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Intro Optical Engineering

by: Manuela Boyer PhD

Intro Optical Engineering EEE 165

Manuela Boyer PhD

GPA 3.66

Russell Tatro

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Russell Tatro
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This 49 page Class Notes was uploaded by Manuela Boyer PhD on Monday October 5, 2015. The Class Notes belongs to EEE 165 at California State University - Sacramento taught by Russell Tatro in Fall. Since its upload, it has received 17 views. For similar materials see /class/218829/eee-165-california-state-university-sacramento in Electrical Engineering at California State University - Sacramento.

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Date Created: 10/05/15
Chapter 3 Semiconductor Science and Light Emitting Diodes Energy Bands The energy of an electron in a crystal is quantized But many states exist very close together These many states with close neighbors is called an energy ban e mmwnmmggmumlmmmummy vsmmcs 7Euhlmbawem a anus e umlmza tmmam mmquks asmmmrmmm mmsxsxbm e u emsu m mammlemaxemt The conduction band is a spread of electron energies where the electron is considered to be free to wander about in the crystal lattice Energy Bands The width ofthe conduction band is called the electron affinity Above this energy level electrons have gained enough energy to leave the crystal OAFhmwnhnmggvmthmll m n mewng vsmuu as 7Euhlmbawem a anus e umlmza tmmam mmquks asmmmrmmm mmsxsxbm xsuuud emsu m mammlemaxemt Section 31 Semiconductor Concepts and Energy Bands Energy Band Diagrams m a A Slmpll edlvm dimensmnal viewufaregun ufthe 5 crystal sheng cumlent bunds b The energy band mam er electmns m the 5 Crystal at absulute zem enempemmre 19w 0 ram amnemempem Ha Energy B ands The states that are tightly bound are called the valence band The valence band represents electron wavefunctions in the crystal that correspond to bonds between the atoms e mmwmmeemmelmmmomma vsmmcs mmmwunsemememmm mmw asxbmmeehmmmumh W as and e mm m mamJVJLMuMl The region between the conduction and valence band is called the an ap There are no allowed electron energies in this forbidden region Effective Mass An electron placed in the conduction band CB is free to move around the stal and also to respond to an external and internal electric eld because there are plenty of neighboring empty energy levels In general we can treat the electron in the CB as if it were free within crystal by simply assigning an effective mequot to it is founu Thus the movement ofthe electron in the conduction band is W e by experiment Excitation of an electron results in the electron freeing itself from a specific locality and then moving thru the c 39 behind an empty electron state called a hole with effective Thermal excitation Thermal generation of electron hole pairs EHP occurs above t 0 K Since the only empty states are in the CB for very low temperatures the excitation of an electron from the VB requires a minimum ener ong This electron absorbs the incident photon and gains sufficient energy to surmount the energy gap Eg and reach the CB When awandering electron in the CB meets a hole in the VB it has found an erupt 39 lower mu 39 it This process of lling a hole trapping an electron is called recombination Any excess energy in the recombination momentum and energy must in either ldLLlL quot 39 L 39 photon Semiconductor Statistics The FermiDirac mction fE is a mdamental property of a collection of interacting electrons in thermal equilibrium 1 f E e E F 1 exp T Where kB is the Boltzmann constant T is the temperature in Kelvin and EF is an electron energy parameter called the Fermi energy k5 13807x10 8 617x10 E T K T K Any Plum 9 Fr 39 rem 39 WUIK input or output per electron Fermi Levels The Fermi energy is de ned as the energy level Where at T 0 K all states below it are lled An equivalent de nition is the Fermi energy is the level Where fE 2 at some temperature T fE 2 means that 2 ofthe states are lled at this energy level Although at EF the probability of electron occupancy is V there may be no states for an electron to occupy What is important is the product of the density of states and the probability of occupation Semiconductor Statistics There are two important concepts Density of states DOS gE represents the number of electronic states in a band per unit energy per unit volume of the c sta The FermiDirac mction fE is the probability of nding an electron in a quantum state with energy E According to quantum mechanics for an electron con ned Within a threedimensional potential energy Well the density of states increases With energy as gE oc JE 73 The energy E r Ec is the electron energy from the bottom ofthe conduction band The density of states mction gives information only on available sates and not on their actual occupation Semiconductor Statistics For a semiconductor system in the dark in thermal equilibrium and with no applied voltage or EMF generated AEF 0 and EF must be uniform across the system r quot v a In quotm39xa uwmmmmmmmmmw milmmm gwtm rmmymwmkmnmcs mermmwm a 11th mmme imxukqwcmmnw mm Fermi Levels The electron concentration n in the conduction band is given by n J gCBEfEdE Whenever Ec 7 EF gtgt kT the fE can be approximated by Boltzmann statistics ErEF fE miK Electron Concentration ever EE 7 EF gtgt kT such semiconductors are called nondegenerate It implies that the number of electrons in the CB is far less that the number of states in the band n NE expiiijri where ND is a temperaturedependent constant called the effective density of states at the CB edge 3 2nmkBT 3 N 2 e hz Intrinsic Semiconductor Thus we see the Fermi energy EF location determines the electron and hole concentrations an intrinsic semiconductor a pure crystal n p and the Fermi level is located at midgap 5 EV 155 lm ml l 2 2 WV where EF is known as the intrinsic Fermi level Thermal Velocity The text brie y shows that the thermal energy imparted to an electron in the CB is about 32 kB T This energy results in a thermal velocity in the semiconductor which is typically about 105 msec Hole Concentration A similar analysis for the concentration ofholes in the VB yields the resu ErEv m PNVSXP7 L we de ne NV r d p d 39 constant called the effective density of states at the VB edge 3 NV 2 2 39Z39thBT Y Mass Action Law There is a use rl semiconductor relation between n and p called the mass action law 7 Eg np NCNV exp k n in which Eg EE 7 Ev is the bandgap energy and n has been de ned by 73 k n NCNV exp 57 and is a constant that depends on temperature and the material properties Extrinsic Semiconductors By introducing small amounts of impurities into an otherwise pure crystal it is possible to obtain a semiconductor in wh39 concentration of carriers of one polarity is much in excess of the other t Such semiconductors are called extrinsic semiconductors For ntype material the electron concentration is larger than that of the holes We track the concentration by the calculation of Nd Ndmm For ptype material the hole concentration is larger than that of electrons We track the concentration by the calculation of Na N accepmr Figure 35 on page 114 As doped Si As is pentavalent With five outer electrons m A A HH HH 1 I W W W V l T V ml immuwmm WWW 1mm mm mm alfwwnmbandmthhsmmhzrm w g mhm z t 15 semi a 1 155 n W 5mm 5 mg mmmmm K etc m mammal w 0 may amulmym Pretendh Conductivity The conductivity 5 of a semiconductor depends on both electrons and holes as both contribute to charge transport The electron dri mobility is given by p The hole dri mobility is given by ph 66nueepuh Electron Conductivity The semiconductor conductivity is given by Which for an ntype semiconductor becomes 2 VI 72Ndt e 39 eNdt a Nd a Electrons in an ntype semiconductor n gt p are majority carriers Whereas holes are minority carriers The electron concen 39 39 Written as nm to refer to the electrons as majority carriers The hole concentration is Wrinen as pm to refer to holes in an ntype material Figure 36 on page 115 Boron doped Si B is trivalent With three outer electrons um um Emmsmxyln snm 2 JllLx mm mm m a Eamn suggest cryml E has mlythxee valence electxms thmt mums m a s amm an uf s bands has an electxanmxssng AndLherme a hale b Endgyband alga fur aptypesx duped mm ppm E nm are acceptm mugylevelsjm abuveEvamundE ates T1125 Mummers accept ale 9115er m vs AndLherme agate bales mm vs 19w 0 Km Wuhmmmwnnm Ha Example Example 312 on page 119 What is the conductivity ofan ntype Si crystal that has been doped uniformly With 1016 cm393 phosphorus P atoms donors if the dri mobility of electrons is about 1350 cm2 V39ls39l Solution Since Nd 10 cm393 gt n Which about 145 x1010 cm3 the electron concentration is n Nd We can ne lert quot contribution A cquot quot since p nfNd ltlt n 216 QL m 2 a 2N4 1 5 x10quot901x10 5m 31350 1 s Thermal Equilibrium for ntype In thermal equilibrium and in the dark the mass action law can be stated as r 2 nu Pm 1 Where nm Nd Thermal Equilibrium for p type Holes in aptype semiconductor p gt n are majority carriers whereas electrons are minority carriers The hole concentration is written as ppm to refer to the holes as majority carriers The electron concentration is written as npu to refer to electrons in a ptype material In thermal equilibrium and in the dark the mass action law can be stated as pa Ppu where ppm Na Section 3 1 F Energy Band Diagrams in an Applied Field Consider the energy band diagram for an ntype semiconductor that is connected to a voltage supply of V volts and is carrying a current The Fermi level EF is above that for the intrinsic case Eh That is closer to EE than Ev 39rVL use appueu I mun u that the electrons in the semiconductor now also have an imposed ener h terminal Section 3 2 Dim and IndirectBandgap Semiconductors Ek diagrams We know from quantum mechanics that when the electron is within an in nite potential energy well of spatial width L its energy is quantized 39ven b Z En ah 7 2mg Where mE is the electron mass and the wavevector kn is a quantum number determined by kni Wheren123 L Section 31 D and E Section D reviews compensation doping where both donors and 41 1 a L l r a Section E reviews degenerate and nondenerate semiconductors In nondegenerate semiconductors the number of states in the CB far exceeds the number of electrons This allows us to use Boltzmann statistics to describe the density of states Please read these sections on your own Energy Band Diagrams in an Applied Field EF must be uniform across the system only for a semiconductor in thermal equilibrium and in the dark When a potential is applied EF follows the electrostatic potential energy behavior Electron concentration in the dem semiconductor is uniform quot quotH E m be constant Thus 1 3 the CB VB andEF all bend by me same amount nzxgybandduugmmafan 714E semlcanductmcannzctzdtna wl esu le Thaw rash durum I because hs39i sm p hiwhashstfaasaaspashaa ihg asvi mvsu hm mama Tammi Dim and IndirectBandgap Semiconductors Ek diagrams The energy increases parabolically with the wavevector kn Th Ehaagyaahd Diagram k aagam ufa duenbandgnp sanicanductm such asGaAs The M many discrete palms with sash paint samspmahg la a Th pm s an sa alas Lhatwe nmma y aaw the Erkrelahanship as a cmhnuwus clxve In Lb endg range 5 ta 5 there an m palms W saimams 19w 0 Kass autumn Mass Ha Direct Bandgap Semiconductors When an electron can gain enough energy to leave the VB and jump to the CB without a change in momentum the material is called a direct bandgap semiconductor The minimum of the CB is directly above the maximum of the VB in terms on energy exchange The also means that an electron can lose energy emit a photon and recombine with ahole EHP recombination without a change in momentum Indirect Bandgap Semiconductors We note that the minimum of the CB is not directly above the maximum ofthe VB but is displaced on the kaxis Such a crystal is called an indirect bandgap semiconductor An electron at the bottom of the CB cannot therefore recombine directly with a hole at the top ofthe