 Today's assignment we are going to look at p n junctions. This is assignment 5. In assignment 4 we looked at metal semiconductor junctions. In assignment 5 we are going to look at junctions form between p and n type. So usually the p and n are of the same material which case it is a homo junction. We have also seen hetero junctions where the junction is formed between 2 different materials. In this assignment we will focus fully on the homo junctions. We will do some calculations on the built-in potential when a junction is formed, the depletion widths and the total that is the total depletion width and also the depletion width on the p and the n side. A p n junction is essentially a diode. It is a rectifier so that it conducts current in the forward bias and does not conduct in the reverse bias. So we will also do some calculations of the forward bias current and the reverse saturation current. So some of this will be similar to what we did in assignment 4 where we looked at a short key junction which is also a rectifier. So later we can compare the properties of a p n junction and that of a short key junction. So let me go to problem number 1. We have a silicon p n junction which has an n region with 10 to the 17 donors per centimeter cube, n region so n d is 10 to the 17 per centimeter cube and there is a p region with acceptor concentration of 2 times 10 to the 17 per centimeter cube, n a. The material here is silicon and it is at room temperature. So temperature T is 300 Kelvin. The intrinsic carrier concentration we have seen this so many times in the past is just 10 to the 10 per centimeter cube. So in part a we want to calculate the built in potential of this junction. So V naught is the built in potential. So it is the potential when the junction is an equilibrium and this forms because we have electrons from the n side moving into the p side. This is a diffusion current. We have holes from the p side moving to the n side. These essentially meet each other and annihilate so that you have a depletion region. So on one side of a depletion region you have a net positive charge. This is the n side. On the other side you have a net negative charge that is your p side and there is a junction potential that develops. So this built in potential is nothing but k t over e ln of n a n d over n i square. So this is just a direct substitution of the numbers n a and n d are given. n i square is also given. If n i is not known can always calculate n i from the band gap and the effective density of states or the effective mass of the electrons and holes. So you can just plug in the numbers and the answer is 0.852 volts. In part 2 we want to calculate the total depletion width. So let me call the W that is the total depletion width. So the depletion width again forms because you have electrons and holes moving across the junction and are recombining. So we have seen this concept of a depletion width earlier when we looked at a short key diode or a short key junction. In that case the depletion width is almost entirely on the semiconductor side. So here the depletion width will be in both the p and the n side. The total depletion width is again just given by a direct formula substitution. So 2 epsilon naught epsilon r n a plus n d times V naught where V naught is your built in potential by e n a and n d and the hole to the one half. So epsilon naught is the permittivity of free space. Epsilon r is the permittivity of silicon. The relative permittivity and epsilon r is 11 point time. So that is a known value for silicon. So once again everything here is known. We just calculated the built in potential V naught. So we can do the substitution and this works out to be 1.3 times 10 to the minus 7 meters or if you want to write in nanometers 130 nanometers. In part C we want to calculate the depletion width on the p and the n side. So the ratio of the depletion widths W p over W n is inversely proportional to the concentration of your dopants. So this is equal to n d over n a. Another way of writing this of course is that W p n a is W d n d and this comes from the charge neutrality. So that the total positive charge due to your positively charged donors on the n side must be equal to the total negative charge due to the negatively charged accepted ions on the p side and those to essentially balance. We also know the total width W is just W p plus W n and total width W has been calculated to be 130 nanometers. So we have all the numbers. It is again a case of doing the substitution and the math. So I will just write down the final values. So W p is 43 nanometers and W n is 87 nanometers. So the total width comes out to be 130 nanometers. The acceptor concentration on the p side is higher. So 2 times 10 to the 17. So the depletion width on the p side is smaller. So let us now go to problem 2. Problem 2, we have a p n junction diode with a concentration of 10 to the 16 acceptor atoms on the p side and 10 to the 17 donor atoms on the n side. So once again you have p side and the n side. So n a is 10 to the 16 and n d is 10 to the 17 per centimeter cube. So we need to know the built-in potential if the material of the semiconductor is