 In today's assignment, we are going to look at transistors. So this is assignment number 6. We are going to look at numerical problems related to transistors. During the course of the lecture, we saw the theory behind the transistor action. We saw that there were 3 main kinds of transistors that we discussed. One was the bipolar junction transistor or BJT. This is essentially a current control device. It had an emitter, base and a collector and transistor action occurred when the minority carriers were injected through the base and then they moved on to the collector. The next kind of transistor that we saw was the junction field effect transistor. In this particular type of transistor, there was an existing channel for carrier conduction. So this could be a channel for electrons which will be an N channel or a channel for holes which will be a P channel and basically the width of the channel and hence the amount of current that could flow through it was controlled by an external field. This is why it was called a field effect transistor. And then finally, we discussed the MOSFET which was the metal oxide semiconductor field effect transistor. In this particular case, the channel did not exist originally in the device but was created by applying an electric field so that the channel was formed. We looked at the MOSFET slightly in more detail than the other two type of transistors and we did some numerical calculations for the MOSFET as part of the course. So in today's assignment, we are going to look at some of the numerical problems related to all three types of transistors. We have already seen problems related to junctions. So simple PN junctions and we can actually treat transistors as essentially having two PN junctions. So some of the concepts that we learned when we looked at PN junctions, we can apply here. So let me now go to problem one. So in problem one, we have a PNP BJT. So we have a PNP bipolar junction transistor. So let me just draw the schematic of that. A transistor has essentially three regions. An emitter region, a base and a collector. The case of a PNP transistor, you have the emitter and the collector to be P type. The base is N type. So the concentrations of the different regions are given. So the emitter region has a concentration of 2 times 10 to the 18 per centimeter cube. The base is 10 to the 16 per centimeter cube and the collector is also 10 to the 16 per centimeter cube. So there are different configurations in which a transistor is connected. So here we are going to look at a connection where we have a common base mode. In this particular case, the emitter base junction is forward biased E, V, E, B and the collector base junction is reverse biased. So in this particular case, holes are injected from the emitter to the base so that there is an emitter current IE. These holes are essentially minority carriers in the base. They will recombine with some of the electrons in the base. But most of the holes will essentially move from the base to the collector and that forms your collector current. Some of the holes which move into the base and get recombined are essentially replenished by the external circuit and that forms the base current. So this is just a simple schematic of how a PNP or how a bipolar junction transistor works in your common base mode. So some of the numbers that are given here says that the hole drift mobility, so mu H in the base is 400 centimeter square per volt per second. So this is in the base. So this is the mobility of the minority carriers in the base. The width of the neutral base region is also given. So the width W of the base region is 2 micrometers and the device cross section is also given. So A is 0.02 millimeter square. The whole lifetime in the base is 400 nanoseconds. That is tau H is 400 nanoseconds. We want to calculate the common base mode transfer ratio alpha. It means the amount of current that goes from the emitter to the collector and also the current gain beta. We also want to know what the emitter base voltage is if the emitter current is 1 milliampere. So we look at the mobility of the holes in the base. From the mobility we can essentially calculate the diffusion coefficient. So dH is kT mu H over E. So this we have seen before in the context of both of extrinsic semiconductors. So dH comes out to be 10.36 centimeter square per second. So once we know the diffusion coefficient, we also know the width of the base. So the width is 2 micrometers from which we can calculate the minority transit time or the time it takes for the electrons to go from the base, for the holes to go through the base and to go from the emitter to the collector. So this is assuming simple two-dimensional diffusion. So this width is equal to square root of 2 times the diffusion coefficient dH times the transit time. So this is the minority carrier. So the width is known. The value of dH we just calculated. So you only need the value of T tau. So the transit time is essentially 1.93 nanoseconds. So within 1.93 