 In today's assignment we are going to look at optoelectronic devices. So this is assignment 8. In assignment 7 we looked at problems dealing with general interaction of light with semiconductors. Today we are going to look specifically at some problems leading to devices. So in the classes we looked at photo detectors, solar cells, LEDs and lasers. We did a bit of solar cell trying to calculate the current voltage characteristics. So some of the assignment problems here will mostly deal with solar cells. We will also look a bit at photo conductors and also a problem on LED. We won't be focusing on lasers because the way the laser works is very similar to how an LED works except that you have a population inversion that is created and you have an incoming photon that stimulates the emission. So let us go to problem number 1. In problem 1 we have a PIN diode. So you have a P type, intrinsic and an N type. So we are asked to draw a qualitative energy band diagram both in equilibrium, forward and reverse bias. So we have a PIN diode. So we have essentially two interfaces, one between the P type and the intrinsic material. The other between the intrinsic and the N type. For simplicity we will just take this to be a homo junction. So all the three materials are the same. The only difference is in the doping concentration and the type. So I will start with the P, I and N. So we can draw the energy band diagram for all three of this. So the material is the same. So the valence band and the conduction band are essentially located at the same point. In the case of a P type material the Fermi level is close to the valence band. For an intrinsic it is close to the middle and for an N type EF is close to the conduction band. So when we have this in equilibrium we know that the Fermi levels must line up. So in equilibrium I have the Fermi levels line up. Far away from the interface your material still behaves as the same. So this is P type. This is intrinsic and this is N type and making sure that the band gaps are approximately the same and then we can join all three. So essentially this is your PIN junction in equilibrium. There are two contact potentials that are developed. One between the P and the intrinsic region and one between the intrinsic and the N type region. So we can again draw this in both forward and reverse biased. The case of a forward bias the P is connected to positive and N is connected to negative. So the carriers are injected into the device and the barriers are lower. In reverse bias it is the other way round P is connected to negative, N is connected to positive so that the barriers actually increase. So let me draw that next. So in the case of a forward bias P is connected to positive and N is connected to negative. So on this side I have a P type semiconductor at the center. I have an intrinsic and here I have an N type. If you draw this slightly more clearly. So the important point is that in the case of a forward bias the barrier is lower so that you can have a current going through the device. Now we have a reverse bias. So P is connected to negative and N is connected to positive. So here the barrier is actually increased so that now it has become more difficult for the current to flow through. So the last part of the question asked for use as a photo detector do we use the PIN junction in forward or reverse bias. So if you look at the working of a photo detector a photo detector is one where light falls onto your material so onto your device and essentially this creates electron and holes which are detected as a current and the intensity or the value of the current is directly proportional to the intensity of the light. So for this to happen we essentially want the current through the device to be very small because if there is a background current that will essentially act as noise to the current that is generated by light shining on the material. So this happens when the PIN junction is essentially connected in reverse bias so that electrons and holes that are typically created in the intrinsic region get separated. So if I have a P, I and N and I have light shining onto the material electron is generated here a hole is generated here these electrons move to the N side and the holes move to the P side because of reverse bias and this gets detected as a current. The current once again is proportional to the number of electrons and holes that are generated which is directly proportional to the intensity of the light that is falling. So for use as a photo detector the PIN junction or the PIN diode must be connected in reverse bias so let us now go to problem 2. So problem 2 essentially deals with an LED, LED is your light emitting diode that based upon the fact that you inject electrons and holes into your material a typical diode is nothing but a PN junction these electrons and holes recombine and basically give you light. The wavelength of the light depends upon the band gap of the material so by tuning the band gap of the material by adding different dopants can essentially tune the light output. So in this particular case we are asked to show that the change in the emitted wavelength lambda so d lambda over dt so the change in wavelength with respect to temperature is approximately given to be minus hc over e g square d e g over dt and then we are asked to do the calculation for a gallium arsenide system where the band gap is given and also the temperature variation is given. So the first part can actually be very easily shown so