you know everyone so today we are going to discuss about Optical sources in the last class we had discussed about or lighting basics of Lighting systems like radiometry photometry and chlorine calorimetry so today now we are going to discuss about Optical sources which are which will be used in Optical wireless communication so the typical Optical sources which we will consider are LED and lasers so let us now understand the typical requirements of your Optical source whether it is for optical fiber communication or it is for optical wireless communication so first that the size should be very small it should be as small as possible the second is you want the power output the optical power output which is coming from the source is proportional to electrical current because ultimately I am going to modulate my electrical signal so I want my power output to be proportional to electrical current or electrical input it should have high electrode to Optical conversion efficiency that means whatever my electrical power is should give corresponding Optical uh Power so you should have that conversion rate between electrical power to Optical power should be high electrical power to Optical power the spectral width that is Delta Lambda should be small what I mean by Spectral weights spectral width is if the light is emitting at a single color then it will have a fixed Lambda suppose this is my Lambda and this is a a single wavelength source so this is a Lambda 0 so we'll say that light is emitting at valence Lambda naught but in normal practice this Lambda 0 is not a single wavelength so it has some spread around Lambda naught so this is spread is called as a spectral words so we want this Delta Lambda to be as small as possible because if the light has many colors in communication this could be a problem because each wavelength will travel in the medium with different speeds and when they travel with different speeds they arrive at the receiver at different times and at different times meaning there will be a you know dispersion in the pulses or the pulses will spread so suppose if you launch a pulse like this if your Delta Lambda is high after say 20 kilometers or 30 kilometers in optical fiber for example this pulse will become something like this so if there is a dispersion in time then we'll start interfering with next pulses so it will result into a limited data rate or it will result into something arbiticus which is called as Intel symbol interference so we want this Delta Lambda should to be very small we want high bandwidth capability which means I should be able to modulate the source at high speed it should not be limited I mean the the wavelength the the modulation bandwidth should not be kilo Hertz or uh even megahertz I should be able to modulate the source at high data rate or it should have high modulation bandwidth of the order of few gigabits per second so these sources should have you know High modulation capability foreign the sources should be energy efficient the kind of power which we apply to LED and Laser should be small and it should give you it should give you adequate Optical outputs uh so the sources are expected to be energy efficient and of course it should be low cost and should have long life so let us first understand the basic uh principle of a LED so LED is basically a PN Junction diode which is forward biased this is p this is n and this is the depletion region and this is forward biased so what happens in a LED uh when you forward bias it sum the electrons from the Lower State they go to the higher state and from from that state they fall onto the Lower State and they release energy and this energy is in the in the form of photons so this is your lower energy level this is your high energy they say E1 this is E2 so some of the electrons which are here they will go up when the current is supplied to the device by giving this forward bias and these electrons come at the higher level and then they fall down and so this may result into two types of radiations one is a radiative the combination and another is this will this result into light output and another could be a non-reditative which may not result or which will not result into any light output not radiatory combination so this is the basic principle of LED and and the the photodiodes are of two types direct bandgap and direct band Gap so for direct band Gap the light which is emitted from a PN Junction is given by HC H is Planck's constant C is velocity of light and EG is the band Gap energy so the Lambda which is emitted uh is given by HC by EG and if I put the value of Planck's Constant and now I also put the value of C here and then the Lambda is actually 1.24 divided by EG and this e g is in electron volts so units are important here so you have to put the value of EG electron volts and then wavelength which you get using this expression will be in micrometer so the Lambda which is emitted is is equal to 1.24 divided by EG in electron volts and the wavelength is in terms of micrometer so if I take different material for example if I take silicon the band Gap energy is 1.12 it will give you a different wavelength if I use gallium arsenide the band Gap energy is 1.43 electron volts and for germanium it is 0.67 for indium phosphide it is 1.35 and then you can have a different combination of gallium and aluminum and then you can get you know different value of band cap energy so basically you can change the doping so this is GAA GA X or rather it will be Alx because you are adding aluminum as a