Chapter 17: Integrated Optical Detectors
Source: Hunsperger, sixth edition (2009), Problems 17.1–17.11, printed pages 368–371.
Problem 17.1 — Choose a detector gap
Working. A 0.9 micrometre photon has energy 1.37760 eV. Direct interband absorption therefore excludes the 2 eV material. Both 0.5 and 1 eV gaps can absorb, but the larger qualifying gap generally suppresses thermally generated dark carriers.
Answer. The 1 eV material is the intended compromise when the other stated material properties are equivalent.
Check. Bandgap alone does not determine detector noise, absorption length, or carrier transport. The conclusion uses the question’s equal-quality assumptions.
Problem 17.2 — Reject short-wavelength background
Working. Require \(e^{-(1000\,{\rm cm}^{-1})t}=10^{-4}\), so
At the signal wavelength, transmission is \(T_s=e^{-0.2t}=0.9981596\).
Answer. A roughly 92.1 micrometre filter suppresses the background by 10,000 while transmitting 99.816 percent of the signal.
Check. Signal attenuation factor can mean either \(T_s=0.99816\) or input/output ratio \(1/T_s=1.00184\); specify which is quoted.
Problem 17.3 — Minimum irradiance for a photocurrent
Working. With \(\eta_q=0.8\) at 0.9 micrometres, \({\cal R}=\eta_qe\lambda_0/(hc)=0.5807\) A/W. Required incident power is \(P=I/{\cal R}=1.7220\,\mu{\rm W}\). The sensitive area is \(10\,{\rm mm^2}=10^{-5}\,{\rm m^2}\).
Answer. Required peak irradiance is \(0.17220\,{\rm W/m^2}\), or \(17.220\,\mu{\rm W/cm^2}\).
Check. This assumes the stated quantum efficiency refers to incident photons. If it is only internal yield, include reflection and incomplete absorption separately.
Problem 17.4 — Material selection and silicon photocurrent
Working. Approximate the GaAs laser photon energy by the listed GaAs gap, 1.4 eV. Si, GaAs, GaSb, and InAs have gaps no larger than that value; GaP does not. Adding Sb or In to GaAs can reduce the gap, whereas adding P generally raises it in the relevant composition range.
For 10 mW incident on 1 cm2 and \(\eta_q=0.8\),
Answer. GaAsSb and InGaAs can provide the needed lower gap; GaAsP generally cannot except at the GaAs endpoint or with sufficient photon excess energy. The illustrative silicon current is 5.71 mA.
Check. Gap equality is only a threshold condition, not a guarantee of strong absorption. If an exact wavelength is specified, use it instead of equating photon energy to the approximate gap.
Problem 17.5 — Photon threshold and pair generation
Working. At 0.600 micrometres, \(E_\gamma=2.06640\) eV, the largest qualifying gap. Incident power is \(P=(20)(5\times10^{-6})=10^{-4}\) W. For unit external quantum efficiency,
Answer. (a) Maximum gap 2.0664 eV. (b) \(3.02047\times10^{14}\) pairs/s. (c) Halving intensity halves the rate. (d) Halving wavelength at fixed intensity also halves the rate.
Check. Shorter-wavelength photons carry more energy, so fixed optical power means fewer photons, not more.
Problem 17.6 — Field-assisted absorption length
Working. Average depletion field is \(E=40.5/(3\times10^{-6})=1.35\times10^7\) V/m. The book’s Franz–Keldysh energy-scale estimate gives
for \(m^*=0.067m_0\). Fig. 17.19 gives about \(\alpha=10^4\,{\rm cm}^{-1}\) at 0.9 micrometres for this field. Require \(1-e^{-\alpha L}=0.99\).
Answer. The characteristic absorption-edge shift is about 88.8 meV; \(L=\ln100/\alpha\simeq4.605\,\mu{\rm m}\) for full optical overlap.
Check. The shift is a field-assisted absorption scale, not a rigid change to the crystal’s true gap. With confinement factor \(\Gamma<1\), replace \(\alpha\) by \(\Gamma\alpha\).
