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### Week 5: Electricity from heat: Devices for a green

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### "A request: This is rather silly. Prof. Datta, could you please say your name out loud in one of the lectures in part II, that is, if you haven't filmed them yet, as I wish to learn how to pronounce your name properly. Thanks for the great course and sorry for the stupid request!"

• Unfortunately I have finished recording Part II, but many thanks for your interest and we are very glad to hear that you liked the course. - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "Why is it called q*q/h the quantum of conductance if there can be any fraction of this value?"

• I don't think it is possible to have a fractional conductance value. I thought a hydrogen molecule conductance was the least you could get and it was never less than the conductance quantum... - Student Fri Mar 09 2012 from Hotseat

• q*q/h is defined as the maximum of conductance of a quasi-one-dimensional system when only the lowest sub-band or transmission channel is activated. In other words, q*q/h is the maximum conductance per transmission channel. In general, the transmission is not ballistic, so the conductance is less. - Student Fri Mar 09 2012 from Hotseat

• Possibly the best justification is in the context of the quantum Hall effect (Part II, Week 3) where the Hall resistance is found to be exactly q^2/h (times an integer) up to many places in decimals. This effect is so precise that it is even used as a standard for resistance. - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "We know, from newtonian mechanics, that if we know the position of the particle at every time we can calculate its momentum. Why in hamiltonian mechanics, the momentum is an independent variable? is it because we are interested only in the position and momentum at one instant of time?"

• Classically you would have to know both position and momentum at t=0 in order to predict position and momentum at later times. I believe this is true of both Newtonian an Hamiltonian mechanics. The latter is just a reformulation of the former. - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "In a ballistic transistor, the electrons are being accelerated in the channel. Do they radiate an electromagnetic field? Accordingly with the theory we saw they can't because they would lose energy."

• Is it something like in the hydrogen atom? where, though the electron is being accelerated it does not lose energy - Student Fri Mar 09 2012 from Hotseat

• Note: the electrons are being transported, not accelerated. They are traversing the channel at a specific velocity. - Student Fri Mar 09 2012 from Hotseat

• Look at these two versions of the same article (fig.6 in the first one and fig.9 in the second one where Dr. S. Datta appears as a coauthor). http://ece.ut.ac.ir/classpages/S84/NanoElectronic/princeton/00974760.pdf http://www.ee.iitb.ac.in/uma/~vharihar/CarrierTransportInNanoscaleMOSFETs.pdf - Student Fri Mar 09 2012 from Hotseat

• It is clear that electrons in a potential profile with a negative slope must gain kinetic energy to maintain the same total energy during its travel from drain to source as it has be the case in an elastic resistor (look at the potential profiles vs. position in this two articles). - Student Fri Mar 09 2012 from Hotseat

• It's not an antenna! Sure they do, but what's the power radiated? The EMF sticker on your laptop does not refer to chips. Have a look at a dismounted laptop (even on the internet) and have a look at a 10-yr old one and its EMI shield locations. All your doubts will be dissipated :-D - Student Mon Mar 12 2012 from Hotseat

• Usually we have talked of low voltages where acceleration is minimal. In any case I believe any EM radiation is negligible. But electrons do radiate significant amounts of acoustic waves or phonons and this makes real resistors deviate from our ideal elastic resistor as length/voltage are increased - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "if there is a quantum capacitance, can we define a quantum inductance also?"

• I know from the book "Quantum Engineering" of A.M. Zagoskin that the answer is yes, but I would like to know your opinion. - Student Fri Mar 09 2012 from Hotseat

• You can look for the "kinetic inductance" too. I didn't hear anything about Quantum Inductance before, but it would be interesting to discuss about this. Which chapter can I find this information? - Student Sat Mar 10 2012 from Hotseat

• Chapter 2.5 of the Zagoskin's book "Quantum inductance and quantum capacitance". - Student Sat Mar 10 2012 from Hotseat

• Please take a look at Eq.(7.21) on page 84 of the notes. We did not discuss it in this course, but my thoughts are summarized in Section 7.4. - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "slightly confused about 5 ceu, what r the contents in CEU I wish to understand Coulunb blockade, Single e lectron transister, CN tubes heterostructures r they coveered in CEU "

