Silicon Carrier-Depletion Mach-Zehnder Modulator

1. Theory

• Definition: A silicon carrier-depletion Mach-Zehnder modulator (MZM) is an interferometer on silicon-on-insulator (SOI) in which each arm contains a reverse-biased lateral pn junction in a rib waveguide; the reverse voltage widens the depletion layer, removes free carriers from the mode, and so changes the effective index and the loss, and the interference at the output turns the phase difference into an intensity (Li et al.; Xiao et al.).
• Why not the Pockels effect: Crystalline silicon has no linear electro-optic effect, which I take as background knowledge (the sources do not discuss it), so the index change comes from the plasma dispersion effect, as in the Soref-Bennett equations $\Delta n=-[8.8\times 10^{-22}\Delta N_e+8.5\times 10^{-18}\Delta N_h^{0.8}]$ at 1550 nm (coefficients from the Ansys page; the exponent from background knowledge). Holes change the index more than electrons per carrier.
• Junction and doping: In the Li device the rib is 550 nm wide with a 90 nm slab on 220 nm SOI, the junction is 50 nm off the centre towards the N region, the doping is non-uniform in depth from 0.8 to 1.5 $\times 10^{18}$ $\mathrm{c}{\mathrm{m}}^{-3}$, with intermediate P${}^{+}$/N${}^{+}$ regions of $2\times 10^{18}$ $\mathrm{c}{\mathrm{m}}^{-3}$ at 200 nm and ohmic P${}^{++}$/N${}^{++}$ regions of $1\times 10^{20}$ $\mathrm{c}{\mathrm{m}}^{-3}$ at 1000 nm. The Xiao device (340 nm SOI, 450 nm rib, 80 nm slab) uses about $2\times 10^{17}$ $\mathrm{c}{\mathrm{m}}^{-3}$ in both regions, with the heavy doping 1 $\mu \mathrm{m}$ from the sidewalls and the junction about 50 nm off centre.
• Efficiency and loss: The phase efficiency is quoted as $V_{\pi }L$: 1.4 $\mathrm{V}\mathrm{cm}$ (Li et al., $-4$ V, 2 mm), 1.6 to 2.05 $\mathrm{V}\mathrm{cm}$ (Xiao et al., 750 $\mu \mathrm{m}$), 3.1 $\mathrm{V}\mathrm{cm}$ for a 250 nm SOI prototype at 2 V. Every doped region also absorbs, so the phase shifter loses 16 to 22 $dB/cm$; a long device is more efficient per volt but costs more loss and has a lower bandwidth.
• Travelling-wave electrodes: A length of 0.75 to 2 mm is a sizeable fraction of the electrical wavelength at tens of GHz (2.5 mm at 30 GHz for a microwave index of 4, computed), so the electrode is built as a transmission line terminated in 35 to 50 $\Omega$, and its microwave index, impedance and loss set the bandwidth; Li et al. removed the substrate to lower the microwave index by 0.3 to 0.4 and to match the optical group index of 3.7 to 3.8.
• Figures of merit: $V_{\pi }L$, the insertion loss in dB, the extinction ratio, the $3\mathrm{dB}$ electro-optic bandwidth, the energy per bit $\frac{1}{4}{C}_J{V}_{pp}^2$, and the product $V_{\pi }L\cdot \alpha$ in $\mathrm{V}\mathrm{dB}$ (about 30 for both devices, my computation).
• Limits: The response is non-linear (the depletion width grows as the square root of the voltage), the efficiency falls with the reverse bias, the extinction ratio at high rates is 2.5 to $4\mathrm{dB}$ with the swings used, and the loss is several dB.

2. Working Principle

• Bias: The junction is held at a reverse bias of $-4$ to $-8$ V (Li et al.) or $-5$ V (Xiao et al.), where the depletion layer is wide and the capacitance is lower; the width goes as $\sqrt{V_{bi}+V_R}$ and the capacitance falls with it.
• Phase: A voltage swing widens the depletion layer, the mode loses holes on the p side and electrons on the n side, the index rises, and the arm phase changes by $\Delta \phi =(2\pi /\lambda )\Delta n_{eff}L$; the phase is a concave function of the voltage because of the square-root law.
• Interference: The two arms are combined in a multimode interferometer (MMI) and the output is $\frac{1}{2}{P}_{in}[1+\cos \Delta \phi ]$ for the ideal case. A bias at quadrature (a phase of $\pi /2$ set with a thermal shifter of 1 to 2 mW per $\pi$) puts the operating point on the steep slope.
• Drive: Li et al. use series push-pull, in which one electrode drives both junctions in series, so the capacitance is halved and the voltage is doubled compared with a dual drive. The drive of 5 V peak-to-peak gives $3.6\mathrm{dB}$ at 70 GBd, and Xiao et al. use 6.5 V for $3.9\mathrm{dB}$ at 50 Gbit/s.
• Speed: The bandwidth is set by the RC time of the junction (about 28 GHz for 750 $\mu \mathrm{m}$ with 102 $\mathrm{fF}$) for the short device, and by the microwave loss and the termination for the long one (58 to 60 GHz for 2 mm after the substrate removal).
• Energy: The driver charges the junction with about $\frac{1}{4}{C}_J{V}_{pp}^2$ per bit, which is about 1 pJ per junction for the Xiao values above, and the thermal bias adds 1 to 2 mW.

3. Applications

• Single-lane 100 Gbit/s links: Li et al. send 112 Gbit/s PAM-4 (56 GBd) and 128 Gbit/s PAM-4 (64 GBd) over 1 to 10 km of single-mode fibre, the 10 km case with a dispersion-compensation module, with a sensitivity of about $-15\mathrm{dBm}$ and $-16\mathrm{dBm}$ at the FEC thresholds used.
• High-rate OOK: Li et al. reach 90 Gbit/s OOK and Xiao et al. 60 Gbit/s OOK with a 750 $\mu \mathrm{m}$ device and $3.5\mathrm{dB}$ of total loss.
• CMOS-compatible transceivers: The Xiao devices are made on 200 mm SOI wafers, which is the process of a silicon photonics foundry; the same junction also makes variable attenuators and the arms of in-phase and quadrature (IQ) modulators for coherent links (background knowledge, no source I read describes them).
• Data-centre optics: The compact size (0.75 to 2 mm) and the integration with germanium detectors (previous page) suit transceivers of 100 Gbit/s per lane; microring modulators, planned as the next page, are smaller but narrower in wavelength.

