 In today's assignment you are going to look at the interaction of light with matter. So this is assignment 7 going to look at light-matter interaction. So during the course of the lecture we first studied how generally light interacts with semiconductor materials, we then looked at some specific examples and applications of this. So we looked at LEDs, photodiodes, lasers, solar cells and so on. So in this assignment we will focus on the general interaction and in the next one we will take up problems related to the specific devices which I just mentioned. So when we talked about light interacting with matter the basic thing we said was that the energy of the light E must be greater than some interaction energy within the semiconductor. So in most cases this interaction is essentially the band gap of the material so that when the energy of the light is greater than the band gap electrons are excited from the valence to the conduction band. You could also have situations where there are defect states or trap states located with in the band gap so these could be located either close to the valence band or to the conduction band and again the interaction of light with the semiconductor causes carriers to excite from either the valence band to the trap state or from a trap state to the conduction band. So that is why E-semi can not only refer to the band gap of the material but could also refer to the energy for a defect state and a band gap. So with this brief introduction let me go through the problems as we go through the problems we will again related to the concepts that we dealt with in the lectures. Problem one so we have a sample of gallium arsenide which is 0.35 micrometers thick it is illuminated with a light source and the energy is given so energy E is 2 electron volts this is greater than the band gap of gallium arsenide. So EG of gallium arsenide which is 1.42 electron volts. Determine the percentage of light absorbed through the sample and we want to repeat the same calculation for silicon. So when we look at light absorption through a material we basically go back to the Bier-Lambert law. The Bier-Lambert law says that the change in intensity over a small distance dx and there is a negative sign because the intensity actually goes down is directly proportional to x and the proportionality constant is called µ. µ here is the absorption coefficient of the material. This in turn depends upon the wavelength of the light that is shining through this. We can integrate this and basically put the boundary conditions which gives us I is equal to I0 exponential – µx. So I here represents the transmitted intensity so if you want to find the absorbed intensity and if you want to find it as a ratio it is nothing but 1 – I over I0. So this is essentially a ratio which we can convert to a percentage so which is 1 – exponential – µ. In this particular case we have been given the thickness of the sample so T here refers to the thickness. So we can basically use this expression to calculate the value for gallium arsenide. So for gallium arsenide the absorption coefficient µ is given. So µ is 5 times 10 to the 5 per centimeter inverse. So from this we can calculate the percentage absorbed which is just 1 – I over I0 times 100 and here we substitute all the values. This works out to be 99.9999 and there are few more trailing lines which means when you shine light of 2 electron volts on a sample of gallium arsenide there is only 0.35 micrometer stick. So 0.35 is just 350 nanometers so it is only 350 nanometers and almost all the light is absorbed so that gallium arsenide is essentially opaque to this radiation. We can do the same calculation for silicon. The value of µ is T times 10 to the 4 per centimeter. So silicon actually has a lower band gap than gallium arsenide. So EG of silicon is 1.10 electron volts at room temperature. So even though it has a lower EG the value of the absorption coefficient at 2 electron volts is slightly lower than that of gallium arsenide and this is basically related to how the density of states is distributed in the valence and the conduction band. So taking this value of µ we can calculate the percentage absorbed. You can again plug in the numbers. This is 93.92% so it is still a high number but nearly 7% of the light gets through while the rest is absorbed. So we can actually go beyond and do some more calculations just to get a feel of this value. So instead of 0.35 micrometers I now reduce my thickness to 0.1 micrometers. This is approximately 100 nanometers. My energy is still 2 electron volts so that I can use the same values for the absorption coefficient. If you do that for gallium arsenide the percentage absorbed is 1 times 100 which is 1 minus exponential minus µt times 100 and this is 99.3. So when we reduce the thickness further so we actually take it down by around 3 times you go from 350 nanometers to 100 nanometers still we get a really high percentage of light that is absorbed. For silicon on the other hand the percentage absorbed for 100 nanometers is only 55%. So only half the light is absorbed and the other half gets transmitted. Another variable we can introduce is to change the wavelength or the energy of the light. So if you take energy to be 1.5 electron volts so instead of shining light of 2 electron volts you shine light of only 1.5 electron volts. For gallium arsenide µ is lower so it is 5 times 10 to the 3 per centimeter so 5 times 10 to the 3 per centimeter. So this is still above the band gap but it is very close to the band gap so that the number of available states are small. So correspondingly the absorption coefficient is also small. Now