doping Alx ga1 minus X into a s so you can change the value of this X and then you can get different band Gap energy and when you change the band Gap energy essentially you are changing the wavelength of emission so for example in this case GA is uh GA 0.97 aluminum 0.03 and a s so this gives a band Gap energy of one point uh five one so similarly if you plot for example have different levels of X this is say aluminum mole fraction and this is the wavelength of emission so this will be something like this so this is starting point point three and this could be 0.8 so by changing this value of x that is the doping concentration of aluminum in the material you can get different uh value of band Gap energy and hence Lambda so you can make your Lambda as per your choice because as we know there are certain wavelengths which are useful for communication for example you know the wavelength which are around 850 nanometer or 1310 nanometer or 1550 nanometer they are all good for communication using optical fiber where they have you know low loss so you can generate a wavelength of your choice by changing this value of x of course this gives you uh this is 1.3 I'm sorry about this is 1.3 and 1.8 micrometers so you can basically get a wavelength ranging from 1.3 to 1.8 by changing the value of x so this is another example where you have EG actually is represented in terms of 1.4 plus 1.2 into X Plus 0.2 into x square so this is a generic equation for EG and depending upon what value of x you choose you can find out this EG and corresponding and corresponding to EG there'll be a wavelength formation so now we will study the two important things of source which one of them as I mentioned is the spectral width which we call as Delta Lambda and the other one is the modulation capability of the device modulation bandwidth of the device so these are the two important aspect of any communication any Optical source so first we will study the spectral width part so uh so let us consider a PN Junction of so so this is for example your valence band then this is your you know conduction band so there are electrons here there there will be some holes when you supply the current so there will be some holes and here there are different states so the electron is not limited to one one energy line but it has some distribution here similarly the holes are distributed here and this is the recombination relative recombination part radiative Z combination part and similarly you have the another part which is non-related free combination so both are happening in the semiconductor and this is balance band so this will energy E1 and this E2 and E2 is not a sharp line it can be you know is a broad line for these energy states so let us find out what is the probability of photon generation so it is proportional to any two into p e one any two means the probability of electrons in the conduction band so this is probability of electrons in the conduction band and p e one is the probability of holes in the valence band so the probability of photon generation will be you should have you know electrons and holes when they combine they either give you relative combination recombination or non-native so for Simplicity we will assume that all electron hole pairs they result into some Photon generation so there is no non-related non-reditative recombination so this probability of photon generation is given by this and this expression for any two and pe1 which takes into account the Fermi level also but that Fermi level part is constant so I am absorbing that constant into that proportionality sign and this is proportional to e raised to minus E2 minus E1 divided by ktk K is the boltzmann constant and T is the temperature so this is the probability of photon generation and the total photons if I see here and I'm assuming that all the electron hole pairs are combining relatively they are emitting photons so basically this will be an integration of because you know this electron May combine with this whole for example this electron May combine with with the different hole similarly there are other holes also so basically I need to integrate all possible you know combining of these electron and whole pairs so I will be integrating this from EC to eph and eph is the photon energy which is emitted which is given by between this point and this point so this is eph and this eph is actually varying for you know different pairs so this total Photon and it is uh integrated from EC to EV Plus eph and if I simplify this integration this becomes EV plus eph minus cc into e raised to minus E P eph by KT yeah and If I multiply by E raised to power plus E G by KT and e minus E G by K T I somehow simplify this expression so this is the constant this part is a constant this part so if you see this part this is nothing but you know X into e minus X kind of form so if I plot this function then this will give me something like this so this is for example eph I'm plotting this and this is the power because this is the photon generated so this is a optical power this will give me something like this this is the EG point where EP this power is 0 e p h is equal to e g is zero and it has a maximum at EG plus KT and if I see this distance or this width this is nothing but 2 KD so this is actually the spectral width of the light emitting diode so I have taken so many approximations so that is why it is 2 KT but in actual practice this is 1.5 KT to 3.5 KD so the spectral width is 2 KT as I mentioned here and Delta Lambda for typical you know experimental