Plasma Dispersion and the Depletion Junction

Name	Formula	Related Formulae	Variables/Constants
Soref-Bennett plasma dispersion of silicon at 1550 nm	$\Delta n=-\left[8.8\times 10^{-22}\Delta N_e+8.5\times 10^{-18}{\left(\Delta N_h\right)}^{0.8}\right]$ $\Delta \alpha =8.5\times 10^{-18}\Delta N_e+6.0\times 10^{-18}\Delta N_h$	${\alpha }_{dB}=4.343\Delta \alpha$	$\begin{array}{r}\Delta N_e,\Delta N_h:changesoftheelectronand\\ holedensitiesin\mathrm{c}{\mathrm{m}}^{-3}\end{array}$ $\Delta n:changeoftherefractiveindex$ $\begin{array}{r}\Delta \alpha :changeoftheabsorptioncoefficient\\ in\mathrm{c}{\mathrm{m}}^{-1}(intensity)\end{array}$ $\begin{array}{r}{\alpha }_{dB}:thesameindB/cm,thefour\\ coefficientsareasprintedontheAnsys\\ page(theabsorptioncoefficients8.5\times 10^{-18}\\ and6.0\times 10^{-18}andtheindexcoefficients\\ 8.8\times 10^{-22}and-8.5\times 10^{-18}),theexponent0.8\\ fortheholesin\Delta nandtheexponent1\\ elsewherearefrommybackground\\ knowledge,astatementthatthedependence\\ islinearforelectronsandnonlinearfor\\ holesisinarXiv1504.04409,the\\ equationsarefor1550nmandarethe\\ startingpointofeverycarrier-depletion\\ designinthisfolder\end{array}$ $\begin{array}{r}(computedforequalchangesofboth\\ carriers:at2\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}\Delta n=-7.65\times 10^{-4}and\\ \Delta \alpha =2.90\mathrm{c}{\mathrm{m}}^{-1}=12.6dB/cm;at1\times 10^{18}\mathrm{c}{\mathrm{m}}^{-3}\\ \Delta n=-3.02\times 10^{-3}and\Delta \alpha =14.5\mathrm{c}{\mathrm{m}}^{-1}=63.0\\ dB/cm;at1\times 10^{18}theholesgive71\%of\\ \Delta n(2.14\times 10^{-3}of3.02\times 10^{-3})andthe\\ electrons59\%of\Delta \alpha (8.5of14.5\mathrm{c}{\mathrm{m}}^{-1}))\end{array}$
Nedeljkovic-Soref-Mashanovich update at 1550 nm	$\Delta n=-\left[5.4\times 10^{-22}\Delta N_e^{1.011}+1.53\times 10^{-18}\Delta N_h^{0.838}\right]$ $\Delta \alpha =8.88\times 10^{-21}\Delta N_e^{1.167}+5.84\times 10^{-20}\Delta N_h^{1.109}$	$\frac{\Delta n_{NSM}}{\Delta n_{SB}}$	$\begin{array}{r}symbolsasinthepreviousrow,the\\ abstractofthepapersaysthatterahertz\\ absorptionmeasurementswereusedto\\ refinethecommonlyused1.3and1.55\mu \mathrm{m}\\ equations,butitreproducesno\\ coefficient,soallsixcoefficientsand\\ exponentsherearefrommyrecollectionof\\ thepaperandnotfromasourceIread\\ (backgroundknowledge,tobeverified\\ againstthepaperbeforepublication),the\\ samerecollectiongivesdifferent\\ exponentsat1310nm,whichIdonotuse\end{array}$ $\begin{array}{r}(computed:at2\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}\Delta n=-6.49\times 10^{-4}\\ and\Delta \alpha =2.27\mathrm{c}{\mathrm{m}}^{-1}(9.9dB/cm),whichis15\\ \%smallerinindexand22\%smallerin\\ lossthantheSoref-Bennettvalues;at\\ 1\times 10^{18}\mathrm{c}{\mathrm{m}}^{-3}\Delta n=-2.71\times 10^{-3}and\Delta \alpha =14.35\\ \mathrm{c}{\mathrm{m}}^{-1},10\%smallerinindexand1\%\\ smallerinloss)\end{array}$
Built-in voltage of the pn junction	$V_{bi}=\frac{k_{B}T}{q}\ln \frac{N_A{N}_D}{n_i^2}$	$\frac{k_{B}T}{q}=25.85\mathrm{mV}\mathrm{at}300\mathrm{K}$	$k_B:Boltzmannconstant$ $T:absolutetemperature$ $q:elementarycharge$ $N_A,N_D:acceptoranddonordensitiesin\mathrm{c}{\mathrm{m}}^{-3}$ $\begin{array}{r}{n}_i:intrinsiccarrierdensityofsilicon,\\ Xiaoetal.use{V}_{bi}=0.9Vinsidetheir\\ capacitanceformula(seethecapacitance\\ row),theformulaisthestandard\\ abrupt-junctionresult(background\\ knowledge,nosourcegivesit)\end{array}$ $\begin{array}{r}(computedwithT=300Kand{n}_i=1.0\times 10^{10}\\ \mathrm{c}{\mathrm{m}}^{-3},bothassumed:V_{bi}=0.869Vfor{N}_A=\\ N_D=2\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3},0.941Vfor8\times 10^{17}and\\ 0.952Vfor1\times 10^{18};thevalueof0.9Vin\\ Xiaoetal.agreeswiththe2\times 10^{17}case\\ within4\%)\end{array}$
Depletion width of the lateral junction	$W=\sqrt{\frac{2{\epsilon }_s\left(V_{bi}+V_R\right)}{q}\frac{N_A+N_D}{N_A{N}_D}}$	$W_p=W\frac{N_D}{N_A+N_D}$ $W\propto \sqrt{V_{bi}+V_R}$	${\epsilon }_s={\epsilon }_0{\epsilon }_r:permittivityofsilicon$ $V_R:reversebias$ $\begin{array}{r}{W}_p:partofthewidthonthepside,the\\ carrierdensityisassumeduniformandthe\\ junctionabrupt(alateraljunctionina\\ ribwaveguidehasadopingprofileand\\ fringingthatthisformulaignores),the\\ formulaisthestandarddepletion\\ approximation(backgroundknowledge)\end{array}$ $\begin{array}{r}(computedwith{\epsilon }_r=11.7and{n}_i=1.0\times 10^{10}\\ \mathrm{c}{\mathrm{m}}^{-3},bothassumed:for{N}_A=N_D=2\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}\\ W=106nmat0V,276nmat-5Vand339\\ nmat-8V;for1\times 10^{18}\mathrm{c}{\mathrm{m}}^{-3}W=50,124and\\ 152nm;between0and-5Vthe2\times 10^{17}\\ junctionwidensby170nm,85nmoneach\\ side)\end{array}$
Junction capacitance per unit length	$C^{\prime }=\frac{{\epsilon }_{s}h}{W}(\mathrm{per}\mathrm{unit}\mathrm{length})$	$C_J^{\prime }\approx \frac{330}{\sqrt{0.9-V_{bias}}}fF/mm$	$h:heightofthedepletionsheet$ $W:depletionwidth$ $\begin{array}{r}{V}_{bias}:biasvoltage(negativeforreverse\\ bias),thesecondformisthefitofXiao\\ etal.forthe450nmwideribguidewith\\ 80nmslabandabout2\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}doping,\\ printedas{}^{\prime }fJ/m{m}^{\prime }inthesummaryIread,\\ whichIreadasfF/mm(myreading,because\\ acapacitanceperlengthismeant),the\\ capacitancefallsasthesquarerootof\\ thebiasbecausethedepletionlayer\\ widens,Xiaoetal.:750\mu \mathrm{m}phaseshifter\end{array}$ $\begin{array}{r}(computed:thefitgives348,194,136and\\ 111fF/mmat0,-2,-5and-8V,so102\mathrm{fF}\\ forthe750\mu \mathrm{m}shifterat-5V;the\\ parallel-plateformwiththe276nmwidth\\ at-5Vgives128fF/mmforh=340nm\\ (thefullribheight)and98fF/mmforh=\\ 260nm,sothefitiswithin6\%ofthe\\ full-heightestimate(assumeduniform\\ doping,nofringing))\end{array}$
Effective-index change and phase shift	$\Delta \phi =\frac{2\pi }{\lambda }\Delta n_{eff}L$ $\Delta n_{eff}\approx {\sum }_i{\Gamma }_i\Delta n_i$	$\Delta n_{eff}\approx \frac{\Delta W/2}{w_{eff}}\left(\left|\Delta n_e\right|+\left|\Delta n_h\right|\right)$	$\lambda :vacuumwavelength$ $L:lengthofthephaseshifter$ $\begin{array}{r}{\Gamma }_i:overlapoftheopticalmodewiththe\\ regioniinwhichthecarrierdensity\\ changes\end{array}$ $\Delta n_i:indexchangeofthatregion$ $\begin{array}{r}{w}_{eff}:effectivewidthofthemode,thelast\\ formismyowncrudeestimatefora\\ symmetricjunctionthatwidensby\Delta W\\ (halfofitoneachside)andamodeof\\ uniformintensityover{w}_{eff},thestandard\\ overlapformisbackgroundknowledge\end{array}$ $\begin{array}{r}(computedwith{w}_{eff}=450nm(theribwidth\\ ofXiaoetal.),the2\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}junction\\ andthe0to-5Vswing:\Delta n_{eff}=\\ -1.44\times 10^{-4},so{L}_{\pi }=5.4mmand{V}_{\pi }L=2.7\\ \mathrm{V}\mathrm{cm}forasinglearm;themeasured2.05\\ \mathrm{V}\mathrm{cm}is1.3timessmaller,whichisthe\\ expecteddirectionbecausethemode\\ intensitypeaksatthejunction(my\\ reading))\end{array}$