nanoseconds, your holes essentially sweep through the base and move from the emitter to the collector. Some of these holes will recombine in the base along with the electrons and this recombination depends upon the transit time and it also depends upon the whole lifetime. So based upon this, we can define a base transport factor. This is also the common base current gain or alpha. Another name for this is the common base. This is alpha and alpha is nothing but 1 minus tau T over tau H. So all the values are given. T tau T is 1.9 nanoseconds. Tau S is 400 nanoseconds. So that alpha comes out to be 0.99517. So higher the value of alpha, then greater is the current transfer from the emitter to the collector. So one way to do that is to reduce the transit time. This you can do by making the base thinner. Another way to do that is to have a higher value of tau H, which is the higher value of your minority carrier lifetime. This again you can do by reducing the doping concentration in the base. So alpha is 0.99517. We also want to calculate the current gain that is beta and this is again given by a formula. Beta is defined as alpha over 1 minus alpha. So we can substitute in the values so that beta comes out to be 206.2. So again higher the value of alpha, we will essentially find that the current gain is also higher. In the last part of the question, we have to calculate the emitter base voltage given that the emitter current is 1 milli amperes. So IE is 1 milli amperes. So if you look in the case of an emitter current, your emitter current is generated because you now have a PN junction between the emitter and the base and you basically have minority carriers that are injected across the junction. So IE is nothing but the current that is generated when a PN junction is essentially forward biased. So IE is IE0 exponential EV over kT. So that your emitter current is due to diffusion of minority carriers. In this particular case, these are holes diffusing into the base. So IE0, so if you look at this expression, this is very similar to the expression that we got for a PN junction under forward bias. In that particular case, I was I0 exponential EV over kT where is a minus 1 term but usually that can be neglected. So in this particular case, IE0 is equal to Ni square E times the cross sectional area A times DH over ND times WB. So here ND which is the concentration of electrons in the base. It is much smaller than NA which is the concentration of holes in the emitter. So since it is 1 over ND, this particular term will dominate. So all of the values here are known. WB is the width of the base which is 2 micrometers. So that this is equal to 1.66 times 10 to the minus 14 amperes. So we have the cross sectional area A here. So that the final answer is in amperes. If we did not have A, this will be a current density. So typically amperes per centimeter square or amperes per meter square. So IEB is known. So IE is given to be 1 milliampere which is equal to IE0 exponential EVEB which is the emitter base voltage divided by kT. So all the terms here are known except VEB. So you can rearrange and then VEB is 0.64 volts. So we essentially looked at a PNP bipolar junction in forward bias. We tried to calculate the current gain and also the transfer ratio and how you can calculate the emitter base voltage for a given value of the current. So we essentially treat a bipolar junction transistor as made up of two kinds of PN junctions. One forward bias, the other reverse bias and then we go through these calculations. So this will be more clear when we look at problem 2. So in problem 2, we have an idealized silicon NPN bipolar junction transistor. So you have a silicon NPN. So problem 1 was a PNP bipolar junction transistor. Now we have an NPN. The cross sectional area is given. So A is 10 to the 4 micrometer square. So micrometers is nothing but 10 to the minus 3 millimetre square. So another way of writing is just 0.01 millimetre square. The base emitter forward bias voltage, so VEB is given to be 0.6 volts and the reverse bias base collector voltage is 18 volts. So once again this is in a common base mode, so CB mode. We can go ahead and draw the schematic for this bipolar junction transistor as well. So we have three regions, an emitter, a base and a collector. This is an NPN. So we have N, P and N. The concentrations are also given. So 2 times 10 to the 18 per centimeter cube. This is a majority carrier concentration. Then you have 10 to the 16 per centimeter cube. This is also 10 to the 16 per centimeter cube. So this is in your common base mode. So that this is VEB. This is VBC. You can compare this with problem 1. You will essentially see that the polarities are reversed because now we have an NPN. But once again your emitter base junction is forward bias. So that electrons from the emitter basically move to the base. This gives you your emitter current IE. These electrons are minority carriers. Some of them will recombine with the holes in the base. But most of the electrons