Germanium and Silicon-Germanium Electroabsorption Modulator

1. Theory

• Definition: A germanium or germanium-silicon electroabsorption modulator (EAM) is a reverse-biased p-i-n waveguide of Ge or GeSi grown on silicon, usually on SOI, in which a field in the intrinsic region changes the absorption near the direct gap, so that the power transmitted from a continuous-wave laser follows the drive voltage (Liu et al.; the 1550 nm GeSi paper). Two mechanisms are used: the Franz-Keldysh effect of bulk Ge or GeSi and the quantum-confined Stark effect of Ge/SiGe wells.
• Why germanium: The direct gap of unstrained Ge is 0.80 eV, which is 1550 nm (strain paper). Tensile strain lowers the gap, and the strain fit is $E_{g\Gamma }=0.80-7.31\epsilon -37{\epsilon }^2$ in eV for uniaxial tension; a small amount of silicon raises the gap again, so the pure-Ge device works near 1600 to 1610 nm and a GeSi device with about 0.7 % Si (an estimate) at 1550 nm.
• Franz-Keldysh effect: A field turns the carrier wave functions into Airy functions with tails in the gap, so photons below the gap are absorbed and the edge moves to longer wavelength (Wikipedia). It is a bulk effect without wells and it is sub-picosecond, so the speed is limited by the RC time (Liu et al.). The field in the Ge is about 25 $kV/cm$ at 0 V and about 80 $kV/cm$ at $-2$ V (Liu et al.).
• Quantum-confined Stark effect: In Ge wells of 6.5 to 10 nm between Si${}_{0.15}$Ge${}_{0.85}$ barriers the direct exciton of the well is sharp and shifts to lower energy with the field, so the edge moves from 1408 nm to 1456 nm between 0 and 4 V (Kuo et al.); with thinner wells the band moves to the O-band at 1.31 $\mu \mathrm{m}$ (Chaisakul et al.). The effect needs a stack of wells and a narrower wavelength window (23 to 28 nm).
• Integration: The Ge or GeSi section is grown in a recess of the silicon waveguide and coupled to it by a taper (Liu et al.: two Si tapers of 5 $\mu \mathrm{m}$; the 1550 nm GeSi paper: about $1.0\mathrm{dB}$ of transition loss); in the quantum-well paper the passive circuit is silicon nitride on Si or SOI. The same layer also works as a photodetector, with 0.49 to 0.73 A/W (Liu et al.).
• Figures of merit: The extinction ratio $ER$, the insertion loss $IL$ and their ratio $FOM=ER/IL$, the link power penalty, the energy per bit $\frac{1}{4}{C}_j{V}_{pp}^2$, the bandwidth $1/(2\pi R_s{C}_j)$ and the optical bandwidth.
• Limits: The extinction ratio is 3 to $6\mathrm{dB}$ at 2 to 3 V in 40 to 55 $\mu \mathrm{m}$, the insertion loss is 5 to $10\mathrm{dB}$, the edge moves by 0.6 to 0.76 $\mathrm{nm}{/}^{\circ }\mathrm{C}$, and the photocurrent adds static power.

2. Working Principle

• Bias: The diode is reverse-biased; the built-in field of the p-i-n already puts about 25 $kV/cm$ in the Ge at 0 V, which causes a residual loss, and the field rises to about 80 $kV/cm$ at $-2$ V (Liu et al.).
• Absorption: In the bulk devices the laser sits a few tens of meV below the zero-field edge (background knowledge), where the Franz-Keldysh tail makes the absorption rise steeply with the field (the ratio $\Delta \alpha /\alpha$ is about 1.6 at 42 $kV/cm$ and 3.1 at 75 $kV/cm$, the 1550 nm GeSi paper). In the quantum-well devices the operating point is about 23 meV below the zero-field exciton, and the field moves the exciton towards the laser (Chaisakul et al.).
• Drive: Liu et al. use 3 V peak-to-peak at $-1.7$ V, the 1550 nm GeSi paper a swing of 0 to 2.8 V (1.4 V bias with $\pm 0.7$ V), and the quantum-well paper 2 V at $-1.1$ V for 50 Gbit/s.
• Speed: The process is fast, and the limit is $f=1/(2\pi R_s{C}_j)$ with about 20 to 49 $\mathrm{fF}$ and 9 to 205 $\Omega$ (Liu et al.; the quantum-well paper); a 56 Gbit/s NRZ eye needs 36 GHz.
• Energy: The dynamic energy is $\frac{1}{4}{C}_j{V}_{pp}^2$, from 13 to 147 fJ/bit in the cited table (leaving out the doubtful 5.6 fJ/bit entry), and the static power is the bias times the photocurrent.
• Temperature: The edge moves by 0.6 to 0.76 $\mathrm{nm}{/}^{\circ }\mathrm{C}$, so a temperature change of 40 to 60 K moves it by about the width of the usable window.

3. Applications

• Short-reach 50 to 100 Gbit/s links: A 40 $\mu \mathrm{m}$ Ge modulator carries 56 Gbit/s NRZ at 1610 nm (Liu et al.), and the quantum-well modulator on silicon nitride carries 100 Gbit/s with a six-tap feed-forward equalizer at 1288 nm (the quantum-well paper).
• C-band links: A GeSi modulator carries 28 Gbit/s at 1550 nm with about 60 fJ/bit (the 1550 nm GeSi paper), and the table of Liu et al. lists devices at 1560 and 1566 nm with 50 and 56 Gbit/s.
• O-band CWDM links: Ge/SiGe wells give more than $8\mathrm{dB}$ at 1327 to 1340 nm with 5 to 6 V in 100 $\mu \mathrm{m}$ (Chaisakul et al.), and the quantum-well paper compares with devices at 1321 and 1340 nm.
• Monolithic transceivers: The same Ge layer serves as modulator and detector on silicon, with a responsivity of 0.49 to 0.73 A/W (Liu et al.) and a bandwidth beyond 50 GHz with 10 $\mathrm{fF}$ for a Ge EAM aimed at chip-level optical interconnects (ElectronicSpecifier).
• Ultralow energy modulation: A GeSi modulator with an active area of 30 square micrometres gives $10\mathrm{dB}$ at 1540 nm, 1.2 GHz and 50 fJ/bit (Nature Photonics abstract).