if you have a thickness of 0.35 micrometers so the same 350 nanometers the percentage absorbed using the same formula is only 16.05. So nearly 84% of the light is transmitted while the rest is absorbed. So the absorption coefficient µ plays a really key role in determining the thickness of the sample that you need in order to either get light to completely pass through or light to be absorbed. So if you are trying to build a transparent semiconductor with gallium arsenide we find that if you have light of energy greater than 1.5 electron volts and this is already in the visible region we find that most of the light essentially gets absorbed and only a small percentage of light gets transmitted. So the value of µ and the corresponding at different wavelengths something that plays a very important role in determining the type of material you choose and also the thickness of the material. So let us now go to problem 2. In problem 2 we have a sample of a semiconductor the cross sectional area A is 1 centimeter square and the thickness is 0.1 centimeters. So we want to find out the number of electron hole pairs. So EHP is nothing but electron hole pairs. So we want to find the number of electron hole pairs that are generated per unit volume when you absorb light of 1 watt. So the power is 1 watt and the wavelength lambda is 630 nanometers. In this particular case the band gap of the semiconductor is not given but we are going to take it such that the light has sufficient energy to excite electrons across the band gap so that we can get electron hole pairs. If the excess minority lifetime is 10 microseconds so the minority lifetime tau is 10 microseconds we are also asked to calculate the steady state excess carrier concentration. So we look at the first part of the problem. So we have light of wavelength 630 nanometers so the first thing is to calculate the energy. Energy is nothing but hc over lambda this works out to be 3.16 times 10 to the minus 19 joules. So we also know the intensity of the light. So i is 1 watt which is 1 joule per second. So we can calculate the total number of photons. So this is the number of photons that are incident per second. This is equal to i divided by the energy. So i divided by hc over lambda. So we can plug in the numbers. This works out to be 3.17 times 10 to the 18 photons per second. So we also want to calculate the electron hole pairs per unit volume. So volume is nothing but a times t so 0.1 centimeter cube. So this we can divide by the volume. We also say that each photon gives rise to one electron hole pair. So this means there is a quantum efficiency of 100%. Usually that is not the case. The quantum efficiency would be lower than 100 in which case you will have to multiply by the appropriate fraction. But for this particular problem we will take the quantum efficiency to be 100%. So this is the number of photons that are incident. These will give rise to an equal number of electron hole pairs. So the number of electron hole pairs per unit volume is nothing but 3.17 times 10 to the 19 per centimeter cube per second. So we are just dividing the number of photons by the volume. So in the next part we are asked to calculate the steady state excess carrier concentration. So we do not know if this material is a p-type or an n-type semiconductor. So just for simplicity I will take it to be an n-type so that the excess carriers are holes. You can do the same thing by assuming the material to be a p-type so that the excess minority carriers are electrons but it will not affect the final result. So these are the excess minority carriers. So when we have light illuminating on a sample it is possible to write an equation for the excess carriers. This differential equation just says d delta p over dt. So delta p represents the excess minority carriers that are created. This is equal to the number of electron hole pairs that are generated minus delta p n over tau h where tau h is the minority lifetime. So when we basically have steady state delta p is 0 so d delta p over dt is 0 at steady state which implies the steady state excess carrier concentration is nothing but the number of electron hole pairs that are generated times the lifetime of the minority carriers. So this we can again substitute. We know the lifetime is given to be 10 microseconds. The number of carriers per unit volume is something we just calculated. So this is equal 3.17 times 10 to the 14 per centimeter cube. So these are the excess holes at steady state. So let us now go to problem 3. So problem 3 we have a direct band gap semiconductor with no trap states. So we do not have any defects or traps that are located within the band gap. So this is important because trap states can usually trap the carriers and these have a longer lifetime than the electrons and holes in the band. This will again change the carrier lifetime and that will again affect properties like the conductivity and also the quantum efficiency. So we have a direct band gap semiconductor. This illuminated with light of intensity so I of lambda. So in this case I is essentially a function of the wavelength. This causes photo generation. So once again you are saying that the wavelength lambda or the energy of the light is greater than the band gap so that we have electron hole pairs that are created. The area of the illumination is given so A is length times W and the thickness of the semiconductor is D. If eta is the quantum efficiency, quantum efficiency defines how many electron hole pairs or how many photons are converted to electron hole pairs. In the last problem we assume the quantum efficiency to be 1. So that every photon gets