VB because for the electron to fall down to the top of the VB its momentum must change from kEb to kvb quotquot quot 39 mich in 39 quot via arecombination center at an energy level Erwithin the bandgap These recombination centers may be crystal defects or impurities Section 3 3 pn Junction Principles T Open Circuit Condition Consider what happens when one side of a sample of Si is doped ntype and the other ptype Assume that there is an abrupt discontinuity between the p and n regions which we call the metallurgical junction a in the gure more m twmunmmn mimsu Wzmamm my Indirect Bandgap Semiconductors In the case of Si the diamond structure leads to an Ek diagram that has the essential features shown be ow z z mummy 2quot A Gm b s c Siwltluxemrrbm nnmntnx a In GaAsthemimmum ufthe CBIS directly abuvethemzximum ufthe VB GaAs is MM t an electxun and shale in Si invulves arecumbmatlun cantEr 1999 s o KaEp Optoaltzrvowrs FrmuceHall Indirect Bandgap The electron is first captured by the defect at By The change in the energy and momentum of the electron by this capture process is transferred to lattice vibrations that is to phonons In some indirect bandgap semiconductors for example GaP with nitrogen defect additions the electron transition from Er to Ev may involve a photon emission pn Junction Principles Due to the hole concentration gradient from the pside p ppm to the nside p pm holes di lue towards the right The nside near the junction therefore becomes depleted of majority carriers nm Similarly the electron concentration gradient drives the electrons by diffusion towards the le Electrons diffusing into the pside recombine with the holes majority carrier in ptype b in the figure Depletion Region The regions on both sides of thejunction M in the figure become depleted of free carriers in comparison with the bulk p and n regions far away from thejunction There is therefore a space charge layer SCL around point M on the gure c in the figure This area is more commonly known as the Depletion Region Space Charge Density The net space charge density pnet is negative and equal to 46 Na in the SCL from Wp x g 0 e in the figure pm1 is positive and equal to e Nd in the SCLfromOxWn The total charge on the le hand side of e must equal that on the right side for overall charge neutrality so that N p Nd Wn Depletion wide Builtin Potential The builtin electric field is given by eNdWw anWp 5U gr 5U gr En The builtin potential is given by 1 wan4WD Vur EDWWI 2 25D57NDN4 Where WU Wa WP the total width of the depletion region under a zeroapplied voltage BuiltIn Potential It is clear that there is an internal electric field EU which tries to dri the holes back into the pregion and electrons back into the nregion The electronhole diffusion and dri uxes will be balanced in equilibrium d in the figure This area is more commonly known as the Depletion Region Builtin Potential The potential Vx at any point x can be found by integrating the electric field since by definition E dVdx Vx increases in the depletion region towards the nside as indicated in f on the figure Notice that on the nside the potential reaches VD which is called the builtin potential Builtin Potential We can write the builtin potential in terms of the dopant concentrations 2 n P Na PM 5 ls quotW The builtin potential is then given by VD 7 kET lranNdl K 2 e n The builtin voltage VB is the potential across a pn junction going from p to ntype semiconductor in an open circuit Forward bias Consider what happens when a battery with a voltage V is connected across a pn junction so that the positive terminal of the battery is m A p r bias r mum biasedpn Juncan and the in action afmmnnty carriers 5 c mm concentraan pro les across the dawn under forward bias in The hale pa nus mergywith and without an appliedbus W15 the width mm SCL with forward bias w s 0 may Wilhcm ur Famedh Forward bias The probability that ahole in the pside will surmount this reduced potential barrier and diffuse to the nside is now proportional to 4M m t 5 Probability oc exp T This results in the injection of excess minority carriers holes into the nregion Similarly excess electrons can now diffuse towards the pside and enter this region as injected minority carriers Law ofthe Junction The electron concentration np0 just outside the depletion region is given by the equivalent law ofthe junction equation It is apparent that an electric current can be maintained thru a pn junction under forward bias and that the current ow seems to be due to the al39 un39on of minority carriers There is some dri of majority carriers as well Forward bias by V L Ur Lu uppr The bulk regions outside the SCL have high conductivities due to plenty of majority carriers Thus the applied voltage drops mostly across the depletion width W Consequently V directly opposes VB and the potential barrier against diffusion is reduced to VB 7 V Law ofthe Junction 39Ihe newL 39 pn0just 394 L 4 39 regioni determined by the probability that a hole in the pside will surmount this reduced potential barrier eVU r V 4M m er some math we nd the the following gives the amount of excess holes diffusing into the n region pw0 pm exp mm V in 0 This is called the law ofthe junction p pp eX13 Diffusion Length There will be a minority carrier concentration pro le that arises from these carriers recombining We de ne the length that results from the mean carrier lifetime L11 Dhrh Lh is the hole diffusion length Dh is the diffusion coefficient for holes th is the mean hole recombination lifetime minority carrier lifetime The diffusion length is the average distance diffused by a minority carrier before it disappears by recombination Diffusion Length Likewise for electrons Le lDere Shockley diode equation The total diffusion current density is JMHW J50 exprl Where Jso is the net diffusion current ow from both electrons and holes The Shockley equation represents the diffusion of minority carriers in the neutral regions Diode equation ll 110 exp Reverse LV charactenmcs at a pee Junchanthe nantve and negntrve eurrent ax es hnve different scales 19w 0 Ken amneme heme rur Current Density The total diffusion current density is Jdr Karl Jh J2 f P39mgmn SCL tern Ida e I if Sin mm The total current anywhere in the device is constant Just outside the de letion re Ion it is due Mnn eneran p men39y to the drffusrgon o minority came pr 1999 s o Kasap Uptoeteeneme prenteee Han Diode equation The total current into the diode Will supply carriers for minority carrier diffusion in the neutral regions and recombination in the space charge layer The diode current is given b 110 exprl Where Io is a constant The diode ideality factor n is either 1 or 2 depending on the diffusion mechanism of the particular situation Reverse bias When a pn junction is reverse biased the reverse current is typically very small Mmmlyczmu m Reverse biasedpn Juncan a Mrnanty cmierprufrles and the angn hr the reverse eurrent b Hale FE acrass the Junean enaer reverse hns rm 0 Kenn OWMIlmpvn meme Reverse bias The applied voltage drops mainly across the resistive depletion region which becomes wider than 39 er unbiased or forward biased m 7n There is a small concentration gradient and thus a small electronhole diffusion current This current 715 is called the reverse saturation current density Recombination Lifetime The semiconductor must maintain charge neutrality At the same time electrons and holes are recombining Thus the rate ofrecombination depends on the electronhole concentrations 6AM 7 AMP at r ere IE is the recombination time for an electron in this case The rate of carrier concentration is given by Anp Under high levels of carrier injection conditions the lifetime 1 is inversely proportional to the injected carrier concentration Thus the lifetime is not constant under some conditions Section 34 The pn Junction Band Diagram In part a the energy band diagram is shown for open circuit conditions The Fermi level must be uniform throughout the material under thermal equilibrium and in the dark Thus the bands are bent to re ect the relative position of the Fermi level EnzybzrddlznmsfmAwmcmnnduzwmcmn wound in ma c me in mrdmm39 a mm gmzcmm ekchmhilz yaks mile deplzcmngwnxesuhs m sma mm m in ms u m ammmlemmHH Depletion Layer Capacitance The depletion region ofa pn junction has positive and negative charges separated over a distance W similar to a parallel plate capacitor 2 C 7 A N mm m 2NaNd Where A the depletion region area and 2 2D srpermitivities Typically Cmp under reverse bias is of the order of a few picofarads Examples Example 331 on page 136 The pn Junction Band Diagram 7 Forward Bias In part b the energy band V 7 di ram is shown for forward bias conditions The potential energy barrier has shrunk by V Electronholes can diffuse from the majority carrier side to become minority carriers in Q1 Fi g1 the other side 3 Emigbe mam bx wwwmqu upmcmt b Emaad in ma c we in mrdmms a mm SHAWNelmth yaks mile unamngmnm m ma vehemle in ms u m mwmlmm 39Ihe pn Junction Band Diagram 7 Reverse Bias In part c the energy band diagram is shown for reverse bias conditions The potential energy barrier has grown by Vr The small reverse current arises from thermal generation 39 of minority carriers Within a diffusion length to the SCL Part d Enzybzrd deems er awmamnnasrra Wmumut b roman bus are e lame bus mrdmms a 1mm semamrelermlels Fans mt densamregmnane ma mu reverse ulna e M95 u use GummzllmmHH Light Emitting Diodes For a direct bandgap material the emitted photon energy is approximately equal to the bandgap energy hv 2 Eg Carrier recombination primarily occurs Within the depletion region and Within a volume extending over the diffusion length LE of the electrons in the p side This recombination zone is called the active region This phenomenon oflight emission from EHP recombination as a result of minority carrier injection is called injection electroluminescence nght numut Light Emitting Diodes 7 Device Structures The pside is on the surface from Which light is emitted and is therefore made narrow a few microns to allow the photons to escape Without being reabsorbed lnsulatur renal Epitaxial layer m r r u L absorbed or re ected back at the substrate interface Not all light rays reaching the semiconductorair interface can escape due to TIR e in GaAs n N 36 to air is only 16 Which means much ofthe light suffers TIR a sinquot 1 55d quots sinquot L 15 127 more dense 3 6 Section 35 Light Emitting Diodes A light emitting diode LED is a pn junction typically made from a direct bandgap semiconductor for example GaAs in Which the EHP recombination results in the emission of aphoton nght numut Metal electde schematic lllustranon of typlcal planar surface ernrmng LED devlces a player grown epltaxlally on an n substrate b Flrst n is epltaxlally grown andthen 1 region s orrn dopant dlffuslon into the epltaxlal layer 1999 s o KaEp Optoelacnontcs Prenueellall Light Emitting Diodes 39Ihe emitted photos are random in direction The LED structure has to be such that the emitted photon can escape or be guided to the surface of the device Without being reabsorbed the the semiconductor material Lrgnl uutput lnsulamr EIde er r Epltaxlal lay Light Emitting Diodes 7 Device Structures It is possible to shape the surface of the semiconductor into a dome or hemisphere so that light rays strike the surface at angles less than 9 a me bl ma mum cl LE mnme mm m names a sane llgm sufferslmal nemrl reneeaanrnnemlesune b lnennl xe zconns canbe rerlrserlrnnreree nae llghtcanbe callzctzdb shaping are emeernnlar ruler earn sv am an anglzsafmcldznce mks semlcanduc rnr surface are sma erthanthz annexing b Anecannmlc mmaeemlam mare llgm tn asap mar LED tn enrpe e llmamnspnremplasnc dame 19w 0 may onmrema name Hall An inexpensive solution is to encapsulated the LED Within a tran harem 39 39 th as a ni nel 39 than air as shown above Section 36 LED Materials Read this section for an overview of various materials used in LEDs Fmrxee 552 rivmlzngth 1mg byddr exentaleD mm mammal spmtsnhthdz m m wave 2 ma cammumcaonns an 2 ran lmzs m