Franz-Keldysh Absorption and Strain

Name	Formula	Related Formulae	Variables/Constants
Franz-Keldysh absorption of a direct-gap semiconductor	$\alpha \left(\hslash \omega ,F\right)\propto \frac{\sqrt{\hslash \theta }}{\hslash \omega }\left[{\mathrm{Ai}}^{\prime }{\left(\eta \right)}^2-\eta \mathrm{Ai}{\left(\eta \right)}^2\right],\eta =\frac{E_g-\hslash \omega }{\hslash \theta }$	$\hslash \theta ={\left(\frac{e^2{\hslash }^2{F}^2}{2\mu }\right)}^{1/3}$ $\mu =\frac{m_e^{\ast }{m}_h^{\ast }}{m_e^{\ast }+m_h^{\ast }}$	$\alpha :absorptioncoefficient$ $\hslash \omega :photonenergy$ $E_g:directgap$ $F:electricfield,\mathrm{Ai}:Airyfunction$ $\hslash \theta :electro-opticenergy$ $\mu :reducedmassoftheelectron-holepair$ $e:elementarycharge$ $\begin{array}{r}\hslash :reducedPlanckconstant,theAiryform\\ and\hslash \theta arethestandardbulkresult\\ withoutexcitons(backgroundknowledge;\\ thesourcesdescribetheeffectonlyin\\ words:thefieldturnsthewavefunctions\\ intoAiryfunctionswithtailsinthegap,\\ sophotonsbelowthegapareabsorbed\\ (Wikipedia),andtheeffectis\\ sub-picosecond,sothespeedissetbythe\\ RCtime(Liuetal.)),Liuetal.givea\\ fieldatthecentreoftheGewaveguideof\\ about25kV/cmat0Vandabout80kV/cm\\ at-2V\end{array}$ $\begin{array}{r}(computedforassumedmasses{m}_e^{\ast }=0.038\\ m_{0}and{m}_{lh}^{\ast }=0.043m_0,whichgive\mu =\\ 0.0202m_0:\hslash \theta =22.8,32.2,47.4and49.4\\ meVat25,42,75and80kV/cm,agrowth\\ as{F}^{2/3})\end{array}$
Sub-gap tail of the Franz-Keldysh absorption	$\alpha \propto \exp \left[-\frac{4}{3}{\left(\frac{E_g-\hslash \omega }{\hslash \theta }\right)}^{3/2}\right]$	$\frac{d\ln \alpha }{d\left(\hslash \omega \right)}=\frac{2}{\hslash \theta }\sqrt{\frac{E_g-\hslash \omega }{\hslash \theta }}$	$\begin{array}{r}theexponentialformholdsonlywellbelow\\ thegap,for{E}_g-\hslash \omega muchlargerthan\hslash \theta \\ (backgroundknowledge),thetailmakesthe\\ absorptionatafixedwavelengthrise\\ steeplywiththefield,whichisthe\\ modulationmechanismofbulkGeandGeSi,\\ Liuetal.noteresidualabsorptionfrom\\ thebuilt-infieldat0Vthatraisesthe\\ insertionloss\end{array}$ $\begin{array}{r}(computedwiththeAiryformandthe\\ assumedmassesabove:at40meVbelowthe\\ gaptheabsorptionrisesbyafactorof\\ 3.5between42and75kV/cm(arbitrary\\ units);atthatdetuning\eta isonly1.2and\\ 0.84,sotheexponentialformisnotyet\\ reached)\end{array}$
Direct gap of strained Ge and the absorption edge	$E_{g\Gamma }\left(\epsilon \right)=0.80-7.31\epsilon -37{\epsilon }^2[\mathrm{eV}]$	${\lambda }_g\left[\mathrm{nm}\right]=\frac{1239.84}{E_g\left[\mathrm{eV}\right]}$ $\Delta \lambda =\frac{{\lambda }^2}{hc}\Delta E$	$\begin{array}{r}\epsilon :tensilestrainasafraction,thefit\\ isforuniaxialtensionalong[100]and\\ holdsbelowabout2\%strain(strain\\ paper),0.80eVistheunstraineddirect\\ gap,1550nm,tensionlowersthegapand\\ movestheedgetolongerwavelength,the\\ biaxialcoefficientsofGeonSiarenot\\ giveninthesource(thestrainpaper\\ mentions0.16\%biaxialstrainfrom\\ thermalmismatchforitssample),Liuet\\ al.:pureGeworksat1600to1610nm,the\\ 1550nmGeSipaper:about0.7\%Si(an\\ estimate)worksat1550nm,adesignstudy\\ with1.19\%Sitargets1528to1560nm\\ (ActaPhys.Sin.),Siraisesthegapand\\ somovestheedgebacktoshorter\\ wavelength(backgroundknowledge)\end{array}$ $\begin{array}{r}(computed:E_{g\Gamma }(0.2\%)=0.785eV,thatis\\ 1579nm,asanillustrationonlywithan\\ assumedstrain;theedgeofthepure-Ge\\ deviceat1600nmandthatofthe0.7\%\\ GeSideviceat1550nmdifferby25meV,\\ whichisabout36meVperpercentSi(my\\ reading,twodifferentdevicesandbiases,\\ notameasuredcoefficient))\end{array}$
Relative absorption change	$\frac{\Delta \alpha }{\alpha }=\frac{\alpha \left(F_2\right)-\alpha \left(F_1\right)}{\alpha \left(F_1\right)}$	$ER\approx 4.343\Gamma \Delta \alpha L$	$F_1,F_2:lowandhighfield$ $\Gamma :confinementoftheguidedmodeintheGe$ $\begin{array}{r}L:length,thereferencefieldofthe\\ quotedratiosisnotdefinedinthe\\ extracts(myreading:thelowerfield),\\ the1550nmGeSipaper:about1.6at42\\ kV/cmandabout3.1at75kV/cm,theActa\\ Phys.Sin.pagelists2.8at1640nmand2\\ to3.3over1610to1640nmforearlierGe\\ devices\end{array}$ $\begin{array}{r}(computed:3.1/1.6=1.94forafield\\ increaseof75/42=1.79)\end{array}$

