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22-Elec-B10 Electro-Optical Engineering · May 2015

Question 5 of 6: Receiver Front-End Choice and the Erbium-Doped Fibre Amplifier

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, May 2015 — 07-Elec-B10 Electro-Optical Engineering. Three hours, closed book; one 8.5 in × 11 in double-sided sheet of hand-written notes and one approved Casio or Sharp calculator are permitted. Six questions are printed; any five constitute a complete paper and all questions are of equal value (20 marks each). All six are solved here, because the set is a study resource rather than a timed sitting. The constants used throughout are those printed on page 1 of the paper: $\varepsilon_0 = 8.854\times10^{-12}$ F/m, $\mu_0 = 4\pi\times10^{-7}$ H/m, $c = 2.998\times10^{8}$ m/s, $q = 1.602\times10^{-19}$ C, $h = 6.626\times10^{-34}$ J·s, $k = 1.381\times10^{-23}$ J/K, and the semiconductor data (GaAs: $n = 3.63$, $E_g = 1.41$ eV; Si: $E_g = 1.11$ eV; InGaAsP: $n = 3.5$).

Reference texts. The Engineers Canada syllabus for this examination code draws on the standard fibre-optic and photonics texts, and the solutions below cite them by chapter:

Question 5: Receiver Front-End Choice and the Erbium-Doped Fibre Amplifier (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 photodetector front end at 300 K with a 680 kΩ bias resistor and 5 pF of total input capacitance, and two candidate amplifiers whose parameters are collected below.

Given data for part (a)
QuantitySymbolValue
Temperature$T$300 K
Detector bias resistor$R_b$680 kΩ
Total input capacitance$C_T$5 pF
Amplifier input resistance (both designs)$R_{in}$1 MΩ
Design A feedback resistor$R_f$220 kΩ
Design A open-loop gain$A$500

Find. The unequalised bandwidth and the input-referred noise current density of each design, a reasoned choice between them, the benefit an equaliser brings to design B, and a description of the EDFA with its role in a WDM system.

Design A - transimpedance amplifier +V 680 kΩ light −A out 220 kΩ open-loop gain A = 500 input resistance falls to Rf/(1 + A) Design B - high-impedance amplifier +V 680 kΩ light A out amplifier input resistance = 1 MΩ no feedback: the input node stays high-R
The two candidate front ends. Design A closes a feedback loop through Rf, which divides the effective input resistance by (1 + A); design B leaves the input node at the full parallel resistance of the bias resistor and the amplifier.

Approach. In both designs the bandwidth is set by the resistance seen at the input node working against the 5 pF, and the noise is set by the thermal current of that same resistance; compute both for each design and compare them at equal signal bandwidth.

  1. Design B: the resistance at the input node. The bias resistor and the amplifier input resistance appear in parallel: $$R_B = R_b \parallel R_{in} = \frac{(680)(1000)}{680+1000}\ \text{k}\Omega = 404.8\ \text{k}\Omega .$$
  2. Design B: bandwidth. That resistance charges the 5 pF, so $$B_B = \frac{1}{2\pi R_B C_T} = \frac{1}{2\pi(4.048\times10^{5})(5\times10^{-12})} = \boxed{\;78.6\ \text{kHz}\;}$$ which is useless for any fibre link — three orders of magnitude short of even the 50 MHz receiver of Question 2.
  3. Design A: the effective input resistance. Shunt feedback through $R_f$ around an inverting gain $A$ presents a Miller-reduced resistance at the summing node: $$R_{\text{eff}} = \frac{R_f}{1+A} = \frac{220\times10^{3}}{501} = 439\ \Omega .$$ This sits in parallel with the 404.8 kΩ already computed, which changes it by only 0.1 %, giving $R_A = 438.6\ \Omega$ — the feedback completely dominates.
  4. Design A: bandwidth. Against the same 5 pF, $$B_A = \frac{1}{2\pi R_A C_T} = \frac{1}{2\pi(438.6)(5\times10^{-12})} = \boxed{\;72.6\ \text{MHz}\;}$$ a factor of 923 more bandwidth than design B, obtained without changing a single component value in the detector itself. Equivalently, $B_A = (1+A)B'$ where $B'$ is the bandwidth $R_f$ alone would give — the classic gain–bandwidth exchange of shunt feedback.
  5. Noise: design B. The input-referred thermal noise current density of a resistance $R$ is $\sqrt{4kT/R}$, so $$i_{n,B} = \sqrt{\frac{4(1.381\times10^{-23})(300)}{4.048\times10^{5}}} = 2.02\times10^{-13}\ \text{A}/\sqrt{\text{Hz}} = 0.202\ \text{pA}/\sqrt{\text{Hz}} .$$ The high-impedance front end is quiet precisely because a large resistance is a weak noise-current source.
  6. Noise: design A. The feedback resistor now appears as an additional thermal source at the input, in parallel with the bias resistor and the amplifier input: $$R_{n,A} = R_b \parallel R_{in} \parallel R_f = 142.5\ \text{k}\Omega, \qquad i_{n,A} = \sqrt{\frac{4kT}{R_{n,A}}} = 0.341\ \text{pA}/\sqrt{\text{Hz}} .$$ Design A is therefore $0.341/0.202 = 1.69$ times noisier per root hertz, a penalty of $$20\log_{10}(1.69) = \boxed{\;4.5\ \text{dB}\;}$$ in input-referred noise current at equal bandwidth. Note it is $R_f$, not the reduced $R_{\text{eff}}$, that sets the noise — the feedback lowers the impedance without adding the noise a physical 439 Ω resistor would.
  7. Make the choice. Design A, the transimpedance amplifier, is the correct choice. It buys a factor of 923 in bandwidth for a 4.5 dB noise penalty, and no realistic amount of equalisation can recover three decades of bandwidth from design B without a matching noise cost of its own. The transimpedance stage also has a far wider dynamic range, because the feedback holds the detector node near a virtual earth so the diode sees an almost constant bias and cannot be driven into a nonlinear region by a strong signal. Design B would be preferred only in a genuinely sensitivity-critical, low-data-rate application — a photon-counting or instrumentation receiver — where the last decibel of noise matters more than speed.

