Chapter 9: Electro-Optic Modulators
===================================

Source: Hunsperger, sixth edition (2009), Problems 9.1--9.11,
printed pages 195--198. Use :math:`\delta n=-n^3rE/2` for the
appropriate principal polarization, with sign selected by crystal orientation.

Problem 9.1 — Compensated GaP phase modulator
---------------------------------------------

**Working.** The carrier difference is
:math:`\Delta N=2.999\times10^{18}\,{\rm cm}^{-3}`.
Using the stated :math:`m^*=0.013m_0` in the carrier-cutoff relation,

.. math::

   t_{\min}=\sqrt{\frac{\pi^2c^2\epsilon_0m^*}{4e^2\Delta N}}
   =0.549582\,\mu{\rm m}.

The breakdown field is :math:`5\times10^7` V/m, so
:math:`V_{\max}=E_ct=27.48` V. Convert
:math:`r_{41}=5\times10^{-11}\,{\rm cm/V}` to
:math:`5\times10^{-13}\,{\rm m/V}`. With :math:`n=3.2`,

.. math::

   |\delta n|_{\max}=4.096\times10^{-4},\qquad
   L_\pi=\frac{\lambda_0}{2|\delta n|}=769.04\,\mu{\rm m}.

**Answer.** Limiting thickness 0.5496 micrometres, limiting voltage
27.48 V, and ideal uniform-field pi-shift length 0.769 mm.

**Check.** Operate slightly above cutoff and below breakdown in practice.
The estimates assume full optical overlap with the uniform field; the
poorly confined cutoff mode would not actually have unit overlap.

Problem 9.2 — Voltage needed to turn on a GaAs guide
----------------------------------------------------

**Data.** The drawing supplies :math:`t=3\,\mu{\rm m}`,
:math:`N_2=10^{15}`, :math:`N_3=3.80\times10^{17}\,{\rm cm}^{-3}`,
and :math:`n_3=3.4`. Use the chapter's GaAs coefficient
:math:`r_{41}=1.4\times10^{-12}\,{\rm m/V}`.

**Working.** At 1 micrometre the Drude contrast with mass
:math:`0.08m_0` is 0.000624917. The asymmetric fundamental
requires approximately
:math:`\Delta n_c=1/[32(3.4)(3^2)]=0.00102124`.
The electro-optic effect must supply the difference:

.. math::

   V_c=\frac{2t(\Delta n_c-\Delta n_{\rm Drude})}{n^3r}
   \simeq43.22\,{\rm V}.

**Answer.** Approximately 43 V is the cutoff estimate; slightly more
is needed to produce a bound fundamental.

**Check.** The existing carrier contrast must be subtracted. This
calculation uses the requested uniform voltage-drop approximation.

Problem 9.3 — Schottky modulation and the Franz–Keldysh effect
--------------------------------------------------------------

**Working.** Reverse bias changes the depletion field, which changes
both refractive index and near-edge absorption. Index change directly
produces phase modulation; birefringence can alter polarization.
Interference or an analyzer converts phase/polarization changes into
intensity changes. A time-varying phase produces instantaneous frequency
change :math:`\delta\omega=-d(\delta\phi)/dt` for the chosen
complex-field convention. A binary drive can encode pulses.

**Answer.** All listed signal formats are possible with appropriate
device and readout arrangements; intensity is not automatically
produced by a purely phase-modulating element alone.

**Check.** Franz–Keldysh electroabsorption is field-assisted sub-gap
absorption associated with band tilting and tunnelling. It is not
a uniform permanent reduction of the material's zero-field bandgap.

Problem 9.4 — Polarization dependence
-------------------------------------

**Working.** The electro-optic response is tensorial:
:math:`\Delta(\epsilon^{-1})_{ij}=r_{ijk}E_k`.
The optical polarization samples particular principal-axis changes.
Rotating polarization can select different coefficients, reverse the
phase response, or select an eigenpolarization with little response.

**Answer.** Polarization generally strongly affects modulation depth.
Crystal cut, electrode direction, and optical polarization must be specified.

**Check.** An isotropic scalar-index approximation cannot predict
polarization switching. Phase depth and detected intensity depth also
depend on the analyzer or interferometer.

Problem 9.5 — Two conditions for synchronous transfer
-----------------------------------------------------

**Working.** The interacting modes need matching longitudinal phases,
:math:`\beta_0=\beta_1`, so their incremental transfers do not cancel.
They also need a nonzero coupling integral, normally supplied by overlapping
evanescent tails with compatible field symmetry and polarization.

**Answer.** Phase matching and spatial/modal overlap.

**Check.** Sufficient interaction length is then required:
:math:`\kappa L=\pi/2` for complete first transfer. Identical
propagation constants in widely separated guides do not transfer appreciable power.