essentially move through the base and go to the collector. This gives you the collector current IC. So those electrons that essentially recombine in the base have to be replenished so that you also have a base current IB. So if you look at it, we have one PN junction that is forward biased. This is your forward biased PN junction. You have another PN junction that is reverse biased. We can also mark the depletion regions. So for the emitter base junction, the base is lightly doped while the emitter is heavily doped. So most of the depletion region will essentially lie in the base. The other hand for the base collector junction, we will have the depletion region both in the base and the collector. And since the concentrations are the same, the depletion regions are essentially of the same width. So the total width of the base is given. So the total width of the base is 4 micrometers. So this includes the neutral region. At the same time, it also includes the depletion region. So this is the neutral plus the two depletion widths in the base. So the first part of the question, we want to calculate the depletion layer width between the collector base and the emitter base. We also want to calculate the width of the neutral base region. So in part A, we want to calculate the two depletion region widths. And we also want to calculate the width of the neutral base region. So let us look at the emitter base EB. So here, the depletion width is fully in the base and this is because ND in the emitter is much greater than NA in the base. So then we can use the formula. So this is the emitter base region. For the collector base region, the depletion layer or the depletion layer or depletion width is on both sides. This is because NA in the base is equal to ND in the collector side. So let us look at the base and the collector region. This is essentially a region that is in reverse bias. So reverse bias tends to increase the depletion width. So that WBC is given by 2 epsilon naught R NA plus ND times the total voltage VR divided by E NA ND and this is the square root of the whole thing. So this is the same formula that we have used when calculating the depletion width of a PN junction. We are just using that formula. The only difference here is that VR is not the contact potential, but it is the contact potential plus the externally applied reverse bias voltage. So VR is nothing but V naught which is your contact potential plus the reverse bias voltage. So this value is given in this particular problem VBC is equal to 18 volts so that this number is usually much higher than V naught. So this I can approximate as VBC. So we can plug this in the formula. All the numbers are essentially known from which we can calculate WBC to be equal to 2.18 micrometers. So WBC let me just write it here is the total width of the depletion region between the base and the collector and this is equally shared between the base region and the collector region. Therefore WBC on the P side which is your base is equal to WBC on the N side which is your collector is equal to half of 2.18 or 1.09 micrometers. So this represents the depletion width on the base side for the base collector junction. We can use a similar argument to calculate the depletion width for the emitter base region. So for the emitter base region so for the EB side we can calculate the width. So once again W is 2 times epsilon naught epsilon r NA plus ND times VR over E NA ND in the whole square root. For the base and the collector region we said that it is reverse biased so that VR is nothing but the reverse biased potential plus the built-in potential V naught. Now we have VR to be equal to the built-in potential V naught minus V EB and V EB is given as 0.6 volts. The built-in potential we can calculate by simply taking this to be a PN junction so that V naught is kT over E ln of NA ND over NI square. This works out to be 0.830 volts. So in this particular case ND is much greater than NA so that W can be approximated as 2 epsilon naught epsilon r VR divided by NA E whole to the half. So everything else we know VR is V naught minus V BE. NA is known so we can calculate W between the emitter and the base region and this mostly lies in the base and this comes out to be 0.174 micrometers. So this is mostly in the base region. The total base width is 4 micrometers. The total depletion layer width, total depletion width is nothing but 0.174 plus 1.09 so 1 on the emitter and base side, 1 on the base and the collector side. Therefore the total neutral width is this minus this which comes out to be 2.74 micrometers. So we can essentially calculate the total width of the depletion regions and also the neutral width on the base side. Let us now go to part B. In part B we want to calculate the values of alpha and beta so the current injection ratio and also the current efficiency beta. We are going to take unity emitter injection efficiency so that is something we looked at in problem 1. So we will follow a similar argument. So electrons are