We have to make the injection angle below 45 deg to get 1:1 beam splitting ratio.
Images attached to this report
1369_20190527122659_img7475.jpg