17-Phys-B2 Electro-Optical Engineering · December 2017
Question 6 of 7: PIN + Transimpedance Amplifier — Responsivity, Bandwidth and Required SNR
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B2 Electro-Optical Engineering, National
Examination December 2017 — a three-hour closed-book examination (one 8.5×11 inch
double-sided handwritten note sheet permitted). The cover page states any five of the
seven questions constitute a complete paper and only the first five as they appear in
the answer book are marked; every question is nonetheless answered in full below so the paper
remains a complete study resource. The Question 6 heading carries the
stray fragment "rework this one" — almost certainly a candidate's or marker's pencil annotation, reproduced verbatim in the question box below but not a
part of the printed exam text.
Reference texts. G. Keiser, Optical Fiber Communications, 4th ed.
(fiber modes and dispersion, link power and risetime budgets, LED/laser diode output
characteristics, PIN photodiode responsivity and noise, receiver design); B. E. A. Saleh and
M. C. Teich, Fundamentals of Photonics, 2nd ed. (semiconductor laser rate equations,
photodiode quantum efficiency and noise); A. Yariv, Quantum Electronics / E. Hecht,
Optics, 5th ed. (electro-optic modulators and Pockels cells).
Find. (a) responsivity $R$, (b) the diode–TIA bandwidth and whether
equalization is needed, (c) the incident optical power for 55 dB SNR, in dBm.
PIN photodiode driving a shunt-feedback transimpedance amplifier; the feedback resistor $R_f$ and total input capacitance $C_d+C_a$ set the achievable bandwidth, boosted by the amplifier's open-loop gain.
Approach. Part (a) is the standard responsivity formula. Part (b) uses the
shunt-feedback transimpedance-amplifier bandwidth relation, where the open-loop gain $A$
multiplies the naive $1/(2\pi R_fC_{tot})$ bandwidth by $(1+A)$ — the reason the exam
supplies $A$ at all. Part (c) balances shot noise (signal + dark current) and Johnson
(thermal) noise from $R_f$ against the target SNR to solve for the minimum signal power.
Check: the diode resistance $R_d=1\ \text{M}\Omega$ and amplifier input
resistance $R_a=10\ \text{M}\Omega$ are both far larger than $R_f=50\ \text{k}\Omega$, so
neither materially loads the feedback node nor contributes significant thermal noise current
next to $R_f$ — both are carried in the Given table for completeness but dropped from the
bandwidth and noise calculations below.
Part (a) — responsivity.
$$R=\frac{\eta q\lambda}{hc}=\frac{(0.5)(1.602\times10^{-19})(0.83\times10^{-6})}{(6.626\times10^{-34})(2.998\times10^{8})}
=\boxed{0.335\ \text{A/W}}.$$
Part (b) — TIA bandwidth. A shunt-feedback transimpedance amplifier's
$-3$ dB bandwidth is enhanced above the bare $R_fC_{tot}$ rolloff by the open-loop gain,
$$f_{3dB}=\frac{1+A}{2\pi R_fC_{tot}},\qquad C_{tot}=C_d+C_a=1+6=7\ \text{pF},$$
$$f_{3dB}=\frac{1+32}{2\pi(50\times10^3)(7\times10^{-12})}=\boxed{15.0\ \text{MHz}}.$$
Since $15.0\ \text{MHz}>10\ \text{MHz}$, the amplifier already exceeds the desired
post-detection bandwidth, so no equalization is necessary.
Part (c) — power for 55 dB SNR. The photocurrent noise variance is
the sum of shot noise (from signal and dark current) and Johnson noise from $R_f$,
$$\sigma^2=\underbrace{2q(I_{ph}+I_{dark})B}_{\text{shot}}+\underbrace{\frac{4kTB}{R_f}}_{\text{thermal}},
\qquad I_{ph}=RP_{in},$$
using the desired bandwidth $B=10$ MHz. Requiring $\mathrm{SNR}=I_{ph}^2/\sigma^2=10^{5.5}=3.162\times10^5$
gives a quadratic in $P_{in}$; the thermal term alone is
$4kTB/R_f=3.26\times10^{-18}\ \text{A}^2$, and solving the quadratic (shot noise from the
signal itself is comparable in size at this power level) gives
$$P_{in}=\boxed{4.90\ \mu\text{W}}=10\log_{10}\!\left(\frac{4.90\times10^{-6}}{10^{-3}}\right)
=\boxed{-23.1\ \text{dBm}}.$$
