3.6: Fiber amplifiers

In long distance communication, losses in the optical fiber play an important role. Consider that the distance from Amsterdam to New York is 5,880 km, while a typical long haul optical fiber has losses of around 0.17 dB/km, meaning that the total attenuation is about 1000 dB. A typical receiver might be reliable for signals as low as -16 dBm (25 µW), meaning that the input power would have to be 1000-16 = 984 dBm (2.5\(\times 10^{95}\) W). Even if the receiver had a very high sensitivity, of -60 dBm, which corresponds to a single photon per bit at a 10 Gb/s data rate, then the input power still has to be unreasonably high.

A typical DFB laser has an output power of 10 dBm (10 mW), meaning that 26 dB is the maximum acceptable attenuation. Thus, the maximum length of fiber is about 150 km, which does not take into account losses due to connecting (splicing) lengths of fiber together. Giving ourselves some margin, we need 58 amplifiers between Amsterdam and New York.

Amplification can take place in two ways: the optical signal can be detected, converted to an electrical signal, then returned to the optical domain by modulating an optical source, or an amplifier that directly amplifies the optical signal can be used. In practice both are used for reasons that will, hopefully, become clear later.

Optical amplification works by the same process as a laser: a population inversion is achieved by exciting a medium, and the input light is amplified by the same process, including the competition for gain. A typical amplifier consists of an optical fiber that has been doped with the element erbium, called an erbium doped fiber amplifier (EDFA). Erbium has a relatively high gain over the C band, as shown in Figure \(\PageIndex{1}\). Despite a factor of four difference in emission cross section, the resultant gain is relatively flat. Nevertheless, the wavelength dependence of the gain must be compensated for. In order to ensure that each channel in the fiber gets equal gain, the signal must be spectrally shaped before amplification: essentially a filter should be shaped such that the gain-intensity product is constant across the C band: \(G(\lambda)I(\lambda)=k\).

Figure \(\PageIndex{1}\): Emission cross section of Erbium (larger is better). There is significant gain between 1500-1600 nm.

Excitation

EDFAs are optically excited. A pump laser with a wavelength of 980 nm drives erbium ions into the excited state, where emission in the C band can take place. Thus, the gain is proportional to the population inversion as described for laser gain through stimulated emission. There is also a second gain mechanism called Raman amplification. In this case, emission from an excited state population is not involved, instead, a nonlinear optical effect converts pump photons to signal photons via the generation of vibrations in the glass. Since vibrations in solids are low frequency (compared to the frequency of light), Raman amplification requires a pump wavelength much closer the signal wavelength (see Figure \(\PageIndex{2}\)).

Both amplification methods have their benefits: at low signal powers, amplification via Erbium ion emission is much more efficient. However, the gain is intrinsically limited by the number of ions available. At a certain pump power, all ions are available, and no more gain is possible. On the other hand, Erbium has many more energetic states available, so at high pump powers, the ions often get driven out of the excited state into other states that do not contribute to gain (called excited state absorption). Raman gain grows with pump power, but the gain is very low at low signal powers, and, because the pump wavelength is close to the signal wavelength, it can be difficult to filter the pump out of the channel.

Figure \(\PageIndex{2}\): Conceptual energy level diagram for an EDFA (the real energy level diagram is far more complex). The laser emission state (\(|3\rangle\) is a set of states that overlap in energy (grey box), creating a band. At high emission energy (bluer wavelengths), the laser behaves like a three-level laser, while for redder wavelengths, the laser is more like a four-level laser (red arrows). Raman emission is driven by a 1450 nm pump (green arrow), which stimulates Raman emission (red arrows) along with high frequency vibrations in the fiber (grey arrows).