Quantum-Confined Stark Effect in Ge/SiGe Wells

Name	Formula	Related Formulae	Variables/Constants
Quantum-confined Stark shift of a Ge well	$\Delta E_{tr}\approx -C_s\frac{\left(m_e^{\ast }+m_h^{\ast }\right)e^2{F}^2{L}_w^4}{{\hslash }^2},C_s=2.19\times 10^{-3}$	$\Delta \lambda =\frac{{\lambda }^2}{hc}\left|\Delta E_{tr}\right|$	$L_w:wellwidth$ $m^{\ast }:effectivemassalongthegrowthdirection$ $\begin{array}{r}F:field,theinfinite-wellformis\\ derivedontheIndiumPhosphide\\ Quantum-WellElectroabsorptionModulator\\ pageandisalowerestimateforfinite\\ barriers,Kuoetal.:tenstrain-balanced\\ Gewellsof10nmbetween16nmS{i}_{0.15}G{e}_{0.85}\\ barriers,ap-i-ndiodeonarelaxed\\ S{i}_{0.1}G{e}_{0.9}buffer,compressivelystrained\\ wells,low-fieldedge0.88eV(1408nm)\\ against0.8eVofbulkGe,edgemoving\\ from1408nmto1456nmbetween0and4V,\\ excitonsresolveduptoabout80kV/cm\end{array}$ $\begin{array}{r}(computed:themeasurededgeshiftis29.0\\ meV;withassumedmasses0.038and0.20m_0\\ theinfinite-wellformulagives4.4meVat\\ 80kV/cmand6.9meVat100kV/cm,whilea\\ finite-barriersolutionwithanassumed\\ offsetof100meVforbothcarriersgives\\ 23.1and36.1meV(second-ordervalues\\ scaledfrom20kV/cm),whichbracketthe\\ measurement;themassesandoffsetsare\\ assumed,notfitted)\end{array}$
Exciton energy and well width	$E_{HH1-C1}=E_g+E_{e1}+E_{hh1}-E_b$	${\lambda }_{ex}\left[\mathrm{nm}\right]=\frac{1239.84}{E_{HH1-C1}\left[\mathrm{eV}\right]}$	$E_g:straineddirectgapofthewell$ $\begin{array}{r}{E}_{e1},E_{hh1}:electronandheavy-hole\\ confinementenergies\end{array}$ $\begin{array}{r}{E}_b:excitonbindingenergy(background\\ knowledge),Chaisakuletal.:20Gewells\\ of6.5nmbetween10nmS{i}_{0.15}G{e}_{0.85}barriers,\\ HH1-C1excitonat0.945eVandLH1-C1\\ transitionat0.989eVat0V,calculated\\ gap0.93eV,atargetof0.97eVfor1.31\\ \mu \mathrm{m}operationneedswellsof5.5to6nm,\\ thinnerwellsmovetheexcitontohigher\\ energy,thewellsarecoherentwitha\\ S{i}_{0.1}G{e}_{0.9}buffer\end{array}$ $\begin{array}{r}(computed:0.945eVis1312nm,0.989eV\\ is1254nmand0.97eVis1278nm;the\\ operatingpoint23meVbelowtheexciton\\ is1345nm;the10nmwellsofKuoetal.\\ (0.88eV)lie64meVbelowthe6.5nm\\ wells(0.945eV),aroughcomparison\\ becausethestrainandbarriersdiffer)\end{array}$
Exciton absorption and contrast of the quantum-well stack	${\alpha }_{eff}=\frac{-\ln \left(1-A\right)}{t_{MQW}},CR=\frac{T_{\max }}{T_{\min }}$	$C{R}_{dB}=10{\log }_{10}CR$ $CR=\exp \left[\Delta {\alpha }_{eff}\Gamma L\right]$	$\begin{array}{r}A:absorbedfraction(theformof{\alpha }_{eff}is\\ backgroundknowledge;thesourceonlysays\\ thatthestackistreatedasonelayer)\end{array}$ $\begin{array}{r}{t}_{MQW}:thicknessofthewellandbarrier\\ stack,0.26\mu \mathrm{m}inKuoetal.,Kuoetal.:\\ lowestexcitonabout6300\mathrm{c}{\mathrm{m}}^{-1},halfwidth\\ athalfmaximumabout8meV,largest\\ change2800\mathrm{c}{\mathrm{m}}^{-1}between0and3V,peak\\ contrastratio4.69at1461nm,28nmof\\ bandwidthwithacontrastabove3,\\ absorptionderivedfromthemeasured\\ photocurrent,thesharpnessofthe\\ excitonsissaidtoexceedthatof\\ InGaAs/InPwells,thedefinitionofthe\\ contrastratioisnotintheextract\end{array}$ $\begin{array}{r}(computed:10{\log }_{10}4.69=6.7dB;asingle\\ passof0.26\mu \mathrm{m}at2800\mathrm{c}{\mathrm{m}}^{-1}givesonly\\ exp(0.0728)=1.08,sothecontrastof\\ 4.69mustcomefromalongerpath;ln4.69\\ /2800\mathrm{c}{\mathrm{m}}^{-1}=5.5\mu \mathrm{m}ofequivalent\\ well-filledpath,thatisabout55\mu \mathrm{m}for\\ aconfinementfactorof0.1(assumed))\end{array}$

