A transmitter has an output power of 0.1 mW. It is used with a fiber having NA= 0.25, attenuation of 6 dB/km, and length of 0.5 km. The link contains two connectors with a 2 dB average loss. The receiver has a minimum acceptable power (sensitivity) of – 35 dBm. The designer has allowed a 4 dB margin. Calculate the link power budget.

Solution:

Source power Ps

$=0.1 \mathrm{~mW}$

$ P_s=-10 \mathrm{dBm}\\ $

Since $\quad N A=0.25$

Coupling loss

$=-10 \log \left(N A^2\right)$

$ =-10 \log \left(0.25^2\right)\\ $

$ =12 \mathrm{~dB}\\ $

Fiber loss $=\alpha_f \times L$

$ I_f=(6 \mathrm{~dB} / \mathrm{km})(0.5 \mathrm{~km}) \mathrm{I}_{\mathrm{f}}=3 \mathrm{~dB}\\ $

Connector loss $=2(2 \mathrm{~dB})$

$ I_c=4 \mathrm{~dB} \text { Design margin } P_m=4 \mathrm{~dB}\\ $

Actual output power Pout = Source power $-(\Sigma$ Losses $) P_{\text {out }}=10 \mathrm{dBm}-[12 \mathrm{~dB}+3+4+4]$

$ P_{\text {out }}=-33 \mathrm{dBm}\\ $

Since receiver sensitivity given is $-35 \mathrm{dBm}$.

$ P_{\min }=-35 \mathrm{dBm}\\ $

As $P_{out} \gt P_{\min }$, the system will perform adequately over the system operating life.