Efficiency, Loss and Doping

Name	Formula	Related Formulae	Variables/Constants
Voltage-length product	$V_{\pi }L=\frac{\lambda }{2\left(d\Delta n_{eff}/dV\right)}$	$V_{\pi }=\frac{V_{\pi }L}{L}$ $\Delta \phi \propto \sqrt{V_{bi}+V_2}-\sqrt{V_{bi}+V_1}$	$\begin{array}{r}d\Delta n_{eff}/dV:changeoftheeffectiveindex\\ pervolt\end{array}$ $\begin{array}{r}{V}_{\pi }:voltageforaphasedifferenceof\pi ,\\ thesquare-rootformisthedepletion\\ approximationappliedtoaswingfrom{V}_1\\ to{V}_2(reversebiascountedpositive;\\ backgroundknowledge),reportedvalues:Li\\ etal.1.4\mathrm{V}\mathrm{cm}(7Vtimes0.2cm,DC,-4\\ Vbias,2mm,seriespush-pull),Xiaoet\\ al.2.05\mathrm{V}\mathrm{cm}inthetableand1.6to2.0\\ \mathrm{V}\mathrm{cm}intheabstract(750\mu \mathrm{m},-5V),the\\ 250nmSOIdesignpaper3.1\mathrm{V}\mathrm{cm}at2V\\ (prototype)and1.8\mathrm{V}\mathrm{cm}at6V\\ (optimised,theoretical),thetableof\\ Xiaoetal.alsolists0.85,1.3,2.4,\\ 2.7,2.8,3.5,4and14\mathrm{V}\mathrm{cm}forearlier\\ devices\end{array}$ $\begin{array}{r}(computed:1.4\mathrm{V}\mathrm{cm}at1.55\mu \mathrm{m}meansd\Delta \\ n_{eff}/dV=5.5\times 10^{-5}pervolt(singlearm,if\\ the7Visthetotalacrossthe\\ interferometer);theswing0to-2Vis\\ 1.83timesasefficientas-4to-6V,and\\ thesmall-signalslopeat-2Vis1.30\\ timesthatat-4V(assumed{V}_{bi}=0.9V))\end{array}$
Insertion loss of the Mach-Zehnder	$IL=2{\alpha }_{sp}+\left({\alpha }_{dp}+{\alpha }_{wg}\right)L$	$I{L}_{\pi }\approx 2{\alpha }_{sp}+\left({\alpha }_{dp}+{\alpha }_{wg}\right)L_{\pi }$	${\alpha }_{sp}:lossofonesplitterorcombinerindB$ ${\alpha }_{dp}:doping-inducedlossindB/cm$ ${\alpha }_{wg}:lossoftheundopedwaveguideindB/cm$ $\begin{array}{r}{L}_{\pi }:lengthforaphasedifferenceof\pi ,\\ Xiaoetal.:splitter0.35dBmeasured\\ (0.1to0.2dBsimulated),{\alpha }_{dp}about7\\ dB/cm,{\alpha }_{wg}about9dB/cm,on-chiploss1.9\\ dB,Lietal.:waveguide0.6dB,twoMMIs\\ 0.4dB,phaseshifter4.4dB,total5.4dB\\ (thesummarydoesnotsaywhetherthe4.4\\ dBincludestheheater,whichIreadas\\ separate,myreading),fibrecouplingis\\ notincluded\end{array}$ $\begin{array}{r}(computed:Xiaoetal.2x0.35+(7+9)\\ x0.075=0.7+1.2=1.9dB,equaltothe\\ reportedvalue;Lietal.0.6+4.4+0.4\\ =5.4dB,equaltothereportedvalue,and\\ 4.4dBover2mmis22dB/cm)\end{array}$
Optimum doping for minimum loss	$I{L}_{\pi }\approx 2{\alpha }_{sp}+AB\sqrt{N}+\frac{{\alpha }_{wg}B}{\sqrt{N}}$ $N_{opt}=\frac{{\alpha }_{wg}}{A}$	${\alpha }_{dp}=AN,L_{\pi }=\frac{B}{\sqrt{N}}$	$N:dopingdensityin\mathrm{c}{\mathrm{m}}^{-3}$ $A:slopeofthedopinglossindB/cmper\mathrm{c}{\mathrm{m}}^{-3}$ $\begin{array}{r}B:constantofthelength(L_{\pi }fallsasthe\\ inversesquarerootofthedoping,Xiaoet\\ al.),theoptimumfollowsfromsettingthe\\ derivativetozero(myderivation),Xiao\\ etal.:optimumdoping5\times 10^{16}\mathrm{c}{\mathrm{m}}^{-3}for{\alpha }_{wg}=\\ 2dB/cmand3.3\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}for15dB/cm,\\ simulatedloss6.5dBat15dB/cmfalling\\ to2.6dBat2dB/cm(at-5V),1.9dB\\ measuredwith9dB/cm,chosendopingabout\\ 2\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3},a3mmdeviceat2dB/cmbelow\\ 2dB(projected)\end{array}$ $\begin{array}{r}(computedwithA=7dB/cm/2\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}=\\ 3.5\times 10^{-17}(assumedconstant):N_{opt}=\\ 5.7\times 10^{16},2.6\times 10^{17}and4.3\times 10^{17}\mathrm{c}{\mathrm{m}}^{-3}for{\alpha }_{wg}=\\ 2,9and15dB/cm;thepapergives5\times 10^{16}\\ and3.3\times 10^{17},sotheratioofthetwo\\ optimais6.6against7.5forthe\\ proportionallaw)\end{array}$
Loss-efficiency product	$FO{M}_{\alpha }=V_{\pi }L\cdot {\alpha }_{ps}$	$I{L}_{ps}=\frac{FO{M}_{\alpha }}{V_{\pi }}$	$\begin{array}{r}{\alpha }_{ps}:totallossofthephaseshifterin\\ dB/cm(dopingpluswaveguide)\end{array}$ $\begin{array}{r}I{L}_{ps}:lossofthephaseshifteratthe\\ lengthofa\pi shift,thefigureofmerit\\ ismyownconstructionfromthequoted\\ values,withXiaoetal.{\alpha }_{ps}=16dB/cmand\\ Lietal.22dB/cm(thelatterfrom4.4dB\\ in2mm)\end{array}$ $\begin{array}{r}(computedin\mathrm{V}\mathrm{dB}:Xiaoetal.2.05x16=\\ 32.8,or28.8with1.8\mathrm{V}\mathrm{cm};Lietal.1.4\\ x22=30.8;sothetwoverydifferent\\ dopings(2\times 10^{17}against0.8\times 10^{18}to1.5\times 10^{18}\\ \mathrm{c}{\mathrm{m}}^{-3})sitonnearlythesametrade-off\\ line,whichiswhattheoptimisationof\\ thepreviousrowwouldpredict(my\\ reading))\end{array}$