Extinction, Loss and Link Power

Name	Formula	Related Formulae	Variables/Constants
Static extinction ratio	$ER=10{\log }_{10}\frac{T\left(V_{ref}\right)}{T\left(V_a\right)},T=10^{-IL/10}$	$ER\approx 4.343\Gamma \left[\alpha \left(V_a\right)-\alpha \left(V_{ref}\right)\right]L$ $E{R}_L=\frac{ER}{L}$	$V_{ref}:referencebias$ $\begin{array}{r}{V}_a:biasoftheabsorbingstate,the\\ quantum-wellpaperprintsER=10{\log }_{10}(ILat\\ V_a/ILat{V}_{ref})withILinlinearunits,\\ Liuetal.:3.8dBat1600nmand3.6dB\\ at1610nmfora3Vswing,4.8and4.6dB\\ for4V,40\mu \mathrm{m}ofGe,the1550nmGeSi\\ paper:6dBat1550nm,55\mu \mathrm{m},the\\ quantum-wellpaper:3.55dB(-2V/0V,\\ Si),3.5dB(SOI),50\mu \mathrm{m},Chaisakulet\\ al.:morethan8dBover1327to1340nm\\ at5V,100\mu \mathrm{m}by4\mu \mathrm{m},TE,theNature\\ Photonicsabstract:10dBat1540nmwith\\ anactiveareaof30\mu {\mathrm{m}}^2\end{array}$ $\begin{array}{r}(computed:theratiosperlengthare\\ 0.090,0.109,0.071and0.080dBper\mu \mathrm{m};\\ theneededvaluesof\Gamma \Delta \alpha are207,251and\\ 163\mathrm{c}{\mathrm{m}}^{-1}forthefirstthreedevices)\end{array}$
Insertion loss and figure of merit	$IL=I{L}_c+I{L}_{tr}+4.343\Gamma \alpha \left(V_0\right)L+{\alpha }_{i}L$	$FOM=\frac{ER}{IL}$	$I{L}_c:couplingloss$ $\begin{array}{r}I{L}_{tr}:transitionlossbetweentheSiguide\\ andtheGeorGeSisection\end{array}$ $\alpha (V_0):absorptionatthelow-lossbias$ $\begin{array}{r}{\alpha }_i:otherloss,Liuetal.:6.2dBat1610\\ nmafterremoving6.7dBpergrating\\ coupler,the1550nmGeSipaper:5dBat\\ 1550nm(2.9to5.7dBover1546to1581\\ nm)withabout1.0dBoftransitionloss,\\ thequantum-wellpaper:9.57dBat1290nm\\ onSiand16.88,12.09and10.31dBat\\ 1272,1290and1300nmonSOI,itstext\\ saysthecouplinglossisbelow1dB,its\\ tablelists1.69and2.08dB,Chaisakulet\\ al.:about5dBabove1345nmandabout10\\ dBat1327nm,afigureofmeritabove1.6\\ isthetargetforshortlinks(ER5dB,\\ loss3dB)\end{array}$ $\begin{array}{r}(computedFOM:0.58forLiuetal.\\ (3.6/6.2),1.2forthe1550nmGeSipaper\\ (6/5),0.37forthequantum-wellpaperon\\ Si(3.55/9.57),1.67forthetarget(5/3);\\ inthetableofLiuetal.theother\\ devicesgive2.04,0.94and0.95)\end{array}$
Link power penalty and optical modulation amplitude	$LPP=10{\log }_{10}\frac{2P_{in}}{P_1-P_0}$	$OMA=P_1-P_0$ $P_1=P_{in}10^{-IL/10},P_0=P_{1}10^{-ER/10}$	$P_{i}n:inputpower$ $\begin{array}{r}{P}_1,P_0:outputpowersoftheonandoff\\ states,thequantum-wellpaper:17.5dBat\\ 1291nm(Si),16.66dBat1305.4nmand\\ 17.69dBat1290nm(SOI),the1550nm\\ GeSipaper:aminimumof8.5dB,the\\ relationofthepenaltytotheoptical\\ modulationamplitudeisfromthe1550nm\\ GeSipaper\end{array}$ $\begin{array}{r}(computed:theidealcase,zerolossand\\ infiniteratio,gives3.01dB;fromthe\\ staticILandERofonewavelength:9.27\\ dBforthe1550nmGeSidevice,11.7dB\\ forLiuetal.,15.1dBforthe\\ quantum-wellpaperonSi;withthedynamic\\ ratiosof2.73and1.8dBofthatpaper\\ thevalueis15.9and17.3dB)\end{array}$
Photocurrent and static power	$I_{ph}=\mathcal{R}{P}_{abs},P_{static}=V_b{I}_{ph}$	$P_{dyn}=E_{bit}B$	$\mathcal{R}:responsivityoftheGeasadetector$ $B:bitrate$ $\begin{array}{r}{V}_b:bias,Liuetal.:darkcurrent6nAat\\ -1Vand12nAat-4V,responsivity0.49\\ A/Wat0Vand0.73A/Wat-2Vat1610\\ nm,staticpowerabout1.2mWat-1.7V\\ (0.7mAfor1mWinput),the1550nmGeSi\\ paper:responsivityabout0.6A/W,dynamic\\ power1.7mWat28Gbit/s,totalabout6.7\\ mWincludingphotocurrent,inputpower\\ below6mW,bias1.4V\end{array}$ $\begin{array}{r}(computed:0.7mAtimes1.7V=1.19mW;\\ 60fJ/bittimes28Gbit/s=1.68mW;the\\ remaining5.0mWat1.4Vis3.6mA,which\\ equals0.6A/Wtimes6mW)\end{array}$