• CEU is not anything about topic; chk out the course registration page, https://nanohub.org/groups/purdue - Student Fri Mar 09 2012 from Hotseat

• i'm confused about the question. CEUs are Continued Education Units, they are like college credits. you can definitely get CEUs for taking courses in other places on other topics. - Student Fri Mar 09 2012 from Hotseat

• A CEU is a Continuing Education Unit and is a measure used in continuing education programs or by a corporation, particularly those required in a licensed profession in order for the professional to maintain the license. - NanoHUB-U Admin Fri Mar 09 2012 from Hotseat

• thanks to all now I understand - Student Sat Mar 10 2012 from Hotseat

### "My only suggestion is it would be nice to indicate which HW go with which lectures. I usually waited until seeing all the lectures to do the HW, but I think it would have been better to do HW after the associated lecture was viewed. Many thanks for this class, enjoyed it thoroughly!"

• Thank you very much, we appreciate your feedback. Also I appreciate your suggestion, will try to implement it for Part II. - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "Ref M5.3, In the lecture heat current was explained as (e-mu) per each electron going from left to right. So it is intuitive to think there should be a electron current for heat current to exist. apart from the algebra, can you pls explain the existence of a heat current with I=0 in the quiz5.3?"

• Please take a look at the discussion at the bottom of page 133 to the top of page 134. - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "The text indicates (page 146) that the quantum of thermal conductance is 325 pW/K. Since this is given by pi*pi*k*k*T/3*h (units are (J*J/K*K*J*sec)*K, what T is used to get this value?"

• T=300 Kelvin or 27 Centigrade, usually assumed for room temperature. - Supriyo Datta Thu Mar 08 2012 from Hotseat

• But T=300 K does not yield even close to 325 pW/K according to the expression of page 146 in the lecture note. - Student Fri Mar 09 2012 from Hotseat

• it gives 283 pF, would u please chk? - Student Sat Mar 10 2012 from Hotseat

• You are right, it should be ~ 284 pW/K and not 325 pW/K. - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "What are the Continuing Education Units for? Where can I use them? Do you recommend to pay for it?"

• A CEU is a Continuing Education Unit and is a measure used in continuing education programs or by a corporation, particularly those required in a licensed profession in order for the professional to maintain the license. .. - NanoHUB-U Admin Fri Mar 09 2012 from Hotseat

### "In Tutorial 5.1 the use of the delta function is emphasized but a general value of G is used to derive the results. Since the delta function essentially makes the integral single valued at E=epsilon, thus defining a single G0 value, why is G0 not used in place of G? It might be simpler to understand"

• The coefficients G,GS,GP,GQ each involve an integral over energy with G(E) in the integrand. I should have used a different symbol say Gtilde(E) for the function G(E) to separate it from the number G obtained from 10.4a by integrating. Gtilde(E) is a delta function, not a constant number. - Supriyo Datta Thu Mar 08 2012 from Hotseat

• OK. Bad choice of terminology on my part. Doesn't the delta function "pluck out" a single value of the integral at the point E=epsilon as shown here: http://en.wikipedia.org/wiki/Dirac_delta_function In the "Definitions" - As A Measure section? - Student Fri Mar 09 2012 from Hotseat

• Yes, I agree with what you say about the delta function. This principle needs to be applied individually to each of the four integrals for G,GS,GP,GQ. I think that is what the tutorial did. So I am not clear what you have in mind when you say "why is G0 not used in place of G." - Supriyo Datta Mon Mar 12 2012 from Hotseat

### "I believe there's a typo in the last paragraph on page 139. "The key difference with PHONONS is that unlike electrons ...""

• Got it, many thanks. - Supriyo Datta Thu Mar 08 2012 from Hotseat

### "The picture in the slide M5.1, and Fig10.1 in the book state that the electron current moves from cold to hot in n-type conductors. But this is the opposite of what I grasped from the lecture. Is there a mistake, or were the lectures describing conventional current? Thank you. "

• I think the confusion is due to the difference between the current inside the channel and that in the external circuit. For n-type, electrons flow from hot to cold in the channel, but from cold to hot in the external circuit as shown in Fig.10.1 - Supriyo Datta Thu Mar 08 2012 from Hotseat

### "Quiz 2, Week 5: Why should the coefficient GS (for the material with conductance G= Go) be zero? Could you elaborate?"