different. So in this particular case, we want the built-in potential for the semiconductor materials, germanium, silicon and gallium arsenide. So if you go back to the formula for the built-in potential, skt over e ln of n a n d over ni square. So if you change the material and if you keep the dopant concentrations the same, the temperature is also same. It is typically room temperature. The only thing we are changing is ni. We have seen earlier that ni depends upon the band gap square root of n c n v exponential minus eg over 2 kt. So as the value of the band gap increases, ni essentially goes down because it has an exponential with a negative term. If the value of ni goes down, then the built-in potential will essentially increase. So in this case, we have 3 materials. So I will write down the table that is given in the problem. So we have germanium, silicon, gallium, arsenide. The band gap values are given in eg and these are mainly used just for comparison. We do not need the band gap values. As far as this problem is concerned, germanium is 0.7, silicon is 1.1, gallium arsenide is 1.4. What we do need is the values of ni and ni is again given per centimeter cube. So germanium is 2.4 times 10 to the 13, silicon is 1 times 10 to the 10 and gallium arsenide is 2.1 times 10 to the 6. So ni essentially decreases as the value of the band gap increases. So we now need to calculate the built-in potential for these 3 materials. We can make use of this formula, just substitute na and nd and then the values of ni for the different materials. So we can go through and work out the math. I will just write down the final answer. So v0 which is your built-in potential is nothing but 0.372. The units is volts for germanium. It is higher for silicon 0.775 and it is even higher for gallium arsenide 1.213. So as the value of ni goes down because you have a higher band gap, the built-in potential at the junction essentially increases. So this information is especially useful when you are trying to build devices with materials apart from silicon because once you know the built-in potential, you will also know what kind of current that needs to be applied through the circuit for a particular kind of application. Let us now go to problem 3. In problem 3, we have a silicon abrupt junction which is in thermal equilibrium at room temperature. So temperature T is 300 Kelvin. It is doped in such a way such that EC-EF is 0.21 electron volts in the n region. So you have an n region and you have a p region. So in this question, the doping concentrations are not given but the position of the Fermi level is. So here EC-EF is 0.21 electron volts and EF-EV is 0.18 electron volts. The material is silicon. So some of the parameters of silicon ni which is 10 to the 10 for centimeter cube, the band gap of silicon EG is 1.10 electron volts. The position of the intrinsic Fermi level EFI, it is not exactly at the center but it is very close to the center. So we can take this as 0.55 electron volts. So these are some of the parameters of intrinsic silicon that we can use. So the first part of the question says draw the energy band diagram for the p-n junction. So before we do that, we have to first draw the energy band diagram for the 2 regions separately and then we can put them together to draw the energy band diagram for the junction. So on the n side, this is my conduction band. This is my valence band. This gap is nothing but EG which is 1.1. EFI is at the center of the gap. So that is 0.55 and EG is 1.1. So this question says EC-EF is 0.21. So the Fermi level on the n side EF-N is 0.21 electron volts. So this height which is nothing but 0.55-0.21 is 0.34. So all the energies are in electron volts. I am just not writing the units where everything is in EV. We can now do the same for the p region. So the material is the same. So the band gap is the same. We just draw this slightly better. So it is the same silicon. So EC and EV are located in the same place. EFI will also be located in the center. EFI. In this particular problem, EF-EV is given to be 0.18 electron volts. So EFP. So this is 0.18. So EFI to EV is 0.55 so that this height is nothing but 0.37. So this is 0.55. This is 0.18. This is 0.37. So we have the energy band diagrams of the n and the p region separately. We can put them together and draw the energy band diagram of the p-n junction. But before we do that, I would like to calculate the concentration of electrons in holes on the n and the p side. So that we can do. We are basically using the formula EFn-EFI is kT ln n over ni and EFP-EFI is equal to minus kT ln of p over ni. So the position of the Fermi level is related to the concentration of the majority carriers. On the n side, it is your donors. On the p side, it is the acceptors. So here this term is known and this is the unknown. Same way here this is known and this is the unknown. So EFn-EFI is 0.34 and EFP-EFI is minus 0.37. So we can substitute in the values so that we get n equal to nd which is equal to 5.1 times 10 to the 15 centimeter cube. p is nothing but nA is slightly higher 1.62 times 10 to the 16 centimeter cube. So even without doing the numbers, we could have predicted that