Operation of PIN photo Diode - Optiwave

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This topic has 4 replies, 2 voices, and was last updated 7 years, 3 months ago by K. Esakki Muthu.

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- December 10, 2016 at 5:19 am #41803
  
  Md. Mahfuzul Azam Ghuri
  Participant
  
  Hello. When we transmit 2 optical side-bands having a frequency spacing of 60 GHz through a fiber, at the end how the PIN photo diode detects this signal as 60 GHz Rf signal?
- December 10, 2016 at 12:42 pm #41804
  
  K. Esakki Muthu
  Participant
  
  Io(t)= R |Eo(t).Eo(t)*|, Where R is the reponsivity , Eo(t) is the optical field and Eo(t)* is the complex conjugate of the field.
- December 10, 2016 at 3:00 pm #41805
  
  Md. Mahfuzul Azam Ghuri
  Participant
  
  thanks for the answer. but it can’t makes me clear . isn’t it represent the amplitude of the RF spectrum. If an optical spectrum contains two sidebands at 193.07 and 193.14 THz, then when PIN detects it Why it place the rf spectrum at 60 GHz ? this i want to know , why in 60GHz? not in 193.07 and 193.14 THz ? Thanks in advance .
- December 10, 2016 at 3:05 pm #41806
  Md. Mahfuzul Azam Ghuri
  Participant
  this is the output .
  Attachments:
  Capture2.jpg
- December 31, 2016 at 3:50 pm #41888
  
  K. Esakki Muthu
  Participant
  
  can you please show your circuit diagram? The frequency separation between the 193.07 and 193.13 THz is 60 GHz. thats why you are getting 60 GHz RF signal.
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