1
\$\begingroup\$

I want to check the thermal noise spectrum of a photodiode (PD) at different frequency response but I do not know how to calculate it. Below graph is an example which shows noise power (dBm) as a function of frequency. I want to calculate and plot similar graph using our PD. I have tried to use a laser beam with different frequency with same power and detect the signal with the PD. The output signal of the PD is connected to an oscilloscope to record the voltage Vpk. Is this the way to calculate noise power (dBm) using this formula (P(dBm)-10)/20=log(Vpk)? It seems I am using wrong procedure. Can anyone please tell how I can do this? Your answer will be appreciated.enter image description here

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

The noise power quoted in most photodiode data sheets is actually noise equivalent power and the "equivalence" is related to the incident light power on the device. Incident power gets converted to photodiode current so you have to have your PD in situ in a circuit to make sense. If you are using a TIA then that current becomes a voltage and then you can more readily measure it on an oscilloscope.

\$\endgroup\$
3
  • \$\begingroup\$ Incident optical power and photodiode current are related by a parameter called "quantum efficiency" which should be in the datasheet, allowing you to work back from a (measured) current to the equivalent optical power. \$\endgroup\$
    – user16324
    Commented May 5, 2018 at 10:18
  • \$\begingroup\$ I measured the Vpk voltage by an oscilloscope at same incident optical power but at different frequency. In this case the Vpk voltage doe not change. How will I measure noise power (dBm) at different frequency? Can you please explain it in details? \$\endgroup\$
    – Subir Das
    Commented May 7, 2018 at 3:35
  • \$\begingroup\$ @SubirDas you should use a spectrum analyser to measure the noise power over the bandwidth you require. \$\endgroup\$
    – Andy aka
    Commented May 7, 2018 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.