mamm mf m g m9 0 K1511 amtumm Pam rur InternalExternal Ef ciency External efficiency 7 def n r the fraction of optical output power versus the electrical input power This is sometimes called the wall plug efficiency Em UPW Pm 0PMl Um T Rnelecmtal WhereI is the diode current and V is the diode voltage Response Time The response time of the semiconductor is due mainly to the lifetime I of the injected minority carriers The rise time is 1 seconds with typical rise times for LEDs of 1 Long lived carriers have a greater chance in escaping the semiconductor and achieve higher internal efficiencies 11quot mm quot d in ani rm 39 39 mm butwith lower 11ml InternalExternal Ef ciency Internal efficiency 7 def n r the fraction of electronhole recombination39s under forward bias that results in a photon emission PUWECWM 7rateufradianvephutunrecumbmatmns7 Phutunsemrttedpersecund 7 mm T 7 39 Tutal rate ur recumbmatmns 39 Tutal camarslust persecund 39 Current 1 echarge 2 As stated in the previous section not all ofthese photons reach the surface and escape due to reabsorption and TIR Responsivity Not in the text at this point are two other items of common interest Relpomiviry is the ratio of emitted optical power PM to the injected current m Pm0ptml ener per hotontimesofphotons hv DWW hv I Current 2 Units are WattsAmp Section 37 Heterojunction High Intensity LEDs A junction between two different bandgap semiconductors is called a heterojunction A semiconductor device structure that has junctions between different bandgap materials is called a heterostructure device quotquot 39 39 of a emiLuuuuLtur 39 dep 4 its bandgap A wider bandgap sernicondcutor has a lower refractive index By constructing LEDs from heterostructures we can engineer a dielectric waveguide within the device and thereby channel photons out from the recombination region Heterojunction High Intensity LEDs The Wide bandgap ofthe MW pAlGaAs layers con ne hoif quot nht i o the CB electrons to the eds mnMEvuks pGaAs Note the barrier gt g p height of AEE Ir gamma Since the bandgap of the 0 mmam nAlGaAs is greater than v the pGaAs the emitted photons do not ge my reabsorbed as the escape 393 5 5 the active region and can reach the surface D ms u no methntptdmmmn LED Characteristics The transitions that involve the largest electron and hole concentrations occur most frequently See b and c below dam pd ramwrth dd madman ths hm mslnbmmnaf ereem eglnh ppte 5 ya Th moste tmeme ometmw shave e cThzrelahvelrgm mymrmondpopnemgypoedmh dRzlauve metedytermoonornmthgmmhedptepetmbdedentbmdte rm 0 pop opponent modem LED Characteristics Note that in c there is a turnon or cutin voltage from which point the current increases sharply With voltage Just like for the ideal diode the extemal circuit must insure the current is appropriately limited d A typical output epeem eddtye rntmstyvswavzlengl rm deeded LED b Typed wutyullrm pow vs rowdd uent c Typed rev ehddaeersoee uf d redLED Th ammv lug rs armmdl 5v 19w 0 Koo anathema Pumice run Section 38 LED Characteristics The energy ofan emitted photo from an LED is not simply equal to the bandgap energy Eg because electrons in the conduction band are distributed in energy and so are the holes in the valence band a Ems r electmns m the and lwlzs m the VB The highest e nmncemrnonn is my abnve 5 eme relative lrghnntznsr asafuncnunafplwtmrenzrgybasedan b d Relative mtenmyas d moan afwmlzngth m the a put spchmbased anGa and c rm 0 K199 anthem memo ypmd mm wth pussrblz recambmanan utls b Enzrgydlslnbmmna ca ctm LED Characteristics Since there is a spread of possible Wave rntions there is a spread in the recombination energies This results in a lineWidth of the output spectrum Av or A d A typical output epeem tddtye memoMeow rm aredGaAsP LED to Typed output up pow vs Totodd nxrmt c Typed rev ehddaeersoee uf d redLED 39l39henmmvnltagus armmdl 5v 19w 0 Koo ahedmonndpohod Ha Example 381 on page 149 Given that the Width of the relative light intensity versus photon energy spectrum of an LED is typically around m 3 kET What is the lineWidth All in the output spectrum in terms of Wavelength c E hoton Solution 5 1v gt 2 and Emmy hv gt v P v c c 115 s a 0 V E photon E h photon Now differentiate With respect to the photon energy 115 he 3 E2 AE E2 photon photon photon photon Examples 381 m 7 AE E2 1mm photon Now rearrange the terms to separate out the A 115 A m i AE W Eim W It was given that AE N 3 kET We can calculate the energy ofthe photon given either the wavelength or the frequency The text went on to substitute back in for the wavelength rather than calculate a photon energy directly ht E hv 1 ho ho AE 1402 A m 7 AE 3 E12 om mm E 2 mm M 1 Example 382 on page 149 Given GaAs LED Bandgap at 300K 142 eV Bandgap changes with temperature as given below dT is the incremental change in tempera ure 01E 45gtlt10quot K dT What is the change in the emitted wavelength if the temperature change is 10 C AimLIE Ehv gtt 115 2 E 2 ho AE 115 d 115 dE A m 2 2 EEP hc EEPEP dT EEPEP dT Example 382 0 277m dT K Thus the change in wavelength for 10 C is d1 mm mm mm 0277 A1 0277 AT 0277 10K277nm dT K K K Note that Egap decreases with temperature and the spread of wavelengths increases with temperature Examples 381 Mm ET 115 Thus at 870 nm we nd 3 13807X103923 1 Ms 870x10 9m2 K 6621x10 3 Js 3x108 1 S 300K 4729nm Example 382 Thus the change in wavelength per degree change in temperature is a t he dT E2 dT gran 52p di 6 621x10 3 Js3x105 V J S 2 745x10 1602x1049 K 2V d7 1 422 V1 602x10quot9 i 2V a 2 769x10quot 1 0 277 dT K K Example 383 on page 150 The ternary alloy InlrxGaxAsyPlry grown on an InP crystal substrate is a quot 4 39 uriLuuuuLLur urarcriar Jul 39 LED and laser diode applications The device requires that the InGaAsP layer is lattice matched to the InP crystal substrate to avoid crystal defects in the InGaAsP layer Experience shows that this requires y m 22x The bandgap Eg of the ternary alloy in eV is then given by Eg m135072y 012y2 for 0 x 047 Calculate the compositions of InGaAsP ternary alloys for peak emission at awavelength o 13 pm Example 383 Solution We need the required bandgap Eg at the wavelength of interest The photon energy at peak emission is hcA Eg kBT an experimentally found value Then in eV h M l le J new 138U7xlu391 3 l I may Er A J J z 2 l EIle mi lixl quotm l EIleEI39Wi 42V 42V The InGaAsP then must have y satisfying 0928 1357072y 012y2 012y2 7072y04220 Section 39 LEDs for Optical Fiber Communications For short haul applications like local area networks LANs LEDs are 1 Simpler to drive Cheaper 3 Longer life N150000 hours than for laser diode 4 Provide the necessary output power The type of light source suitable for optical communications depends not only on the communication distance but also on the bandwidth requirement But even with the LED virtues listed above the output spectrum is much wider than that ofa laser diode The situation will dictate which compromises can be made Edge Emitting LED The light is guided to the edge ofthe crystal by a dielectric waveguide reu u LII wiuer or 39 quot a double 1 m e heterostructur mam mm m WWWm mm m mm mat nxzwamm My 41x mm w 125w Winged w WWW Schema mum mm m 5mm Ufa mu hlemjumnmr 5m suntan edge mm LED 9 0 Km Willu39mzl hm Ha Example 383 012y27072y04220 Solving the quadratic gives y 06583 Recall that y m 22 x or x y22 0658322 0299 03 The ternary alloy is thenan 7Ga 3Asu MP 34 LEDs for Optical Fiber Communications There are essentially two types of LED devices If the emitted radiation emerges from an area in the plance of the recombination layer as in a then the device is a surface emitting LED Ligmt a Suface amtqu LED in Edge mung LED arm 0 may Opymlmmp Remandh If the emitted radiation emerges from an area on an edge ofthe crystal as in b then the LED is an edge emitting LED ELED Coupling Methods The light from the LED must be launch into the fiber to do use il work Below is shown a Burrs device in a and microlens in b mm mm mmmq mam Maw sm lm mm a n Lg cmpLaammLxe mm LED A quotnodal 5mm mmmmmmm mm mime Renting mmx matching mm LED mm mime uphczl 5m my 11L men burdzd m LL LED Wm 19w 0 Km amtumm Pam run These devices are used for surface emitting LEDs Coupling Methods Edge emlmng LEDs pronde a greater mtensrty lrght heterostrueture devlces and also abeam thatrs more eolhmatedthan surfaee emlmng LEDs EM mm W at o nglmmn meda mmgm n swam Ithrlypxl NWulxgl mm nnmnuum emm m Memeth Example not m the text A slmple LED flasher wrth an ae power supply Let ywy 9 Vm The forward eurrentrs requrredto be M 20 mA at Va 1 9 V where Va ls the LED dlode voltage drop oluuon use a dlode so that the negatwe supply voltage ls shunted away from the LED The senes resrstorwas found before Example not m the text Flnd the power eoupledmto afrber by an LED emrtter er 1mw wlth surfaee dameter 00 um r e um Let ED ow Step mdex ber wrth eore damete 148 an 146 NA148271462 en 24 The eoupled energy effleleney l5 glyen by emxgvcmyladmdrlzmu NA umg emrgyemtuuhymu t mm Thus the power eoupled vaucmylzdz n Aquot A l ltvn unl Example notm the text Deslgn a slmple de power supply for an LED 2 Let me 9 V The forward eurrent ls requlred to be M g 20 mA at Va 1 9 V here les the LED dlode voltage drop Choose R ZUmA So ehoose the next hlgher eommon R value of 360 or 390 o What powerraung must the reslstorhave7 r W71VaUmAEIMZW 18 Watt 0125 w 14 Watt 0 25W So mustuse a AWattreslstor More sophrstreated LED dnyer erreurts Flgure a prondes for a slmple onoff modulatron ma a swrteh The voltage Vb ls suffrerent to swrteh the transrstor on There ls l p d p be e eoll etor demrttertermm s urrent wmg throu the LED ls determmed b the voltage V andthe senes resrstanee R5 Ln frgure b the dlode output may be modulated by yoltage y The resrstors R ut h oth the txanslstor andthe LED are then blased wtll lnto the llnear reglons lt Fquot o 5 Example notm the text The area ofthe LED emrtter ls engj quotHTMJ7556gtltID The area ofthe ber eore r 5quotTM mennltax Rauo ofthe two areas 7 4 Thus the power eoupled 392 slur Av Fuwercuuled p p m tee Am 391 u058564lmWl41464xlu w nut Wlth no re eedon losses eonsldered Example not in the text Example not in the text We will assume a fiber is spaced a short di stance away from the The total power transmitted across this interface and into the fiber as a emitter with air as the medium betwe em proportion of the emitted power 395 The Fresnel losses due to re ections is 2 R quot1 39 quot2 So including re ection losses the LED successfully launches quot1 quot2 1464 uW 0655 96 uW of power into the fiber The power transmitted into the next medium is equal to 1 r Power transmitted 17 Rmmlr RM 170 319517 0 0375 0 555 So ifwe let the emitter be GaAs LED with n 36 then 2 2 3571 1 4871 R 03195 3 61 RM I 48 0 0375 End of Chapter 3 What Wavelength Goes With a Color spectrum labeled vlslble llght Thls vlslble llght corresponds to awavelength range of n eapable of seelng radlatlon VJth wavelenng outslde me vl ble spectrum The vlslble colors from shortesttolongest wavelenglln are vlolet blue green yellow ange and red TwFr r d the colors othe vlslble speeaurn Blaekrs atotal absenee ofllght re eleeaornagnene speeaurn Gamma my x my 400 nm J Uhmvmkt Vth mdvarinn hu res m mu 1 pm Vellow mum 39 rmbk Mg OW m 39 lnfnm 700 nm I000 pm 1 mm radlanun lOmm I cm Mimowavzs n m IOU m m m m r Wave cngm Radm waves IOU km W39avelengvh mug mm Umwvse by Freedman and Kaufm am esu as 5m 5m Wavelengfh nm 5x 7n ar 5 Wave mm m Indigo Light The visible indigo light has a wavelength of about 445 nm Violet Light The visible violet light has a wavelength of about 400 nm Within the visible wavelength spectrum violet and blue wavelengths are scattered more efficiently than other wavelengths The sky looks blue not violet because our eyes are more sensitive to blue light the sun also emits more energy as blue light than as violet Energy with wavelengths too short for humans to see Energy with wavelengths too short to see is quotbluer than bluequot Light with such short wavelengths is called quotUltravioletquot light How do we know this light exists One