• Please look at the answers to the question from Ms. Satpathi below, might help answer your question. If not, please let me know if you are looking for a mathematical answer or a physical one. - Supriyo Datta Tue Mar 06 2012 from Hotseat

### "It would be appreciated if the admin gives a little effort to concoct the most replied discussions and upload them as downloadable format for future references and clarifications of the topics. The discussions were highly attractive and informative."

• I wish all discussions easily downloadable. Forum is remarkable, lots of excellent questions and discussions, highly informative replies and explanations from Professor, collaborators, students --- and no chatter or stupidities. Would also nice to have "Expand all questions" button at top of page. - Student Tue Mar 06 2012 from Hotseat

• Great suggestion! We will let you know as soon as we have a solution for this. - NanoHUB-U Admin Tue Mar 06 2012 from Hotseat

• In the meantime, big thanks for allowing photo uploads for the future :) - Student Tue Mar 06 2012 from Hotseat

### "Thank you very much for the course.Looking forward to Part 2"

• We are glad you enjoyed the course. We will be sending out a course evaluation with the Proofs of Completion and hope you will help us improve our service by completing the short survey. - NanoHUB-U Admin Tue Mar 13 2012 from Hotseat

### "Would someone please explain the correct ans of Quiz 5.6? Is there any details on this point in the lecture note?"

• Sorry, for the late posting. I was kinda busy this week, so could not take part in regular discussions. Thanks to Supriyo datta Sir for presenting such interesting topics of nano-electronics through distant learning. - Student Mon Mar 05 2012 from Hotseat

• Sorry specifically I wanted the Quiz 5.6 question 2 answer explanation. - Student Mon Mar 05 2012 from Hotseat

• The correct answer is (a). Not sure if the notes address this question anywhere other than at the top of page 20 right before the equation. - Supriyo Datta Mon Mar 05 2012 from Hotseat

### "In Quiz M5.2a) Gs is zero for G(E) a constant,G0, but in T5.1, Gs is non-zero and proportional to G0, for a single level device, what is the difference between the two, a single level device and a device with constant G(E)?"

• Key point: Levels above and below the electrochemical potential mu, contribute oppositely to Gs. Non-zero Gs requires difference between the two. For constant G(E) there is no difference. For single level device, Gs depends on whether level is above or below mu. If level is right at mu , Gs=0. - Supriyo Datta Sun Mar 04 2012 from Hotseat

• Are you stating that if G(E) is constant, D(E) is also constant with only a single value so it is a horizontal line (in the E vs G(E) or E vs D(E) plot) and not a curve? - Student Tue Mar 06 2012 from Hotseat

• If G(E) is a constant, it has the same value for all E. So it should be a VERTICAL line in a E versus G(E) plot. By contrast, the single-level device is represented by a horizontal line, G(E) is a delta function: zero everywhere except for a particular energy. - Supriyo Datta Tue Mar 06 2012 from Hotseat

• So, in Week 5 Quiz, Question 1 why does the fact that G(E) = a constant, G0, result in Gs being zero? There is nothing in the derivation of the expression for Gs that would lead to such a conclusion nor is there anything in the expression for Gs, unless mu=E and/ or df/dE=0. - Student Tue Mar 06 2012 from Hotseat

• Sorry, I meant Quiz 2, Question 1. - Student Tue Mar 06 2012 from Hotseat

• Consider the integral in Eq.(10.4b) on page 121. df/dE is a symmetric function around E=m0. (E-mu0) is an antisymmetric function around E=mu0. The overall integrand being a product of the two is antisymmetric. Integral of an antisymmetric function is zero: positive parts cancel the negative parts. - Supriyo Datta Wed Mar 07 2012 from Hotseat

• The above is the mathematical answer. Physically every level above mu0 has a corresponding level below mu0 that contributes an equal and opposite amount to Gs. Overall they cancel. - Supriyo Datta Wed Mar 07 2012 from Hotseat

• Thanks. That helps me understand the solution. - Student Thu Mar 08 2012 from Hotseat

### "Can you pls ans further Q on week 3 exam Q2? We always integrate from -infty to +infty, then in T=0 case we may cutoff integration extremes and only consider interval, but if we do not (e.g. computer program with T variable), same numerical results. I picked "right", but seems to me both options OK."