the minority carriers. So mu E is 1250 centimeter square volts per second. So the first thing is to calculate de kt mu E over E. This is 3.2 times 10 to the minus 3 meter square per second. You can also write this in centimeter square from which you could calculate a diffusion length de tau E. This is equal to 36 micrometers. So from this we can also calculate the transit time. The transit time is the time it takes for the electron to move through the neutral base region. So tau t and we saw the expression last time. It is nothing but WB square over 2 DE. So last time we wrote it as WB equal to square root of 2D times tau just writing it the other way. So this is the transit time and this has a value 1.161 nanoseconds. Alpha is then 1 minus tau t over tau E. So tau t is the transit time. Tau E is the lifetime of the electron in the minority in the base region. So this is nothing but 1.161 nanoseconds by 400 nanoseconds is equal to 0.9971. So beta which is your current gain is nothing but alpha over 1 minus alpha which is equal to 343. In part C we want to calculate the emitter, collector and base currents. So we want to know the values of IE, IB which is your base current and then IC. So once again in the case of an emitter region you have ND to be 2 times 10 to the 28 per centimeter cube that is your doping concentration. Mu H which is the mobility of the holes is 100 centimeter square per volts per second. So the first thing to do is to calculate DH. So DH is KT over E mu H is equal to 2.59 times 10 to the minus 4 meter square per second. So we can calculate the length and this length works out to be 1.61 micrometers. So this is much smaller than the emitter width. So you can basically treat it as a simple PN junction that is essentially in forward bias. So that IE which is the emitter current is IE due to the electron flow. So the electron flow is due to recombination. So the electron flow is due to the carrier injection plus IE which is the contribution to the hole which is because of recombination. So IE which is the electrons that are diffusing from the emitter to the base is just given by I S0 for the electrons exponential E V EB over KT. So this is similar to the equation for a PN junction in forward bias. I S0 is nothing but ni square E over A DE over NA WB. So again all the numbers are essentially known. So we can calculate the value for IE. So IE the hole component is basically the holes that are diffusing into the emitter. So we can write a similar expression I S0 H exponential E V EB over KT. In this particular case I S0 H will have a similar value except this will have ND and will have DH. So we can take these two components and add them together so that your IE which is the emitter current I S0 E plus I S0 H exponential E VB over KT. So we can substitute all the values all the different components are essentially known. So I will not do the substitution here but just write the final answer. So the emitter current is 0.513 milliamps. So let me just write that down here. So I will separate a small section make it easier. So IE is 0.513 milliamps. So we want to also calculate the collector current that is given by alpha. So I C is nothing but alpha times IE. Your alpha is the value of 0.9971. So that this is 0.5 115 milliamps. So almost all the emitter current is nearly transferred to the collector. Small portion of the current is essentially lost in the base due to recombination. So that I B is nothing but I C over your current gain beta which is 1.49 micrometers. So by again just treating your bipolar junction transistor as a series of two PN junctions one in forward and one in reverse can essentially go ahead and calculate all the current and the voltage parameters. Let me now move to problem three. In problem three you essentially have a junction field effect transistor. So you have an N channel JFET. So let me draw the schematic. So you have a source and you have a drain. So this is the drain that is my source. This is the gate. The total width of the channel is essentially A. So this is A and you also have a certain width of the depletion region. So that this is W. So the channel depth is 2A. So let me just re-draw this. So the channel width goes from here to here and that is A. So the total channel width is 2A. So half the channel width is A and the depletion width is W. So we want to calculate basically the potential at which pinch of occurs and how this is related to the built-in potential and also the doping concentrations. So basically in the case of a JFET you have a heavily doped P plus region when you have an N channel. So that the width or the depletion width is almost entirely in the channel. So you have a P plus N junction. So that the depletion width is in the channel. This again depends upon the reverse bias. So that W is nothing but 2 epsilon R V0 minus VGS over E times ND whole to the half. So VGS is actually a negative number so that when you do V0 minus VGS you are essentially adding. This is basically a case of a reverse bias