Question 2 of 6: 850 nm LED/PIN Link — Loss Budget and SNR

Question 2: 850 nm LED/PIN Link — Loss Budget and SNR (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) Tolerable system loss for SNR = 13 dB (thermal-noise-limited). (b) Maximum fiber length given 12 dB connector/coupling loss and 3.5 dB/km fiber loss. (c) Confirm shot noise $\ll$ thermal noise at the part (a) operating point. (d) SNR for AM modulation ($m=0.5$) at 26 dB system loss.

Approach. Compute the receiver's thermal-noise current from Johnson-noise theory, work backward from the required SNR to the minimum detectable power, then convert to a loss budget and a fiber length; separately compute the shot-noise current at that operating point for comparison, and re-apply the SNR formula with an AM signal term for part (d).

1. Thermal (Johnson) noise current. $$\overline{i_{th}^2}=\frac{4kTB}{R_L}=\frac{4(1.381\times10^{-23})(300)(50\times10^6)}{50} =1.657\times10^{-14}\ \text{A}^2,\qquad i_{th}=1.287\times10^{-7}\ \text{A}.$$
2. (a) Minimum received power for SNR = 13 dB. With thermal noise dominant, $\mathrm{SNR}=(RP_r)^2/\overline{i_{th}^2}$, and $13\ \text{dB}\to\mathrm{SNR}=10^{1.3}=19.95$: $$P_r=\frac{\sqrt{\mathrm{SNR}}\;i_{th}}{R}=\frac{\sqrt{19.95}\,(1.287\times10^{-7})}{0.65}=8.85\times10^{-7}\ \text{W}.$$ The tolerable system loss is $$\text{Loss}=10\log_{10}\!\frac{P_{LED}}{P_r}=10\log_{10}\!\frac{5\times10^{-3}}{8.85\times10^{-7}} =\boxed{37.5\ \text{dB}}.$$
3. (b) Maximum fiber length. Of the 37.5 dB budget, 12 dB is fixed (connector + coupling), leaving fiber attenuation at 3.5 dB/km: $$L_{\max}=\frac{37.5-12}{3.5}=\boxed{7.29\ \text{km}}.$$
4. (c) Shot noise at the part (a) operating point. The signal photocurrent there is $I_p=RP_r=0.65\times8.85\times10^{-7}=5.75\times10^{-7}$ A, so $$\overline{i_{sh}^2}=2q(I_p+I_d)B=2(1.602\times10^{-19})(5.75\times10^{-7}+10^{-8})(50\times10^6)=9.37\times10^{-18}\ \text{A}^2.$$ Comparing to the thermal term, $$\frac{\overline{i_{sh}^2}}{\overline{i_{th}^2}}=\frac{9.37\times10^{-18}}{1.657\times10^{-14}}\approx5.7\times10^{-4}\ll1,$$ confirming shot noise is indeed negligible next to thermal noise, consistent with the part (a) assumption.
5. (d) SNR under AM modulation, 26 dB system loss. The received average power is $$P_r=\frac{P_{LED}}{10^{26/10}}=\frac{5\times10^{-3}}{398.1}=1.256\times10^{-5}\ \text{W}, \qquad I_p=RP_r=8.16\times10^{-6}\ \text{A}.$$ The sinusoidal AM signal current has peak amplitude $mI_p$ and mean-square value $(mI_p)^2/2$: $$\overline{i_{sig}^2}=\frac{(mI_p)^2}{2}=\frac{(0.5\times8.16\times10^{-6})^2}{2}=8.33\times10^{-12}\ \text{A}^2.$$ Total noise now includes both shot and thermal terms at this (larger) $I_p$: $$\overline{i_{sh}^2}=2q(I_p+I_d)B=1.31\times10^{-16}\ \text{A}^2,\qquad \overline{i_{th}^2}=1.657\times10^{-14}\ \text{A}^2\ (\text{unchanged, same }R_L,T,B).$$ $$\mathrm{SNR}=\frac{\overline{i_{sig}^2}}{\overline{i_{sh}^2}+\overline{i_{th}^2}} =\frac{8.33\times10^{-12}}{1.31\times10^{-16}+1.657\times10^{-14}}=\boxed{499\approx27.0\ \text{dB}}.$$