way is that this kind of light causes sunburns Our skin is sensitive to this kind of light If we stay out in this light without sunblock protection our skin absorbs this energy After the energy is absorbed it can make our skin change color quottanquot or it can break down the cells and cause other damage Energy with wavelengths too long for humans to see Energy whose wavelength is too long to see is quotredder than redquot Light with such long wavelengths is called quotInfraredquot light The term quotInfraquot means quotlower thanquot How do we know this kind of light exists One way is that we can feel energy with these wavelengths such as when we sit in front of a campfire or when we get close to a stove burner Scientists like Samuel Pierpont Langley passed light through a prism and discovered that the infrared light the scientists could not see beyond red could make other things hot Very long wavelengths of infrared light radiate heat to outer space This radiation is important to the Earth39s energy budget If this energy did not escape to space the solar energy that the Earth absorbs would continue to heat the Earth Responsible NASA Officialquot Bruce R Barkstrom Ph D Site AdministrationHelpquot NA SA Langley ASDC User Services acheos asa go v Privacn Securitz Notices Last Updated39 Mon Dec 6 063322 PS T2004 Responsible NASA O 39iciax Bruce R Barkstrom Ph D Site Administrationquot NA SA Langez ASDC User Services Last Updatequot December 3 2004 Chapter 7 Polarization and Modulation of Light Polarization The eld vibrations and the direction of propagation z in this case de ne a plane ofpolarization so it is also called plane polarized Any arbitrary wave linear can be described in terms of vectors EX and E v Y Ex E cosat rkz and a cosattrkz o Where 1 is the phase difference between EX and Ey In this gure the E eld is 45 relative to the x axis End 180 z i This assertion can be con rmed by plugging these values into the equations for EX and Ey above Polarization Ifthe magnitude ofthe electric eld is a constant but the vector traces out a circle by rotating we have circular polarization Right circularly polarized r the vector traces out a circle in the clockwise sense with time from the perspective of the light approaching the observer EaEyaA and 2 Ex Acosarrkz and Ey iAcosaJtikZ Le circularly polarized r the vector traces out a circle in the counterclockwise sense with time from the perspective of the light approaching the observer Section 7 1 Polarization State of Polarization A propagating electromagnetic EM wave has its electric and magnetic eld at right angles to the direction ofpropagation E l B The term polarization of an EM wave describes the behavior of the electric eld vector as it propagates through a medium If the oscillations ofthe electric eld at all times are contained within awell de ned line then it is said to be linearly polarized as shown to the right Polarization Thus for this linear polarization we nd E EX In other words the vectors have the same magnitude but are out of phase by 180 E tax hm tam eosax imam cosatr k2 E cosmr k2 where E 191139 There are also many other behaviors ofthe electric eld besides the simple linear polarization just discussed A in link is below httpwww ee buffalo edufacultycartwrightjavaiappletsindex htzml Figure 72 A 71th Czrcularlypulanzed lzght The eld vectorE is always at right ang1estozrotatese oc wisearoun z wit time and traces out afull circle over one wavelength of distance propagated 1999 s o Kasap Optamwams Prentice Hall Flgure7 3 y y v 712 7 2 Exam zs arllnslrly l and b and slrsularlypnlanseel light c andd c ls right slrsullrlylna a ls lell slrsullrlypnlanzeel light ls seen whzn the wave alrestlylppnlshss l Vlswer my s u lslamplaaaaml Puma lull Flgure74 y y y la DI E lcl f E 2 z El 2 2 if v n t 14 PM l Llnslrlypnlanleul llgn wuh E 25 ard n b thlw 4445 the llghhs Ugh elllplslllypnlanseu wuhl ulteullnalurlns c thn 42 am the light ls Ugh elllplslllypnlanseu n andinwexe equal thls wmlldbe right slrsularly pullnleel light my s u Kusprmkkmlmme run Malus39s Law There are vanous opueal devlces that operate on the polanzatlon state of a Wave passmg through lt and thereby modlfy the polanzatlon state llnear polanzerwlll only allow eleetne eld oselllatlons along some preferred alreetlon ealledthe transrnlsslon alrls to pass through the deVlee DlehrOle erystals sueh as tourrnallne are good polarlzers beeause they are opueally anlsotrople and attenuate EM Wave Wlth elds that are not osclllatmg along the optleal alrls DlehrOle refers to arnatenal ln whleh llghtln dlfferentpolanzatlon states travellng through lt elrpenenee avarylng absorpuon Fulanzaum Ifthe magmde othe eleetne eld veetors are not equal the veetor traees out an elllpse as the wave propagates Thls ls ealled elllpueally polanzedllght The phase Qmustnotbe zero or any rnuluple ofnforEXO E Th re Em wa e ye ls a speelal ease where or the phase pun4 orrnultlples of 4 Example7 l l unpage 27a Elllpml and ureularpularlzatlun shuwtlatlrl A eugute kz andlLIE eugutelm thatlrthe arnplltuuesA a ande phase daffamce 7lZ then the wavels Elllpumlly pulanzed Suluuun cosaik and ensuen9 slnlalzetu Tnslasnaty sln1arkcos1arlxl 5 5 A a Thelast equahunls that ura nrele rurA 3 and ruran elllpse whenA a Alsu nutethatyuu En setz n andlett surne values we seethatthe eleetrle l39lelulutates dunkwlseand ls thus rlght elllptluullypularlzeu Mall nan mama Dithroil Sheet Polarizers Malus39sLaw EuameLuusMalusluly23 l77seFebruary24 1812 Unpolanzed lrglnt ls by de muon random m dreetron and random m tlne onentauon ofthe E andE elds wl eonstrarnt ofbelng perpendlcular to tlne dreetron ofwhl rt propagates ofpropagatlon tlun tlne Howeyerlrglnt emerglng from apolanzens also by deflm eonsrstent m rts on ntatron E andE elds andtlne dlre propagatron Ifwe rnte r te oyerzero to 21t or tlne random onentauon tlne unpolanzedlrgntrnerdent onto tlne polanzertlnen we see tlne output from tlne polanzens uon now euon of M J Inaye usedtlne plcket fencequot analogy m tlne pastto desenbet e rntensrty yersus angle relatronslup The analogy falls wh s yeral po ow en dlscusslng lanzers m a r For example let unpolarrzedlrglnt fall on polanzer l mum Tne outputrs1 mm A secondpolanzens at45 wrt to tlne rst The lrglnt output wlll be 1 I eusims In eusimsj my 24 Malus39s Law The rntensrty ls propomonal to tlne E eld squared so tlne rntensrty ls I I I now quot 2 If we now pass tlus llnearly polanzed lrglnt tlnrougn a seeond polanzerwluen nas atransmrssron axls at an angle 6wth respeet to tlne rstpolanzerwe flndthatthe outputrntensrty ls Malus39s La 155050525 Glare rlum water sur ee r Glare from water surfaee Glare bloeked by vertleal polanzer polanzauon eomwater Malus39s Law Nowlet tlus lrgnt fall on atlurdpolanzer also at45 wrt to tlne seeond polanzer Thus we have now gone tlnrougln 90 What ls tlne lrgnt output from tlne tlurd polarlze 3415551955 44mm any Ir hus note tlnat wlule tlne lrgnt ls conslderably dlmlnlshed rt has not been bloekedl un way to perform tlus enpenmentrs to set up two polanzers at 90 to eaeln otlner and note tlne lrglnt ls bloeked Tnen rnsert atlurd polarlzer between tlne flrsttwo Rotate tlus thrd polanzer and note that tlne lrglnt rntensrty wlll vary The lrglnt beeomes unblocked Ernally note that we have assumed an attenuatron ofenergy of supposedly slngle pnotons A pnoton nas rts energy or not Quan um mee amesmustre ly be usedto solyetlus analyueal problem The concluslon from tlne quantum analysls ls strll tlne same Malus39s Lawl Seettun7 2 LtghtpmpagauuntnAn AntsutmpteMedum Blre 39mgmce An tnnpottant ehataetenstte of etystals ts that many ofthett ptopetttes depend on the etystal dtteetton thatts they are antsottopte The dlelecmc eonstant adepends on eleetxonte polanzauon whteh tnyolyes the dtsplaeennent of electrons wtth tespeetto postuye atomte nuelet Electromc polanzatton depends on the etystal dtteeuon tnasmueh as ttts eastet to dtsplaee electrons along eettatn etystal dtteeuons hts means that the teftaettye mdex n ofa etystal depends on the dtteetton othe eleetne eldm the ptopagaung ltghtbeam Consequently the yeloetty ofltght tn a etystal depends on the dtteetton of ptopagauon d on the state oftts polanzauon thute7 a Altne ytewedthtough aeubte sodmm ehlonde haltte etystal opueally lsotxoplc and aealute etystal opueally amsotxoplc Tahle7 l 0y MWquot e e Blrefnngmce Mostnonetystalltne matenals sueh as glass llqulds and all eubte etystals are apncallyxsavapxc Thatts the teftaeuye mdex ts the same tn all dtteeuons Sutpnsed thatwe satdglass ts nonetystalltne7 Glass ts an example ofa l qutdt frozen state Medleval ehutehes haye ptesetyauon b m wt ent atned glass The glass ows very slowly andts thtnntng at the tops ofglass seettons Thus ey are mote suseepuble to on age ete are othetnnote destructive forces at wotk on these statned glass Works of art 12 a E Othet etystals those not eubte tn suuetute are anlsotxopl The tesult ts that along eettatn sp etal dtteettons tn the etystal any unpolanzed ltght tay entenng breaks tnto two dlfferent tays wtth dlfferent polanzauon and phas lo tes Sueh etystals ate blrefnngent ot doubly refracted Blrefnngmce We ean desenbe ltght ptopagauon tn tenns of thtee teftaettye tndtees ealledpnnstpal refractive mdtses n n2 and n along thtee mutually otthogonal dtteettons tn the etystal ctystals thathaye thtee dtsunetpnnetpal tndtees also have two opue axes and are ealled blaxlal etystals new ctystals thathaye two dtsttnetpnnetpal tndtees have one opue axls and etystals e ealled umaxlal us gt n ts ealled apnsthvs Crystal quot1 2 quot3 n3 W1 ts ealled ansgattw etystal Unlaxlal ctystals Any EM waye entenng an antsottopte etystal spltts tnto two otthogonal ltneatly polanzed waye whteh travel wtth dlfferent phase yeloetttes These two onhogonally polanzed wayes tn aumaxlal etystal ate ealled the ardmary a and Exvaardmmy 2 wayes These oewaye has the same phase yeloetty tn all dtteeuons andbehaves llke an otdtnaty waye tn whteh the eldls petpendteulatto the phase ptopagauon dtteetton Unlaxlal Crystals erwave has aphase yeloetty that depends on tts dtreetton of propagatton and on tts state ofpolanzauon The eleetrte eldm the er e ts notneeessartly perpendteularto the phase propagauon dtreeuon These o wayes propagate wtth the sarne yeloetty only along a speetal dtreetton ealled the opue axls T oewaye ts always perpendteularly polanzedto the optte axls and obeys Snell39s law Elre 39mgmce urCalute Constder a ealette erystal CaCO3 whteh ts anegauye unlaxlal erystal and well known for tts double refractlon When an unpolanzed natural ltght enters a ealette erystal atnorrnal tnetdenee andthus also norrnal to aprtnetpal seetton to thts surface but at an angle to the opue axls the ray breaks tnto an ordmary and entraordtnary ray see gure 7 11 e rwa ally orthogonal tn polanzauon However they wtll not travel at the same yeloetty and attertrayeltng through the ealette erystal there ts aphase dlfference between the o and e wayes when not propagaung along the optte axls see gure 712 15 sees a dlfferent n and refracts tnto the array El propagates lto the surface and ts the orray a Abmfnngem crystal plate wtththe opus axts pm el tn the date surfaces b A btrernngenterystalplate wtththe opus axtsperpendteular tn the date surfaces 19w 0 ltasan oumrmnet Pram Han thure77 Twu pularutd analyzers are plaeed wlth thetrttansnntsstun axes alung the lung edges at nght angles tu ach ether The urdtnary tay undetleeted gues thmugh the le pulanzer Wheras the mamdmary wave detleete uesthmugh thenght pulanzer The twu waves thenerure have u ugunal pulanzattuns thure 7 ll can use mwdsemm urnan a rurmrrntau M wzve rm ts untlu untn arts era calntta crystal sylns tntu wowAves callatl nvelwnh dx 39exem valentaes nu WM has a Pulmnnon rm ts a s Perpendxmhx te tlu 0Mch arts 19w 0 Max oumrmnet Hamelh Dtehrotsrn 1n addttton to the yartatton tn the refractlve mdex sorne antsotropte erystals also enhtbtt dtehrotsrn aphen enon tn whteh the opueal absorpuon tn a substanee depends on the dtreetton ofpropaganon and the state of polartzatton ofthe ltghtbearn