I just describe the working of an LED and said that the wavelength of the light is approximately equal to the band depends upon the band gap of the material. So lambda this is the wavelength of the light is equal to hc over e where e is the energy of the photon this in turn depends upon the energy of the electron and hole that are involved in recombination. This electron and hole is usually very close to the valence band edge and the conduction band edge so that this can be written approximately as hc divided by e g. So usually the electron or the hole is not located exactly at the band edge but there is some thermal energy which is typically of the order of k t but we can ignore the thermal energy and say that lambda is approximately hc over e g. If you look at this expression h and c are essentially constants the only variable is e g. So if you differentiate this we get d lambda over dt is nothing but minus hc over e g square since you are differentiating 1 over e g times d e g by dt. So we are asked to do this calculation for gallium arsenide gallium arsenide has an e g of 1.42 electron volts this typically lies in the infrared region and the value of d e g or d t is also given 4.5 times 10 to the minus 4 electron volts per kelvin. So we can calculate the change in wavelength with respect to temperature. So d lambda over dt hc square over e g times d e g over dt. Here we can substitute all the values and this gives you a value 0.277 nanometer per kelvin. So if you look at it d e g over dt with respect to temperature is negative which means the band gap essentially decreases with rise in temperature corresponding to a decrease in band gap the wavelength will increase because lambda is inversely proportional to energy. So d lambda with respect to temperature is a positive quantity. So the question also says your delta t is nothing but 10 degrees. So in kelvin this is the same 10 kelvin. So the corresponding change in the wavelength delta lambda is nothing but 2.77 nanometers. So all we are doing is substituting delta t here and then just multiplying. So let us now go to problem 3. So problem 3 relates to a solar cell. So we have a solar cell at room temperature and it is under illumination so the intensity of the light is given. So i is 500 watts per meter square. The short circuit current is also given. So isc is 150 milliamps and also the open circuit voltage voc is 0.53 volts. So the question asks what is the short circuit current and also the open circuit voltage when we double the value of the illumination. So illumination goes from 500 to 1000 watt or 1 kilowatt per meter square. Similarly what are the values when the illumination is essentially halved. So the short circuit current is essentially the current when this voltage is equal to 0 and isc is proportional to the intensity of the light that is falling. So typically it is equal to some constant times iph. So iph if you write the subscript here iph is nothing but the intensity of the photons that is falling onto your solar cell. So isc is essentially some constant k times iph voc which is your open circuit voltage. So this is the voltage when the current through the system is essentially 0 is equal to kt over e ln of iph by is0 where is0 is your reverse saturation current. So this depends upon the material of the solar cell and also the kind of dopants the concentration of the dopants the width of the depletion regions and so on. So we have simple expressions that relate both isc and voc. So the short circuit current and the open circuit voltage to your incident photon radiation. So when you change the radiation intensity you can basically calculate the change in these values. So now the new value of iph is doubled. So let me write this as iph prime. So it is 2 times the original value 1000 watts per meter square. So the new short circuit current so isc nu so I am going to write it as isc prime divided by the original isc it is nothing but iph prime by iph which is equal to 2. Therefore the new short circuit current is essentially the double of the original short circuit current. So this is nothing but 300 milli amperes. Here I can write the value of voc so voc prime by voc again you are taking the ratio of these two quantities kt over e ln of iph prime by is not iph by is not. So kt over e will essentially cancel this we can simplify because the values of iph prime iph are known is not is not known but we can calculate is not by knowing the open circuit voltage when the current intensity is 500 watts per meter square. So from this we can calculate the value of voc. So voc prime is essentially 0.55 volts so we now have a situation where instead of doubling the photon intensity we are reducing it by half. So iph is nothing but 250 watts per meter square so we have reduced it by half so the new short circuit current isc double prime will also be half so this will be 75 milli amps and the new voc so let me write this as double prime is 0.48 volts. So this we can get by following the same procedure where instead of having it as 1000 we now have it as 250. Let me now go to problem 4. Problem 4 is also related to a solar cell so we have a solar cell of area A is given and is connected to drive a load R of 20 ohms so R is essentially 20 ohms. So we can actually draw how the solar cell connection looks so this is given as part of the question so you have the solar cells essentially connected to a resistor so R is the resistor so light falls on the solar cell so you have some light I which drives a