converted to an electron hole pair. We take the quantum efficiency to be 0.5 or 50%. If you have two photons coming in, only one of them will get converted to an electron hole pair. The quantum efficiency is given and tau is the recombination lifetime of the carriers. So when we shine light in a material, we generate excess electrons in holes. These electrons in holes can basically take part in conduction so that there is a change in conductivity when we expose light. This is essentially called photo conductivity and the difference in conductivity delta sigma is defined as sigma in the presence of light minus sigma in the absence of light. Typically that is a dark state and we have to show that delta sigma is equal to E eta i lambda tau mu E plus mu H over Hc D. So in some ways problem 3 is similar to problem 2 except that we are using symbols instead of numbers and in the last part of problem 3 we will actually plug in some numbers to get this expression. So once again the first thing we need to do is to calculate the number of photons or the photon flux that is incident on the sample. So phi ph is your photon flux. This is given by i is the intensity divided by Hc over lambda which is the energy. So this is the intensity and then this is the energy. So this is i lambda over Hc. So we define quantum efficiency as the number of electron hole paths that are generated for a certain photon flux. So gph which is the number of electron hole paths generated nothing but eta which is the quantum efficiency times gamma ph. So this is i lambda eta over Hc. So in this particular problem the intensity is given per second. So intensity is given per unit area. So to calculate the volume change also to calculate the number of electron hole paths that are generated per unit volume we basically need to multiply by the area and also divide by the volume. So the number of electron hole paths per second per unit volume. So again I am going to use the same symbol gph but now this is per unit volume is nothing but eta times gamma ph times a divided by the volume. So a gets cancelled we can plug in the value of gamma ph. So this is nothing but eta i lambda over Hcd. So again we have a steady state situation. So we can write the continuity equation. So if you take the excess carriers to be electrons then d delta n over dt is nothing but gph minus delta n over tau and at steady state this change is equal to 0. So delta n is gph times tau you can substitute that expression. So it is tau eta i lambda over Hcd. So this is the excess electrons that are generated this must be the same as the extra holes because electron hole paths are generated. So every time an electron is generated a hole is also generated. So the change in conductivity delta sigma is nothing but delta n e mu e plus delta p e mu H these two terms are the same and equal to this. So this you could take it out and when you rewrite you get the final expression delta sigma is e eta i lambda tau mu e plus mu H divided by Hcd. So in some ways it is very similar to the previous problem except that instead of putting in numbers we have divided a more general expression that takes into account the lifetime of the carriers and also the excess carriers that are generated. So now we try to put some numerical values to this. So now we say that the material we have is cadmium sulphide the dimensions are essentially given. So 1 millimeter by 1 millimeter by 0.1 millimeter the wavelength of the light is given. So lambda is 450 nanometers and intensity per unit area is 1 milli watt per centimeter square. So we need to calculate the number of electron hole pairs per second the quantum efficiency lambda is equal to 1. So it is the same application of the formula. So number of electron hole pairs gph per second is nothing but a eta i lambda over Hc a is the cross sectional area can substitute all the values. So this gives you 2.26 times 10 to the 13 per second if you divide this by the volume you can get the number of electron hole pairs per unit volume. In part b we need to calculate the photo conductivity. So delta sigma is essentially E eta i lambda tau mu E plus mu H divided by Hc and D. So this is the expression that we derived all the values are essentially known mu E and mu H are also given. So delta sigma works out to be 1.30 ohm inverse centimeter inverse. In part c we need to calculate the photo current delta j when you apply a potential of 50 volts. So delta j is nothing but E times delta sigma so E is 50 volts. So we can calculate delta j this works out to be 6.50 times 10 to the 4 amperes per meter square. To calculate the current so delta i is nothing but delta j times the cross sectional area a this is 6.5 milli amperes. So let us now go to problem 4. So problem 4 you have a gallium arsenide sample is illuminated with a helium neon laser beam. So you have a helium neon laser the wavelength of the laser light is 632.8 nanometers and the intensity is 50 milli watts. We want to calculate how much power is dissipated as heat in the sample due to thermalization. So once again we can calculate the energy. Energy E equal to head c over lambda is 1.96 electron volts. So this is greater than E g of gallium arsenide which is 1.42 electron volts. So what happens is that the excess energy of the photons basically gets translated into excess energy of the electrons and holes. This excess energy is lost as heat to the surrounding material and this is essentially your thermalization process. So to calculate the energy lost or the power lost to thermalization we first need to calculate the number of photons. This is again i over head c over lambda which gives you the number