Question 6 of 7: pin diode with a transimpedance front end — responsivity, bandwidth and required optical power

Question 6: pin diode with a transimpedance front end — responsivity, bandwidth and required optical power (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. A silicon pin diode feeding a transimpedance (shunt-feedback) preamplifier.

Find. The responsivity, the bandwidth of the diode-plus-amplifier combination and whether equalisation is needed, and the incident optical power in dBm that achieves the specified SNR.

Approach. Responsivity comes straight from the quantum efficiency; the bandwidth from the Miller-reduced input resistance working into the total input capacitance; and the required power from equating the signal-to-noise ratio, with shot and thermal noise, to the specified 55 dB and solving the resulting quadratic for the photocurrent.

1. Responsivity. Each absorbed photon of energy $hc/\lambda$ yields $\eta$ electrons of charge $q$, so $$R_{0} = \frac{\eta q \lambda}{hc} = \frac{(0.50)(1.602\times10^{-19})(0.83\times10^{-6})} {(6.626\times10^{-34})(2.998\times10^{8})}$$ $$\boxed{R_{0} = 0.335\ \text{A/W}}$$ The convenient shortcut is $R_{0} = \eta\lambda[\mu\text{m}]/1.24 = 0.5(0.83)/1.24$, which gives the same value.
2. Effective input resistance. Shunt feedback around a gain of $A$ makes the summing node look like $R_{f}$ divided by the loop gain: $$R_{\text{in}} = \frac{R_{f}}{1+A} = \frac{50\times10^{3}}{33} = 1.52\ \text{k}\Omega$$ The diode and amplifier bias resistances sit in parallel with it, $R_{d}\parallel R_{a} = (1\ \text{M}\Omega)\parallel(10\ \text{M}\Omega) = 909\ \text{k}\Omega$, which is so much larger that it barely matters: $$R_{\text{eq}} = R_{\text{in}}\parallel(R_{d}\parallel R_{a}) = 1.513\ \text{k}\Omega$$
3. Combination bandwidth. The total capacitance shunting that node is $C_{T} = C_{d}+C_{a} = 1 + 6 = 7\ \text{pF}$, so the front-end pole sits at $$B_{\text{amp}} = \frac{1}{2\pi R_{\text{eq}} C_{T}} = \frac{1}{2\pi (1512.7)(7\times10^{-12})}$$ $$\boxed{B_{\text{amp}} = 15.0\ \text{MHz}}$$ Since $15.0\ \text{MHz} \gt 10\ \text{MHz}$, the front end already passes the required post-detection band, so no equalisation is necessary. (Had the same 50 k$\Omega$ been used as a plain high-impedance load, the pole would have sat at $1/[2\pi(47.4\ \text{k}\Omega)(7\ \text{pF})] = 480\ \text{kHz}$ and heavy equalisation would have been unavoidable — that factor of 31 is exactly the loop gain $1+A$, and it is the reason the transimpedance topology is used.)
4. Noise sources at the required bandwidth. Thermal noise is generated by the resistances shunting the input; taking the parallel combination $R_{\text{th}} = R_{f}\parallel R_{d}\parallel R_{a} = 47.4\ \text{k}\Omega$, $$\langle i_{th}^{2}\rangle = \frac{4kTB}{R_{\text{th}}} = \frac{4(1.381\times10^{-23})(295)(10^{7})}{47.38\times10^{3}} = 3.44\times10^{-18}\ \text{A}^{2}$$ The dark-current shot noise is $\langle i_{d}^{2}\rangle = 2qI_{d}B = 1.60\times10^{-21}\ \text{A}^{2}$, three orders of magnitude smaller, and the signal shot noise is $2qI_{p}B$.
5. Solve for the photocurrent. Writing the signal-to-noise ratio as a power ratio, $\text{SNR} = 10^{55/10} = 3.162\times10^{5}$, and $$\text{SNR} = \frac{I_{p}^{2}}{2q(I_{p}+I_{d})B + \langle i_{th}^{2}\rangle}$$ Rearranging gives a quadratic in $I_{p}$, $$I_{p}^{2} - \text{SNR}\,(2qB)\,I_{p} - \text{SNR}\left(2qI_{d}B + \langle i_{th}^{2}\rangle\right) = 0$$ $$I_{p}^{2} - (1.013\times10^{-6})I_{p} - 1.088\times10^{-12} = 0$$ whose positive root is $$\boxed{I_{p} = 1.67\ \mu\text{A}}$$ Substituting back reproduces $\text{SNR} = 3.162\times10^{5}$ exactly, confirming the root.
6. Convert to optical power. Dividing by the responsivity, $$P_{\text{in}} = \frac{I_{p}}{R_{0}} = \frac{1.666\times10^{-6}}{0.3347} = 4.98\ \mu\text{W}$$ $$P_{\text{in}}[\text{dBm}] = 10\log_{10}\!\left(\frac{4.978\times10^{-6}}{10^{-3}}\right)$$ $$\boxed{P_{\text{in}} = -23.0\ \text{dBm}}$$ The signal shot-noise term contributes about two-thirds of the total noise at this operating point, so the receiver is close to the crossover between the thermal-limited and shot-limited regimes — which is why the quadratic, rather than the simpler thermal-only approximation, was worth solving.