Chapter 4: Waveguide Fabrication Techniques
===========================================

Source: Hunsperger, sixth edition (2009), Problems 4.1--4.12,
printed pages 79--80. Carrier concentrations are per cubic centimetre.

Shared material models
----------------------

The Drude index difference for a depleted core is

.. math::
   :label: integrated-drude

   \Delta n=\frac{(N_3-N_2)e^2}{2n\epsilon_0m^*\omega^2}.

Convert the concentrations to :math:`{\rm m}^{-3}` when using SI constants.
For AlGaAs, Table 4.2 supplies the Sellmeier model

.. math::
   :label: integrated-algaas-index

   n^2=10.906-2.92x+
   \frac{0.97501}{\lambda_0^2-C(x)}
   -0.002467(1+1.41x)\lambda_0^2,

where wavelength is in micrometres,
:math:`C=(0.52886-0.735x)^2` for :math:`x\le0.36` and
:math:`C=(0.30386-0.105x)^2` above that composition.

Problem 4.1 — Carrier interval for a compensated GaAs guide
-----------------------------------------------------------

**Working.** Equate the Drude contrast to the strongly asymmetric cutoff
and eliminate :math:`\omega^2\lambda_0^2=4\pi^2c^2`:

.. math::

   (N_3-N_2)_{c,m}
   =\frac{(2m+1)^2\pi^2c^2\epsilon_0m^*}{4e^2t^2}.

For :math:`t=3\,\mu{\rm m}` and the book's GaAs electron-mass
approximation :math:`m_e^*=0.08m_0`, the interval is
:math:`6.19\times10^{17}<N_3-N_2<5.57\times10^{18}\,{\rm cm}^{-3}`.
Taking a representative hole mass :math:`m_h^*=0.5m_0` multiplies
both bounds by 6.25.

**Answer.** For an almost fully compensated core, the substrate bounds
are the above electron values, or
:math:`3.87\times10^{18}<N_3<3.48\times10^{19}\,{\rm cm}^{-3}`
under the stated hole-mass model.

**Check.** If residual core carriers are significant, add :math:`N_2`
to both bounds. Heavy doping, absorption, and multiple valence-band
masses limit the quantitative accuracy of the simple p-type estimate.

Problem 4.2 — Repeat at 1.06 micrometres
----------------------------------------

**Working.** Both the Drude-produced contrast and the required weak-step
cutoff contrast scale as :math:`\lambda_0^2`. Their ratio is unchanged,
so the concentration interval in 4.1 is unchanged within this model.

**Answer.** The same electron and hole concentration bounds apply.

**Check.** This cancellation assumes fixed effective mass and the
ideal Drude approximation. It does not predict wavelength-independent
loss, evanescent penetration, or refractive-index dispersion.

Problem 4.3 — Contrast the wavelength dependences
-------------------------------------------------

**Working.** Problems 2.1--2.2 prescribe an index step independent of
its fabrication mechanism. Their required step decreases with
:math:`\lambda_0^2`. In a carrier-depleted guide, however, the step
generated by a fixed carrier difference decreases by exactly the same
factor when wavelength is reduced.

**Answer.** The ordinary index-design interval changes with wavelength;
the ideal carrier-concentration interval does not.

**Check.** The cancellation concerns cutoff only. A material absorption
edge can still prevent guiding at a wavelength that satisfies the
geometric condition.

Problem 4.4 — Modes in an epitaxial AlGaAs guide
------------------------------------------------

**Working.** At :math:`\lambda_0=0.90\,\mu{\rm m}`, Table 4.2 gives
:math:`n_2(x=0.1)=3.49708` and :math:`n_3(x=0.83)=3.12344`.
Take air cover and treat the 10 micrometre lower confining layer as
optically thick. For the 3 micrometre core, apply the exact cutoff:

.. math::

   3>t_{c,m}(3.49708,3.12344,1;\,0.90).

**Answer.** The TE family contains 11 orders, :math:`m=0,\ldots,10`.
The TM family contains 10 orders, giving 21 modes if both polarizations
are counted.

**Check.** The lowest effective index is approximately 3.13066, still
above the lower-cladding index. The actual GaAs substrate behind a
finite barrier makes these approximate guided modes potentially leaky;
the thick barrier suppresses that tunnelling.

Problem 4.5 — Repeat the epitaxial count at 1.15 micrometres
------------------------------------------------------------

**Working.** Re-evaluate material dispersion rather than holding the
previous indices fixed:
:math:`n_2=3.38890,\ n_3=3.03969`.
The same 3 micrometre geometry exceeds cutoffs for
:math:`m=0,\ldots,7` only.

**Answer.** Eight TE and eight TM orders in the thick-barrier
approximation: 16 vector modes altogether.

**Check.** The normalized optical thickness decreases at longer
wavelength; fewer orders are expected. The last root is
:math:`N\simeq3.07318>3.03969`.

Problem 4.6 — Tantalum-pentoxide mode count and last bounce
-----------------------------------------------------------

**Working.** Use :math:`n_2=2.09,\ n_3=1.5,\ n_1=1`,
:math:`t=2\,\mu{\rm m}`, and :math:`\lambda_0=0.6328\,\mu{\rm m}`.
The exact TE phase equation has nine roots, the last at
:math:`N_8=1.615909`.