current I. So this in turn generates a voltage V that opposes the inbuilt voltage in the solar cell so the intensity of the light is given so that is 1 kilowatt per meter square and we want to calculate the voltage in the circuit when the intensity is 1 kilowatt per meter square the corresponding value of current for that is given. So in the question along with how the solar cell is connected we are also given its IV characteristics so let me plot this this IV characteristic is for 500 watts per meter square so we have current I that is in milli amps and we have voltage V this is 0 we have 10 we have 20 this is 0.2 0.4 let me just extend this a bit and you have 0.6 so the IV characteristics so this is the IV characteristics of the solar cell so this is drawn when the illumination is 500 watts per meter square so we are asked to calculate the current for a particular voltage or the voltage for a particular current when the illumination is now changed to 1000 watts per meter square so we can go to the problem let us go to part A so when a solar cell is essentially driving an external resistor R there is some current that is flowing through so this current at 500 watts per meter square illumination I am going to call this as I 500 so I 500 is given by the intensity of the light plus a voltage due to the potential barrier that is created by the resistor this is given as I0 exponential EV over kT minus 1 so this is a general expression that relates the current at a particular intensity to both the open circuit current the short circuit current and also the voltage when you have some external load are sitting on the system so from the graph we can essentially mark the values of these various points so when V is equal to 0 this is a short circuit voltage the corresponding current I is minus 16.2 milliamps this is nothing but I 500 comma pH and just for simplicity I will write this as I pH so this is the short circuit current you can also calculate or you can also look at the open circuit voltage so that when the current is 0 the voltage is exactly 0.49 volts this is the open circuit voltage the question also says that the device at 500 milliamps or 500 watts per meter square illumination is essentially operating at point P and for this we can calculate the current and the voltage so V is 0.45 volts and the corresponding current I is minus 13.1 milliamps so this represents the operating point for the circuit so we now have to use these to calculate the values when your solar cell is illuminated with 1 kilowatt per meter square so once again let me write the expression so I is equal to minus I pH plus I naught exponential EV over kT minus 1 so I naught we do not know but I pH we know so when voltage V is equal to VOC which implies current is equal to 0 this is minus I pH plus I naught exponential EV OC over kT minus 1 so in this I pH is known VOC is known the only unknown is essentially I naught so we can make the substitutions and I naught is calculated to be 2.58 times 10 to the minus 11 amperes so this represents the reverse saturation current in the system so now the new illumination I nu so let me call it I pH nu so nu illumination is 1000 watts per meter square so in the previous problem we saw that when the illumination is doubled the short circuit current is essentially doubled so I pH nu is 2 times I pH at 500 so this is minus 32.4 milliamps so the corresponding current is also given so I of 1000 is nothing but I pH nu plus I naught exponential EV over kT minus 1 so this current is given and this is equal to 0.024 amperes so I pH nu we have just calculated I naught is something we calculated from the data for 500 watts per meter square so this is known this is known the only unknown is the new operating voltage so we can make the substitutions so that the new voltage V nu is equal to 0.475 volts we can also calculate the power in the system so power is nothing but V prime times the current so the current is I 1000 this is equal to 0.011 watts we are also asked to calculate the efficiency of the system so the input radiation is 1000 watts so P in is 1000 watts per meter square and the area of the solar cell is 1 centimeter square so that is 10 to the minus 4 meter square so this gives you 0.1 watt so 0.1 watt is the input power the output power in the form of electrical current is 0.011 watt so the efficiency in terms of percentage 0.011 divided by 0.1 times 100 that is equal to 11% so let us now go to part B so in part B we are asked to calculate the load so to have a maximum power transfer from the solar cell to the load so once again the illumination I is 1 kilowatt per meter square or 1000 watts per meter square so in problem A we essentially looked at the load for a given value of current and voltage we now have to find out the load at which the power transfer is essentially maximum so power P is nothing but I times V which can be written as V times minus I of 1000 so we have I 1000 plus I0 exponential EV over kT minus 1 so this is the expression for the power you look at it everything here is a constant except for the voltage so to have maximum power so at maximum power dp over dv must be essentially 0 so we can differentiate this expression for P so dp over dv and that is given by I0 kT by E V times exponential EV over kT minus I 1000 plus I0 exponential EV over kT minus 1 this is equal to 0 so this you can get by essentially treating this as the differential of two functions so you differentiate one keeping the other constant then you differentiate