of photons. So excess energy which is the electrons and holes possess are essentially lost to the lattice and when the energy is lost the electron comes close to the conduction band edge or the hole goes close to the valence band edge. There is always a certain thermal energy which the electron will possess. So the final energy of the electron after thermalization is nothing but E g which is the band gap plus 3 over 2 kT. This is with respect to the top of the valence band so that the top of the valence band is taken as 0. So the energy of the electron after losing the excess energy to the lattice is just E g plus 3 half kT. So this is the initial energy of the electron, this is the final energy. So the total energy lost delta E is nothing but head c over lambda minus E g plus 3 half kT. This energy lost to do the numbers is 0.503 electron volts. So this energy is lost by every electron that is generated which is equal to the total number of photons that are incident. So the total power that is lost is nothing but gamma ph times delta E which is nothing but the incident power pl divided by head c over lambda times delta E. So this is nothing but the ratio of the incident power to the final energy or the energy that is lost which depends upon the band gap. So all these values are known pl is 50 milliwatts. So if you substitute this works out to be 12.76 milliwatts lost to thermalization. If you increase the energy of the incident right so instead of helium neon with 1.96 electron volts you have a higher energy light the value of delta E will be higher which means more amount of power will be essentially lost to thermalization. So this is important when we decide what kind of incident radiation we want in order to generate your electron hole pairs. So if there is a large mismatch between the incident energy and the band gap of the material most of the heat will essentially be lost as thermalization. If you are trying to operate this in the form of a device this heat loss can basically increase the temperature of the device and cause the device to be more inefficient. Let us now go to the last problem. Problem 5 you have a silicon sample with 10 to the 15 donors so ND is 10 to the 15 donors per centimeter cube is illuminated with light to create 10 to the 19 electron hole pairs so 10 to the 19 per centimeter cube per second. So we need to find the separation of the quasi Fermi levels talk about it in a minute and the change in conductivity upon shining the light. So these are the number of electron hole pairs that are generated. So we need to calculate the excess carriers at steady state. This we have seen before is nothing but gph times tau the value of tau is also given so tau is 10 microseconds so that this is equal to 10 to the 14 per centimeter cube. So ND is 10 to the 15 this is silicon so ni is 10 to the 10. So before shining light we can basically calculate the position of the Fermi level. So we can calculate position of ef before shining light so this is nothing but an n type semiconductor so that efn-efi is kT ln of ND over ni this works out to be 0.298 electron volts and this will be above efi. So we now shine light and the light generates excess electrons in holes. So your new electron concentration n is nothing but ND plus delta n which works out to be 1.1 times 10 to the 15 per centimeter cube. So if you have an excess concentration of electrons this will again cause a slight shift in the Fermi level this is your quasi Fermi level. So the new efn-efi is kT ln n over ni where the value of n is now 1.1. So this gives you 0.3004 the shift is very small because the increase is not that much but there is still a small shift in the Fermi level can also do a similar calculation for the holes efp-efi is minus kT ln of p over ni where p is delta p which is the excess carriers that are generated this is equal to 10 to the 14 per centimeter cube. So if you do this efp-efi is minus 0. through T8 electron volts this is basically below efi. So when we shine light on to the material we generate excess electrons in holes we can define a quasi Fermi level for these excess electrons in holes and we just did the calculation taking the n separately and taking the p separately. These do not reflect the real Fermi levels in the material because these are excess that are generated during illumination but we can treat them as n and p type semiconductors and get an idea of where the Fermi level will be located. We now want to calculate the excess conductivity delta sigma. So we find that there is only a small increase in the electron concentration but there is a drastic increase in the hole concentration. So the change in conductivity will be more or less driven by this excess hole concentration. So delta sigma is nothing but delta p e mu h. So we ignore the change in minority carriers and only look so they ignore the change in the majority carriers and only look in the minority carriers. So mu h is not known but we do know the value of dp. dp is 12 centimeter square per second and this is nothing but kt mu h over e. So from this we can calculate the value for mu h. Mu h works out to be 463.8 centimeter square per volt per second. So this we can substitute here and we can calculate the change in conductivity and this is driven by the excess holes 7.42 times 10 to the minus 3 ohm inverse centimeter inverse. So we can treat a semiconductor which has a non equilibrium concentration of electrons and holes as essentially an n and p type separately and then we can basically calculate the increase in conductivity and this increase in conductivity is mainly due to the increase in the minority carrier concentration.