.. math::

   \theta_8=\cos^{-1}(N_8/n_2)\simeq39.36^\circ.

**Answer.** (a) Nine TE and nine TM modes, or 18 vector modes in total.
(b) The highest-order TE ray makes
about 39.36 degrees with the guide plane.

**Check.** The limiting substrate-critical bounce angle is
:math:`\cos^{-1}(1.5/2.09)\simeq44.14^\circ`; the last discrete
mode need not sit exactly at this limit.

Problem 4.7 — Carrier and alloy confinement together
----------------------------------------------------

**Working.** Evaluate the undoped indices at 1.06 micrometres using
Table 4.2, then subtract each layer's Drude carrier contribution with
:math:`m^*=0.08m_0`. The approximate indices are

.. math::

   n_1=3.33621,\qquad n_2=3.47255,\qquad n_3=3.47073.

For the 5 micrometre core, solve the asymmetric TE phase equation.

**Answer.** One TE mode, with :math:`N\simeq3.47168`, and one TM mode.
There is only one guided spatial order per polarization.

**Check.** Alloying strongly lowers the top-cladding index, while carrier
depletion supplies only a small substrate contrast. Ignoring the carrier
correction to the GaAs substrate would incorrectly remove the guide.

Problem 4.8 — Repeat the asymmetric TE count
--------------------------------------------

**Working.** This is the same parameter set as 3.7. The exact cutoffs
begin at :math:`0.165518\,\mu{\rm m}` and increase by
:math:`0.63/[2\sqrt{1.65^2-1.52^2}]`.
At 1.18 micrometres, orders zero, one, and two are bound.

**Answer.** Three TE modes, with effective indices
1.63494, 1.59010, and 1.52251.

**Check.** The last effective index barely exceeds the substrate index,
so rounding the cutoff phase too early can change the apparent count.

Problem 4.9 — Electro-optically add an order
--------------------------------------------

**Working.** Retain the lower-cladding index from 4.4. Solve
:math:`t_{c,11}(n_2',n_3,1;0.90)=3\,\mu{\rm m}`.
The threshold core index is :math:`n_2'=3.55615`, an increase
:math:`\delta n\simeq0.05907`. With
:math:`|\delta n|=n_2^3|r_{41}|E/2`,

.. math::

   E\simeq\frac{2(0.05907)}{3.49708^3(1.4\times10^{-12})}
   =1.97\times10^9\,{\rm V/m},
   \qquad V\simeq Et=5.92\,{\rm kV}.

**Answer.** This is not a practical electro-optic way to add the order.

**Check.** The required field is far beyond an ordinary operating field,
and extrapolating the linear electro-optic law that far is itself invalid.
An index just above the calculated threshold would be needed for binding.

Problem 4.10 — Why carrier cutoff loses its wavelength dependence
-----------------------------------------------------------------

**Working.** Substitute :eq:`integrated-drude` into the approximate
fundamental contrast cutoff:

.. math::

   \frac{\Delta Ne^2\lambda_0^2}
        {8\pi^2n\epsilon_0m^*c^2}
   >\frac{\lambda_0^2}{32nt^2}
   \quad\Longrightarrow\quad
   \Delta N>\frac{\pi^2c^2\epsilon_0m^*}{4e^2t^2}.

**Answer.** Wavelength cancels because both sides contain the same
quadratic wavelength factor.

**Check.** This derivation does not remove material dispersion or
absorption. It establishes the stated approximate cutoff property only.

Problem 4.11 — Lattice-matched InGaAsP at room temperature
----------------------------------------------------------

**Working.** The question uses
:math:`{\rm In}_{1-x}{\rm Ga}_x{\rm As}_{1-y}{\rm P}_y`.
On Fig. 4.12, intersect 1.1 eV with the **300 K** bandgap curve,
read :math:`y`, then use the solid lattice-match curve to read
:math:`x`. The graph gives approximately
:math:`y\simeq0.6\text{--}0.65,\ x\simeq0.17\text{--}0.20`.

For interband transparency, the photon energy must lie below the gap:

.. math::

   \lambda_0>\frac{hc}{1.1\,{\rm eV}}=1.12713\,\mu{\rm m}.

**Answer.** A graph-level composition estimate is roughly
:math:`{\rm In}_{0.82}{\rm Ga}_{0.18}{\rm As}_{0.38}{\rm P}_{0.62}`,
with a band-edge wavelength about 1.13 micrometres.

**Check.** Treat these as graphical estimates, not an exact alloy
calibration. A practical low-loss wavelength needs margin beyond the
absorption edge; equality is not strong transparency.

Problem 4.12 — Alloy composition at 77 kelvin
---------------------------------------------

**Working.** Use the **77 K** curve of Fig. 4.12 at 0.9 eV. Read its
phosphorus fraction at the intersection, then read the gallium fraction
from the lattice-match curve at the same horizontal coordinate.

**Answer.** Approximately :math:`y\simeq0.15\text{--}0.20` and
:math:`x\simeq0.38\text{--}0.40`, corresponding roughly to
:math:`{\rm In}_{0.61}{\rm Ga}_{0.39}{\rm As}_{0.83}{\rm P}_{0.17}`.

**Check.** Fractions on each sublattice sum to one. Using the 300 K
curve for this question produces a different composition; the temperature
is an essential datum, not a unit conversion.