Speed, Capacitance and Energy

Name	Formula	Related Formulae	Variables/Constants
Junction capacitance	$C_j=\frac{{\epsilon }_0{\epsilon }_{r}A}{d}$	$C_j=\frac{{\epsilon }_0{\epsilon }_{r}Lh}{w_i}(\mathrm{lateral}\mathrm{junction})$	$A:junctionarea$ $\begin{array}{r}d:thicknessoftheintrinsicregion\\ includingspacers\end{array}$ $L:length$ $h:heightofthejunction$ $w_i:intrinsicwidth$ $\begin{array}{r}{\epsilon }_r:relativepermittivity,takenbetween\\ 11.9(Si)and16(Ge)(quantum-well\\ paper),the1550nmGeSipaper:junction\\ 55\mu \mathrm{m}long,2.5\mu \mathrm{m}high,0.75\mu \mathrm{m}wide,30\\ \mathrm{fF}stated,Liuetal.:Geguide2\mu \mathrm{m}wide,\\ 0.5\mu \mathrm{m}high,40\mu \mathrm{m}long,C-Vabout60\mathrm{fF}\\ at0Vand24\mathrm{fF}at-2V,fitted20\mathrm{fF},\\ thequantum-wellpaper:2.5\mu \mathrm{m}by50\mu \mathrm{m},\\ eightwellsof10nmand12nmbarriers\\ withtwo15nmspacers,estimated77.48\mathrm{fF}\\ andfitted49\mathrm{fF}(Si)and40\mathrm{fF}(SOI)\end{array}$ $\begin{array}{r}(computed:26.0\mathrm{fF}forthe1550nmGeSi\\ junctionat{\epsilon }_r=16against30\mathrm{fF};22.7\mathrm{fF}\\ forLiuetal.atanassumedd=0.5\mu \mathrm{m}\\ and20\mathrm{fF}ford=0.57\mu \mathrm{m};77.4\mathrm{fF}forthe\\ quantum-wellstackwithd=206nmand{\epsilon }_r\\ =14.4,sotheestimateof77.48\mathrm{fF}needs\\ {\epsilon }_r=14.42)\end{array}$
RC-limited bandwidth	$f_{3dB}=\frac{1}{2\pi R_s{C}_j}$	$f_{3dB}=\frac{1}{2\pi \left(R_s+R_{load}\right)C_j}$	$R_s:seriesresistance$ $C_j:junctioncapacitance$ $\begin{array}{r}{R}_{load}:loadresistance(background\\ knowledge),Liuetal.:R_s=205\Omega ,C_j=\\ 20\mathrm{fF},measured36GHzat-1V,unchanged\\ withbias,thequantum-wellpaper:Si49\\ GHzat-1Vandabove65GHzat-3V\\ (limitedbythebiasteeandprobe),SOI\\ 19GHzat-2Vand24GHzat-2.5V,R_s\\ fitted9\Omega orestimated60\Omega (Si),180or\\ 160\Omega (SOI),the1550nmGeSipaper:20.6\\ GHzat-1V,31.6GHzat-1.9Vand40.7\\ GHzat-2.8V,thetableofLiuetal.\\ listsabove50GHzfortwootherdevices\end{array}$ $\begin{array}{r}(computed:38.8GHzforLiuetal.;361\\ GHzfor9\Omega and49\mathrm{fF},54GHzfor60\Omega and\\ 49\mathrm{fF}and34GHzfor60\Omega and77.48\mathrm{fF};22\\ GHzfor180\Omega and40\mathrm{fF};for30\mathrm{fF}the\\ resistancewouldbe258\Omega at20.6GHzand\\ 130\Omega at40.7GHz)\end{array}$
Dynamic energy per bit	$E_{bit}=\frac{1}{4}{C}_j{V}_{pp}^2$	$C_j=\frac{4E_{bit}}{V_{pp}^2}$	$\begin{array}{r}{V}_{pp}:peak-to-peakswing,thefactor1/4is\\ usedbyLiuetal.,the1550nmGeSipaper\\ andthequantum-wellpaper,Liuetal.:45\\ fJ/bit(20\mathrm{fF},3V),the1550nmGeSi\\ paper:about60fJ/bit(30\mathrm{fF},2.8V),the\\ quantum-wellpaper:49fJ/bit(Si)and\\ 62.5fJ/bit(SOI),theNaturePhotonics\\ abstract:50fJ/bit,1.2GHz,30\mu {\mathrm{m}}^2,1539\\ to1553nm,thetableofLiuetal.lists\\ 147,12.8,13.8and5.6fJ/bitforfour\\ otherdevicesat3,2,2and4V,\\ ElectronicSpecifier:aGeEAMwith10\mathrm{fF}\\ andabandwidthbeyond50GHz\end{array}$ $\begin{array}{r}(computed:45.0fJ/bit,58.8fJ/bitand\\ 49.0fJ/bitforthefirstthree;theSOI\\ valueis40\mathrm{fF}timesa2.5Vswing,62.5\\ fJ/bit;theestimated77.48\mathrm{fF}wouldgive\\ 77.5fJ/bit;thetableimplies65,12.8,\\ 13.8and1.4\mathrm{fF}forthefourother\\ devices)\end{array}$