Problem 9.6 — Directional-coupler switching voltage
---------------------------------------------------

**Working.** With :math:`\kappa=1\,{\rm cm}^{-1}`,
:math:`L=\pi/(2\kappa)=1.5708` cm. The first detuned zero needs
:math:`|\Delta\beta|=\sqrt3\pi/L=2\sqrt3\,{\rm cm}^{-1}`.
Thus :math:`|\Delta N|=\sqrt3\lambda_0/\pi=5.844\times10^{-5}`
for wavelength in centimetres.

Using the Table 4.2 index :math:`n\simeq3.47255` at 1.06 micrometres,
full electro-optic overlap, and electrode separation 5 micrometres,

.. math::

   V=\frac{2(5\times10^{-6})|\Delta N|}
           {n^3(10^{-12})}\simeq13.96\,{\rm V}.

**Answer.** (a) 1.5708 cm. (b) Approximately 14 V under the
stated uniform-field/material-index assumptions.

**Check.** Only one channel is tuned. Push–pull tuning of both
channels changes the voltage relation, and a nonunit overlap increases
the required voltage.

Problem 9.7 — Compare plasma and electro-optic index changes
------------------------------------------------------------

**Working.** With the GaP parameters of 9.1, the Drude relation gives
:math:`\Delta n_{\rm carrier}=0.0128326`. Matching that magnitude
by a uniform electro-optic field across the limiting thickness requires

.. math::

   V=\frac{2t\Delta n_{\rm carrier}}{n^3r}
   \simeq860.9\,{\rm V}.

**Answer.** (a) About 0.01283.
(b) About 861 V in the ideal uniform-field calculation, already far
beyond the 27.48 V breakdown estimate.
(c) A uniformly doped Schottky depletion region has a triangular,
not nearly uniform, field; the same numerical voltage cannot simply
be reused as an exact answer.

**Check.** At fixed depletion width and fixed peak field, a triangular
field drops half the voltage of a uniform field. For the actual guided
mode, the relevant change is a field-weighted overlap, so a self-consistent
depletion and mode calculation is needed.

Problem 9.8 — Field profiles for compensated and uniform material
-----------------------------------------------------------------

**Working.** Poisson's equation gives
:math:`dE/dx=eN_D/\epsilon`. A nearly compensated layer has small
slope, so its field is almost constant across the layer and then drops
rapidly in the heavily doped substrate. A uniformly doped depletion
region instead has :math:`E(x)=eN_D(W-x)/\epsilon`.

.. figure:: /_static/knowledge_base/worked_exercises/integrated_optics_6e/electro_optic_fields.svg
   :alt: Nearly uniform field in a compensated layer compared with a triangular depletion field in uniformly doped material.
   :width: 100%

   Normalize both examples to the same depletion width; voltage is area
   under each field curve.

**Answer.** Draw a nearly flat profile for the compensated guide and a
linear triangular profile for uniform doping. Increasing doping increases
the field gradient and its influence on the index profile.

**Check.** Index perturbation follows the tensor-weighted local field,
while depletion width itself depends on voltage and doping.

Problem 9.9 — Electrical drive power for a phase swing
------------------------------------------------------

**Working.** Let voltage denote peak sinusoidal amplitude:
:math:`\Delta\Phi=\pi Ln^3rV/(\lambda_0t_g)`.
With :math:`C=\epsilon A/t_g` and
:math:`R_L^{-1}=2\pi C\Delta f`, the average load power is
:math:`P=V^2/(2R_L)=\pi C\Delta f\,V^2`.
Substituting for :math:`V` yields

.. math::

   P=\frac{(\Delta\Phi)^2\lambda_0^2\epsilon A\Delta f\,t_g}
            {\pi L^2n^6r^2}.

**Answer.** This is the requested result.

**Check.** :math:`A` must be the electrode-overlap area used by the
capacitance model. Using RMS voltage without adjusting the peak phase
definition creates a factor-of-two discrepancy.

Problem 9.10 — Bandgap for a near-edge absorption modulator
-----------------------------------------------------------

**Working.** The 1.15 micrometre photon has energy
:math:`E_\gamma=1.239841984/1.15=1.07812` eV.
Choose a zero-field gap slightly above this energy, leaving low
off-state interband absorption while allowing the applied field to
increase the absorption tail at that photon energy.

**Answer.** A material with gap near, and modestly above, 1.08 eV.

**Check.** A gap far below the photon energy gives excessive off-state
absorption; a gap far above it may require an impractically large field.

Problem 9.11 — Insertion loss and extinction
--------------------------------------------

**Working.** The high-transmission and low-transmission states have
:math:`T_{\rm on}=e^{-0.2(2)}` and :math:`T_{\rm off}=e^{-2(2)}`.
Therefore

.. math::

   {\rm IL}=\frac{10}{\ln10}(0.2)(2)=1.73718\,{\rm dB},
   \qquad
   {\rm ER}=\frac{10}{\ln10}(2-0.2)(2)=15.6346\,{\rm dB}.

**Answer.** Insertion loss 1.737 dB; extinction ratio 15.63 dB.
For 500 mW incident, outputs are approximately 335.16 and 9.158 mW.

**Check.** The input power cancels from both decibel ratios.
Extinction uses the difference of absorption coefficients.