Transfer Function, Extinction and Chirp

Name	Formula	Related Formulae	Variables/Constants
Intensity transfer with unequal arm loss	$P_{out}=\frac{P_{in}}{4}\left[e^{-{\alpha }_{1}L}+e^{-{\alpha }_{2}L}+2e^{-\left({\alpha }_1+{\alpha }_2\right)L/2}\cos \Delta \phi \right]$	$P_{out}=\frac{P_{in}}{2}\left[1+\cos \Delta \phi \right](\mathrm{ideal})$ $\Delta \phi ={\phi }_0+\Delta {\phi }_1-\Delta {\phi }_2$	$P_{i}n,P_{o}ut:inputandoutputpower$ $\begin{array}{r}{\alpha }_1,{\alpha }_2:powerlosscoefficientsofthetwo\\ armsin\mathrm{c}{\mathrm{m}}^{-1}\end{array}$ $L:phaseshifterlength$ $\begin{array}{r}{\phi }_0:staticphasedifferencesetbyarm\\ lengthandtheheater\end{array}$ $\begin{array}{r}\Delta {\phi }_1,\Delta {\phi }_2:carrier-inducedphasesofthe\\ arms,idealsplittersof50\%are\\ assumed,theidealformandthepush-pull\\ relationarederivedontheLithium\\ NiobateMach-ZehnderModulatorpage,the\\ unequal-lossformismyownderivation\\ (backgroundknowledge),insiliconthe\\ depletionchangesthelosstogetherwith\\ theindex,whichiswhythelosstermis\\ kept\end{array}$ $\begin{array}{r}(computedcheck:forequallossesthe\\ bracketreducesto{e}^{-\alpha L}(1+\cos \Delta \phi ),theideal\\ formtimesthetransmission{e}^{-\alpha L};the\\ maximumofthetransferisreachedat\Delta \phi =\\ 0andthenullat\pi ,whereonlytheloss\\ imbalanceleaveslight)\end{array}$
Extinction ratio of a limited phase swing	$ER=\frac{1+\sin \left(\Delta {\phi }_{pp}/2\right)}{1-\sin \left(\Delta {\phi }_{pp}/2\right)},\Delta {\phi }_{pp}=\pi \frac{V_{pp}}{V_{\pi }}$	$ER\left[\mathrm{dB}\right]=10{\log }_{10}ER$	$\begin{array}{r}\Delta {\phi }_{pp}:peak-to-peakphasedifferenceofthe\\ interferometer\end{array}$ $V_{pp}:peak-to-peakdrive$ $\begin{array}{r}{V}_{\pi }:voltageforaphasedifferenceof\pi at\\ theoperatingbias,theformassumes\\ quadraturebias,aphaselinearinvoltage\\ andlosslesssplitters(myderivation,\\ backgroundknowledge),Xiaoetal.:6.5V\\ peak-to-peakat-5V,3.9dBatquadrature\\ at50Gbit/s(text)and4.4dBat60\\ Gbit/s(table),7.5dBbelowquadrature,\\ Lietal.:5Vpeak-to-peakat-6V,3.6,\\ 2.7and3.3dBat70,80and90GBd(OOK),\\ thefinalswinginthetransmission\\ experiments3.8V\end{array}$ $\begin{array}{r}(computed:Xiaoetal.with2.05\mathrm{V}\mathrm{cm}(V_{\pi }\\ =27.3V)gives3.33dBandwith1.8\mathrm{V}\mathrm{cm}\\ (24.0V)3.81dB,so1.8\mathrm{V}\mathrm{cm}matchesthe\\ 3.9dB;Lietal.with{V}_{\pi }=7Vgives12.8\\ dB,withtheefficiencyloweredto0.84\\ timesat-6V(square-rootlaw)9.8dBand\\ with3.8Vswing6.8dB,allwellabove\\ thereported3.6dB)\end{array}$
Chirp (linewidth-enhancement) factor of carrier depletion	${\alpha }_H=\frac{2k_0|\Delta n_{eff}|}{\Delta {\alpha }_{eff}},k_0=\frac{2\pi }{\lambda }$	$\Delta {\phi }_1=-\Delta {\phi }_2\Rightarrow \mathrm{no}\mathrm{net}\mathrm{chirp}(\mathrm{ideal}\mathrm{push}-\mathrm{pull})$	$\begin{array}{r}{\alpha }_H:ratioofthephasechangetohalfthe\\ relativechangeofthepower\end{array}$ $\begin{array}{r}\Delta n_{eff}and\Delta {\alpha }_{eff}:changesoftheeffective\\ indexandoftheintensityabsorption\\ causedbythesamecarriers,the\\ definitionfollowstheHenryfactorofthe\\ IndiumPhosphideQuantum-Well\\ ElectroabsorptionModulatorpage,no\\ sourceIreadmeasuresthechirpofa\\ siliconmodulator,thepush-pullstatement\\ isbackgroundknowledge\end{array}$ $\begin{array}{r}(computedfromtheSoref-Bennettvaluesat\\ 1\times 10^{18}\mathrm{c}{\mathrm{m}}^{-3}forbothcarriers:{\alpha }_H=16.9;\\ fromtheupdateofthepreviousgroup\\ 15.3;soremovingcarriersisalmosta\\ purephasechange(large{\alpha }_H),butthe\\ smallabsorptionpartstillleavesa\\ residualamplitudechange,whicha\\ single-armdriveturnsintochirp)\end{array}$