Note that the two amplification techniques complement each other to great extent. It is, therefore, common to combine both in a single amplifier. The basic geometry of a double pumped EDFA is shown in Figure \(\PageIndex{3}\). The Erbium doped gain fiber is ~100 m long and double clad, with the 980 nm pump traveling in the same direction as the signal. This provides optimum small-signal gain, but, as the signal grows, the gain drops because the Erbium concentration is not large enough. The remainder of the gain fiber is glass without doping. The Raman pump (at 1450 nm) is injected in the opposite direction to the signal, so that the pump power is highest where the signal is most intense, ensuring that the gain grows as the signal intensity grows.

Designing a fiber amplifier involves a balance between pump intensity, ion doping concentration, and fiber length.

Figure \(\PageIndex{3}\): Double pumped EDFA. The 980 nm pump (for stimulated emission) is guided in the cladding in the same direction as the signal. The Raman pump is injected in the reverse direction and provides gain for the signal once it has already been amplified by stimulated emission. In the configuration shown here, the Raman pump is guided in the core.

Amplified spontaneous emission

The amplification process that we have described so far relates to stimulated emission, where the presence of the signal light wave drives an excited ion to emit light, amplifying the wave. However, if left to its own devices, an excited emitter will eventually emit a photon on its own, generating a new light wave. This light wave can then be amplified by other excited ions, in a process called amplified spontaneous emission (ASE).

ASE is generated in parallel to signal amplification, creating a growing noise floor that the signal must overcome. Figure \(\PageIndex{4}\) shows how ASE grows after each amplifier stage. The growth of ASE and optical losses are used to calculate the optimum spacing between amplifiers, and the maximum number of stages before the signal must be converted back to the electrical domain. We begin by stating that over the length of the link, the amplification must exactly compensate for the losses (the total optical power in the signal band is, on average, constant over the length of the link)

\[\begin{equation}G= \alpha L_s\label{eqn:gain}\end{equation}\]

where \(G\) is the gain in dB, \(\alpha\) is the per-km loss (dB/km) and \(L_s\) is the length of fiber between amplifiers. The ASE is given by

\[\begin{equation} P_{n} = 2\eta_{sp}hfN(G-1)\Delta f\label{eqn:Pase}\end{equation}\]

where \(\eta_{sp}\approx 2\) is the inversion factor, \(h\) is Plank's constant, \(f\), \(\Delta f\) are the optical center frequency and bandwidth, and \(N\) is the number of amplifiers. Since the total power is a constant, the signal to noise ratio is

\[\begin{equation}\textrm{SNR} = \frac{P-P_n}{P_n}\label{eqn:snr}\end{equation}\]

In a communications link, the specified bit error rate defines the minimum acceptable SNR, while the sensitivity of the detector defines the minimum optical power at the detector. Thus, we can combine Equations \(\eqref{eqn:gain}-\eqref{eqn:snr}\) to find the maximum number of amplifiers.

Figure \(\PageIndex{4}\): Noise growing in a fiber link. In this calculation, the noise power for a single channel is calculated. It is assumed that each amplifier stage is set to compensate exactly the total losses between each amplifier stage.

Example \(\PageIndex{1}\)

An optical fiber link (center wavelength: 1550 nm) with a single channel data rate of 10 Gb/s is to be deployed between Amsterdam and New York. The total length of passive fiber is 6000 km, with an average attenuation of 0.17 dB/km. The input power 10 dBm, and the amount of input noise is negligible. To obtain a bit error rate (BER) of 10^-9 SNR ≥ 7 is required. The detectors require an average flux of 20 photons per bit to reliable detect a signal. The chosen optical amplifiers have a gain of 50 dB

Calculate the number of optical amplifiers and electronic repeaters required for the link.

Solution

First step: calculate the power (in dBm) at the receiver: 20 photons per bit = \(\frac{20\times10^9}{2}\) photons per second. The energy per photon is \(E_p=\frac{hc}{\lambda} = 1.3\times10^{-19}\) J. So, the average signal power at the end detector is \(E_p\times (5\times 10^9)\) = 0.64 nW (-62 dBm). This, as can be seen below is not the limiting factor.

