0
\$\begingroup\$

I try to calculate the Q value of APD photodetector with these parameters:

$$R \hspace{0.1in} (response) = 1$$ $$M \hspace{0.1in} (magnification) = 10$$ $$k \hspace{0.1in} (ionization \hspace{0.1in} constant) = 0.1$$ $$N_{th} \hspace{0.1in} (noise \hspace{0.1in} density \hspace{0.1in} of \hspace{0.1in} thermic \hspace{0.1in} current) = 12 \hspace{0.1in} pA/\sqrt{Hz}$$ $$P_1 \hspace{0.1in} (optical \hspace{0.1in} power \hspace{0.1in} of \hspace{0.1in} '1' \hspace{0.1in} bit) = 0.5 \hspace{0.1in} \mu W$$ $$f \hspace{0.1in} (frequency)= 193.1 \hspace{0.1in} THz$$ $$r_e \hspace{0.1in} (extintion \hspace{0.1in} ratio) = 0.05 \xrightarrow \quad r_e = P_0/P_1$$ $$B_R \hspace{0.1in} (bitrate) = 10 \hspace{0.1in} Gbits/s$$ $$B_e \hspace{0.1in} (electrical \hspace{0.1in} bandwidth) = B_R$$

I use as reference the explanation of "Principles of Optical Communications" of José Capmany and it applying the following equation for Q value:

Q equation

The possible values of Q are:

  • 77.499
  • 2.9592
  • 0.019739
  • 0.3947

I search more information about how I can calculate the Q value in APD photodetectors. However, I haven't found any way to get one of the above results. Somebody could help me?

Thanks!

Improve this question
\$\endgroup\$
4
  • \$\begingroup\$ How did you arrive at 4 possible values? \$\endgroup\$ – Andy aka Apr 29 '20 at 10:24
  • \$\begingroup\$ @Andyaka It’s an exercise. These values are options of the statement. \$\endgroup\$ – Adrián Marín Villar Apr 29 '20 at 11:19
  • \$\begingroup\$ What value did you arrive at? \$\endgroup\$ – Andy aka Apr 29 '20 at 11:20
  • \$\begingroup\$ @Andyaka Q = 1.8425. I don’t found the error. \$\endgroup\$ – Adrián Marín Villar Apr 29 '20 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.