nA will be higher than nd simply because the Fermi level on the p side is located much is closer to the valence band. It is only 0.18 compared to the Fermi level on the n side which is 0.21 electron volts below the conduction band. So we have drawn the energy band diagram separately. We also have the concentration of the electrons and holes. So let me draw the energy band diagram when the junction is formed. To do that we know that the Fermi levels must essentially line up at equilibrium. So this is EFn, this is EFP. Far away from the junction you still have an n-type semiconductor and you still have a p-type semiconductor. Let me just arbitrarily mark an interface between these two and we can show the bands bending so that these two join. So this is EV, this is EC, this is EC, this is EV. So you have the Fermi levels lining up and there is a built-in potential, this is a straight line and there is a built-in potential V0 formed at the junction. So part B we need to determine the concentration of the impurities. So we actually just did that. So this is essentially part B. Just by looking at the shift in the Fermi levels we can calculate the concentration of the impurities. Part C we want to calculate the built-in potential. So V0 is nothing but kT over E ln of Na Nd over Ni square. We can do all the substitution and the numbers. So this works out to be 0.71 volts. We can also calculate the built-in potential by looking at the energy band diagram. So in this particular case the distance between EFN and EFP, so this distance is essentially 0.34 plus 0.37. So this distance delta is 0.34 plus 0.37 which is 0.71 electron volts. So when the junction is formed we know that the Fermi levels have to line up. So we can think of as either the n side shifting completely by 0.71 or the p side shifting up by the same 0.71 so that they line up. So the built-in potential or the built-in voltage is nothing but the difference between the Fermi level positions. So this is 0.71 electron volts. If you divide by E it is 0.71 volts. So instead of using the formula we can also calculate the built-in potential by just looking at the energy band diagram. So let us now go to problem 4. So in problems 1 to 3 we looked at the PN junction in equilibrium so that there was no external potential that was applied and no current that was flowing through the junction. In problem 4 which is slightly a long problem we are going to look at a PN junction that is essentially biased and we are going to calculate some values for the current in the forward and the reverse bias. So problem 4 you have an abrupt PN plus junction. So when we say a PN plus or a NP plus N the plus essentially denotes that this is heavily doped. So when one of the carriers or when one of the sides of a PN junction is heavily doped then the depletion region lies almost entirely on the other side. So one way to see that is to go back to this equation. So NA WP is equal to ND WN. So when ND is much greater than NA this implies WN is much smaller than WP. So that the depletion width is almost entirely on the P side. I will also just write the reverse when NA is much greater than ND then you have WP much smaller than WN and the depletion is almost entirely on the N side. So we have an abrupt PN junction. The cross sectional area A is 1 millimeter square. We will use the cross sectional area to calculate the current. The acceptor concentration of 5 times 10 to the 18 boron atoms on the P side. So NA is 5 times 10 to the 18 per centimeter cube and there is a donor concentration. So this is boron. There is a donor concentration ND is 10 to the 16 per centimeter cube and this is arsenic. So in this problem NA is much higher than ND. So this should actually read P plus N not PN plus. It is my mistake because NA is much greater than ND. So we have 5 times 10 to the 18 boron and 10 to the 16 arsenic atoms on the N side. The whole lifetime values are also given. So the lifetime of the whole tau h in the N region. So these are your minority carriers. This is equal to 417 nanoseconds. Similarly the lifetime of the electrons in the P region is only 5 nanoseconds and this difference is because of the difference in concentration of the dopants. The thermal generation lifetime is also given. So tau g is 1 microsecond. Some other values are also given for this problem. So mu e which is the mobility of the electrons. So 120 centimeter square per volt per second. Mu h is 440 centimeter square per volts per second and e g is 1.1 e v. The length of the P and the N regions are also given. So you have a P region is 5 micrometers and the N region with this 100 micrometers. So these are a whole set of data that is given about the silicon P in junction. So the first thing we need to calculate is the minority diffusion length and to determine what type of diode this is. So we want to calculate the minority diffusion lengths. So in the case of a P N junction under equilibrium you have a dynamic equilibrium so that electrons and holes are moving across the junction and constantly get annihilated. When we apply a forward bias the P side is connected to the positive. The N side is connected to the negative. The Fermi