Question 4 of 7: Receiver SNR — Thermal and Shot-Noise Limits, APD Gain and Photon Counting

Question 4: Receiver SNR — Thermal and Shot-Noise Limits, APD Gain and Photon Counting (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.

Find. (a) SNR, (b) thermal-limited SNR, (c) shot-limited SNR (quantum limit), (d) APD gain $M$ for SNR = quantum limit − 5 dB, (e) photons/bit at 100 Mbps.

Approach. Compute the unmultiplied signal current $I_s=RP$, then the shot and thermal noise variances $\sigma_{\text{shot}}^2=2q(I_s+I_d)B$ and $\sigma_{\text{th}}^2=4kTB/R_L$; the actual SNR is $I_s^2/(\sigma_{\text{shot}}^2+\sigma_{\text{th}}^2)$ and each limiting case keeps only one noise term. For an ideal ($F=1$) APD of gain $M$, signal and shot noise both scale with $M$ while thermal noise does not, so raising $M$ pushes the SNR toward the shot-noise (quantum) limit.

1. Part (a) — actual SNR. The signal current is $$I_s=RP=(0.5\ \text{A/W})(1\times10^{-6}\ \text{W})=0.50\ \mu\text{A}.$$ The shot-noise variance (signal + dark current) and thermal-noise variance are $$\sigma_{\text{shot}}^2=2q(I_s+I_d)B,\qquad \sigma_{\text{th}}^2=\frac{4kTB}{R_L},$$ which evaluate to $\sigma_{\text{shot}}^2\approx8.07\times10^{-17}\ \text{A}^2$ and $\sigma_{\text{th}}^2\approx1.66\times10^{-13}\ \text{A}^2$ — thermal noise already dominates by more than three orders of magnitude with this small $50\ \Omega$ load. The total SNR is $$\mathrm{SNR}=\frac{I_s^2}{\sigma_{\text{shot}}^2+\sigma_{\text{th}}^2} =\boxed{1.51\ (\,1.78\ \text{dB}\,)}.$$
2. Part (b) — thermal-noise-limited SNR. Keeping only $\sigma_{\text{th}}^2$, $$\mathrm{SNR}_{\text{th}}=\frac{I_s^2}{\sigma_{\text{th}}^2} =\boxed{1.51\ (\,1.79\ \text{dB}\,)},$$ essentially identical to the total SNR above — confirming this receiver is thermal-noise limited as designed (unmultiplied, $50\ \Omega$ load, no gain).
3. Part (c) — shot-noise-limited SNR (quantum limit). Keeping only $\sigma_{\text{shot}}^2$, $$\mathrm{SNR}_{\text{shot}}=\frac{I_s^2}{\sigma_{\text{shot}}^2} =\boxed{3096\ (\,34.91\ \text{dB}\,)}.$$ This is roughly $10^3$ times better than the thermal-limited value — the gap an APD's internal gain is meant to close.
4. Part (d) — APD gain for SNR = quantum limit − 5 dB. As $M\to\infty$ an ideal APD's SNR approaches $\mathrm{SNR}_{\text{shot}}$ from part (c) (both signal and shot noise scale as $M^2$, so $M$ cancels, while the fixed thermal term becomes negligible) — that is exactly the "quantum limit." The target is $$\mathrm{SNR}_{\text{target}}=\mathrm{SNR}_{\text{shot}}\times10^{-5/10} =3096\times0.3162\approx979.$$ With gain $M$, $I_s(M)=MI_s$ and $\sigma_{\text{shot}}^2(M)=M^2\cdot2q(I_s+I_d)B$ (negligible excess noise, $F=1$) while $\sigma_{\text{th}}^2$ is unchanged, so $$\mathrm{SNR}(M)=\frac{M^2I_s^2}{M^2\sigma_{\text{shot}}^2+\sigma_{\text{th}}^2} =\mathrm{SNR}_{\text{target}}.$$ Solving this quadratic in $M^2$ gives $$M=\boxed{30.8}.$$
5. Part (e) — photons per bit at 100 Mbps PCM. The photon energy at $1.3\ \mu\text{m}$ is $$h\nu=\frac{hc}{\lambda}=0.954\ \text{eV},$$ and at 100 Mbps the bit period is $T_b=1/(100\times10^{6})=10\ \text{ns}$, so the energy delivered per bit at the (unmultiplied) received power $P=1\ \mu\text{W}$ is $E_{\text{bit}}=PT_b=1\times10^{-14}\ \text{J}$, giving $$N_{\text{photon/bit}}=\frac{E_{\text{bit}}}{h\nu}=\boxed{65443\ \text{photons/bit}}.$$