device experimentally it is found that this Delta Lambda is between 1.5 KT and 3.5 KT so now let us find out the value of Delta Lambda so we know Delta Lambda Lambda is equal to SC by eph this is the wavelength which is generated because of this eph and if I sort of differentiate Delta Lambda with respect to Delta eph I get Delta Lambda is equal to minus SC EPS Square Delta eph and if I replace you know right in a different fashion that is Delta Lambda by Lambda is nothing but 2 KT by eph so we can get some values or Delta Lambda for different wavelengths so if I put Lambda is equal to 850 nanometers uh what I get Delta Lambda as 30 nanometers and if I increase the wavelength to 1310 nanometers then I get 17. nanometers for 1550 it is about 100 so typical LED gives you you know 30 depending on the wavelength 30 to 100 nanometer so it's quite High actually for communication from communication perspective in optical fiber particularly it is quite High so because it lead to lot of dispersion right now we have understood about the spectral width let us understand about the LED power how much power is emitted and also we will try to understand the efficiency part so let us take the first the efficiency so there are two types of efficiency which is defined one is internal Quantum efficiency which is defined as number of photons which are generated inside the device divided by total number of electronical pairs are which are generated inside the device so if you know number of electron hole pairs are all converted into photons the efficiency is one but that is not the case most of the time so this is how internal Quantum efficiency is defined external Quantum efficiency is actually ratio of number of photons which are emitted out of the device and number of photons which are generated inside the device so this is called as the external to the external world or external Quantum efficiency so these are the two important parameters what is your internal Quantum efficiency and what is your external because ultimately the photons which are generated they are going outside to either getting coupled to the fiber or traveling to uh different destination at some point B so that is also important so these two quantities both are important for sources so let us try to understand these efficiency parameter so as We Know in in in in a photodiode in a light emitting diode there's a p Junction there's an N Junction and there is a depletion region here so this rate minus d n by DT because the N is the carrier density charge carrier density it is getting depleted so minus DN by DT is actually proportional to n this is what what is happening inside the LED and if I solve this so This n is equal to n0 e minus t by Tau where tau is the average lifetime or against recombination so n is equal to n0 and 0 becomes an initial charge carrier density is e raised to minus t by Tau and tau is the average lifetime against recombination and if I differentiate this uh this expression this becomes DN by DT minus n 0 by Tau e raised to minus t by top and I can replace this with n so this becomes DN by DT is equal to minus n by Tau so what is a net rate you are externally supplying the current and there is a depletion happening because of the recombination the net rate is externally supplied current plus depletion rate due to the recombination so externally supplied is J by Q D J is the current density and Q is the charge and D is the width of the depletion region plus the depletion is happening at the rate of DN by DT that is a depletion rate due to recombination so this becomes J plus Q D plus d n by DT and under equilibrium state this is 0 and I have replaced here this is DN by DT by n minus Tau using this expression so this becomes J minus Q D minus n by Tau and net rate will be 0 under a steady state so this gives me the steady state electron density in the active region when the constant current is Flowing across the device which is given by n is equal to J Tau divided by Q D so that was the the n and the current current part so now let us see the internal efficiency of the device the internal efficiency is defined by neeta internal this is how I am designating is designating this so this is a ratio of radiative recombination rate divided by the total recombination rate so this is a useful light which is getting generated inside the device so that is called the internal Quantum efficiency and this can be represented as RR that is a relative recombination rate and total combination which consists of relative as well as non-reditive so it will be RR plus RNR and this is equal to d d n by DT radiative this is the rate and divided by d n by d t total including both relative as well as non-reditative so as from earlier expression d n by DT is equal to minus n by top and for relative it will be minus n by Tau r so this is the average career lifetime against relative recombination and the DN by DT will be total so this will be minus n by Tau minus n by Tau NR so this can be little simplified here and gets canceled and what you get an internal ninja internal that is efficiency internal as Tau by Tau NR so Tor is the actually bulk recombination time this is the bulk Z combination time and this is the non-reditative recombination time non-reditative time also we know D N by DT another expression which will connect Tau and Tau R and Tau NR so d n by DT total is com is addition of DN by DT radative plus d n by DT non relative so this we know minus n by tau is equal to minus n by Tau R minus n by Tau n r and this n will get