The benefit of an equaliser after amplifier B. The high impedance at the input of design B does not destroy information; it integrates it. The front end behaves as a single-pole low-pass filter with a corner at 78.6 kHz, so signal components above that corner are attenuated by a known $20$ dB/decade — they are still present, merely small. An equaliser placed after the amplifier is a filter with the complementary response, a zero at 78.6 kHz, which lifts those components back to their proper amplitude and flattens the overall channel out to whatever bandwidth the amplifier itself can support. The crucial point is that the equaliser is after the first stage: it amplifies signal and noise equally and so does not degrade the signal-to-noise ratio the front end achieved, whereas simply lowering the bias resistance to gain bandwidth would raise the noise as the inverse square root of the resistance. The high-impedance-plus-equaliser combination therefore keeps design B's superior sensitivity while restoring usable bandwidth. Its remaining weaknesses are a limited dynamic range — a strong signal saturates the integrating front end before the equaliser can act — and the need for the equaliser response to track component tolerances and temperature drift in $R$ and $C$.

Part (b) — the erbium-doped fibre amplifier. An EDFA is a length of silica fibre whose core is doped with erbium ions, pumped by a semiconductor laser and spliced directly into the transmission path, so that a weak signal is amplified in the optical domain without ever being converted to an electrical signal.

Principle of operation. The trivalent erbium ion has a ground manifold, a metastable level about 0.8 eV above it, and a higher pump band. A pump laser at 980 nm lifts ions from the ground state into the pump band, from which they decay non-radiatively within microseconds into the metastable level; alternatively a 1480 nm pump excites the metastable level directly. Because the metastable lifetime is around 10 ms — extraordinarily long — even a modest pump power accumulates more ions there than remain in the ground state, and the medium is inverted. A signal photon in the 1530–1565 nm band then stimulates a transition back to the ground state, producing a second photon identical in wavelength, phase, direction and polarisation. The pump is combined with the signal through a wavelength division multiplexing coupler; optical isolators at each end prevent reflections from turning the amplifier into a laser, and an output filter removes residual pump light and out-of-band spontaneous emission.

Main characteristics. A typical EDFA delivers 20–35 dB of small-signal gain over the 1530–1565 nm C-band (extendable to the L-band with a longer, differently pumped fibre), saturated output powers of $+15$ to $+23$ dBm, and a noise figure of 4–6 dB against the 3 dB quantum limit. The gain spectrum is not flat — it peaks near 1532 nm with a shoulder around 1550 nm — and the metastable lifetime of 10 ms is far longer than a bit period at any practical line rate.

Advantages in a WDM system. The gain bandwidth of tens of nanometres covers dozens of wavelength channels at once, so one amplifier replaces one regenerator per channel — the economic fact that made dense WDM viable. Amplification is transparent to modulation format and bit rate, so the line can be upgraded without touching the amplifiers. The device is all-fibre, giving low coupling loss and negligible polarisation sensitivity, and the 10 ms upper-state lifetime is so long compared with a bit interval that the gain cannot follow the data and therefore introduces no pattern-dependent distortion or inter-channel crosstalk.

Disadvantages. The gain spectrum is intrinsically non-uniform, so channels at different wavelengths receive different gain; cascading amplifiers multiplies that tilt, and gain-flattening filters or dynamic equalisers become mandatory in long chains. Amplified spontaneous emission accumulates amplifier by amplifier, degrading the optical signal-to-noise ratio and eventually limiting the reach even though the power is adequate. Because an EDFA amplifies without reshaping or retiming, it lets dispersion and nonlinear distortion accumulate too, so dispersion compensation must be provided separately. It operates only in the erbium bands, so nothing near 1310 nm can be amplified this way. Finally, in a system where channels are added and dropped, the total input power changes abruptly and the shared gain medium transfers that transient to the surviving channels, so fast automatic gain control is required.

signal in isolator WDM coupler pump 980 or 1480 nm erbium-doped fibre 10-30 m isolator filter amplified out EDFA as a line amplifier in a WDM system Erbium ion energy levels pump band (4-I-11/2) metastable level (4-I-13/2), lifetime ~10 ms ground state (4-I-15/2) pump fast non-radiative decay stimulated emission, 1530-1565 nm
Configuration and energy levels of an erbium-doped fibre amplifier. The pump creates the population inversion; signal photons in the 1530-1565 nm band stimulate emission from the long-lived metastable level.
Question 5 — final results
QuantityDesign A (transimpedance)Design B (high impedance)
Resistance at the input node438.6 Ω404.8 kΩ
Bandwidth (unequalised)72.6 MHz78.6 kHz
Noise resistance142.5 kΩ404.8 kΩ
Input noise current density0.341 pA/$\sqrt{\text{Hz}}$ 0.202 pA/$\sqrt{\text{Hz}}$
Relative noise penalty+4.5 dBreference
Bandwidth advantage× 923reference
Recommendation Choose design A; design B is viable only with an equaliser, and then only where sensitivity outweighs dynamic range.