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.
Problem 4.1 — Carrier interval for a compensated GaAs guide
Working. Equate the Drude contrast to the strongly asymmetric cutoff and eliminate \(\omega^2\lambda_0^2=4\pi^2c^2\):
For \(t=3\,\mu{\rm m}\) and the book’s GaAs electron-mass approximation \(m_e^*=0.08m_0\), the interval is \(6.19\times10^{17}<N_3-N_2<5.57\times10^{18}\,{\rm cm}^{-3}\). Taking a representative hole mass \(m_h^*=0.5m_0\) multiplies both bounds by 6.25.
Answer. For an almost fully compensated core, the substrate bounds are the above electron values, or \(3.87\times10^{18}<N_3<3.48\times10^{19}\,{\rm cm}^{-3}\) under the stated hole-mass model.
Check. If residual core carriers are significant, add \(N_2\) 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 \(\lambda_0^2\). 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 \(\lambda_0^2\). 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 \(\lambda_0=0.90\,\mu{\rm m}\), Table 4.2 gives \(n_2(x=0.1)=3.49708\) and \(n_3(x=0.83)=3.12344\). Take air cover and treat the 10 micrometre lower confining layer as optically thick. For the 3 micrometre core, apply the exact cutoff:
Answer. The TE family contains 11 orders, \(m=0,\ldots,10\). 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: \(n_2=3.38890,\ n_3=3.03969\). The same 3 micrometre geometry exceeds cutoffs for \(m=0,\ldots,7\) 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 \(N\simeq3.07318>3.03969\).
Problem 4.6 — Tantalum-pentoxide mode count and last bounce
Working. Use \(n_2=2.09,\ n_3=1.5,\ n_1=1\), \(t=2\,\mu{\rm m}\), and \(\lambda_0=0.6328\,\mu{\rm m}\). The exact TE phase equation has nine roots, the last at \(N_8=1.615909\).
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 \(\cos^{-1}(1.5/2.09)\simeq44.14^\circ\); 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 \(m^*=0.08m_0\). The approximate indices are
For the 5 micrometre core, solve the asymmetric TE phase equation.
Answer. One TE mode, with \(N\simeq3.47168\), 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 \(0.165518\,\mu{\rm m}\) and increase by \(0.63/[2\sqrt{1.65^2-1.52^2}]\). 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 \(t_{c,11}(n_2',n_3,1;0.90)=3\,\mu{\rm m}\). The threshold core index is \(n_2'=3.55615\), an increase \(\delta n\simeq0.05907\). With \(|\delta n|=n_2^3|r_{41}|E/2\),
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 (1) into the approximate fundamental contrast cutoff:
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 \({\rm In}_{1-x}{\rm Ga}_x{\rm As}_{1-y}{\rm P}_y\). On Fig. 4.12, intersect 1.1 eV with the 300 K bandgap curve, read \(y\), then use the solid lattice-match curve to read \(x\). The graph gives approximately \(y\simeq0.6\text{--}0.65,\ x\simeq0.17\text{--}0.20\).
For interband transparency, the photon energy must lie below the gap:
Answer. A graph-level composition estimate is roughly \({\rm In}_{0.82}{\rm Ga}_{0.18}{\rm As}_{0.38}{\rm P}_{0.62}\), 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 \(y\simeq0.15\text{--}0.20\) and \(x\simeq0.38\text{--}0.40\), corresponding roughly to \({\rm In}_{0.61}{\rm Ga}_{0.39}{\rm As}_{0.83}{\rm P}_{0.17}\).
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.