channel. So pinch of occurs when this depletion width starts to increase and essentially reaches the center of the channel. So we have pinch of when W is equal to A. So that A is nothing but 2 epsilon naught epsilon R can rest write it as epsilon times V0 minus VGS divided by E and D whole to the half. So this we can basically rearrange so that VP which is minus VGS is your pinch of voltage. So all I am doing is taking this expression and rearranging is equal to A square E ND by 2 epsilon minus V naught where epsilon is nothing but epsilon naught and epsilon R. So we can do a simple calculation that relates your pinch of voltage to both the width of the channel and also the concentration of the dopants in the end channel. So for this particular numerical problem NA and ND are given. So we can essentially calculate V naught. V naught is KT over E ln of NA ND over Ni square. So the material is silicon. So V naught is 0.8936 volts. We need to calculate the pinch of voltage A is 1 micrometer. So everything else is known. V naught is known A is known from which we can calculate VP to be equal to 6.71 volts. So this represents the pinch of voltage to essentially close the end channel and basically stop conduction in your JFET. So let us now go to the last problem. So problem 4, you have an NPN silicon MOSFET. Now you have a MOSFET. So we are trying to create an end channel in a p type material NA is 10 to the 18 per centimeter cube. We will again take the material to be silicon. So we can use all the silicon parameters. So part A, we need to determine the position of the Fermi level. So Fermi level position we can just calculate. It is a p type dope material. So E of p minus E of i is minus KT ln of NA over Ni. So for silicon Ni is 10 to the 10. NA is known. So that E of p minus E of i is essentially minus 0.48 electron volts. So the Fermi level is located 0.48 volts below the intrinsic Fermi level. Next is part B. We need to calculate the applied voltage to achieve strong inversion. We also want to calculate the width of the depletion region and the width of the end channel at strong inversion. So strong inversion is defined as basically the voltage at which the channel is as much N type as it was originally p type. So this occurs when the Fermi energy is 0.48 electron volts above the intrinsic level. Since originally it was 0.48 electron volts below the intrinsic level. So Phi S for strong inversion is essentially 2 times Phi B. Phi B is the bulk potential which is the difference in the Fermi levels in the bulk. So Phi S is 0.96 volts. This width is essentially related to the depletion width and it is given by the formula Phi S is equal to E times NA WD square by 2 epsilon not epsilon R. So Phi S is a surface potential that is equal to 2 times Phi B. NA is known. Everything else is known. The only thing we do not know is WD which is the depletion width at strong inversion. So you can substitute all the values and essentially calculate WD. WD works out to be 50.3 nanometers. So the next thing you want to calculate is the width of the end channel. To know the width of the end channel we need to know how the potential varies as we go from the surface to the depth. So in this particular case this is given by the expression Phi of X is equal to Phi S. So it is your surface potential minus 1 over X by WD the whole square. So WD is the total width of the depletion region and Psi of X is the potential at some depth X from the surface. So we define the end channel to be in a region where the potential goes from 2 times Phi B which is the surface potential to 1 times Psi B. So that now you have a channel which has a higher value of electrons than holes and if you go deeper you have a higher concentration of holes than electrons. So we basically calculate X when Phi of S is equal to Phi B. So we can use expression Phi B the surface potential is 2 times Phi B. This is something we have seen in class 1 minus X over WD the whole square. So WD is known. So we can calculate the value of X and X is essentially 14.73 nanometers. So this is the width of the end channel. In the last part C we need to plot the energy bands as a function of distance starting from the bulk and then moving on to the surface. Let me just draw the section corresponding to part D. This is the surface. At the surface you have an n type material. So EC, EV, E of N. The Fermi levels should essentially be constant. So within the bulk where you have a P type material this is essentially E of P. This is EV, this is EC. You can also draw E of I. E of I should be at the center E of I. So this represents how your band diagram changes as you go from the bulk to the center, from the bulk to the surface. Part C is something that I missed. So part C wants to calculate the depletion width when Phi is 0.5 volts. So once again you have to just use the formula. So Phi S is equal to ENA WD square by 2 epsilon naught epsilon r. So all the numbers are essentially known. You only want to calculate WD. So WD is 26.2 nanometers.