lhaye already shown examples of dtehrote rnatenals when dtseusstng Malus39s Law Section 73 Birefringent Optical Devices Retarding Plates Consider a positive uniaxial crystal such as quartz ne gt nu Letalineall p normal 39A m rl face Aretarder plate The optic axis is parallel to the plate face The e and Erwaves travel in the same direction but at different speeds 1999 s o KaEpOptolzllzmomc FremceHall Retarding Plates The phase difference 1 expressed in terms of full wavelengths is called the retarda on of the plate We know that depending on the phase difference 1 ofthe orthogonal components of the field the EM wave can be linearly circularly or elliptically polarized 9pm ees A halfwave retarder has a thickness L such that m E the phase difference 1 is 7 or 180 corresponding to a halfwavelength of retardation 1 m7 27 nenL e 1 quun x In this case the plane of oscillation has rotated le ccw by 90 by vector addition of E L and EN halfwave retarder Example 73 1 on page 288 Quartz halfwave plate Given the incident wavelength A 590 nm Quartz plate with nu 15442 and nE 15533 What should be the thickness ofa halfwave plate 7r27 nlrrliL 22 1 a T e n 1 03m x103quot 5 32418x10 m 2 n en 21 5334 5442 20 0091 Retarding Plates Given L thickness ofthe plate 39Ihe owave will experience aphase change of k L 27mg L The ewave will experience a phase change of krwwL 2 A L The phase difference 1 between EL and EU is 2 1P 7 V le Relative Phase Retarding Plates A quarterwave retarder has a thickness L such that the phase difference 1 is 7r2 or 90 corresponding to a quarterwavelength of retardation 27 nrmL e emerging lightwill be was a of Elliptically polarized for 0 lt a lt 45 I is 7 quarterwave retarder Circularly polarized for a 45 Birefringent Prisms Prisms made from birefringent crystals are use ll for producing a highly polarized light wave or for polarization splitting of light The Wollaston prism is apolarization splitter in which the split beam has orthogonal polarizations So ifwe take a polarizing sheet we can cancel one beam at a time by rotating the sheet b 90 39Ihe prism is made by joining two blocks with orthogonal orientations of the optical axis arbitrary wave splits into an owave E2 and ewave E For calcite negative uniaxial nu gt n3 we have El sees nu then nE which is a decrease in the index of refraction along the optical path 39Ihus El refracts away from the normal EZ sees nE then nu which is an increase in the index of refraction along the optical path 39Ihus EZ refracm toward the normal Figure 7 15 may array The and orthogonal to E l999 s o KEEP Oprocllzcrmmc FrentlceHall Optical Activity and Circular Birefringence 39 l I 1 Dim m This rotation increases continuously With the distance traveled through The rotation ofthe plane of polarization by a substance is called optical activity It is a very important effect in liquid crystal displays LCD as a voltage can control the rate per distance of this rotation This effectively tums onoff a pixel in the display The speci c rotatory power is de ned as the extent of rotation per unit length of distance traveled in the optically active substance This property depends on Wavelength Rotatory power Section 7 5 ElectmOpu39c Effects Electrooptic effects refer to changes in the refractive index of a material induced by the application ofan external electric eld This eld modulates the optical properties ofthe device The electrooptic effects are classi ed according to rst and second order effects Let the refractive index n be a function of the applied electric eld n nE Expand this by a Taylor series in E n nqEt17E2 The coefficient al is called the linear electrooptic effect The coefficient a7 is called the second order electrooptic effect The even higher order terms have been found to be negligible for the highest practical electric elds and are ignored Section 7 4 Optical Activity and Circular Birefringence When a linearly polarized light Wave is passed through a quartz crystal along its optic axis the emerging Wave has its E vector plane of polarization rotated An optlcally actlve rnatenal such as quartz rotates the plane ofpolarlzatlor ofthe tnctdentwave The optlcal fleld E rotated to E Ifwe re ect the Wave back into the material E39 rotates back to E l999 s o Kasap Oproalacnomc FrentlceHall Figure718 Rotation Angle 5 gm ennL y E Input x y 5 E Output x Veraeally polanzedwave at the input can be thought of as two nght and left handed circularly polanzed waves that are symmetrical 12 at any instant ct lfthese travel at different ueloetaes through arnedurn then atthe output they are no longer symmetne with respect toy ct a and the result ts a vector E at an angle a to y 1999 s o Kasrp moclcmom premcellall ElectroOptie Effects The change in n due to the linear term is the Pockelr e ect An 15 The change in n due to the second order term is the Kerr e ect An 25 ugly2 All materials exhibit the Kerr effect But only noncentrosymmetric material exhibit the Pockels effect ofWhich GaAs is an example that has a Kerr effect FuckelsE ect The expresslon for the Poekels effect stated prevlou v ls an e strnpltfteatton The e eetrnust conslder the applled fleld along a parueular ervstal dreetton on the refraeuve mdex for ght vvtth a glven propagauon dreetton and polarlzatlon Thls ls atensor type of analysls Wlth coef clents dependlng on s pararn t ovvever lt ls elear that the eontrol ofthe refra by an external applledfleld and henee a voltage ls a dlstln advantage ansuatvotttvmnrent The phase ehange m aPoekels ervstal I r 1 1 can be controlled or modulated such as P2 aphase modulator calledaPockels call I a pneltelsmeet Md1 dn v stnee the phase ehange ls proporuonal to Ld when the devlce ls fabncated vve ean reduee d andlncrease L to bulldrup rnore phase ehange for a glven voltage V From the phase rnodulator ofthe last sllde add polanzers before and after the Poekels eell Thls 15 now an lntenslty rnodulat r In many applteauons ltls advantageous to have alaser operate as a eonunuous vvave devlce and externally modulate a slgnal e transverse eell vvtth polanzers ean do thls atverv hlgh frequencles FuckelsE ect Ifwe have the polanzers at45 and glven E915 the amplltude ofthe tnerdentvvave then amplltudes at the ervstal face wlll be Ev E cusw ycuswA Atter sorne tngonornetne rnanlpulauon we have E E sanAd smn 3m Let Ia be the llght lntenslty under full txansmlsslon Then the lntenslty of the deteeted beam ls then deteeted at the farnght tn the gure 2 Where VA 15 the applted vnltage ever etlve lndex t Flgure 7 m V E d m nevus dunner manner A monomeme n unn nu rs rnrunxynumn uh Lents pnnntnu Pmlhlwy am lama Manson mt m V 1 gm wng r zRzkds madam rare t YL39YHZJ Why a r zPuckels paranan 4 a Flgure 7 ll e elelseeurnensltvnauurtnr mpunnrprnurnuyrern have hernnsnenonrrnn ht lesnrup unnsnrn leASNn inn ht Tnnsrnsennnerertm lfppl wluge ethsenstss lF l nrervie pute gm erreereurttermnenneensaenmuhthensnuem e my s e nonhuman them up Example7 s l unpage 299 Pnekels Cell Mudulatnr 3le a husva39se LlNloZ phase mndulatu opemung at lKEIEI nm se shl Applled Vulta e no 2 table7 2 dues nethaveths dataat lKEIEInm 4 x ln urnv frumtahle7 2 r2 Wlat shnuld the meet mhun Ld be7 2v 1 7 t 7V 4 9 a a 2 2 t X n 17quot humans m 71w mmcr39 Tahle7 2 nuts n Mes unetxtnmr murmur mum Cmm mm mm rnrmrnns x mm x ut quotmv we meet unurtu nu nu ma Unluul l x mm um name a s mum Ivnntp u muv lelvcmru tumult t x N l Ken meet 5 me hqulds sueh as mtrotoluene C5117N02 andmtrobenzene CEHSNOZ exhlbltvery large Kerr eonstants A glass eell fllled Wth one ofthese llqulds ls ealled a Ken 5211 These are frequently usedto modulate hght slnee the Kerr effect responds very qulckly to ehanges ln eleetne eld nght ean be modulated wlth these demees at frequencles as hlgh as 10 Hz The polanzatlon modulator andlntenslty modulator we exammed before ean be extendedto the Kerr effect b Kzncz phase madulator Example 7 s 2 un fags m l Kerr Erreet Mudulatur ek The ught has bean pulauzed parallel tu the applled eld Ea dlreeuun What ls the applled wltage that mdueesa phase change uh halfiwavelengh 2m 251 m ZW 7 1 l quot id mm a a 7 2y 4 M 22 72 2 We aw hum equatlun 6b that the rmmmumlntmsltyls when V 11sln 39Vquot 1 when V 7 any V Kerr Erreet Suppose we apply a strong eleetrle flel 30ka to an otherwlse opueally d up to ple matenal sueh as glass or llquld lsotxo mymnmtmuwH n M The ehange tn the refractlve lndex wlll be due to the Kerr effectwhlch ls a seeond or er effect et orthe quadratic Electrnrnptic Effect QED elreet ls a ehange tn the refractlve lndex ofamatenal ln response to an eleetne e o ofto the magmde ofthe All matenals show aKerr effect but eertatn llqulds dlsplay the affect more strongly t an other matenals o K 187 The err effect was dlscoveredln 5 by John Kerr a Seotush physlelst Kerr Erreet E ls the applledfleld the ehange tn the refraetave lndex for b If a polanzauon parallel to the applledfleldls glven An Am where K ls the Kerr cmef nem Example7 s 2 unpage ml KerrErreetMuunlatur Gwen a glass reetangular hluek Thlelmess a um um I Thus a lEIEIXlU m e 9le 7V 72 2 rauxl 39mj xl quot The Kerr erreet ls st hut eumes wlth a eustly pnee m the rum urhlgh vultages th1em7 17 en page 319 150 Phase Madmamr th1em7 17 en page 319 150 Phase Mudulamr LZEImmWmeD2mm L2UmmW2mmD 2mm 11zunhm 11zunhm 2 12 2 12 nN36 nN36 Suppuse we suddmly app1y a s1epxm11age mm a supp1y wnh an Hulle reasahee Calculate 1he hme eemam and harm me in K sun r 1 rhsdsmce he LWEDA SUDZCMEIZM885gtlt1U mm The ume reqmredm emrge and mseharge the Efa 39ance e1 between 7 4quot e 4L 01062 edmdensdelmmnedby39heums mmnt unhee1ee1ne menu 71W 0 WXW A v V LWEDE re TRlcml D hilwgh 20x10 quot006 024quot 3x10 7 11g111a1se has 1 have1 1hmugh 39helengvh enhe crystal The hunt ume1s 1h 1h1s demee 1he 11m pmpagaueh 1s s1ewer1hah 1he nrcmlrespunse an m z 7 th1em7 17 en page 319 150 Phase Madmamr th1em 7 17 en page 319 150 Phase Mudulamr Ifwe app1yahae aga1 Vs fruma seureea1se Themlpedance ufths hemmed ureums wnh empmmpeaahee ufsn Q Lhanthe 1 mammumm du aum frequmcyu Z J C 0 i JhL Ru 7 71 1 e r 155112 1JhRcm quot 1JRr Cm L 27411 1u62x1n39 ML Nuwweanempl 1a 1mpmvelhehg1 frequmcypafunnance enhe mndulamr hy Mammmpewens dehvered when eehheenhg ah mpedanceL wnhapara11e1 essance as shnwnbeluw 1 1 m n 2110 Fmblem7 17 an page 319 E0 Phase Madmatur Semun7 mitigated Opuml Madmalurs If m GHzxsme amed mum men Integrated opucs refers 10 the mtegranon components on a smgle common substxat f ofvanous opucal devmes and 1 e Where 51472124 r 2 Low p Le rating devmes sueh as 1aser d1odeswavegu1des sphtters 1 1 modulators photodetectors on the same substrate lead 1 z 01M m1n1atunzat1on and overall performance and usab1l1ty Zne Cm 4721nx1nsz2124x1n4 Spual MZEMS Lman Inductor err r her 1 nHmductors m r 1r 1 7 pm WW I W Umvemty umenessee Flgure7 23 ms serum lntegnted hunverse Packzls cell phase maduhmxm whtsh uwrmgtuds ls durused mm un electmrapnc Eo substrute Co m stnp electmdss upplyu mususe thrv hths wavegmdz The suhshute lsanxrcleNbOund typrcrllythsre thrch slog between the sunuse electrodes my hon the wmgmdz my s u K s110lpckrmn kme x Macthehnder Modulator Induced phase shrlt by applled voltage can be convertedto an amphde vanauon by uslng an lnterferometer n rnterferorneterrs a devrce Lhatlnterferes two waves othe sarne A frequency but drlTerent phases Conslder the structure shown m the next sllde whrch has rrnplanted slnglecmode waveguldes m a LrNbo3 substrate m the geornetry shown Macthehnder Modulator The wavegulde at the rnput branches out rnto two arrns A andB These two arrns are later recornbrned atporntD to constrtute an output case the pow rls equally splrt atpornt c so that the eldls scaled by afactor ofsquare root of2 golng rnto each arrn The two waves travelrng through the arrns A and B lnterfere at the outpu port D andthe output arnplrtude depends on the p ase dlfferenceophca1 path dlfference between the two branches Two backctocback rdenual phase rnodulators center of the gure m black ena e the phase changes m an be rnodulated e devrce ls conflguredto apply uppustte applled flelds Lucent E0 