the other keeping the first one constant this is now equated to 0 for maximum power so in this case I0 is known I 1000 is known this is the short circuit current at 1000 which is two times the short circuit at 500 and we just calculated it so the only unknown is essentially the voltage so if you solve this this gives the voltage Vm this is the voltage at maximum power and this is equal to 0.436 volts the corresponding current I m is 0.031 milli amps and the resistor R is Vm over I m which is 14.07 so this is essentially the load at which the power is maximum we can also do this for the illumination at 500 watts per meter square so if I is 500 watts per meter square we can repeat this I will just write the answer I m is 0.015 amps Vm is 0.42 volts and the load Rm is essentially 28 ohms so the last part of the question part C so we need to connect solar cells in order to drive a calculator the calculator is to be used at a light intensity of 500 watts per meter square so when we have solar cells each having a particular voltage and they should essentially be connected in order to give the voltage of 3 volts these solar cells should be connected in series so the solar cells should be connected in series so if you look at the initial operating condition that was specified in part a the voltage for each solar cell is 0.48 volts so voltage per cell is 0.48 volts we basically want a total voltage of 3 volts so the number of cells is essentially 3 divided by 0.48 and you round off to the highest integer so this gives you 7 so we need 7 solar cells all connected in series to basically supply enough voltage to run a calculator let us now go to the last problem so problem 5 is related to a photo conductor so we saw some variation of this problem in as the previous assignment assignment 7 so you have a photo conductor that is placed under uniform illumination it says the absorption of light basically causes an increase in current the increase is given when you have a particular voltage then the radiation is cut off so that the excess carriers start to recombine and when that happens the carrier concentration drops and the current also drops so in this particular case we are going to take electrons to be the minority carriers so the increase in current is always due to the increase in the minority carriers this is especially true when we have weak illumination where we assume that the illumination intensity is smaller compared to the concentration of the majority carriers so electrons are the minority carriers and we have also given the electron mobility so mu e to be 3600 centimeter square per volt per second so the first part of the question we need to calculate the equilibrium density of the electron hole pairs generated under radiation so we can write an expression for that so the current I which is your photo current is due to the excess electrons that are created so this is equal to delta n this is the excess electrons times the mobility times the electric charge e of the electron times an electric field e that is applied across the material and since it is the current we have to multiply by the area so this expression is got from the original expression for conductivity conductivity is n e mu e plus p e mu h conductivity sigma is related to the resistivity as 1 over rho and r is nothing but rho l over a so this is 1 over sigma l over a so this is the expression for the current due to the increase in the electron concentration in this particular case ip is given so this value is 2.83 milliamps the electron mobility is given so this should actually be mu e so electron mobility is given the dimensions of the device are given e is the electric field this is nothing but the voltage v divided by the length l so the only unknown in this expression is delta n so delta n is 1.47 times 10 to the 13 per centimeter cube so this is the excess electron whole pair concentration that is created when light is shining part b we want to calculate the minority carrier lifetime so when the light is switched off the concentration of the minority carriers essentially decrease and this is given by the value at a particular time t divided by the minority carrier lifetime tau so since we are dealing with electrons I will call this tau e so you can write this in terms of current and if you do this dip over dt is nothing but e mu e v over l times wd d delta n over dt d delta n over dt is nothing but delta n at time 0 divided by tau e so delta n at time 0 is nothing but what we calculated in part a dip over dt which is the rate at which the current falls is given in the question this value is 23.6 ampere per second so this is known this is known the only unknown is tau e so tau e is essentially 119.6 microseconds part c we want to calculate the excess density of electrons and holes 1 millisecond after the radiation is turned off so part c you want to calculate delta n 1 millisecond later so delta n at time t is nothing but delta n at time t equal to 0 exponential minus t over tau. So t is 1 millisecond delta n at time t equal to 0 is what we calculated in part a so is the excess equilibrium concentration tau is what we calculated in part b so delta n is nothing but 3.44 times 10 to the 9 centimeter cube once again we are assuming a condition of weak illumination so that any change in conductivity is driven by the minority carriers so that in this particular case the change in concentration of the majority carriers which are holes is not significant enough to affect the conductivity.