canceled so we get a simple equation 1 by tau is equal to 1 by R Tau R plus 1 by Tau NR so generally what happens star R is equal to Tau NR these two times are same so if I calculate the efficiency by putting these values are tall and R in the neeta initial expression I get 0.5 or 50 percent that means 50 percent of the energy is only relatively emitted the other part is there is no radiation it is absorbed or lost in the device so internal efficiency for a typical diode this is the order of 50 percent ah let us also try to understand the relationship between the internal efficiency and the power which is generated that is p i n t or P internal so this this is an internal is defined as neeta internal is defined as relative recombination divided by total recombination and this is i i r r divided by total recombination will be I by Q and RR that is a relative is the P internal that is a power generated because of this relative recombination divided by H Nu H Nu is the photon energy divided by I by Q so the total becomes P internal into Q divided by I into H into Nu and so the P internal the power which is generated inside the device is neeta internal into I into H Nu divided by q and it can be little written in a different form so this is the efficiency part this is the Planck's constant this has units Joule second so while calculating you should be very careful with the units and C is in velocity of light meter per second Lambda is in meters and Q is a charge in coulomb and I is the current in amps and you will get the power in terms of Watts so if you are using MK system units these are the units you have to follow so let us do an example and try to get some you know idea about the typical values so if I assume a in gas phosphide LED which is emitting at say Lambda 1310 nanometer and it has uh relative recombination time as 30 and 2 and are non-derivative is 100 millisecond here it is shown different but normally these times are you know similar and the drive current which is the ie is actually is 40 milliamperes and we will try to calculate what is the bulk recombination time which is nothing but top and what is the internal efficiency of the device and how much power is generated inside the device so if you put all these values in the expression which I have discussed what you get tall as 23 nanosecond what you get efficiency as 0.77 and the power which is generated inside the device is the order of 29.2 millivolts so these are the typical values which are used in you know Optical communication using optical fiber so now let us try to understand the con external Quantum efficiency so for understanding the quantum efficiency external Quantum efficiency uh suppose I have this interface I mean this is my device and this has some interface and this is air so this is say N2 reflective index here is N2 and inside the device it is N1 this N2 can be anything but it if it is air then N2 will be 1 actually so there's the interface and this is suppose the photon is generated here and then you know there is only certain cone here and let me put this as angle as 5c so if the your angle is greater than 5c then the light gets reflected back into the medium it gets reflected back into the media which is not useful and if your angle is less than 5c the light is reflected out what I have shown here is a critical angle where the light is just grazing the interface so reflected out right so all the light or all the photons which are in this cone only will come out the other will be reflected back and will be absorbed so if I you know put mathematically this this is a solid angle where if the photon is is confined to this solid angle only then it is reflected out so this can be expressed as this is this part is from the solid angle and the limits are from 0 to 5 C and 4 Pi is a total solid angle so the external efficiency will be you know integral 0 to 5c 2 pi sine Theta into D Theta divided by the total solid angle which is 4 pi and I can write this sine Theta uh I I can you know solve this integral and limits from 0 to 5 C and also so this gives me half into 1 minus cos Phi C and we know from Snell's law that sine Phi C is equal to N2 by N1 uh the other angle is sine 90 which is one so sine Phi C can be converted into cos Phi C and put in this expression by putting this expression using cos Phi C is equal to root 1 minus n Square Phi C and doing some approximation here assuming that 5c is very small so this half can come here what we I get as an external efficiency as this so this is my an external one minus half into square by N1 Square so as the outside medium is air so N2 will be 1 and effectively I will get an external one as one by four and one square assuming N2 is equal to 1 and N1 for a typical semiconductor material is 3.6 and if I put this 3.6 value in here then n external one is 0.0193 which is roughly two percent so the external efficiency because of the light which is going back to the medium because of the total internal reflection and only part of the light is coming out of the device which is about two percent of the light which is generated inside so this part gives you two percent efficiency so there's another thing which is happening here at the interface not everything which goes here and is inside the solid angle will go out of it some some part will be reflected back right because the refractive index is changing at the boundary so even if the light is inside that cone but when it is trying to come out of the device there is some part which will be reflected back so so we we