• I believe none of the options given include f1 - f2 in the integrand. If f1 - f2 is included in the integrand then I agree it does not matter whether limits are (mu1, mu2) or (-inf, +inf). But without f1-f2, integrating from (-inf, +inf) would give infinity since G(E) is not zero at high energies. - Supriyo Datta Fri Mar 02 2012 from Hotseat

• But the limits (-inf, mu1) might give the correct answer anyway if we assume G(E) goes to zero at low energies and G(E=mu2) = 0, which is needed for the current to saturate. - Supriyo Datta Fri Mar 02 2012 from Hotseat

• Thank you for clarifying this particularly insidious mix-up I had. - Student Mon Mar 05 2012 from Hotseat

### "Thanks Prof. Datta and the team for nice course. Waiting the II part of the course.. Thanks "

• Thank you very much. - Supriyo Datta Fri Mar 02 2012 from Hotseat

• I agree - Student Thu Mar 08 2012 from Hotseat

### "Thank you so much for the awesome lectures . I really enjoyed with this course. Will you provide certificate for this course?"

• Yes, we will be providing a certificate or “Proof of Completion” for successful completion of the course. See the course FAQ's for more details. - NanoHUB-U Admin Thu Mar 01 2012 from Hotseat

• thank you for responding sir. - Student Thu Mar 01 2012 from Hotseat

• Thank you very much, we are very glad you liked the course. - Supriyo Datta Fri Mar 02 2012 from Hotseat

### "Thank you very much for such a nice course which provides a very good platform in this field. Due to some personal problem I could not go through the last module properly and still attempted the exam as I will be out of station for 10 days. Sir,will you be providing any certificate for this course?"

• Yes, we will be providing a certificate or “Proof of Completion” for successful completion of the course. See the course FAQ's for more details. - NanoHUB-U Admin Thu Mar 01 2012 from Hotseat

• . - NanoHUB-U Admin Thu Mar 01 2012 from Hotseat

• Many thanks, we are glad you liked it. - Supriyo Datta Fri Mar 02 2012 from Hotseat

### "What a great class, can't wait for the next one. Thank you "

• I can't wait for the next part either (Actually, I was eager to see new material today, just to find out the next part starts in 3 weeks). Thanks Prof Datta for this great course. I really enjoyed your lessons. - Student Mon Feb 27 2012 from Hotseat

• It would be great to have material posted earlier, leaving in place only the deadlines. This way it would also be helpful to those who have a problem and are away for a week or two. - Student Tue Feb 28 2012 from Hotseat

• Thank you very much, we greatly appreciate the feedback. The material is still in production, but we will try to have it posted as soon as it is ready. - Supriyo Datta Fri Mar 02 2012 from Hotseat

### "There are 2 versions of the World Scientific book available for pre-order, one is \$75, the other \$36. What is the difference? "

• One is paperback the other is hard cover is my guess. - Student Sat Feb 25 2012 from Hotseat

• Yes as Mustafa says. See "pbk" remark. How does it work with preorder? Book is supposed to be published in the summer, so it says on World Scientific website. - Student Sat Feb 25 2012 from Hotseat

• I presume they collect your order and ship it to you when it is available. I would assume that they charge you only when the book is actually shipped, but I could check with World Scientific, if you are interested. - Supriyo Datta Mon Feb 27 2012 from Hotseat

• Thank you! I have just ordered. I was not immediately charged. I have asked them and when they answer, I will post the answer here for everyone else. - Student Mon Feb 27 2012 from Hotseat

• For anyone who wishes to order: World Scientific records our pre-orders - provided we give them a credit card with validity beyond August 2012 - and will charge when book is ready and shipping. - Student Tue Mar 06 2012 from Hotseat

### "Thank you Professor Datta, for a wonderful course. I attend college now and I have several professors that know you, have worked with/for you, and that know of your work in this field. I look forward to the secound course. "

• Thank you very much. I greatly appreciate your feedback. - Supriyo Datta Sun Feb 26 2012 from Hotseat

### "Thank you for the wonderful course Prof Datta. I enjoyed the course and learned more than I imagined. Very well done. Many Kudos and Congratulations!"