The total loss due to the fiber absorption is 0.17\(\times\)6000 = 1020 dB, while the gain of the amplifier is 50 dB, indicating that we need 21 amplifier stages at 286 km intervals. The ASE, given by Equation \(\eqref{eqn:Pase}\) is 126 µW per amplifier (-9 dBm). After 21 stages, the ASE is 2.6 mW, thus the SNR is (from Equation \(\eqref{eqn:snr}\)) \(\frac{10-2.6}{2.6}=2.8\), which implies a BER>10^-9.

Setting \(P_n = NP_{amp}\) where \(P_{amp}\) is the ASE from a single stage, equation \(\eqref{eqn:snr}\) can be rearranged to \(N = \frac{P}{P_{amp}(\textrm{SNR} + 1)}\). Filling in the numbers we get \(N\approx 10\). Thus, we will need a single electronic repeater, which replaces amplifier number 10.

Note that this leaves no margin for error and does not take into account other losses. This includes losses when fibers are spliced together, losses for dispersion compensating fiber, losses due to spectral shaping filters, and more. Thus, three electronic repeaters at positions 5, 12, and 19 is a better choice.

Example \(\PageIndex{1}\), illustrates how noise grows in a fiber link. This idea is illustrated in Figure \(\PageIndex{4}\), and is a good place to start when thinking about a link budget.

Link Budget

When planning to deploy a fiber link, the requirements will generally give a data rate and a maximum acceptable BER. The data rate, \(B\) and the average number of photons per bit \(\bar{n}\) required to meet the BER requirement lead give us the minimum acceptable power at the receiver in dBm.

\[\begin{equation}P_r = 10\log\left(\frac{\frac{hc}{\lambda}\bar{n}B}{1\times10^{-3}}\right)\label{eqn:minPow}\end{equation}\]

It is never a good idea to operate at the minimum, so some margin should be added, \(P_m\). Attenuation losses can be split into two components: power loss per slice/coupling \(P_c\), power loss due to attenuation of the fiber, \(\alpha L\). Thus, the link budget for absorption losses is given by

\[\begin{equation}P_s - N_cP_c - \alpha L = P_r\label{eqn:linkabsBudget}\end{equation}\]

where \(N_c\) is the number of splices and couplers.

On top of that the pulse will spread out due to the dispersion of the material (we consider only single model fibers), such that

\[\begin{equation}LB = \frac{1}{4|D_\lambda \sigma_\lambda}\label{eqn:disp}\end{equation}\]

where \(D_\lambda\) is the dispersion coefficient (ps/km.nm), and \(\sigma_\lambda\) is the the bandwidth of the pulse. If we assume the best case scenario (the minimum bandwidth for a pulse of a given duration), then \(\sigma_\lambda = 0.75\frac{\lambda^2}{c}B\). After a number of substitutions related to the dynamics of pulses that have a Gaussian shape, equation \(\eqref{eqn:disp}\) becomes

\[\begin{equation}LB^2 = \frac{\pi}{4}\frac{c}{|D_\lambda \lambda^2}\label{eqn:tldisp}\end{equation}\]

From equations \(\eqref{eqn:linkabsBudget}\) and \(\eqref{eqn:tldisp}\), it is possible to graphically see how desired bit rate and link length are related to each other (see Figure \(\PageIndex{5}\)).

Figure \(\PageIndex{5}\): How dispersion and attenuation limit the maximum bit rate of a link. Without dispersion compensation, high-speed, long distance links are impossible.

The dispersion can be compensated by using two different fibers such that \(D_\lambda L + D'_\lambda L' = 0\), where \(D'_\lambda\) is the dispersion coefficient of a second fiber that has anomalous dispersion in the transmission band. These fibers have higher loss, but their dispersion is typically much stronger, so only short lengths are required to compensate for the dispersion of the normally dispersive fiber.

In a typical communication system, the entire C and L band might be used, while the dispersion compensating fiber is chosen to only exactly compensate in the middle of the communication band. Thus, there is residual positive and negative dispersion for channels adjacent to the center channel. However, this residual dispersion is still smaller than the uncompensated dispersion, and allows for much longer links.