Speed and the Travelling-Wave Electrode

Name	Formula	Related Formulae	Variables/Constants
Lumped RC bandwidth	$f_{RC}=\frac{1}{2\pi R_{eff}{C}_J}$	$C_J=C^{\prime }L$	$\begin{array}{r}{R}_{eff}:totalresistanceseenbythejunction\\ (diodeseriesresistanceplusthesource\\ orloadshare)\end{array}$ $\begin{array}{r}{C}_J:junctioncapacitanceofthephase\\ shifter,theformulaisexactonlyfora\\ lumpedloadandisusedhereasafirst\\ estimateforthe750\mu \mathrm{m}shifter\\ (backgroundknowledge),Xiaoetal.:27.8\\ GHzelectro-optic3dBand27.7GHz\\ electrical6dBat-5V,500\mu \mathrm{m}above40\\ GHz,aresonancenear34GHzfromthe\\ impedancemismatch\end{array}$ $\begin{array}{r}(computed:C_J=102\mathrm{fF}at-5V,sothe\\ measured27.8GHzneeds{R}_{eff}=56\Omega ;50\Omega \\ wouldgive31.2GHzand25\Omega wouldgive\\ 62.4GHz)\end{array}$
Velocity-mismatch bandwidth of a travelling-wave electrode	$H\left(f\right)=\frac{\sin x}{x},x=\frac{\pi f|n_{\mu }-n_g|L}{c}$	$f_{3dB}=\frac{1.392c}{\pi |n_{\mu }-n_g|L}$ ${\beta }_{\mu opt}=\frac{{\omega }_m}{c}\left(n_{\mu }-n_{opt}\right)$	$n_{\mu }:microwaveindexoftheelectrode$ $n_g:opticalgroupindex$ $L:electrodelength$ $c:speedoflightinvacuum$ $\begin{array}{r}{\omega }_m:modulationangularfrequency,thesinc\\ formisthelosslesscase(background\\ knowledge;seetheLithiumNiobate\\ Mach-ZehnderModulatorpageforthe1.9c/\pi \\ conventionusedthere),Lietal.:\\ removingthesubstrateloweredthe\\ microwaveindexby0.3to0.4tomatchthe\\ opticalgroupindexof3.7to3.8,andthe\\ characteristicimpedancerosebynearly4\\ \Omega \end{array}$ $\begin{array}{r}(computed:foramismatchof0.3and2mm\\ f_{3dB}=221GHz,for0.4itis166GHz,for\\ 0.3and0.75mm590GHz;with1.9c/\pi \\ insteadof1.392c/\pi the0.3and2mmvalue\\ wouldbe302GHz;sothevelocitymismatch\\ aloneisnotthelimit,thelossofthe\\ electrodeis)\end{array}$
Electro-optic response with electrode loss	$EO\left(f\right)=10{\log }_{10}\frac{{|S_{21}|}^2-2|S_{21}|\cos \left({\beta }_{\mu opt}l\right)+1}{{\left(\ln |S_{21}|\right)}^2+{\left({\beta }_{\mu opt}l\right)}^2}$	$\left|S_{21}\right|=e^{-{\alpha }_{m}l}$	$S_{21}:electricaltransmissionoftheelectrode$ $l:electrodelength$ ${\alpha }_m:microwaveamplitudelossperlength$ $\begin{array}{r}{\beta }_{\mu opt}:asinthepreviousrow,Lietal.\\ printthisform(partlygarbledinthe\\ summaryIread,soIrestoredthestandard\\ form,myreading),Lietal.:removingthe\\ substrategaveabout1dB/cmless\\ microwaveloss(simulatedfor1000\Omega cm\\ silicon),thesimulated3dBelectro-optic\\ bandwidthrosefrom47to55GHzat5V,\\ themeasuredresponsewas1.2dBdownat\\ 50GHz(-4V)with58GHzpredicted,and\\ 60GHzat-8V\end{array}$ $\begin{array}{r}(computedcheck:for{\alpha }_{m}lgoingto0and\\ {\beta }_{\mu opt}lgoingto0theargumenttendsto1,\\ soEO(0)=0dB;thegainofthesubstrate\\ removalis8GHz,whichis17\%of47\\ GHz)\end{array}$
Reflection of the electrode termination	$\Gamma =\frac{Z_L-Z_0}{Z_L+Z_0},S_{11}=20{\log }_{10}\left|\Gamma \right|$	$Z_0=\sqrt{\frac{L^{\prime }}{C^{\prime }}}$	$Z_L:loadimpedance$ $Z_0:characteristicimpedanceoftheelectrode$ $\begin{array}{r}{L}^{\prime },C^{\prime }:inductanceandcapacitanceper\\ length(thecapacitanceincludesthe\\ junction),theformulasareforalossless\\ lineina50\Omega system(background\\ knowledge),Xiaoetal.:agapof40\mu \mathrm{m}\\ gives50\Omega andagapof5\mu \mathrm{m}about30\Omega \\ (chosen,S_{11}about-10dB,below-9dBup\\ to40GHz),Lietal.:a35\Omega on-chip\\ termination,S_{11}below-15dBupto40GHz\\ andbelow-10dBabove50GHz,aload\\ below50\Omega givespeakingatthecostof\\ power\end{array}$ $\begin{array}{r}(computed:a30\Omega lineina50\Omega system\\ gives|\Gamma |=0.25,thatis-12.0dB,\\ againstthereportedabout-10dB;35\Omega \\ gives0.176,thatis-15.1dB,closeto\\ thereported-15dBbelow40GHz)\end{array}$