• I agree with Mustafa, the course is really wonderful! For a short time It gives a lot in understanding of physics of most important carrier transport phenomenas. I enjoyed it a lot! - Student Thu Feb 23 2012 from Hotseat

• Many thanks, Mustafa and Oleg, I appreciate the feedback. This is a new experiment on our part, and I am really glad you liked it. - Supriyo Datta Thu Feb 23 2012 from Hotseat

• I second my colleagues' admiration, amazing course provided us with original and thought-provoking overview, presented lots of concepts in few hours, for all of us non-specialists to apply lessons from nanoscience in our professional life. Looking forward to part 2. Big thanks to prof Datta and team - Student Fri Feb 24 2012 from Hotseat

• Thank you very much. Really glad you liked it. - Supriyo Datta Sun Feb 26 2012 from Hotseat

• Thank you for this great course, just what i needed to get familiarized with some basic concepts of nanoelectronics before i begin my Msc this September :). I will definitely apply for the second part. - Nick Matthaiakakis Thu Mar 01 2012 from Hotseat

### "From previous lectures G is proportional to M*lambda/(L+lambda). Would it be true for Gs (10.4b), Gp (10.13a), Gq (10.13b) in the CHANNEL (i.e. if energy transport is considered in channel)? Should lambda be same for these coefficients?"

• I'm asking because some time ago we had a look on a HD model (DD + div(Iq)=Rhs) problem in MOSFETs, one of conclusions at that time was that carrier thermal conductivity model should be improved. Note, it was for relatively long channels (50-100 nm) where lambda should not affect much. - Student Thu Feb 23 2012 from Hotseat

• Yes, I believe it should be the same G(E). For Gs this seems clear since it comes from linearizing the same equation (10.1), same as (3.3). For Gp and Gq too this is true because the heat current equation (10.11) comes from extending the same equation (3.3). - Supriyo Datta Thu Feb 23 2012 from Hotseat

• lambda is average length of "free flight" up to carrier will scatter. I'm not sure that lambda should be same for energy transport. Not each scattering event will change the carrier energy. If so, then lambda for energy transport coefficients might be higher (D(E) should be same, of course). Or not? - Student Thu Feb 23 2012 from Hotseat

• Good point. Note, however, that our result is obtained for an elastic resistor model which has no energy relaxation at all. So I would argue that the energy relaxation length is not relevant here. - Supriyo Datta Thu Feb 23 2012 from Hotseat

• Yes, I agree that energy relaxation length should not be used. We saw that HD model started to fail (mostly div(Iq) equation) for longer L in comparison of just DD (I'd say critical L /beginning of problems/ is 30nm for DD and 100nm for HD). It might be related to lambda difference, but not sure. - Student Thu Feb 23 2012 from Hotseat

### "I am falling behind. Will this material remain available for a while, or will access be shut down at the end of this week? Thanks!"

• No, the course materials will be available to you for about a year after the course ends. Just log back into the course content page. - NanoHUB-U Admin Thu Feb 23 2012 from Hotseat

• There is still some time to do the exams, if you plan to try catching up in the next few weeks. - Student Fri Feb 24 2012 from Hotseat

### "I discussed spin transport in semiconductors with a researcher and he asked me why the electro-potential splits into spin up and down splitts and does that affect the spin transport. In other words, does electrons having spin up will be more transported because of the splitting."