Mudulatur Tl d usedllthlurn nluhate dedmrnpuc Fuckels effect rnuunlaturs hlgmeed Upuml Mammum u u batcummummhuns uptu 16 GHz operatesat lssnnrn rnuunlauun vultage ls 2W Cuurtesy quucmt Technuluges Figure 724 An rntegntedMashZendsropussl rntensrtynrodulutor The In splumm m cahzxem wmsA and upphs mv su who Macthehnder Modulator By rnodulatrng the electrode the opposrte phase wlll reduce the output amphde when the The output power ls propomonal to E2output whl ch ls re The last equauon ls over slmpllfled and the cancelled above resu gm hght ls E whlchue phase shrrted yths dvoltrge und than the Wm ure cam rnsd rgnn ut the nu ut 9 mm Puma nun rnaerrrnurn phase dlfference Qbetween the two arrns ls ro we ma 4 me output wlll notbe totally However the general form of the response follows the lt Coupled Waveguide Modulators Figure 7 25 When two parallel optical waveguides are sufficiently close together and not blocked by the material in between then power Crasssecum may be transferred between the waveguides by the evanescent camiiveguiues wave Spacing Li 1 where c the coupling ef ciency c The efficient transfer of energy back and forth between the two 4 A waveguides requires that the two modes be in phase to allow the a ciuss seetiun urtwu elusely maeeu waveguidesl andB sepamteu lay a angerer ampllLUde to bulld39up along the dlrecuon 239 emb eddedln a submte The mnescent eld ruml extends lntu Sandiice versa Nute in anal n gt n suhstmte index hTup Viewurthetvm guideslandBthatarecuupledalungthezrdirecnun Light B the use ofan lied volt e ockels effect we can ad 1151 the is reuintuA atzn anuitis gauuallytmnsrmeutu Balungz Atz w allthe Y aPP 3 l lightheentmnsrmeutu B Beyunu thispuintlightheginstu hetmnsieneuhaektu amount of power transfer The mismatch of the propagation A mmsamemy constants 3A and BB is given by to i999s o Kasap0ptoalactmmcsKennella A AnM 2 n3r2 wharerlsthePockels eoe fieient Figure 7 26 Figure 7 27 PELoPA0 Tiansmission power ratio from guide A to 100 guide 5 over the transmission length L as a function of mismatch AS v3 A 30 V Cuupleuwaveguiues 0 7r VLO 1999 s o Kasap Optoelectronic FrentlceHall coupler l two guides and changes the strength ofcoupllng 1999 s o Kasap Optoelectronic Prentice Hall Exam 1e 76 1 on a e 307 Modulated Directional Cou lei Coupled Waveguide Modulators P P g P Two optical guides are embedded in a substrate are coupled The transmission len th 10 mm A sw1tchlng voltage can be determined for a given wavelength The can 1i S Main dis 10 which will turn onoffthis transfer ofpower n 2 2 p g ep Pm The operating wavelength is 1300 nm 574 The Pockels coefficient r 10 x103912 mV anrL What is the switching voltage 7 Jim J3 1300nm1om V 7 3 3 712 Zn rL 22 2 10x10 10mm 10 57 Volts Seettun7 7 Acumerpu Mudulatur An lndueed stratn m a erystal wlll alter the refraettve lndex eps Where p ls the photoelastre eoemerent and s ls the rndueed stratn s ls atensor relauonshrp Thls stratn ean be rndueed by travelrng aeoustre or ultrasonre waves The Llnk heluwls a guud 2n fags uvervlewrurthuse rnterested In rnure rnrurrnauun him wwwaavptudeetmnte cmnDummemaumAoiDUE pdr Seettun7 7 Acumerpu Mudulatur In the last gure we see that rnterdrgrtal eleetrodes have applled F voltage The result ls the plezoelecmc effect Modulatrng vt generates surface aeousue waves SAW whreh leads to penodre vanauonr the lndex ofrefracuon n Thus for lrght rner dent at the Bragg angle the lrght ean be H eted t t t penodreally de e M Mansel Flgure7 3n Puhxnu39 snrr The sense a moan orthe opueul eld E depends only on the dueeuan orthe nugneue eld ror u gwen medtum glven Verdet eonstent lrhght rs reneeted e tes u tEurther a m the e to back rnto the Fendeynt dmm the eld mu eome out ts Equot wnh u 23 moan wnh respeet tu my s u KAsptwkmmn mm Hm Flgure 7 28 Acmlsncabmx uex mmmnenn smut nuneted ltght tnte nut eleetnueu vuuunng Rletxge m ifm Y Tnveung mete wee humane vnuaonnute rerneuve nuer nu heregy erene nunetnn gnang tun mm the when been hnug a my s u renonmmrm nut Seeuun 7 a MaguslmOph Erreets When an opueally rnaeuve rnatenal ls plaeed m a strong rnagneue eld and then plane polarrzedlrght ls sent along the drreetron of the rnagneue eld the emerglng llght s plane ofpolanzauon has been rotated Thls ls known as the Faraday effect The amount ofrotatlon ls glven y a VBL Where B ls the rnagnetre fleld ux densrty Lls thelength othe rn d A v ls the sorcalled Verdet eonstant A rnagnetre eld of0 l Telsa eauses arotauon of about 1 through a glass rod oflength 20 rnrn MagumnrOpu meets one applreatron othe Faraday effect ls m very hlgh voltage powe tran slo Faraday rotatron does provlde a handy way to rneasu e the eunentrn EHV or UHV power lln ean be qurte long ls plaeed nearthe power eable and apolarlzedlaser ls usedto rneasure the rotatron and thus rnferthe voltage from the assoerated rnagneue eldthe power llne produees The drreetron ofrotauon ofthe plane ofpolanzatlon depends only on e r e agne e ux s long as the lrght propagauon ls parallel to these llne of ux the rotauon ls the same he ergolng leftto nghtquot or nghtto lettquot Thls ls the basls for s w th optreal rsolator Section 7 9 NonLinear Optics and Second Harmonic Generation For linear dielectric material the polarization is P an 11239 where is the electric susceptibility However the linearity fails at high elds The induced polarization P becomes a higher order inction ofthe E eld P 1 E5u12E213E3 These nonlinear effects nan at about 107 Vm and require a light intensity of about 1 x 106 Wcmz So a laser is invariably required Figure 731 eld oseillaaons between 5 resultln polanzauon oseillaaons betweenP andP c frequencies a fundamental 2n seeondhannonie and a smalch componmt 99 s o Knsnn Uptoelecnomc menace Hall Figure 732 Second harmonics S gt k2 WS2 gt v W x 1 Fundamental Crystal As the fundamental wave propagates it periodically generates I the amplitude of the second harmonic light builds up 1999 s o Kasap Optoelectronic FrentlceHall Second Harmonie Generation One the most important consequences of the nonlinear effect is the second harmonic generation SHG When an intense light beam of angular frequency oopassing through an appropriate crystal for example quartz generates a light beam of double the frequency 2m SHG is based on a nite xi coef cient in Which the effect of X is negligible IfWe Write the optical eld as E EU sin out then P SaltEn 5mm s l lesi 005415n1253 These terms are called the indamental rst term the second harmonic and the dc term third term Which a constant polarization term Second Harmonie Generation There is a condition for constructive interference or the process Will die out This phase matching condition requires that the index of refraction for both the indamental and second order term be quot60 quot260 This is not possible for most crystals as they are dispersive index of refraction depends on Wavelength AWork around is to use birefringent crystals that match as closely as possible to the desired indamental and then the second harmonic This is called index matching and a measure of the t is the phase matching angle Figure733 Laser KDF Clemens Filter 5 1 532nm 1 mam 1 lEI Anm 1 532m at fr KT IM 1 quot angle a about 35 to the optic axis along which man may The focusing o the aserbeam on e Perystal and the collimation ofthe light emerging from the aystal are not shown to 1999 s o KaEpOptolzhzcramcrPrmt1ce Hall Opuml Ken39Effed The ppm Km a crlsthe use lhwhlehlhe eleeule eldls duetu lhellghz llself Thls musesa vanauuhlhlhdeh urlalaeuuhwhlehls pm mluhal m lhe lna Thl a hdex all huhllhear uphml e eels ur self rueuslhg and selfrphase mudllanuh an ls hasls f rkm39lms mudeluelllhg Thls effed uhly heeumes Sg mnl wnh very lhlmse hams sueh as 39huse um lasers Sel phade Imduhu39nn 5PM ls a huhllhar upuml effed urllglmaud lhlaae uh Ah uluashull pulse urllgll when uavdllhg lh a medlum Wlll lhduee a le ae velhdex unhe medlum due lu lhe uphml Km Efrem Thls vanauuhlh le ae ve lhdex Wlll pruduce a phase shl lh lhe pulse ladlng lu a ehahge unhe pulse s eduehey wemum Opuml Ken39Effed a amuse unhe huheuhlrulm pUWa39 dalle dlsulhuuuh lh a Gausslah hm lhe le ae ve lhdex ehahges auuss lhe hm plu le The refmmve lhdex Expmmcedbylhebmmls Wlennlhe cema39 uf39hebmm39hanauhe edge Therefure a lud urah ae ve Km medlumvmrks llke a lens fur hlgl lhlmslly llgll Thls ls alled Selfrfucusng and lh meme mses lads lu halmal desuueuuh lhe lasa39 mully shun bursts urllghz Wlll lheh he rueused amllylu eumllluuuswavesmm lamw thlem7 zl Ophml Kme ed Cuhslda a malmal that dues hm have lhe seeuhd urdenemllh the pulauzauuh edua uh P Emma15 2 p1 squot The m lemlapplles unda39luw elds The seeuhd lemlwnh E2 ls lhe lnadlahee unhe llg hmahd ws lhanhe lhdex urle ae uh depends unLhB lhlehslly ghz hm ufthsh WW 4 1 Buy where p lZEIrr3770 whlehlslhelmpedaheeuf eespaee Selrphase mudulauuh llllmlll m ml n u lullu Wm nuu l A Dlagam by a uh Melllsh opml Km Effed e Appllmhuns hee Kenlens mudelueklhg ls an Efrem lhal direle mas uh lhe eleeule dd sl lhe lepuhse ume ls hlgh ehuugh m pluduee llgll pulses lh lhe ulslhle and ha lh ared wlthlmglhs ufless llah s fantusecunds Due in he hlgl deanml eld sumglh rueused uluashull lass hams the lhreshuld ur In W mlz whleh sulpasses lhe eld s mglh ufthe EIEETXEIYHEIn uhd lh alums These shnnpulses umehB hew eld uruluam upues whehlsa eld mm W hen menallkemasuran l upueslhal gvesaccesslu a eu lelel newelass urp u an Elemmn muvanmts lh an 2mm anuseeuhd phehumeha cuhermt hluadhahd nght geneauuh ulua hluad 12535 and lhelehy glues llse lu rmny nEWapphEthS lh Elfum sehslhg e g cuherml lasa39 ladar uluahlgh resuluuun epual euhemlee mmugzphy halmal plueesslhg and ulhla elds llke melmlugy mahely Exam eduehey and me msuremehls thlem 7 zl Opnml Kan Efrem leeh mulls suseepuhlllues 13 Glasses 13 1EI 21mm Deped Glasses 13 ln H mm Sume uhgahle suhslahees 1 Inquot mZW hdudselh lu ah dyzssvchasblseMSE s euhdueluls 13 a n n mwz a use ps nquot 4 WM 1 M Calculate h1 and lhe lmehslly urllghl heeded lu ehahge h by In 3 n 1fur ach use nu H4104quot Y I m 1 En wemume ur Problem 7 2l opu39eal Kerr effect 3 7 quota GIVenVanoussuscepubllmesxz En 7 3 An Glasses x e 1072 mZw Doped Glasses 13 e 1048 mIw 39 39 m10397m2VJ A u Semiconductors X e 10quot4m2V End of Chapter 7 39 39 39 39 M w and a spot size ofl 5 mm This is equal to 2 25 x 106 wm2 39 ofoptical much powel in he same spot size or about IOkW Chapter 5 Photodetectors Seetaon 5 l Pnnerple of the pn Junetron Photodmde Photooleteetors eonyert alrght slgnal to an eleetneal slgnal sueh as ayoltage or eurrent eleetron hole pans EHPs my the oly th gen any photooleteetors sueh as photoeonoluetors anol photodwdes oleyree whreh ehanges permlmvlty bythe In oleyrees sueh apyroeleetne oleteetors the es e eratron ofheat whreh rnerea Pnnerple ofthe pn Junetaon Photodroole When aphoton wth energy ogenerate or eol anol phot the olepletaon layer the E fl separates o pesl Th pho nu ls separatron and ow of toeurrent 1W eN where rnber of eleetxons hv gt Egap ls s an EHP m lolthen the EHP anol dnlt thern m pposrte dlrectlons erto nrslde and hole to de eharge ls a N ls the total typreally aehreyeol by the ereatron offree absorptaon ofphotons Thls ls ataon ofeleetrons m the eonoluetron banol and holes m the banol energy eonverslon ses the ternp rts polarrzatron anol henee rts relataye erature of Photodmde closerup The rnoleseent square blue thes bo wrres are prote teol wrth an eneapsulant Th p l ty devl ls marked wrth a srnall rnolentata on m the lower nght Pnnerple othe pn Junetron Photodwde The pnjuncuon baseolphotoohoole type oleyrees are srnall and have hlgh speed and goool sensrtwrty v Thls gure ls a sehernatre of areyerse blased pnjunctlon photodwde r ps he hlghly resrstaye olepletaon layer w Net spaee eharge olensrty aeross the ohoole m the olepletaon reglon Npl and N are the olonor anol aeeeptor eoneentratrons m the n and p sldes The E eldln the olepletaon reglon The p so the E eldls rnostly m the n type rnatenal as shown Seetlon 5 2 Ramo39s Theorem and External Photoeurrent Conslder a sernreonduetorrnatenal wrth anegllglble dark conductle that ls