Temperature and Wavelength Window

Name	Formula	Related Formulae	Variables/Constants
Temperature drift of the absorption edge	$\Delta {\lambda }_{edge}=\frac{d{\lambda }_{edge}}{dT}\Delta T$	$\frac{d{E}_g}{dT}=-\frac{hc}{{\lambda }^2}\frac{d\lambda }{dT}$	$\begin{array}{r}the1550nmGeSipaper:0.76\mathrm{nm}{/}^{\circ }\mathrm{C}over30\\ to70{}^{\circ }Candabandwidthof35nm(1546to\\ 1581nm)fora1dBlinkpenalty,the\\ quantum-wellpaper:about0.6\mathrm{nm}{/}^{\circ }\mathrm{C}(Si)\\ and0.58\mathrm{nm}{/}^{\circ }\mathrm{C}(SOI)over20to80{}^{\circ }C,the\\ extinctionratioat-2V/0Vis3.95dB\\ (Si)and4.82dB(SOI)atthepeak,andat\\ 40,45and50{}^{\circ }Citis3.55,4.13and3.61\\ dB(Si)and4.96,5.09and4.46dB(SOI),\\ withalargestdropbelow0.65dB,Liuet\\ al.:morethan30nmofopticalbandwidth\\ atroomtemperatureonly,Chaisakulet\\ al.:23nmformorethan8dB,Kuoetal.:\\ 28nmforacontrastabove3\end{array}$ $\begin{array}{r}(computed:0.76\mathrm{nm}{/}^{\circ }\mathrm{C}at1550nmis0.39\\ meV/K,and0.6and0.58\mathrm{nm}{/}^{\circ }\mathrm{C}at1290nm\\ are0.45and0.43meV/K;0.76\mathrm{nm}{/}^{\circ }\mathrm{C}times\\ 40Kis30nm,closetothe35nmwindow,\\ and0.6\mathrm{nm}{/}^{\circ }\mathrm{C}times60Kis36nm,more\\ thanthe28nmwindow)\end{array}$

Important Questions

Ques: Is the extinction ratio of a Ge modulator $3.6\mathrm{dB}$, $6\mathrm{dB}$, $8\mathrm{dB}$ or $10\mathrm{dB}$?

Each value belongs to a different device, length and swing. Liu et al. give a static ratio of $3.8\mathrm{dB}$ at 1600 nm and $3.6\mathrm{dB}$ at 1610 nm for a 3 V swing (4.8 and $4.6\mathrm{dB}$ for 4 V) in 40 $\mu \mathrm{m}$ of Ge, and a dynamic ratio of $2.7\mathrm{dB}$ at 56 Gbit/s. The 1550 nm GeSi paper gives $6\mathrm{dB}$ for 55 $\mu \mathrm{m}$ and $5.9\mathrm{dB}$ dynamic at 2.8 V. Chaisakul et al. give more than $8\mathrm{dB}$ at 5 V for a 100 $\mu \mathrm{m}$ quantum-well guide, and the Nature Photonics abstract $10\mathrm{dB}$ at 1540 nm. Per micrometre the first three are 0.09, 0.11 and $0.08\mathrm{dB}$ (computed), so the spread is mostly length and voltage. The dynamic value is what a receiver sees, and 3 to $6\mathrm{dB}$ is the more likely range for 50 Gbit/s links (my reading). Reference: Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary); High speed GeSi electro-absorption modulator at 1550 nm wavelength on SOI waveguide, Optics Express 20(20), 22224 (2012) (read through a ResearchGate fetch-tool summary; authors not recorded); Chaisakul et al., J. Appl. Phys. 116, 193103 (2014) (re.public.polimi.it PDF; read through a fetch-tool summary); Waveguide-integrated, ultralow-energy GeSi electro-absorption modulators, Nature Photonics (2008), nphoton.2008.99 (abstract and figure captions only; the article is behind a paywall; authors not recorded).

Ques: Why does the pure-Ge modulator work at 1610 nm and the GeSi modulator at 1550 nm?

The direct gap of unstrained Ge is 0.80 eV, or 1550 nm, and the strain paper fits $E_{g\Gamma }=0.80-7.31\epsilon -37{\epsilon }^2$ for uniaxial tension. Ge on Si keeps a small tensile strain (background knowledge) and the Franz-Keldysh tail reaches further, so the pure-Ge device of Liu et al. peaks near 1600 nm with more than 30 nm of bandwidth. The 1550 nm GeSi paper estimates 0.7 % Si, and a design study uses 1.19 % for the C band. The 25 meV between 1600 and 1550 nm is about 36 meV per percent Si (computed from two devices, my reading). One value is an estimate for a built device and the other a design target, so I take 0.7 % as more likely for a real device and read 1.19 % as including margin (my reading). Reference: Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary); High speed GeSi electro-absorption modulator at 1550 nm wavelength on SOI waveguide, Optics Express 20(20), 22224 (2012) (read through a ResearchGate fetch-tool summary; authors not recorded); Design of an evanescent-coupled GeSi electro-absorption modulator based on Franz-Keldysh effect, Acta Physica Sinica (2013) (ResearchGate landing page only); Germanium under high tensile stress: nonlinear dependence of direct band gap vs. strain, arXiv 1606.01668 (ar5iv; read through a fetch-tool summary).

Ques: Which capacitance belongs in the energy per bit, 77 fF or 49 fF?

The quantum-well paper estimates a junction capacitance of 77.48 $\mathrm{fF}$ from the parallel-plate formula and fits 49 $\mathrm{fF}$ (Si) and 40 $\mathrm{fF}$ (SOI), and it reports 49 and 62.5 fJ/bit. With $E=\frac{1}{4}{C}_j{V}_{pp}^2$ and a 2 V swing, 49 $\mathrm{fF}$ gives 49 fJ/bit and 77.48 $\mathrm{fF}$ gives 77.5 fJ/bit (computed); the SOI value of 62.5 fJ/bit needs 40 $\mathrm{fF}$ and a 2.5 V swing (computed), which the paper does use at its SOI bias. Liu et al. show the same spread: 60 $\mathrm{fF}$ at 0 V, 24 $\mathrm{fF}$ at $-2$ V and a fit of 20 $\mathrm{fF}$, which gives their 45 fJ/bit at 3 V. The fitted values reproduce the reported energies, so they are the more likely ones, and the estimate leaves out the fringing and the depletion (my reading). Reference: A 100 Gb/s quantum-confined Stark effect modulator monolithically integrated with silicon nitride on Si (PMC12046052; read through a fetch-tool summary; authors not recorded); Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary).

Ques: Is the printed 5.6 fJ/bit for a 4 V device credible?

