Q & A
A Collection of Questions & Answers Asked by Current and Past ECE 3040 Students
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Q:I am having a difficult time visualizing what’s going on when a band gap is created. The band gap is created when the individual Si atoms are brought close together, but what actually happens to those valence electrons (aside from them A: Basically, if you take the y-axis as energy and x-axis as the distance, the energy states split and form multiple energy states. As the # of atoms (,say, N atoms) increases, the split states may overlap for certain energy levels and start to fill up the energy space. As N approaches infinity, the spacing between levels becomes indistinguishable. Energy “bands”, instead of discrete “energy states”, are formed. For simplicity, we can say a semiconductor forms 2 groups of “bands” (i,e, C.B. and V.B.) and in between these bands exists the “forbidden gap”, or the “bandgap” |
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Q:Is there any material you can refer me to (for topics on the formation of energy band and bandgap)? A:Introductory semiconductor books usually do not cover this topic. You may check other advanced solid-state physics books on this topic. You may read the following book chapters if you wish: 1. Chapter 1, “Fundamentals of semiconductor devices,” Anderson&Anderson, McGraw Hill, 2005 2. Chapter 15-17, “solid state physics,” Ashcroft & Mermin,1976 |
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Q: How can you find the mobility of an intrinisic material? How do I find the intrinsic carrier concentration for different semiconductor materials? A:You may assume constant mobility for electrons and holes, respectively, at low doping concentration (<10^14 cm^-3) from Fig. 3.5 and read the intrinsic ni values form Fig. 2.20. The mobility at low impurity level is dominated by lattice scattering. |
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Q:I was wondering whats the best way to study for the upcoming exam? A:Please read the exam-related section in the syllabus carefully. The textbook, lecture notes, your classnotes, homework, and practice exams are good resources for your study. Please focus on the understanding of physical concepts and physical implications of equations, not on the memorization of equations. You also need to know what equations to use under what questions, of course. |
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Q: How can you find the mobility of an intrinisic material? How do I find the intrinsic carrier concentration for different semiconductor materials? A:You may assume constant mobility for electrons and holes, respectively, at low doping concentration (<10^14 cm^-3) from Fig. 3.5 and read the intrinsic ni values form Fig. 2.20. The mobility at low impurity level is dominated by lattice scattering. |
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Q: What is the intrinsic carrier concentration at T=0k? A: 0 |
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Q: I had a small question regarding sample problem no. 2 on page 128 of the pierett textbook. They ask us to find delta(P) as a function of x. But at what time t should we find this delta(P(x))….at time t=0 or t=infinity? The text book assumes steady state conditions which means that they do not A: At steady state, the recombination and diffusion process for minority carrier in a semiconductor reach balanced status so the spatial distribution of minority carriers are “frozen” in the time domain. Therefore the “time derivative” term is set to zero. |
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Q:My understanding of a hole was it is simply the absence of an electron in a crystal lattice. The hole could move around the lattice and later be filled by an electron. The analogy I used in my mind was this: at an amusement park (ie Six Flags), you have long waits for a ride. If you watch people move up in a big burst, you can see a “hole” propagating the opposite way that all of the people are moving. Is it? A: Good analogy. |
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Q: But then, I read page 105 of Pierret. The book says “An electron and hole moving in the semiconductor lattice stray into the same spatial vicinity and zap! – the electron and hole annihilate each other. The excess energy released during the process typically goes into the production of a photon (light)”. How can an electron-hole collision create a photon? This makes it sound like electron-positron annihilation. Are holes the same as positrons? A: This is exactly what happens in a band-to-band recombination process. Remember that band-to-band recombination process takes a electron in the CONDUCTION BAND to recombine with a hole in the VALENCE BAND. The actual process in the energy space is that the electron falls back to valence band, lose energy in the form of photon (radiative recombination), phonons(non-radiative recombination), or energy transfer (Auger recombination), and recombine with the hole at the valence band in the lower energy state. |
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Q: My question is about 5.2. I don’t understand how the fermi-level can be at E_v – 2*k*T on the p-side. That would mean that the level is below the E_v. If you can explain this to me, I would appreciate it. Also, is it possible to have the electric field at two different values at x=0, for example on Problem 5.4(c) The Electric Field at x = 0. And finally, for problem 6.1(e) I don’t seem to understand the question. If you explain it to me in another way, that would be great. Thank you A: 1. The fermi level can be below the valence band. Please note that the fermi level represent the energy level where one can find 50% of electron occupancy (by definition) 2. There must be a unique solution for electric field at any given points in a PN junction diode The elecrostatics law states that the electric flux density (D) is continuous through out the material. Unless the p-side has different material than that for the n-side, the electric field is continuous across x=0. 3. rephrase 6.1 (e): Why minority carrier diffusion equations are not applicable in the depletion region? |
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Q:I have a question about the voltage at a pn junction. The equation is V = kT/q * ln(Na*Nd/ni^2). In Exercise 5.2 on page 214, the book equates Kt/q = 0.0259. This isn’t true because kT = 0.0259, which should then be divided by 1.6*10^-19. If you do not divide by q, then your answer will be in eV and not V. Could you explain why the book does not divide by q? A: Please note the units of Boltzmann constant “k”. k = 1.38 E-23 (J/K) or 8.617 E-5 (eV/K). Also note that 1 eV = 1.6 E-19 J. If you apply 1 volt on an electron, the electron gain energy with the amount of 1.6E-19 Joule (E=q*V), which is equal to 1 eV. |