eleetrooleol anol blased as shown below The eleetrooles olo notlnject earners but allow excess e nersln e sarnpletoleaye and ar beeorne eolleeteol by the battery The elecmc fleldD m the sample ls unlform and ls VL Suppose that a slngle photon ls absorbeol at aposmon n e Lfrom the left eleetroole anolrnstantly ereates an EHP Seetaon 5 2 Ramo39 s Theorem and External Photoeurrent Ramo39s Theorem and External Photoeurrent The eleetron andhole olrrtt rn opposrte drreetrons wrth respeetrye Let the external photoeurrent due to the motron othls eleetron be lea olrrtt yeloertres ye h E and yh nhE The transrttrme ofa earner rs the trme rt takes for a earnerto dntt from rts generatron porntto the eolleetrng eleetroole The eleetron photoeurrent rs lzv 10 L rnr rltr The eurrent eontanues to ow as long as the eleetron rs dn mg rn the sample 1t lasts for a duratron te atthe enol ofwhlch the eleetron reaehes the battery Thus although the eleetron has been photogenerateol rnstantaneously the external photoeurrentrs notrnstantaneous and has atrme spread Ramo39s Theorem and External Photoeurrent Ramo39s Theorem and External Photoeurrent The photoeurrent for both the eleetron and hole rs shown below The eolleeteol eharge Q rs QM Ixzdzrzdt 2 n n Mr M The hole photoeurrentrs m The eolleeteol eharge rs not 2e buttust one eleetron tU fur m These three equatrons eqn 2 3 and 4 form Ramo39s Theorem The total external eurrentrs the sum ofleO andrhtt In general lf a eharge qls berng dnfted wrth ayeloerty vdt by an n W W M elecm fleldbetween two braseol eleetrooles separated by L the thrs M motron ofqgenerates an external eurrent gryen by r a h fur t Um Semun S 3 Absurpuun Cuef nemand thudmde Malmals Table 5 1 un page 222 The photon absorption proeess forphotogenerauon requlres that the photon energy to be at least equal to the bandgap energy Em of the semreonoluetor matenal to exerte an eleetron from the yalenee mtrr nnlnprmn 1quotng anhtanstsrrmtnnquotWWW unborn band VB 0 the conduction band CB tatmtnwsnrtuarrrrrrrmramr mm 4 Arlwnv 1w The upper rubywavelenth or the threshold wavelength Ag for 3 H if 391 g photogenerataye absorptron rs therefore determrneol by the bandgap v m m x m aathurn lm H o energy P othe semreonoluetor when M75 lo p M lm t M has i n E 1 r as m 7 tr W 1 39 Eu For example 7 5r E 12 ev thus A51 11mm 1t rs elear that 5r photodrooles eannotbe usedln opaeal eommunreataons at l 3 and l 55 pm Problem 5 1 a Find the bandgap energy Egap for a semiconductor whose upper cut offwavelength Ag 600 nm Solution 11 a Z xlU z Jsec3xllxl E M25 hv 1 1 3313 l quot 1 xl wm X w 3313x10 1206786V ISOZXIO WC Problem 5 1 b Example where detector area is less than beam spot size Laser power 5 mW and spot size 2 cm with radius 1 cm Detector area 1 cm Solution Power SmW Arm 7mm2 PM per area Awakka 7r0 5m2 0 785m2 Power received by the detector is 13 ch Detector AreaPower per area 0 785 m2l 59 1 249mW Absorption Coef cient and Photodiode Materials Most of the photon absorption 63 occurs over a distance 101 This distance lcz is called the penetration depth B m 1 t 2 wherex gt2 2 0368 a So the amount absorbed a er traveling to this distance x is 1 r equot 1 7 0368 0632 In direct bandgap semiconductors the photon absorption is a direct process that excites an electron to the CB with essentially no change in momentum This process corresponds to a vertical transition on the Ek diagram Problem 5 1 b A photodetector whose area is 5 x 10392 cm2 is irradiated with yellow light whose intensity is 2 mW cm39z Assuming that each photon generates one EHP calculate the number of EHP per second Solution Pam AreaIncident Intensity 5 x10 25m22 04 IUUXIU W 0 lmW What ifthe area of the photodetector is lessmore than the beam size of the source If more then all the source power is received the the detector Ifless Absorption Coef cient and Photodiode Materials Incident photons with wavelengths shorter than Ag which can excite an electron up to the CB become absorbed as they travel in the semiconductor The light intensity which is proportional to the number of photons decays exponentially with distance into the semiconductor I x I Ue m Where x is at the surface ofthe semiconductor and In is the incident light intensity The absorption coef cient a depends on the photon energy and is a material property Figure 5 3 The absorption eoef eientversus wavelength anurw I I I I my us n7 Indirect bandgap Si and Ge IXIIIH the p II 1mm 1A mm Direct bandgap m I GaAS W InAs InP mu GaSb 2 IIIIIII nnnsnxinuim Ahsm nme 39mmt m vs Waugh arms smrductms Dzhselzcowlycdlzctad ma mrrbnd summers sources rams Figure 5 4 Photon absorption for dim and indirect bandgap semiconductor k 4 a Gm Dmctbandgap b s Irduec mndgap a Fhutun absurptmn m a ma bandgap sanicunaucmr h Fhutun absurptmn m an mama bandgap sanicunaucmr VB valence band cs Eunduchun band 1999 s o Kassp Qmelemovnc PrenhceHa Section 54 Quantum Efficiency and Responsivity Not all the incident photons are absorbed to create free EHPs that can be collected and give rise to a photocurrent The ef ciency of the conversion process of received photons to free EHPs is measured by the quantum e iciem QE q of the detector 7 number offree EHIP generated and collected 77 number ofincident photons I h The number of electrons per second Iphe 77 P The number of photons per second PUhu Quantum Ef ciency and Responsivity m R of r A A in terms of the photocurrent generated Iph per incident optical power PU at a given wavelength R Fhutu current A 1 Incident Optical Power W F From the de nition of QE it is clear that Rzi ii ggm39Uv 39Uc m L 1 39 39 39 dc end uu the avelength R is also called the spectral responsivity or radiant ty W sensitiv1 Absorption Coef cient and Photodiode Materials The choice of materials for a photodiode must be such that the photon energies ofthe anticipated incident light be greater than 33939 Further at the wavelength of radiation the absorption should occur over a depth covering the depletion region so that e photogenerated EHPs can be separated by the electric eld and collected at the electrodes us the absorption coef cient must not be too large surface recombination nor too small few photons absorbed Quantum Ef ciency and Responsivity The device QE is always less than unity It depends on the absorption coef cient of the semiconductor at the wavelength of interest and on the structure of the device The internal quantum ef ciency is the number of free EHPs quot 39 uitehi h But the extemal circuit only ows if the electron is collected and an EHP is created only when the photon is absorbed Figure 55 Responsivity versus wavelength Respunsivity AW Madmanad U77 u mum n a mu ADD uu arm innninn Wavelengthmm Rewunsvnymws WavelenglhD urantdal phumdmdewtm QE1UU 7land furatypim cummemai s phutudmde y su mwmemmmmmlm Problem 5 3 on page 247 responsivity mammalVa Photosensitive area o 008 m2 i 7 VRZIOVwithdarkcun39entofOIi uA 35 Junction capacitance c 4 F N Rise time ofthe diode o 5 h ha m m i Calculate its quantum ef ciency at850 1300 and 1550 mi 5 wiwhxeiiiim 116 R 116 R q 1241x10 A e A R t850 7033 a 33 1 241xl quot7 482 a m 7 mmiXSEleEl39y 7 I 58 rl Rat 13oohmoss mm mxmtrmlxm M554 Rat1550nm073 an 73 llAlxlEIquot584 lSSEIXlEl39y Section 5 5 The pin Photodiode The simple pn junction photodiode has two major drawbacks Its junction capacitance is not sufficiently small to allow photodetection at high modulation frequencies This is an RC time constant limitation Secondly its depletion layer is at most a few microns wide This means at at long wavelengths where the penetration depth is greater than the depletion layer width the majority of the photons are absorbed outside the depletion layer where there is no electric field to separate EHPs and dri them ii m We photodiode r r The pin Photodiode The pin refers to a semiconductor device that has the structure ptintrinsicn The intrinsic layer has much smaller doping than both p and n regions and it is much wider than these regions typically 5750 um depending on the particular application When the structure is first formed holes diffuse from the p side and leave behind a thin layer of exposed negatively charged acceptor ions Electrons diffuse from the n region into the intrinsic region as well leaving behind exposed positively charged donor ions Problem 5 3 on page 247 2 Consider a commercial Ge ph junction photodiode which has the g following responsivity a Photosensitive area 0 008 mm2 0 008 10393 m28 x 10399 m2 w mJZJ 39 VR 10 v with dark current of0 3 uA Junction capacitance c 4 pF Rise u39me ofthe diode 0 5 his emehn R at1550nm0 73 Dark current0 3 uA 1ph 1 as DetectedPower 537 U 3 A411x1u4w R 173 P Pu Area1ncident Light Intensity Lighz Intensity A quot rea 4 11x10 7W 1 Imusijmii i Light Intensity m m irvm m 7 5 mm Figure 56 mt Schematic structure of an idealized pin photodiode m The net charge density across the diode m f if r l l l l The builtin field across the diode 5W The pin photodiode in photodetection is 39 e reverse bias d I Note that VB is the detection voltage 39Ihepin Photodiode The two charges are separated by the iSi layer of thickness W There is a uniform builtin electric field EU in iSi layer from the exposed positive ions to the exposed negative ions In contrast the builtin field in the depletion layer ofa pn junction is not uniform and has its maximum at the metallurgical junction The separation of two very thin layers of negative and positive charges by a fixed distance width W of the iSi is the same as in a parallel plate capacitor The pin Photodiode The junction or depletion layer capacitance of the pin diode is given by 55A Gwen v Where A is the cross sectional area and 2 g is the permittivity of the semiconductor Since the Width W of the iSi layer is xed by the structure the junction capacitance does not depend on the applied voltage in contrast to the pn junction Cm3p is typically of the order of apicofarad in fast pin photodiodes so that With a 50 Q resistor the RCD Ep time constant is about 50 ps The pin Photodiode The response time ofthepr39n photodiode is determined by the transit times ofthe photogenerated carriers across the Width W Idquot K where vd the dn39ft velocity ofthe particle vi The reverse bias Vr increases the builtin voltage to VD VI The electric field E in the iSi layer is still uniform and increases to At high electric fields vd does not follow the expected pd E behavior Where pd is the drl mobllity but instead tends to saturate at vs vsat is ofthe order of 105 ms Problem 5 4 on page 248 km Annrlhm Rlnth In if la 0 125 rnrn2 4 rnrnin diarneter Differences nsmrutyo m u 5 m reshmmtyerue warmth m rlmenenes 39Ihepr39n Photodiode When a reverse bias voltage Vr is applied across thepin device it drops almost entirely across the Width ofthe iSi layer The reverse bias Vr increases the builtin voltage to VD V The electric field E in the iSi layer is still uniform and increases to E En K since V gtgtVn W W The pin structure is designed so that photon absorption occurs over the iSl layer Thus in the uniform E field as shown above the charges are separated by dri and give rise to an external photocurrent Figure 5 7 Dri velocity Versus electric eld Drl veloclty rn squot 105 Electron 4 10 Hole 103 102 104 06 107 105 l Electrlc field or mquot Drift velocity vs electric eld for holes and electrons in Si 1999 s o Kasap Opmelecvomcs Prentiee Hall Problem 5 4 Caleulate the photocurrent horn eaeh when they are illuminated it A with blue light ofwavelength 450nm and light intensity 1 mW ean 37 What is the QE of eaeh device For type A Responsivle 1eA 0 2 AM at 450 nrn wavelength light from the graph From the given data intensity 101 uw cm392 10 wrn 2 and 71 rn areahol25rnrn2 Then the power PD 7 QA 104 Wmm397gtlt0125 rnrni l 25 x 1075 w Photocun enthhZRAPDZQ 2AWx125 x 10395W2 5 x 106A 2 5 uh The quantum etrieieney is 7 he 6 626x 3 Jsx3xlozrnsquotx0 2AM A 77A 21 l 60218x10quot9c x450x10 3rn 551 Problem 5 4 2 Problem 5 4 2 Calculate the photoeunent from each when they are illuminated 1 A The same results are tabulated below foi three wavelengths