doping profile and refractive index profile of optical fiber amplifier

Many people know. It's in all the textbooks, having been worked out between

For communications fibers, multimode is ~parabolic, while for single-mode it's step index, with perhaps a few stepped rings.

For carriage of gross amounts of optical power, it's usually a larger stepped-index fiber than for communications.

The difference in refractive index between core and cladding is 1% to 2%.

Joe Gwinn

I suspect you'll agree, however, that when you look at the actual (i.e., measured) index profiles of real fibers, they often display horrible looking spikes and irregular small-scale variations about these idealized parabolic or step index profiles -- irregularities that lead, in most cases, to very little or practically no real difference in actual mode propagation.

Absolutely. In practice, the desired refractive-index (RI) profile is approximated by some number of constant-RI layers. The number of layers varies between fiber makers, and I recall one boasting about using at least 100 layers, to better approximate the desired profile.

Now I remember. It is Draka: .

It should be noted that many practical multimode optical fibers exhibit a centerline RI dip or bump and that unless avoided this central defect has a large effect on the bandwidth (MHz-Km product) of the fiber. Avoidance of such central defects was a major motivation for the donut-launch requirement in IEEE

Joe Gwinn

Joe, I've been looking at the analysis of fiber modes (or more often, just TE planar waveguide modes), motivated by some fundamental questions as to the effects of loss and especially gain on these modes. (For example, an elementary step-profile fiber with even a small amount of uniform loss in the core has _no_ stable modes; all of its modes are perturbation-unstable to any kind of random deviations in the fiber. With uniform gain in the core, however, the modes become stable.)

In the process, I've observed that most of the texts and published literature tackle the problem of finding the modes in an optical waveguide by specifying an index profile, then attempting to solve for the modes that propagate in this waveguide -- which is often hard.

But if you turn this process around and start by specifying the modal function (most any arbitrary modal function) that you want, then the index profile that you need to produce or obtain this modal function falls out trivially (I've come to refer to this as "designer waveguides" or "designer fibers") -- though of course fabricating that desired index profile may be a "whole 'nother problem".

Have you encountered this "designer fiber" concept in the published literature anywhere?

Why would uniform gain yield stable modes, while uniform loss would yield unstable modes?

But this arrangement is very common, as almost all fibers have some loss, so somehow it isn't a problem. What immediately come to mind is that the fiber is usually coiled up. Will curvature break the symmetry and allow stable modes to exist?

I think that fiber designers do exactly this (design fiber for the desired modes), but fiber manufacture is a very competitive industry. A good example would be designing fibers to well-match the donut launch mentioned above.

Manufacture of such a designer RI profile would be no problem with current computer-controlled profile generators. And Draka's original patents are long since expired.

Not in the literature, but I have not looked very hard either. All the fiber design groups I knew of (from working in standards groups) had company-developed proprietary full-fidelity simulation codes to predict fiber performance from RI profile. But the details were very closely held, if for no other reason than that such codes are very expensive to develop and validate, and would give much detailed process information away.

Joe Gwinn

Short answer: Higher-order modes have lower filling factors in the core (i.e., they're 'larger', with more of their energy outside the core and in the cladding) than do lower-order modes. The core is lossy. Therefore, higher-order modes have _lower_ losses than higher-order modes, and therefore survive longer (i.e., propagate further down the waveguide) than do lower order modes.

Any kind of random scattering along the waveguide scatters energy among the modes; or any kind of deviation from the lowest-order mode in the excitation at an input to the waveguide puts some energy into the higher order modes -- and these modes last longer and eventually dominate.

They're more vulnerable to microbending, though, because the guiding is weaker. That's particularly true at short wavelength--if you run UV through a MMF, the high order modes are much lossier than the low order ones.

And that business about lossy fibre being perturbation-unstable intrigues me--in one way of speaking, lossy fibre has no stable modes at all, simply because of the loss. Since that's sort of trivial, I gather you're talking about lossy fibres being more vulnerable to scattering loss, is that right? What's the physics there?

Cheers

Phil Hobbs

So, the key is that because gain is by stimulated emission, the new photons are going in the same direction as the original photons (and all are phase coherent), so there is little mode mixing; while loss causes scattering into other often higher-order modes, so there is considerable mode mixing? This seems to require loss by scattering versus loss by absorption.

Joe Gwinn

Not sure I ever answered this one. Still looking for an answer?

--AES