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:
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!