The table of Liu et al. lists a GeSi modulator of 40 by 50 $\mu \mathrm{m}$ at 1566 nm with a 4 V swing, 44 GHz and 5.6 fJ/bit. With $E=\frac{1}{4}{C}_j{V}_{pp}^2$ the energy needs only 1.4 $\mathrm{fF}$ (computed), whereas the other rows of the table imply 65, 12.8 and 13.8 $\mathrm{fF}$ and the paper itself 20 $\mathrm{fF}$, and ElectronicSpecifier quotes 10 $\mathrm{fF}$ for a fast Ge EAM. A device of that footprint is unlikely to have only 1.4 $\mathrm{fF}$. A lost digit, that is 56 fJ/bit and 14 $\mathrm{fF}$, would be in line with the other rows, so I take 56 fJ/bit as the more likely value, although a different definition of the energy, such as only part of the swing, is also possible. I could not check the original (my reading). Reference: Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary); Germanium waveguide EAM demonstrated at OFC (ElectronicSpecifier news item).

Ques: Which bandwidth is realistic: 20, 36, 49, 65 or more than 100 GHz?

All are measured, for different devices and biases. Liu et al. measure 36 GHz at $-1$ V, close to the 38.8 GHz that $1/(2\pi R_s{C}_j)$ gives for 205 $\Omega$ and 20 $\mathrm{fF}$ (computed). The quantum-well paper measures 49 GHz at $-1$ V and above 65 GHz at $-3$ V on Si, limited by the bias tee and probe, so those are lower bounds, and 19 to 24 GHz on SOI. The 1550 nm GeSi paper rises from 20.6 to 40.7 GHz between $-1$ and $-2.8$ V. A fitted 9 $\Omega$ with 49 $\mathrm{fF}$ would give 361 GHz and the estimated 60 $\Omega$ gives 54 GHz (computed), so the fitted value cannot describe the whole circuit. For 30 $\mathrm{fF}$ the series resistance would fall from 258 to 130 $\Omega$ with bias (computed), more likely because the depletion grows (my reading). Reference: Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary); A 100 Gb/s quantum-confined Stark effect modulator monolithically integrated with silicon nitride on Si (PMC12046052; read through a fetch-tool summary; authors not recorded); High speed GeSi electro-absorption modulator at 1550 nm wavelength on SOI waveguide, Optics Express 20(20), 22224 (2012) (read through a ResearchGate fetch-tool summary; authors not recorded).

Ques: Is the coupling loss below $1\mathrm{dB}$ or 1.7 to $2.1\mathrm{dB}$, and why does the SOI loss reach $16.9\mathrm{dB}$?

The quantum-well paper says in its text that the coupling loss is below $1\mathrm{dB}$, but its Table 1 gives 1.69 $\pm$ $0.49\mathrm{dB}$ on Si and 2.08 $\pm$ $0.1\mathrm{dB}$ on SOI; the table has uncertainties and is the more likely value (my reading). The insertion loss at 0 V depends strongly on wavelength because the zero-field exciton absorbs: $9.57\mathrm{dB}$ at 1290 nm on Si, and 16.88, 12.09 and $10.31\mathrm{dB}$ at 1272, 1290 and 1300 nm on SOI. Liu et al. quote $6.2\mathrm{dB}$ at 1610 nm after removing $6.7\mathrm{dB}$ per grating coupler, and the 1550 nm GeSi paper $5\mathrm{dB}$ with about $1.0\mathrm{dB}$ of transition loss. The ratios $ER/IL$ are 0.58, 1.2 and 0.37 (computed), below the target of 1.67 for short links, so loss is the weak point of all three. Reference: A 100 Gb/s quantum-confined Stark effect modulator monolithically integrated with silicon nitride on Si (PMC12046052; read through a fetch-tool summary; authors not recorded); Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary); High speed GeSi electro-absorption modulator at 1550 nm wavelength on SOI waveguide, Optics Express 20(20), 22224 (2012) (read through a ResearchGate fetch-tool summary; authors not recorded); Chaisakul et al., J. Appl. Phys. 116, 193103 (2014) (re.public.polimi.it PDF; read through a fetch-tool summary).

Ques: How large is the field in the germanium, and does it agree with the capacitance?

Liu et al. give about 25 $kV/cm$ at 0 V and about 80 $kV/cm$ at $-2$ V at the centre of the Ge guide, the 1550 nm GeSi paper uses 42 and 75 $kV/cm$, and Kuo et al. resolve excitons up to about 80 $kV/cm$. The change of 55 $kV/cm$ over 2 V implies a width of 0.36 $\mu \mathrm{m}$ and a built-in voltage of 0.91 V (computed). The fitted 20 $\mathrm{fF}$ with ${\epsilon }_r=16$ (assumed) and the area of 80 square micrometres implies 0.57 $\mu \mathrm{m}$ (computed), and the Ge is only 0.5 $\mu \mathrm{m}$ high. The field is quoted at the centre of the guide, where it peaks, while the capacitance includes fringing and the whole area, so the two need not agree. I take the fields as orders of magnitude, because the extract does not say how they were obtained (my reading). Reference: Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary); High speed GeSi electro-absorption modulator at 1550 nm wavelength on SOI waveguide, Optics Express 20(20), 22224 (2012) (read through a ResearchGate fetch-tool summary; authors not recorded); Kuo et al., Ge/SiGe quantum-well electroabsorption, Nature 437, 1334-1336 (2005) (Stanford file 326.pdf, read through a fetch-tool summary; the Stanford abstract page gave no numbers).

Ques: Does the infinite-well formula explain the 48 nm shift of the Ge/SiGe edge?