Energy and Thermal Tuning

Name	Formula	Related Formulae	Variables/Constants
Dynamic energy per bit	$E_{bit}=\frac{1}{4}{C}_J{V}_{pp}^2$	$C_J\to \frac{C_J}{2},V_{pp}\to 2V_{pp}(\mathrm{series}\mathrm{push}-\mathrm{pull})$	$C_J:capacitanceseenbythedriver$ $\begin{array}{r}{V}_{pp}:peak-to-peakswing,thelumpedform\\ withthefactor1/4followstheGermanium\\ andSilicon-GermaniumElectroabsorption\\ Modulatorpageandisonlyafirst\\ estimateforaterminatedtravelling-wave\\ line(inwhichthepowerismostly\\ dissipatedinthetermination,background\\ knowledge),Lietal.donotreportan\\ energyperbit,Xiaoetal.donoteither,\\ theseriespush-pullstatement(halfthe\\ capacitance,doublethedrivevoltage\\ comparedwithdualdrive)isfromLiet\\ al.\end{array}$ $\begin{array}{r}(computedforXiaoetal.,assumingone\\ junctionof102\mathrm{fF}drivenwith6.5V:1.08\\ pJ/bit;twosuchjunctionsdriven\\ separatelytake2.15pJ/bit,andthesame\\ twoinseriespush-pull(51\mathrm{fF}at13V)\\ alsotake2.15pJ/bit;thesearefarabove\\ the45to60fJ/bitofthegermanium\\ devicesofthepreviouspage,becausea\\ silicondepletionshifterofthislength\\ needsvoltsandnotfractionsofavolt)\end{array}$
Thermal phase tuning	$\Delta {\phi }_{th}=\frac{2\pi }{\lambda }\frac{dn}{dT}\Delta TL$ $\Delta T=R_{th}P$	$P_{\pi }=\frac{\lambda }{2L\left(dn/dT\right)R_{th}}$	$\begin{array}{r}dn/dT:thermo-opticcoefficientof\\ silicon,takenas1.86\times 10^{-4}{\mathrm{K}}^{-1}(assumed\\ frommybackgroundknowledge,nosourceI\\ readgivesit)\end{array}$ $\Delta T:temperaturerise$ $\begin{array}{r}{R}_{th}:thermalresistanceoftheheaterto\\ thesurroundingsinK/W\end{array}$ $\begin{array}{r}P:heaterpower,Lietal.:a\\ doped-siliconresistorononearm,1to2\\ mWper\pi ofphase,thesamesubstrate\\ removalthatservestheedgecoupler\end{array}$ $\begin{array}{r}(computedwiththeassumedcoefficientand\\ aheateralongthewhole2mm:\Delta T=2.1K\\ foraphaseof\pi ,so1to2mWmeans{R}_{th}=\\ 1.0to2.1K/mW;theheaterlengthisnot\\ stated,sothethermalresistanceisonly\\ indicative)\end{array}$

Important Questions

Ques: What is the voltage-length product of a silicon carrier-depletion Mach-Zehnder: 1.4, 1.8, 2.05 or 3.1 V cm?

The numbers belong to different devices and biases. Li et al. quote 1.4 $\mathrm{V}\mathrm{cm}$ from 7 V times 0.2 cm at $-4$ V with series push-pull, so it is not clear whether the 7 V is per arm or across the interferometer (my reading). Xiao et al. give 2.05 $\mathrm{V}\mathrm{cm}$ in the table but 1.6 to 2.0 $\mathrm{V}\mathrm{cm}$ in the abstract; the measured $3.9\mathrm{dB}$ at 6.5 V fits 1.8 $\mathrm{V}\mathrm{cm}$ ($3.81\mathrm{dB}$, computed) better than 2.05 ($3.33\mathrm{dB}$, computed), so I judge 1.8 more likely for that device. The 250 nm SOI paper gives 3.1 $\mathrm{V}\mathrm{cm}$ at 2 V for a prototype and 1.8 at 6 V for a theoretical design. I treat 1.4 to 2 as the realistic range. Reference: Li, Wang, Li, Xiao and Yu, ‘Silicon intensity Mach-Zehnder modulator for single lane 100 Gb/s applications’, Photonics Research 6(2), 109 (2018) (researching.cn PDF; read through a fetch-tool summary, in which the first equation was partly garbled); Xiao et al., ‘High-speed, low-loss silicon Mach-Zehnder modulators with doping optimization’, Optics Express 21(4) (2013) (ResearchGate landing page, read through a fetch-tool summary; the loss equations were garbled; full author list not recorded); ‘Design of a carrier-depletion MZM in 250 nm SOI technology’, Advances in Radio Science 15, 269 (2017) (ars.copernicus.org; abstract and metadata only; authors not recorded).

Ques: Can a 0.5 mm silicon modulator really switch $32\mathrm{dB}$?

The abstract of the 250 nm SOI design paper lists a phase shifter of 0.5 mm, 3.1 $\mathrm{V}\mathrm{cm}$ at 2 V, and an extinction ratio of about $32\mathrm{dB}$. A $\pi$ shift in 0.5 mm needs 3.1 divided by 0.05 cm, that is 62 V (computed), or 36 V for the optimised 1.8 $\mathrm{V}\mathrm{cm}$ (computed), far outside a depletion bias. So the $32\mathrm{dB}$ cannot be a dynamic swing of the voltage; it is most likely the static depth of the interferometer null between a balanced pair of arms, set with a heater or a long-arm bias (my reading, the abstract does not say). I would not quote $32\mathrm{dB}$ as the modulation depth. The insertion loss of $4.2\mathrm{dB}$ and the 30 GHz cut-off at 2 V are the usable numbers. Reference: ‘Design of a carrier-depletion MZM in 250 nm SOI technology’, Advances in Radio Science 15, 269 (2017) (ars.copernicus.org; abstract and metadata only; authors not recorded).