need to calculate this part also which will be reflected back so this is a reflection coefficient this tau is the reflection coefficient which is given by N2 minus n 1 divided by N2 plus N1 whole Square and we need to find out the transmission coefficient so transmission coefficient will be 1 minus Tau uh the transmission coefficient will be 1 minus top and if I put this value of door here and do some calculation what I get is four one one and two and one plus N2 square and again assuming N2 is equal to 1 I get 0.68 so this is a transmission coefficient I mean the light which is going out of the device so n external 2 earlier was n external one which we saw was 0.2 or 2 percent 0.02 rather and N external 2 is 0.68 so the total power emitted will be multiplication of these two efficiencies into the power generated inside the device so if I do this so basically it comes out to be point zero one one percent so only one percent of the light which is generated inside the device is able to come out so now we will study about the frequency response and bandwidth of the device so the optical bandwidth let us first try to understand the optical bandwidth first and then there's other another concept of electrical bandwidth for the device so we will try to understand the relationship between the optical bandwidth and the electrical bandwidth so as we know or you can find in some standard books of optical fiber communication when you are modulating your source with a light of frequency Omega so as you increase the Omega the power output decreases so which is given by this expression P Omega is equal to P naught p 0 is actually at DC power at DC divided by 1 plus Omega square into Tau Square tau is the average lifetime of the photon so P Omega is the power at frequency Omega so if we if I plot this so this is for example the my frequency or Omega and this is the power say PW so as the frequency decrease increases it follows this trend so this decreases with change in frequency so what is the the how do you define the bandwidth of the device so where the power has fallen by a factor of half so if somewhere here say this is one this is 0.5 corresponding to this Whatever frequency you get that is the bandwidth of the divides Optical bandwidth of the device so this is how we Define so this P Omega is equal to P naught plus 1 by Omega Square Tau Square so as we know this bandwidth depends on this Factor which is the bulk recombination time so for this to happen P Omega I am trying to calculate the optical bandwidth so P Omega divided by p 0 will be half when this is you know three then one plus three is equal to 4 square root of 4 is 2 so this will be half so at this point the we will have 3 DB bandwidth so this can be so w 3 DB will be root 3 by top and you can write in terms of frequency which is root 3 by 2 pi 2 pi Tau so this is the 3 DB bandwidth Optical bandwidth of the device root 3 by 2 pi by 2 pi Tau now let us try to calculate the electrical bandwidth ah as we know the current which is produced in the device to calculate the electrical bandwidth actually see this is your Optical Source Optical source and then you have the photodiode which actually converts this Optical energy to electrical energy and then there is some processing so the current which is produced in this photodiode as a result of this Optical input is proportional to the light falling onto it so there is a linear relationship so I Omega is proportional to P Omega and electrical power if I calculate will be I Square R which is this is electrical power P Omega I Square Omega r and if I take the ratio of p e Omega divided by P E naught that is electrical power at Omega and electrical power is 0 this will be ratio of I Square Omega by I Square 0 and this can be related to Optical power P Square Omega P Square 0. so as you see the I and P they are directly proportional if it is Optical bandwidth and for electrical power the this power is proportional to I Square so this becomes equal to 1 by 1 plus Omega Tau square and this will be half when Omega tau is equal to 1. so the electrical 3 EDP bandwidth is whenever this Omega T is equal to half and by doing the calculation in similar way we find out that the 3 DB Electrical bandwidth is nothing but 1 by 2 pi Tau 1 by 2 Tau so if I plot just to explain you so this is your Omega and this is how the power is falling with Rising Omega so this is the power or let me now write in terms of current because that is a you know quantity which is common there so I let me plot this as I Omega I is 0. so from the the Expressions which I have mentioned here if you take there will be two points one for the electrical bandwidth other will be for the optical bandwidth the optical bandwidth will be somewhere 0.2 and this will be for the electrical bandwidth 0.7 because this becomes 1 by 1 by root 2 so this will be 0.7 points and the corresponding bandwidth is depicted here so this is the electrical bandwidth this is the electrical bandwidth and this is the optical bandwidth so so there's a difference between the electrical bandwidth and the optical bandwidth and this is how they are related so earlier I had shown you the expression for laughs the earlier I had shown you the expression for electrical bandwidth and similarly for the optical bandwidth and this is how it is the difference is shown in this diagram so I'll stop at this point and from next class uh we will work we will discuss about the laser and we will try to calculate similar quantities for the laser also so thank you very much foreign