What 1 A with blue light ofwavelength 450nm and light intensity 1 mW em2 7 eonelusions ean you draw 7 Whatis the QEofeach deviee7 i i I For type B Responslvlty RE 0 12 NW at 450 nm wavelength light Photoeunent1PhRglgtu0 12AWx125 x 1075110 15 x 10395A1 5 uh The quantum efficiency is p 55ni ME 6 626xan Jsx3xlxmsquotx12AW 121 lauzlixlu cx45mluym Example 5 5 1 on page 223 opeiation and speed ofapin photodiode Exam 5 5 1 quot 3 228 0PM 3quot SW Wm Ph d The eleetiie eld is mainly across the isi layer and is appioximately si pm photodiode isilayeiwidth2o um E LULV 5xl0 p layer illumination side 0 1 pm W 20m R A From gure 5 7 we see that the eleetion dntt veloeity v is about 9 5 x 10 ms 51 l gh Pm 900 quotm which is Very near its saturation wdue of105 ms 5 The hole dntt veloeityv is a slower7 x 104ms andis the longest tiavel time layerquot h 39 pamele 5 Z xm mzis7xluquot Solution MD the entiie isi layei7 f 39 39 m From gure 5 3 my eyes saw M 4 x 104ml so the penetiation depthis 1 CL 7 en with my number the assumption ofabsoiption ovei the bandwui39h m o ev entire isi layer seems reasonable Exam 5 5 2 quot 3 m WNW D fmswquot quot a P quot Ph Example 5 5 2 on page 223 Photoeaniei Diffusion in a pin photodiode A Solution n in gun 3 39 wavelengths have a absoiption eoeflieient otthat is larger and thus lais a m 5 1 m dd quot 391 P 5 5 d 391 d Pl F n iegion The photogeneiated eleetions have to make it across shorter distance into the photodetecuon to the n side to give nse o a otoeunent The movement across the p is by diffusion The photogeneiated eleetion has to diffuse to the depletion iegion Where it is swept into the ilayer and diitted acmss me 7 an demon an demon on avenge diffuses a I be distaneel given by equation not denved in this xt pm photodiode 2 Imam 39s391 39dth20 167 10 i i aycrWl I I win 7 1 42D tdw39 2D 2mm X 5 p layer illumination sidee 1 um V V Va 120 V Diffusion coef cient D in p giom 3 x 1074 mzs The txanslt time ofthe eleetion onee it reaches the Ilayer is l 1 2xluquot n Zn V 1 Whatis the speed ofiesponse ofthis photodiode7 absorbed near the surface is about tam tum 1 37 ns Example 5 5 3 on page 229 Responsivity of a pin photodiode si pm photodiode Active light receiving area ofdiameter o 4 mm Area o 125 mm Incident light at wavelmgth 700 nm with intensity In 0 1 mwcm2 The incident light generates a photocun39ent 1ph 55 5 nA What is the responsivity and QE ofthe photodiode at 700 um Solution 004 2 02 W Pt Army 0126uW I 55 5m Th b R L o 454 e responslvlty IS given y P 0 IZWW W The quantum ef ciency QE is given by 7 7R 5 ms 1241x10quotm 774571241x10 700nmlt a A 0 7983 7 0 798 A 7 0 798 Avalanche Photodiode Mm The electric eld is given by the integration of the net r L A pm 0L A A subject to an applied voltage VR The absorption ofphotons and hence photogeneration takes place mainly in the long 7 layer The electric eld in the 7 region dri s the EHPs apart When electrons reach the player they experience even greater elds and acquire sufficient kinetic energy greater than Egzp to impactionize some of the Si covalent bonds 39Ihese impact liberated electrons can themselves be accelerated sufficiently to continue the impact ionization process Avalanche Photodiode 39Ihe multiplication of carriers in the avalanche region depends on the probability of impact ionization Which depends strongly on the electric eld in the region and hence on the reverse bias VR Or in other Words the avalanche multiplication is a statistical process Note that only electrons are multiplied in the device M Multiplied photocment i Primary unmultip lied photocment 1W 1 Or in terms ofthe reverse voltages M 1 a Where Vbr is a parameter called the avalanche breakdown voltage and n is a characteristic index to best t experimental data at a speci c temperature Section 5 5 Avalanche Photodiode APD Avalanche photodiodes are Widely used in k W optical communications due their high speed A and internal gain There are four layers n p p7K p On sufficient reverse bias the depletion region Widens and reaches through from the n layer through the 7 layer The 7 layer is nearly intrinsic The n side far le is thin and it is the side that is illuminated through three p layers of different doping levels suitably modify the d39ode No The electric eld distribution across the i te that the electric extends from the n region all the Way to the p region Avalanche Photodiode Thus from a single electron entering the player one can generate a large number of EHPs in an avalanche of impact ionization processes gtE h a 239 39 u 7 region 1 me p E Awlanchzregan 39Ihe photodiode possesses an internal gain mechanism in that a single photon absorption leads to a large number of EHPs generated 39rVL r c l r A effective quantum ef ciency greater than unity Avalanche Photodiode The speed ofthe APD depends on 1 The time it takes for the photogenerated electron to cross the absorption region the 7 layer 2 The time it takes for the avalanche process to buildup in the p region and generate EHPs 3 The time it takes for last hole released in the avalanche process to transit through the 7 region The response time of an APD to an optical pulse is therefore somewhat longer than a corresponding pin structure But e multiplicative gain o en makes up for the reduction in speed ability to detected Weak signals Table 52 on page 233 unzsz lmnhhzmnamlmniwmmyuanMnMdAFDInKphmmuqubmmlunN a c w 1mm rar M Ma N 1 WWW m my mum plum11 ml at 111mm w remhrrrrwtrrmparturmrhrhrahmmmwmurmmhmrrmh 1 pm em rum 1 am 1 martrparaeutmtrr murmur m1 1mm 1 mm 11 A n 1 aquot wowme m an AW nah net 7m WWW m1 quot0 mm w 1 tn gram mm 1m 1154 lt1 mum 51 AVE um um Min 1mm 11 ca Warm 7mm 1mm nun lt1 1n prawn 741mm must e 4 that 111 lanu lul39pm mnm 51wa at he 1 ma 1nmA11nP1PD mm mum 39HK mt mun Seenuh s 7 Helatqun mn Fhmudmdes Th1s seeuoh W111 hotbe eovered and1s hottestable Fhututranasmrs Smce the photon generated pnmary photoeurreht 1W ampl1f1ed as1f1twere abase eurreht IE the photoeurreht hr the extema1 erreurtrs Ia 4 Wm ere 1315 the eurrehtgarh or 33331 g393 hm of the trahsrstor msgnm L131 Example 5 a 1 mpage 234 lnGaAsAPD Respehamty QE EIVaal 15511 rmw1thM1he mmhphmum Thenb1asedfurM 12 Ca1eu1atethe pheteeurrent 1f39he madam upuml puwens 211 M Suluhun if 16x1o 71550x1o 77 A R quotPr 7mm 6626x1039A 3x10 7 707w RF n757 gt2nx1n W 15 For the uhrhu1trp1red photoeurreht answemnm Whatrs the rewunsumy whenthe muluphmuums 127 1 MI au a R Pquot Pquot MR 1z075e9w Seem s a thtettranasturs The phototrahsrstor 1s a b1polarjuncuon trahsrstor BJT that operates as aphotodwde wrth a photoeurreht gam m ii 1 Sam 352331 The base terrhrha1 1s normally open and there 15 avo1tage apphed between the eo11eetor and em1ttertermmals eorhrhor erhrtter BI T Extraet mm Famhld ANmus thutmnsmur Data sheet rhmt Damman mmmtht For the rhu1taphed photoeurreht Harnarmtsu 2829 Phetutranslstur Semun s 9 Phutueunduetlve Daeeturs he photoeondueuve deteetors have tvvo eleetrodes attaehedto a sernreonduetorthathas the deslre absorpuon eoemerent and QE overthe vvavelength oflnterest Inerdentphotons beeorne absorbed and photogenerate EHPs Thls results m an lncrease m the eondueuvrty and henee the external eurrent slow hole drlftrnust be eovered by several faster eleetrons dungn the tlrne lt takes for the hole to drltt to an eleetrode Thus there ls gan m the Fhmucundumve Deteeturs The transrt urnes through length L are glven by L L K5 M5 Let 1 equal the rnean reeornbrnatron trrne of excess eleetrons r r I te 1 r J Gam G J 11 Phototmnsistor Example Applications Appltranuus mmuuumtuwb umntttltw tunnuuutttsuttutuntnvuuu lyludlytlllwlll m must mu le Mm u nta mum mmwmm mnvmmw n tulrtu utna o tun Hymllzmwnumwm Ul unnhwr Dvumlsmnzh up u Wane outu t t autumnw huntst tuutuuwuuusutnutwtt mullrm u M u my u ltunttumruuu mmmlh ulhuttwtmuntnuuturt tummy WWWHull m nut Wm H t t tut l u tun mu M tut nu t t W n Wm Flgure s 18 Eleetruns cuvmng39 slvvvhuletranslt nu D D d let 1L WSW o a a l a a A nhetunundrnturwrthuhrno cuntuots perms nuthrnmns camarentrv nanerhrurtsnn us tle sluv huh dn s tlmsh tle nhetuoundrntue many an ehntruns emerald dnn tluvush dd nhetunundrnturbenarse n anvrnsm the Plrnmcuzldudm mum mud Elechuns dnn Ashe w mans as urn lazveszmkxmsentnx lyyy s o Knxlvavalquotmn Puma Hull Seebun s In Nulseln thlmmnslsturs The lowest slgnal that aphotodeteetor ean deteet the extend ofrandom uetuau deteetor ls deterrnrned by ons m the eunent through the The uetuauons are the result of vanous statlstleal proeesses m the devlee When the deteetor ls reverse blased and ln the darkquot there ls sull a reverse blas eunent ealled the dark eunent TM or Iv1 Thls dark eurrentIv1 exhrbrts shat muse due to the arnval of ds rete eharges of quanth e c ers are eolleeted as dlscrete arnount of eharge e that anlve atrandorn trrnes and not eontrnuously Noise in Photocransistors The rms value ofthe uctuations ofthis dark current is I39M Jzem Where B is the bandwidth ofthe photodetector 39Ihe photocurrent signal must be greater than this shot noise to be detected See the next gure 519 Noise in Photocransistors Just y have an unavoidable random uctuation in the rate of arrival This type of uctuation is called quantum mire i 2211 Where again B is the bandwidth of the photodetector Since we have characterized these noise effects in terms of rms we can add them to nd the total noise 2 7 2 2 In 47de 47qu 139 22I um Noise in Photocransistors 39Ihe mire equivalent power NEP is the optical power required to generate a photocurrent signal Iph that is equal to the total noise current This is at a given wavelength and within a bandwidth of HZ Thus the NEP equates to a SNR 1 The detectivity D is the reciprocal of NEP 1 D 7 NEP Figure 5 19 Shot noise versus photocurrent signal In I W Mummated TimB Inpnjuncnon andpm devices the main source ofnoise is shot noise due to the dark current and photocurrent 1999 s o Kasap Oprochzcvomcr Frames Hall Noise in Phototransistors In our detection circuit we now have the dark current Id the photocurrent Iph and the noise current in Added to this is the thermal noise in the detection resistor and the noise in the input stage of the receiver arnpli er The receiver system is usually characterized by the n gna to Moire rario or SNR SM 7 Signal Power Noi Se Power For the photodetector alone no external detection cimcit considered the S is SNR Noise in Phototransistors Given the responsivity R and the monochromatic light incident optical power PU such that Iph in Iph RPH I 7 ZeUd 113 111 P i 1 2 I I Enid 8 2 P 39Ihus NEP is de ned in units OfWHZ39W as P 1 W NEPE 2e1 1 J R a P2 r Hz Example 5 10 1 on page 244 N39EP ofa 5 pin Photodiode P W NEPE D JEJHZ A 5 pm photodiode N39EP 1 x 10quot3 wHzm quotLL f the bandwidth of operationis 1 5H2 Solution P W Mkii PNEP B JE Hz in J 1043 Lle GHZ JHz 3 162x10 3W Example 5102 on page 244 Noise of an ideal photodetector Consider an ideal photodiode with n 1 and dark current Id 0 and is operating at 1300 nm with a bandwidth ofl GHz Now we can plug and chug with the values above P 7 2th 7 25 525x10 Jx3x103g TE 7 11300x10 3m 3 05x10m 3 06nW 1x109 39 39 i due oiei quantum 39 random arrival of photons It is a unrealizable minimum End of Chapter 5 Example 5 10 2 on page 244 Noise ofanideal photodetector Consider an ideal photodiode with n 1 and dark current Id 0 and is operating at 1300 nm with a bandwidth ofl GHz Calculate the minimum optical power for an SNR 1 Solution 114 zeal up My 1 2213 From the work in section 54 we have 2116 VB The power required at SNR 1 411228 11 Avalanche Noise in the APD In the avalanche photodiode both the photogenerated and Limuuaii 39 39 L L multiplied The shot noise associated with these carriers are also multiplied Let the unmultiplied dark current and photocurrent be given as Idu an I P new M ejle du 11 221dn 11 MZB There is also a randomness to the impact ionization process and this adds another factor to the noise 1H zea 1WM1FB Where F is called the excess noise factor and is a function ofM and the impact ionization probabilities


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