Ques: Is $2\times 10^{17}$ or $1\times 10^{18}$ per cubic centimetre the better doping?

The two reference devices differ by a factor of five to ten: Xiao et al. use about $2\times 10^{17}$ $\mathrm{c}{\mathrm{m}}^{-3}$ (loss 16 $dB/cm$, $V_{\pi }L$ about 2 $\mathrm{V}\mathrm{cm}$) and Li et al. $0.8\times 10^{18}$ to $1.5\times 10^{18}$ $\mathrm{c}{\mathrm{m}}^{-3}$ (loss about 22 $dB/cm$, 1.4 $\mathrm{V}\mathrm{cm}$). Higher doping gives a shorter device but more loss per length, and the optimum $N_{opt}={\alpha }_{wg}/A$ (my derivation from the loss relation of Xiao et al.) is $2.6\times 10^{17}$ $\mathrm{c}{\mathrm{m}}^{-3}$ for 9 $dB/cm$ of waveguide loss (computed), close to their choice. Their own simulation puts it at $5\times 10^{16}$ to $3.3\times 10^{17}$ $\mathrm{c}{\mathrm{m}}^{-3}$ for 2 to 15 $dB/cm$. The product $V_{\pi }L\cdot \alpha$ is 29 to 33 $\mathrm{V}\mathrm{dB}$ for both (computed), so neither is clearly better; the choice then follows from the speed, since higher doping lowers the series resistance. Reference: Xiao et al., ‘High-speed, low-loss silicon Mach-Zehnder modulators with doping optimization’, Optics Express 21(4) (2013) (ResearchGate landing page, read through a fetch-tool summary; the loss equations were garbled; full author list not recorded); Li, Wang, Li, Xiao and Yu, ‘Silicon intensity Mach-Zehnder modulator for single lane 100 Gb/s applications’, Photonics Research 6(2), 109 (2018) (researching.cn PDF; read through a fetch-tool summary, in which the first equation was partly garbled).

Ques: Do the Soref-Bennett and the Nedeljkovic coefficients agree, and which is more likely?

They agree within 10 to 22 % for the density range of modulators. For equal electron and hole changes of $2\times 10^{17}$ $\mathrm{c}{\mathrm{m}}^{-3}$ the Soref-Bennett fit gives $\Delta n=-7.65\times 10^{-4}$ and 12.6 $dB/cm$, the Nedeljkovic fit $-6.49\times 10^{-4}$ and 9.9 $dB/cm$ (computed), 15 % and 22 % lower. At $1\times 10^{18}$ $\mathrm{c}{\mathrm{m}}^{-3}$ the difference is 10 % in the index and 1 % in the loss (computed). The Nedeljkovic paper says it refined the 1.55 $\mu \mathrm{m}$ equations with terahertz measurements, so I judge it more likely to be right, but I used coefficients from my recollection, not from a source I read, and the Soref-Bennett ones are from the Ansys page. The page should be checked against the paper before publication. Reference: Soref-Bennett coefficients for 1.55 um as printed on the Ansys Lumerical help page (optics.ansys.com, article 360034382494; the exponents were not on the page); the original, Soref and Bennett, SPIE 704 (1987), was not read; Nedeljkovic, Soref and Mashanovich, IEEE Photonics Journal 3(6), 1171 (2011) (abstract page eprints.soton.ac.uk/337781 only; the coefficients are from my recollection, not from a source I read); arXiv 1504.04409, Quantum Noise of Free-Carrier Dispersion in Semiconductor Optical Cavities (appendix D.2 only; used for the statement that the dependence is linear for electrons and nonlinear for holes).

Ques: The capacitance is printed as 330 fJ per mm; is that 330 fF per mm?

The summary of Xiao et al. gives $C_J\approx 330/(0.9-V_{bias}{)}^{1/2}$ with the unit ‘fJ/mm’. A capacitance per length has the unit of farad per metre, so I read it as 330 $fF/mm$ at $V_{bias}=-0.1$ V (my reading). At $-5$ V it gives 136 $fF/mm$, or 102 $\mathrm{fF}$ for the 750 $\mu \mathrm{m}$ shifter (computed), and the RC time then requires 56 $\Omega$ for the measured 27.8 GHz (computed), a sensible value. A parallel-plate estimate with the depletion width of 276 nm for $2\times 10^{17}$ $\mathrm{c}{\mathrm{m}}^{-3}$ and the full rib height of 340 nm gives 128 $fF/mm$ (computed, assumed), within 6 % of the fit, which supports the reading as femtofarad. The 0.9 V also matches the built-in voltage of 0.87 V (computed). Reference: Xiao et al., ‘High-speed, low-loss silicon Mach-Zehnder modulators with doping optimization’, Optics Express 21(4) (2013) (ResearchGate landing page, read through a fetch-tool summary; the loss equations were garbled; full author list not recorded).

Ques: Why does the computed extinction ratio of the 2 mm device exceed the measured one?

With $V_{\pi }=7$ V and 5 V peak-to-peak at quadrature the ideal extinction ratio is $12.8\mathrm{dB}$ (computed). Allowing for the lower efficiency at $-6$ V (0.84 times, computed with the square-root law and an assumed $V_{bi}=0.9$ V) it is $9.8\mathrm{dB}$, and with the 3.8 V swing of the transmission experiments $6.8\mathrm{dB}$. Li et al. report $3.6\mathrm{dB}$ at 70 GBd. I cannot close the gap from the summary I read: possible causes are a bias away from quadrature, the $1.2\mathrm{dB}$ roll-off and the equalisation-free eye, or a $V_{\pi }$ that already refers to the series pair (my reading). I therefore trust the reported eye values and treat the ideal formula as an upper bound. Reference: Li, Wang, Li, Xiao and Yu, ‘Silicon intensity Mach-Zehnder modulator for single lane 100 Gb/s applications’, Photonics Research 6(2), 109 (2018) (researching.cn PDF; read through a fetch-tool summary, in which the first equation was partly garbled).

Ques: Is the bandwidth of the 2 mm silicon modulator 58 or 60 GHz, and is it consistent with a $1.2\mathrm{dB}$ roll-off at 50 GHz?