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Q: How does the charge density and E-field diagram for the NPN differ from that of PNP? A: You should use Poisson’s equation to conceptualize this question. In the |
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Q: In Chapter 5, the voltage drop across the DR when an applied voltage exists is given by: V(built-in) – V(applied). However, in Chapter 6, for the case of large forward or reverse bias, the voltage drop across the DR is given by: V(applied) – V(ohmic contacts and QNRs). For the case when there is a large forward or reverse bias can the voltage drop across the DR be written as: V(built-in) – V(applied) – V(ohmic contacts and QNRs)? A: Good Question! The Vbi-VA is the “potential difference” between the p-side QNR and n-side QNR in the band diagram, or the potential barrier that electrons and holes need to overcome if they wish to reach the other side of the junction. In chapter 6, VA – Vohmic is the case where the ohmic loss in the QNR is taken into account. If we refer to the “voltage applied across the DR”, we are not referring to the POTENTIAL DIFFERENCE in the DR, instead, we are referring to the voltage difference applied across the DR. The electron still see a potential across the DR with a value of Vbi-(VA-Vohmic). Think: Electrons see a potential barrier across the DR with a value of Vbi at zero voltage.) |
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Q:Can you explain the directions of the hole and electron currents in an NPN BJT? Is it simply the opposite of that for the PNP? On the practice exam, electron flux is utilized and on the lecture notes it is not. A: I guess you are talking about the PNP and NPN under “active” biasing mode. I will not give you a straight answer and I would like you to follow the thoughts below: 1. The EB junction is forward biased and the BC junction is reverse biased for Active mode 2. In the base, the minority carriers (injected from the emitter) diffuse across the quasi-neutral base and are swept into the collector (due to the electric field inside of the DR for the reverse-biased BC junction), contributing to one of the collector current components 3. The electron flux density and the electron current are directing in the opposite directions because electrons are negatively charged. Once you understand these facts in the BJT, it should not be a problem for you to find the answer. |
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Q:The problem 5.1(f) in the book says that the _Ohmic contacts reduce the built-in voltage drop across a junction, true or false A:The answer is “False” because the built-in voltage is a fixed value that is only dependent on the doping concentration, temperature, and intrinsic carrier concentration. |
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Q:How do the electron and hole behave in a forward-biased PN junction? A: The electron are “injected” from n-type into the p-type region due to the reduction of potential barrier across the space-charge region. Once the electrons are injected into the p-side, they become the minority carriers and start to diffuse further into the p-side QNR. As these electrons diffuse in the p-side QNR, the electrons recombine with the hole simultaneously. — That is how you see the “electron diffusion current” in the p-side under forward bias. Similar analogy can be obtained on the holes at the n-side under forward bias condition.. |
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Q: What about the electron and hole behavior in the reverse biased PN junction? A:In the quasi-neutral region, you will see the minority carrier concentration is depleted as it approaches the depletion edge. This is due to the existence of electric field at the DR with reverse bias. Under steady-state, the minority carrier concentration is “quenched” near the depletion edge and is restored back to equilibrium at a position far away from the DR. The minority carrier will tend to “diffuse and generate” in a region where there is a lack of minority carriers. The resulting “minority carrier diffusion current” components in the depletion edge are used to characterize the total reverse bias saturation current. As you see in the lecture slides, the diffusion current components reach their maximum at the depletion edge. It implies that the corresponding the minority carrier flux will be moving into the DR — note that this is for rev. biased case only. Inside of the DR, the minority carriers experience an electric field and hence the (minority) carrier transport mechanism inside the DR is due the “Drift.” |
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Q: [ Question on MS Junction ] For an n-type semiconductor, when a positivie bias is applied, wouldn’t the Fermi level on the metal side be higher than the Fermi level on the semiconductor side since phi/sub/M A:The difference between phiM and phisS gives you the bandgap alignment guideline at 0 V. Whether it is ohmic or rectifying depends on the bandgap prfile. (please see textbook 14.1, 14.2) It is nothing to do with the applied voltage. Once you apply positive voltage on the metal side, the Fermi level of metal moves downward in the band diagram. |
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Q:1) On qeustion 2 of the practice exam 3, part c , I am confused about the phenomenon that occurs. I understand that in a PMOS, during inversion, holes being the minority carrier concentration will accumulate at the OS interface changing the type from n-type to p-type, but I’m unsure of why this occurs. A:The energy band bending under strong inversion results in the hole accumulation at the O-S interface. Mathematically, one can calculate the hole concentration at the O-S interface from chapter 2 of Pierret. One will find that the hole concentration equals the background donor concentration at the “onset of strong inversion”, i.e., p=ND. Beyond the onset of strong inversion, hole concentration increases exponentially and p>>n. This is why we define such biasing mode as “strong inversion” |
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Q:Can you explain Question 4, part d in practice exam 3? A:In the Pierret book, it assumes Phi/sub/MS =0 (for an ideal MOS), which is not true in reality. If Phi/sub/MS is a nonzero value, one would need to apply an extra voltage V/sub/FB to achieve the “flat band” condition. From MOS band diagram, if one assumes the perfect oxide layer and interface (which is not true either as there is no “perfect oxide” in the world), the flat band voltage will be equal to Phi/sub/MS. The resulting threshold voltage will be shifted by an amount of Phi/sub/MS from that for the “ideal” MOS. |