Problem 17.7 — Audit all optoisolator combinations
Working: wavelengths and transparency. Laser (a) has \(\lambda_a=2(3.6)(0.340)/3=0.816\,\mu{\rm m}\). Lasers (b) and (c) emit at 0.9 micrometres. At 0.816 micrometres the Fig. 4.8 curves show strong absorption in both candidate guide cores, eliminating the long guide routes for (a). Its source efficiency and threshold are also not specified well enough for an independent current-transfer calculation.
Working: electrical feasibility. For (b) and (c), \(\alpha_m=\ln(1/0.4)/0.0935=9.800\,{\rm cm}^{-1}\). Using the supplied spontaneous width and modal thicknesses in (1) gives
Thus (b) cannot operate as the assumed laser at the available drive. For (c), the one-facet photon efficiency at 5 A is approximately
Working: current-transfer ratios. For a guide with transmission \(T\) and a detector absorbing fraction \(A_D\) with internal yield one and avalanche factor \(M\),
For detector (f), using the zero-field 0.9 micrometre absorption \(25\,{\rm cm}^{-1}\) from Fig. 17.19 and length 0.01 cm gives \(A_f=1-e^{-0.25}=0.22120\). With \(M_f=2\), even perfect guide transmission gives \({\rm CTR}_{c,f}\le0.06689<0.1\). If its avalanche field also changes absorption, that new absorption must instead be supplied; multiplication alone does not specify it.
For detector (g), strong field makes \(A_g\simeq1\), hence \({\rm CTR}_{c,g}\simeq0.15119T\). It passes only if
Answer. These inequalities are the explicit acceptance test for the remaining (c,d,g) and (c,e,g) routes. For illustrative book-graph readings \(\alpha_d\simeq2\) and \(\alpha_e\simeq25\,{\rm cm}^{-1}\), their one-facet ratios are only about 0.0556 and 0.00102, so neither meets 0.1. The (c,*,f) routes fail with the stated zero-field absorption model; (b,*,*) fails the drive limit; the (a,*,*) routes are strongly absorbed.
Check. This exercise is sensitive to graph readings, field-modified detector absorption, and one-face versus two-face collection. The book’s Chapter 12 power expression counts both faces. Allocating that total to the single waveguide doubles CTR and can make (c,d,g) appear acceptable. If both faces really are collected, state that additional optical arrangement explicitly and double the one-face results. Unit internal yield does not mean every incident photon is absorbed.
Problem 17.8 — Avalanche photocurrent from an irradiance
Working. Incident power is \((10^{-3}\,{\rm W/cm^2})(10^{-4}\,{\rm cm^2})=10^{-7}\) W. With 0.85 quantum efficiency and avalanche factor 1000,
Answer. Approximately 89.1 microamperes.
Check. Area is required because the supplied optical datum is irradiance. Multiplication affects carriers, not the photon energy.
Problem 17.9 — Infer avalanche factor and gap
Working. At 1.3 micrometres and quantum efficiency 0.8, the unmultiplied current for 1 microwatt is \(I_0={\cal R}P=0.83884\,\mu{\rm A}\). Therefore \(M=20/I_0=23.8431\). The 1.7 micrometre cutoff estimates \(E_g=1.239841984/1.7=0.729319\) eV.
Answer. Avalanche factor about 23.84; absorber gap about 0.729 eV.
Check. Use the cutoff wavelength for the gap and the operating wavelength for responsivity. They serve different purposes.
Problem 17.10 — p-i-n and avalanche advantages
Working. An intrinsic layer gives a p-i-n detector a broad depleted absorption region with a controlled electric field and reduced junction capacitance. It can combine useful absorption with rapid drift collection. An APD adds impact-ionization multiplication before external electrical amplification.
Answer. The p-i-n architecture improves the absorption, capacitance, and speed tradeoff; the APD provides internal current gain.
Check. APD multiplication also adds excess noise and bias sensitivity. Larger internal current is not automatically better signal-to-noise in every receiver.
Problem 17.11 — Silicon p-i-n photocurrent
Working.
Answer. Approximately 44.36 nA.
Check. Negligible dark current means no dark-current subtraction is needed, but shot noise and amplifier noise have not thereby vanished.