Kuo et al. find the edge moving from 1408 nm to 1456 nm between 0 and 4 V, which is 29.0 meV (computed). The infinite-well formula with assumed masses of 0.038 and 0.20 $m_0$ gives only 4.4 meV at 80 $kV/cm$ and 6.9 meV at 100 $kV/cm$ for 10 nm wells (computed), four to seven times too small. A finite-barrier solution with an assumed offset of 100 meV for both carriers gives 23.1 and 36.1 meV (computed, second-order values scaled from 20 $kV/cm$), which bracket the measurement. The barrier is low, so the wave function leaks and the effective well is wider. The exciton Coulomb term adds a further field dependence. Leakage is the more likely main reason, but the offsets and masses are assumed, not fitted (my reading). Reference: Kuo et al., Ge/SiGe quantum-well electroabsorption, Nature 437, 1334-1336 (2005) (Stanford file 326.pdf, read through a fetch-tool summary; the Stanford abstract page gave no numbers); Indium Phosphide Quantum-Well Electroabsorption Modulator page of this vault.

Ques: Is the edge drift 0.76 or 0.6 nm per degree, and does it limit the window?

The 1550 nm GeSi paper measures 0.76 $\mathrm{nm}{/}^{\circ }\mathrm{C}$ between 30 and 70 ${}^{\circ }\mathrm{C}$, and the quantum-well paper about 0.6 $\mathrm{nm}{/}^{\circ }\mathrm{C}$ on Si and 0.58 $\mathrm{nm}{/}^{\circ }\mathrm{C}$ on SOI between 20 and 80 ${}^{\circ }\mathrm{C}$. These are 0.39, 0.45 and 0.43 $meV/K$ (computed). The bulk edge and the exciton need not move equally, and the two sets of measurements use different definitions of the peak, so I do not call one wrong (my reading). A change of 40 K moves the bulk edge by 30 nm, close to its 35 nm window, and 60 K moves the exciton by 36 nm, more than the 28 nm window of Kuo et al. The extinction ratio of the quantum-well device stays within $0.65\mathrm{dB}$ between 40 and 50 ${}^{\circ }\mathrm{C}$, and the InP modulator page gives 0.5 $nm/K$ for comparison. Reference: High speed GeSi electro-absorption modulator at 1550 nm wavelength on SOI waveguide, Optics Express 20(20), 22224 (2012) (read through a ResearchGate fetch-tool summary; authors not recorded); A 100 Gb/s quantum-confined Stark effect modulator monolithically integrated with silicon nitride on Si (PMC12046052; read through a fetch-tool summary; authors not recorded); Kuo et al., Ge/SiGe quantum-well electroabsorption, Nature 437, 1334-1336 (2005) (Stanford file 326.pdf, read through a fetch-tool summary; the Stanford abstract page gave no numbers); Indium Phosphide Quantum-Well Electroabsorption Modulator page of this vault.

Ques: Why does the computed link power penalty differ from the reported one?

The penalty is $10{\log }_{10}[2P_{in}/(P_1-P_0)]$. With the static insertion loss and extinction ratio of one wavelength it gives $9.27\mathrm{dB}$ for the 1550 nm GeSi device, $11.7\mathrm{dB}$ for Liu et al. and $15.1\mathrm{dB}$ for the quantum-well paper on Si (computed). The first paper reports a minimum of $8.5\mathrm{dB}$, found at the best wavelength of the scan, which is lower than a single-wavelength estimate (my reading). The quantum-well paper reports $17.5\mathrm{dB}$ at 1291 nm; with its dynamic ratios of 2.73 and $1.8\mathrm{dB}$ the formula gives 15.9 and $17.3\mathrm{dB}$ (computed), so the reported value is consistent with the dynamic ratio at 100 Gbit/s. Static values understate the penalty, and the dynamic ones are more likely to be the right basis. Reference: High speed GeSi electro-absorption modulator at 1550 nm wavelength on SOI waveguide, Optics Express 20(20), 22224 (2012) (read through a ResearchGate fetch-tool summary; authors not recorded); A 100 Gb/s quantum-confined Stark effect modulator monolithically integrated with silicon nitride on Si (PMC12046052; read through a fetch-tool summary; authors not recorded); Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary).

Ques: Is Franz-Keldysh or the quantum-confined Stark effect the better choice on silicon?

They serve different bands. The Franz-Keldysh devices work in the C and L bands: Liu et al. have $3.6\mathrm{dB}$, 36 GHz and 45 fJ/bit at 1610 nm, and the table lists 4.2 to $10.6\mathrm{dB}$ at 1550 to 1615 nm with 2 to 4 V. The quantum-well devices work in the O-band: $3.55\mathrm{dB}$ in 50 $\mu \mathrm{m}$ at 2 V and 100 Gbit/s with equalization, or more than $8\mathrm{dB}$ in 100 $\mu \mathrm{m}$ at 5 to 6 V. Per length both give 0.07 to $0.11\mathrm{dB}$ per $\mu \mathrm{m}$ (computed). The wells need a stack and give a narrower window of 23 to 28 nm against more than 30 nm, while the bulk devices need no well growth. The choice follows the wavelength, so neither is better in general (my reading). Reference: Liu et al., ‘56 Gbps high-speed Ge electro-absorption modulator’, Photonics Research 8(10), 1648-1652 (2020) (researching.cn PDF; read through a fetch-tool summary); A 100 Gb/s quantum-confined Stark effect modulator monolithically integrated with silicon nitride on Si (PMC12046052; read through a fetch-tool summary; authors not recorded); Chaisakul et al., J. Appl. Phys. 116, 193103 (2014) (re.public.polimi.it PDF; read through a fetch-tool summary); Kuo et al., Ge/SiGe quantum-well electroabsorption, Nature 437, 1334-1336 (2005) (Stanford file 326.pdf, read through a fetch-tool summary; the Stanford abstract page gave no numbers).

Related pages in this vault: Indium Phosphide Quantum-Well Electroabsorption Modulator, Silicon Carrier-Depletion Mach-Zehnder Modulator, Silicon MOS-Capacitor Modulator, Silicon Microring Modulator, Lithium Niobate Mach-Zehnder Modulator, Quantum-Well Laser (Single and Multiple Quantum Well), Silicon-Compatible Lasers (Hybrid III-V, Epitaxial Quantum-Dot and Ge-GeSn), Silicon on Insulator (SOI) MOSFET, Waveguide Photodiode.