Li et al. state a $1.2\mathrm{dB}$ roll-off at 50 GHz ($-4$ V) with 58 GHz predicted, and 60 GHz at $-8$ V. A single-pole response with $1.2\mathrm{dB}$ at 50 GHz would have its $3\mathrm{dB}$ point at 88.6 GHz (computed), and a Gaussian response with $3\mathrm{dB}$ at 58 GHz would be $2.2\mathrm{dB}$ down at 50 GHz (computed), so the three numbers do not describe one simple curve. I read 58 GHz as an extrapolation of the measured curve and 60 GHz as a different bias (my reading), both close to the simulated 55 GHz (47 GHz without substrate removal, a gain of 8 GHz or 17 %, computed). I would quote ‘about 55 to 60 GHz’ and flag it. Reference: Li, Wang, Li, Xiao and Yu, ‘Silicon intensity Mach-Zehnder modulator for single lane 100 Gb/s applications’, Photonics Research 6(2), 109 (2018) (researching.cn PDF; read through a fetch-tool summary, in which the first equation was partly garbled).

Ques: What limits the bandwidth of a long travelling-wave silicon modulator?

Not the velocity mismatch. With a mismatch of 0.3 and a length of 2 mm the sinc response reaches $3\mathrm{dB}$ at 221 GHz (computed with the 1.392 convention; 302 GHz with 1.9c/$\pi$, computed). The limits are the microwave loss of the electrode, which the substrate removal lowered by about $1\mathrm{dB}$/cm (simulated), the resistance of the doped slab, and the termination: Li et al. use 35 $\Omega$ for a gain from peaking, and Xiao et al. use a 30 $\Omega$ line (the resonance near 34 GHz came from the mismatch). For the 750 $\mu \mathrm{m}$ device the junction RC time (31 GHz with 50 $\Omega$, computed) is already comparable to the measured 27.8 GHz. Reference: Li, Wang, Li, Xiao and Yu, ‘Silicon intensity Mach-Zehnder modulator for single lane 100 Gb/s applications’, Photonics Research 6(2), 109 (2018) (researching.cn PDF; read through a fetch-tool summary, in which the first equation was partly garbled); Xiao et al., ‘High-speed, low-loss silicon Mach-Zehnder modulators with doping optimization’, Optics Express 21(4) (2013) (ResearchGate landing page, read through a fetch-tool summary; the loss equations were garbled; full author list not recorded).

Ques: What does ‘series push-pull’ do to the drive voltage and the energy?

In series push-pull the two junctions are connected back to back, so the same current charges both and the capacitance seen is halved, but each junction gets only half of the applied voltage; for the same phase the drive must double compared with a dual drive (Li et al.). The energy $\frac{1}{4}{C}_J{V}_{pp}^2$ then gives $\frac{1}{4}\cdot \frac{C_J}{2}\cdot (2V{)}^2=\frac{1}{2}{C}_J{V}^2$, which equals the total of two separately driven junctions, $2\times \frac{1}{4}{C}_J{V}^2$ (my computation). So the series scheme gains in speed and impedance, because the capacitance is halved, but not in energy. For the Xiao junction (102 $\mathrm{fF}$, 6.5 V) both give 2.15 pJ/bit (computed, assumed). The sources give no measured energy, and the termination dissipates more in a travelling-wave line (background knowledge). Reference: Li, Wang, Li, Xiao and Yu, ‘Silicon intensity Mach-Zehnder modulator for single lane 100 Gb/s applications’, Photonics Research 6(2), 109 (2018) (researching.cn PDF; read through a fetch-tool summary, in which the first equation was partly garbled); Xiao et al., ‘High-speed, low-loss silicon Mach-Zehnder modulators with doping optimization’, Optics Express 21(4) (2013) (ResearchGate landing page, read through a fetch-tool summary; the loss equations were garbled; full author list not recorded).

Ques: How can the insertion loss be $1.9\mathrm{dB}$ for one modulator and $5.4\mathrm{dB}$ for another?

The Xiao device is 0.75 mm long at about $2\times 10^{17}$ $\mathrm{c}{\mathrm{m}}^{-3}$ doping: $2\times 0.35+16\times 0.075=1.9$ dB (computed). The Li device is 2 mm long at higher doping: $0.6+4.4+0.4=5.4$ dB (computed), and the phase shifter alone loses 22 $dB/cm$. The $1.9\mathrm{dB}$ device needs 27 V for a $\pi$ shift (2.05 $\mathrm{V}\mathrm{cm}$ over 0.075 cm, computed) whereas the other needs 7 V. Both lie on one line of loss against efficiency (29 to 33 $\mathrm{V}\mathrm{dB}$, computed), so the loss is the price of the voltage, not a difference in quality. A low-loss modulator is a long, slow, high-voltage one, or one with a lower doping, unless the junction itself is improved. Fibre coupling is not included in either value. Reference: Xiao et al., ‘High-speed, low-loss silicon Mach-Zehnder modulators with doping optimization’, Optics Express 21(4) (2013) (ResearchGate landing page, read through a fetch-tool summary; the loss equations were garbled; full author list not recorded); Li, Wang, Li, Xiao and Yu, ‘Silicon intensity Mach-Zehnder modulator for single lane 100 Gb/s applications’, Photonics Research 6(2), 109 (2018) (researching.cn PDF; read through a fetch-tool summary, in which the first equation was partly garbled).

Ques: Why does the efficiency fall with reverse bias, and does it matter for the chirp?

The depletion width grows as $\sqrt{V_{bi}+V_R}$, so each volt moves the edge less at high bias. With an assumed $V_{bi}=0.9$ V the swing from 0 to $-2$ V is 1.83 times as effective as $-4$ to $-6$ V (computed), and the response is concave, so a large swing is also slightly distorted. Reports at different bias (3.1 $\mathrm{V}\mathrm{cm}$ at 2 V, 1.8 at 6 V for the 250 nm design) point the other way because the geometry and doping differ. The high Henry factor (16.9 from Soref-Bennett, computed) shows that removed carriers change mainly the phase, so a push-pull drive has little chirp (background knowledge); no source I read measures it. Reference: ‘Design of a carrier-depletion MZM in 250 nm SOI technology’, Advances in Radio Science 15, 269 (2017) (ars.copernicus.org; abstract and metadata only; authors not recorded); Soref-Bennett coefficients for 1.55 um as printed on the Ansys Lumerical help page (optics.ansys.com, article 360034382494; the exponents were not on the page); the original, Soref and Bennett, SPIE 704 (1987), was not read; Indium Phosphide Quantum-Well Electroabsorption Modulator page of this vault.

Related pages in this vault: Silicon MOS-Capacitor Modulator, Lithium Niobate Mach-Zehnder Modulator, Silicon Carrier-Injection PIN Modulator, Silicon Microring Modulator, Polymer Electro-Optic Modulator, Silicon on Insulator (SOI) MOSFET, Waveguide Photodiode, Uni-Traveling-Carrier Photodiode (UTC-PD), PN Junction.