• I found out the answer and please correct me if I got it wrong. The splitting is because a mysterious exchange interactions which shift the spin up potential above spin down potential about 2 eV. - Student Thu Feb 23 2012 from Hotseat

• Also, the higher spin level will be easily injected if a suitable channel was chosen so that the conduction band matches with the higher spin level and falls a little bit above the lower spin level. - Student Thu Feb 23 2012 from Hotseat

• I got the answers from watching lesson 4 but I would like you to check my understanding. I'm proceeding with less speed compared the course progression. - Student Thu Feb 23 2012 from Hotseat

• Important to distinguish between splitting of electrochemical potentials (Module 4.3) and splitting of density of states (Module 4.4). The latter occurs inside magnets (the contacts) and is due to exchange interaction while the former occurs in channel (non-magnetic) due to different interface R. - Supriyo Datta Thu Feb 23 2012 from Hotseat

• Thank you for clarifying that. - Student Thu Feb 23 2012 from Hotseat

### "I have heard Dr Lundstrom refer to your basic bottoms up approach as the Landauer-Datta model. That is indeed an honor. How exactly did you modify the Landauer model? I am sorry to put you on the spot. Can you direct me to your original paper?"

• Not sure of context. Historically Landauer introduced "elastic resistor" in 1960, became popular in 1980's after Imry/Buttiker showed relevance to understanding experiments in small conductors. But common belief still is that it is of no relevance to large conductors which require different approach - Supriyo Datta Thu Feb 23 2012 from Hotseat

• We have argued (as in this course) for ~ 10 years that they are useful for large conductors as well. The results you see in this course are typically obtained from Boltzmann equation, but we obtain them from Landauer approach making them more transparent and accessible. - Supriyo Datta Thu Feb 23 2012 from Hotseat

• I have stressed this connection more in the context of quantum transport. For example, Phys. Rev.B40, p.5830 (1989) relates the Landauer approach to the Keldysh formalism (quantum version of Boltzmann) or the NEGF method discussed in Part II. This is probably more than you wanted to know ! - Supriyo Datta Thu Feb 23 2012 from Hotseat

• Thanks for the info! It's interesting that Landauer introduced elastic resistor. Very long time ago (probably more than 25 years ago...) I tried to read Landauer-Lifshitz books, but it was not so easy... Your lectures at this point are much more understandable. Thank you very much for that!!! - Student Thu Feb 23 2012 from Hotseat

• Thanks. One thing: There may be a potential source of confusion here between Lev Landau and Rolf Landauer. In this course I am generally referring to the latter, except possibly in Part II when we talk about Landau levels in the context of Quantum Hall effect. - Supriyo Datta Thu Feb 23 2012 from Hotseat

• Oh, I do not know much about Rolf Landauer, sorry, it's confusion for me.... :) I'll try to learn it more. Thanks for clarification! - Student Thu Feb 23 2012 from Hotseat

• Thanks for the clarification. I was wondering if you were the originator of explaining concepts with the "bottoms-up" approach in nano electronics. - Student Fri Feb 24 2012 from Hotseat

• I believe so, I do not know of any other place where you would see this approach. Of course, Lundstrom has contributed much to its development. Week 3 (Nanotransistors) is largely his work and Week 5 (Thermoelectricity) is work done in collaboration with him. - Supriyo Datta Sun Feb 26 2012 from Hotseat

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### "Current flows from T2 to T1 in a p-type semi-conductor. I understand electrons are flowing and not heat -- but is the second law of thermodynamics still preserved in all this? "

• Good point. I believe there is no problem with second law as long as the two conductances (G = Current/Voltage and GQ=Heat current/Temperature difference) are both positive. The other two GS, GP can be negative. I discuss second law issues a little in Lectures 16,17 of notes, discussed in Part II. - Supriyo Datta Tue Feb 21 2012 from Hotseat

### "Is the temperature gradient through the channel linear going from T1 to T2?"

• That is what is usually assumed. - Supriyo Datta Tue Feb 21 2012 from Hotseat

### "these particles near to this device without damaging it.I heared about mitigation techniques to prevent radiation.Is there any way,we can regulate the incident radiation on the device and generate electricity?"

• You'd have more than enough electricity from charged particles. Your problem is get rid of it. Ask designers of satellites in Molniya orbit, Tundra orbit, Sirius radio orbit, and similar :) Idea is not feasible: large and vastly varying currents + how to power the s/c when outside Van Allen belts? - Student Mon Feb 20 2012 from Hotseat

• ^^^ I mean, electrically charged particles effect is much, much larger than spin effect in that context. - Student Mon Feb 20 2012 from Hotseat