levels no longer line up but essentially gets shifted and when this happens the barrier comes down. So V naught is the built in potential or the barrier during equilibrium. When you apply an external potential the barrier is V naught minus V external. When the barrier goes down we basically have minority carriers moving across the junction. So we have electrons from the N side moving to the P side where they are minority carriers. We also have holes from the P side moving to the N side and there they are the minority carriers. So it is this minority carrier diffusion that essentially causes current to flow in a P N junction. So the first thing you want to calculate is the diffusion length. To know the diffusion length we need to know the diffusion coefficients. So D e which is the diffusion coefficient of electrons is nothing but k t mu e over e. So it depends upon the mobility and D h is k t mu h over h. So the values of mu e are given. Mu h is given. Everything else is a constant. So we can plug it in. So the D e is 3.10 centimeter square per second. D h is 11.39 centimeter square per second. So D h is higher than D e because mu h is higher than mu e and this is because the holes are diffusing on the N side and the concentration on the N side is two orders of magnitude less than the P side. So because you have less concentration of your dopants, the diffusion coefficients are higher. From the diffusion coefficients we can calculate the length. So L is nothing but square root of D times tau. So L e is D e tau e. L h is D h and tau h. So once again D e and D h we have calculated. Tau e and tau h are given. So we can substitute the numbers. So L e works out to be 1.2 micrometers. I am not doing the math. So all your units are in centimeters. So the answer will also be in centimeters. You can just convert that into micrometers. So L e is 1.2 and L h is larger. It is 20.8 micrometers. So if you looked at the length of the diodes on the P and the N side, on the N side the diode length is 100. On the P side, the diode length is 5 micrometers. So L e is smaller than the 5 micrometers. So this is the length on the P side. L h is smaller than 100 micrometers which is the length on the N side. So that this is essentially a long diode. So a long diode is 1 in which the diffusion lengths are smaller than the physical lengths of the P and the N region. So this is part A. Let us go to part B. So in part B, we want to calculate the built-in potential across the junction. So this is the potential when the junction is in equilibrium. So this is fairly straight forward. So just kT over E ln of Na Nd over Ni square. So we have all the numbers. We just need to substitute that. This works out to be 0.877 volts. So this particular problem does not ask you to calculate the depletion widths. But you can go ahead and do the calculation and you will find that the depletion width is almost entirely on the N side and that there is a very small depletion width on the P side. Part C, what is the current when there is a forward bias of 0.6 volts across the diode? So now you have the diode to be forward bias. The external potential V is 0.6 volts. So when you apply an external potential, the barrier height is lowered so that there is an increase in current due to the minority carriers diffusing across the junction. In this particular case, the current density is given by a constant J s0 times exponential EV over kT minus 1. Usually the exponential term is much higher than 1 so that this can be written as J s0 exponential EV over kT. J s0 is your reverse saturation current and this is given by ni square E dh over lh and d plus de over le na. So we saw the derivation for this during the course notes but this is your reverse saturation current and this is something that plugs in here. So if you remember the assignment from the short key junction, so the metal semiconductor junctions, we had a similar expression to this except that the constant out front had a different value which depended upon the thermionic emission but now here we have a P n junction so the constant here is your reverse saturation current. In this particular problem, na is much higher than nd and since they are in the denominator, this term will essentially dominate over the other term. So this is the reverse saturation current density. To calculate the current, we just need to multiply this by the area. So all the numbers are here, we calculated dh and lh in part a, na and nd are known, ni is also known so that J s0, so instead of J s0, I will directly write i s0 which is J s0 times the area. So this is 8.36 times 10 to the minus 14 amperes. So the reverse saturation current is essentially a really small value. Once you know i s0 or J s0, you can calculate the current during forward bias. So i nothing but J times a which is J s0 times a times exponential e v over k t, v is 0.6 that is given. So the current works out to be 0.96 times 10 to the minus 3 amperes or 0.96 milli amperes. So the current in the forward bias is 0.96 milli amperes. So that is nearly 10 to the 11 orders of magnitude or 10 to the 10 orders of magnitude higher than the reverse saturation current. This is why we essentially