Question 2 of 6: LED Link Budget, Noise and Modulated SNR

Question 2: LED Link Budget, Noise and Modulated SNR (20 marks)

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. An 850 nm LED feeding a multimode fibre and a PIN receiver whose electrical parameters are collected below.

Find. The system loss the link can absorb at the specified SNR, the fibre length that loss buys, confirmation that shot noise is negligible beside thermal noise, and the SNR that survives when the source is sinusoidally modulated at $m = 0.5$ through 26 dB of loss.

Approach. Fix the noise floor from the load resistor, invert the SNR definition to get the photocurrent the detector must deliver, convert that to a received optical power, and express the ratio of launched to received power in decibels; the remaining parts re-use the same noise floor with a modulated signal.

1. Establish the thermal noise floor. The mean-square Johnson noise current of the load resistor over the receiver bandwidth is $$\langle i_T^{2}\rangle = \frac{4kTB}{R_L} = \frac{4(1.381\times10^{-23})(300)(50\times10^{6})}{50} = 1.657\times10^{-14}\ \text{A}^{2},$$ that is, an rms noise current of $i_T = 128.7$ nA. Every subsequent part of this question is measured against that number.
2. Invert the SNR specification for the required photocurrent. Thirteen decibels is a power ratio of $10^{1.3} = 19.95$, and with thermal noise dominant $\mathrm{SNR} = I_p^{2}/\langle i_T^{2}\rangle$, so $$I_p = \sqrt{19.95 \times 1.657\times10^{-14}} = 5.75\times10^{-7}\ \text{A} = 0.575\ \mu\text{A}.$$ Note this is 4.5 times the rms noise current — a modest but perfectly usable margin.
3. Convert to a received optical power and a loss allowance. Dividing by the responsivity, $$P_r = \frac{I_p}{R} = \frac{5.75\times10^{-7}}{0.65} = 8.85\times10^{-7}\ \text{W} = 0.885\ \mu\text{W},$$ and the loss the link may absorb between the LED and the detector is $$\text{Loss} = 10\log_{10}\frac{P_t}{P_r} = 10\log_{10}\frac{5\times10^{-3}}{8.85\times10^{-7}} = \boxed{\;37.5\ \text{dB}\;}$$ Equivalently the launched $+6.99$ dBm may fall to $-30.5$ dBm.
4. Part (b): spend the budget on connectors and fibre. Taking 12 dB for connectors and coupling leaves $37.52 - 12 = 25.52$ dB for the fibre itself, and at 3.5 dB/km that is $$L_{\max} = \frac{25.52\ \text{dB}}{3.5\ \text{dB/km}} = \boxed{\;7.29\ \text{km}\;}$$ The fixed 12 dB of connector loss costs almost 3.5 km of reach, which is why a link of this class is engineered with as few demountable joints as the layout allows.
5. Part (c): size the shot noise against the thermal noise. Shot noise accompanies the signal photocurrent and the dark current together: $$\langle i_s^{2}\rangle = 2q(I_p+I_d)B = 2(1.602\times10^{-19})(5.75\times10^{-7}+10\times10^{-9})(50\times10^{6}) = 9.37\times10^{-18}\ \text{A}^{2},$$ an rms current of only 3.06 nA. The ratio to the thermal term is $$\frac{\langle i_T^{2}\rangle}{\langle i_s^{2}\rangle} = \frac{1.657\times10^{-14}}{9.37\times10^{-18}} = \boxed{\;1.77\times10^{3}\ \ (32.5\ \text{dB})\;}$$ Thermal noise is more than three orders of magnitude larger, so the assumption made in part (a) is amply justified; ignoring the shot term changes the answer by less than 0.03 %.
6. Part (d): received power and mean photocurrent under 26 dB of loss. The average optical power reaching the detector is $$P_r = (5\ \text{mW})\,10^{-26/10} = 12.56\ \mu\text{W}, \qquad I_p = R\,P_r = 8.16\ \mu\text{A}.$$ This is fourteen times the photocurrent of part (a), because 26 dB of loss is 11.5 dB less than the link could have tolerated.
7. Form the modulated signal power and the total noise. With a sinusoidal modulation index $m$, the signal component of the photocurrent has amplitude $m I_p$, so its mean-square value is $$\langle i_{\text{sig}}^{2}\rangle = \frac{(m I_p)^{2}}{2} = \frac{(0.5\times8.16\times10^{-6})^{2}}{2} = 8.33\times10^{-12}\ \text{A}^{2}.$$ The noise is the same thermal floor plus the (now larger) shot term: $$\langle i_n^{2}\rangle = \frac{4kTB}{R_L} + 2q(I_p+I_d)B = 1.657\times10^{-14} + 1.31\times10^{-16} = 1.670\times10^{-14}\ \text{A}^{2}.$$ Even at fourteen times the photocurrent the shot term contributes under one percent of the noise, so the receiver is still firmly thermal-noise limited.
8. Evaluate the modulated SNR. $$\mathrm{SNR} = \frac{8.33\times10^{-12}}{1.670\times10^{-14}} = 498.8 \quad\Longrightarrow\quad \boxed{\;\mathrm{SNR} = 27.0\ \text{dB}\;}$$ The 11.5 dB of unspent loss budget would by itself have bought 23 dB of electrical SNR (electrical power goes as the square of optical power), but half of that is handed back by the modulation index: sinusoidal modulation at $m = 0.5$ costs $20\log_{10}(m/\sqrt{2}\,/\,1) $ relative to full on-off keying, about 9 dB. The net 14 dB improvement over part (a) is exactly what the arithmetic returns.