call a p-n junction to be a rectifier because it conducts very well during forward bias and the reverse bias current is very small. So let us now go to part d. Part d, we want to estimate the forward current at 100 degrees. So the temperature is now increased. You can write this in Kelvin so that is 373 Kelvin. So voltage is the same. So v is 0.6 volts. The question also says that assume the temperature dependence of Ni dominates over everything else. So over the diffusion lens, the diffusion coefficients itself also the mobilities. So if you only take Ni into consideration, so we can see that the current or if you write this down, J is J s0 exponential Ev over kT. So ratio of J at 373 Kelvin to that at 273 or 293 Kelvin, this should be 373. So 373 by 293 Kelvin or 297 Kelvin which is room temperature. So let me just draw this, write this as 297 is nothing but J s0 prime or J s0 at 373 Kelvin divided by J s0 297 Kelvin. So the potential is the same. So the ratio of the currents or the ratio of the current densities is nothing but the ratio of the reverse saturation currents. This is directly proportional to Ni square. So that this is Ni square at 373 Kelvin divided by Ni square 297 Kelvin. Ni square is nothing or Ni square root of NcNv exponential minus EG over 2 kT. So for this problem we can take Nc and Nv to be independent of temperature. So the ratio of Ni is just given by the exponential term. So if you do this, this ratio works out to be approximately 100. So that the reverse saturation current is increased by 100 when we go from room temperature to 100 degrees C. So the new values of I s0 will be 100 square. So the new values of I s0 at 373 Kelvin, just me write down the final answer is 8.36 times 10 to the minus 10 and current I at 373 Kelvin is 0.10 amps. So that the current essentially increases by 4 orders of magnitude. In part E we want to calculate the reverse current when you have a voltage of 5 volts. So we want to know the reverse current when Vr is 5 volts. To calculate the reverse current we first need to calculate the new width when you apply a reverse bias. So the width W is 2 epsilon 0 epsilon r V0 plus Vr and we said that the depletion region lies almost entirely on the N side. So I only have E and B whole to the half. So this formula is something we used before to calculate the width of the depletion region when you have a p-n junction under equilibrium. So you only modified it to add the reverse bias voltage and we also removed the contribution due to N A because N D is much smaller than N A. So we can plug in the numbers. The new depletion width comes out to be 0.88 micrometers and most of this is in the N side. So when you have a reverse bias you have thermal generation of carriers within this depletion region and this thermal generation of carriers is responsible for your reverse current. So I gen which is the current due to thermal generation of carriers is E times the cross sectional area times depletion width times N I divided by tau g where tau g is the thermal lifetime of the carriers and that value is also given. So everything here is known we can substitute the numbers and I gen works out to be 1.41 times 10 to the minus 9 amperes. So this number is again much smaller than your forward bias current which is of the order of milliamps. So once again even if you take thermal generation of carriers into account we essentially have a rectifying action in a P N junction. So let us now look the last question. So problem 5 you have a germanium P N junction diode. So it is germanium P plus N the values are given. So N A is 10 to the 18 per centimeter cube N D is 10 to the 16 D H. So D H I will write on this side. So D H is 49 centimeter square per second D E because we are looking at minority carriers is 100 centimeter square per second tau E is equal to tau H is equal to 5 microseconds in the cross sectional area is 10 to the minus 4 per centimeter square. So we want to calculate the diode current. So we want to calculate I when we have a forward bias V of 0.2 volts N I is 2.4 times 10 to the 13 per centimeter cube. So this is very similar to the previous question. So J is J is not exponential E V over K T and then J is not this N I square over E D H over L H N D plus D E over L E N A. So all the numbers are given. So J is not I will just substitute and write the final answer but you can directly do the substitution. J is not is 2.94 times 10 to the minus 5 ampere per centimeter square. So the current I is not is 2.94 times 10 to the minus 9 amperes. Once you know I is not you can substitute here and you can get the current. The current I is nothing but 6.687 times 10 to the minus 6. So in this particular case the difference between the current and I is not is not as high as in the case of silicon. One particular reason is because your applied voltage is very small. It is only 0.2 volts. Another difference is that the material is germanium. So that the band gap is smaller. So the built in potential is also smaller. At the same time N I is larger. So that J is not is also larger. So these are some of the factors you have to take into account when choosing materials for forming p-n junctions.