Summary of Lecture Notes on Subbands, Density of States, and Current in Ballistic Nanowires and Nanotransistors

Transcription

1 Mani Vaidyanathan 1 Summary of Lecture Notes on Subbands, Density of States, and Current in Ballistic Nanowires and Nanotransistors Overview 1. We began the final part of the course by recapitulating the main topics we have covered up to this point. The topics, and what we covered, are as follows: (a) The Basics of Electron Flow in Nanodevices: We took an initial view of all the elements that are typically required to understand electron flow in a structure that is very small, i.e., a so-called nanostructure. These include ideas such as Fermi function, electrochemical potential (or quasi-fermi level), channel energy levels, density of states, and channel self-consistent potential. (b) The Schrödinger Equation: We became acquainted with the basics of quantum mechanics, the governing physics of nanostructures and nanoparticles, and the the timeindependent Schrödinger equation (TISE), which must be solved to gain energy information for electrons in the channel materials of transistors, such information being a crucial factor in determining electron flow. We also saw how to convert the TISE from a differential equation into a matrix equation using the method of finite differences, leading to a matrix representation [Ĥ] of the Hamiltonian operator, and an associated vector representation {φ} of the wave function. (c) Basis Functions: We learned another way, i.e., that of basis-function expansion, to convert the TISE into a matrix equation, this time involving the matrices [Ĥ] u and [S] u, and an associated vector representation {φ} (different from the earlier case) of the wave function. (d) Band Structure: We learned how to use the method of basis-function expansion to solve for the energy-versus-wave-vector or E-k information of a solid that has a regular, periodic array of atoms. 2. We then stated the final goals of the course: (a) To introduce the idea of a subband, which refers to E-k information derived as a subset of the overall E-k information for a material; such a subset results from a physical constraint, such as when graphene is rolled into a tube. (b) To see how E-k information can be used to get the density of states (DOS), which we ve already used to predict current with the rate-equation model at the beginning of the course. (c) To see how E-k information can be used to predict current in a nanowire and nanotransistor, using an approach that is different from the rate-equation model. 3. The key point is that collectively, this course then provides an introductory story to answer the question, When a voltage is applied across a nanostructure, how does one compute the current through it? If I have been successful, then by the end of the course, you would be able to provide a credible answer to this question, even if you can t remember all the details.

2 Mani Vaidyanathan 2 4. We ended our introductory remarks by considering a few quotes on future technologies, which I supplied to you on a handout. Nobody knows what future information-processing technologies will be like. However, at least from an electrical-engineering point of view, this course has introduced you to the fundamentals required to investigate and understand potential successors to present-day technology. Subbands: Graphene to Nanotubes 1. To illustrate the idea of subbands, we considered what happens to the E-k picture of graphene when it is rolled into a tube, commonly called a carbon nanotube We covered the idea of a circumferential vector c. Such a vector connects two lattice (not atomic) points L 1 and L 2 when graphene is rolled over into a tube. The vector c must hence be a lattice vector: c = ma 1 + na 2 = (m + n)aˆx + (m n)bŷ (1) where a 1 and a 2 are lattice basis vectors, m and n are integers, and a and b are lattice dimensions in the ˆx and ŷ directions, respectively. 3. By reconsidering what had happened when we had a one-dimensional array of atoms connected in the form of a ring, we argued that once graphene was rolled into a tube, one would require e ik c = 1. Look at that argument again, and make sure you are happy with it. We then showed that the requirement would ultimately lead to the following condition: or, equivalently, k c = 2πν, ν an integer (2) k x a(m + n) + k y b(m n) = 2πν (3) Equation (3) defines a series of parallel lines in the k x, k y -plane, one for each ν. The point you should note is that the vector k is now constrained to lie on one of these lines; that s what rolling graphene into a tube does. Also notice from (3) that the allowed lines are determined by the kinds of things you d expect: m and n, which are integers in the circumferential vector c, and which hence describe how the tube has been rolled up; and the dimensions a and b, which are properties of graphene s lattice structure. 4. We drew a sketch illustrating some hypothetical allowed lines of k in k-space, superimposed on the Brillouin zone. We then stated that a subband is just an E-k relation along one of 1 If needed, please review the band structure information for graphene itself; reacquaint yourself with the main results, especially the idea of a Brillouin zone and the fact that the energy E is really a surface on the k x, k y -plane.

3 Mani Vaidyanathan 3 these lines, which then represents a subset of the overall E-k relation; each such allowed subset is said to define a subband or subrange of allowed energies With the idea of a subband established, we considered metallic versus semiconducting tubes. We considered a collective plot of E(k) versus k, where k is now a (one-dimensional) scalar quantity along any allowed line of k. Such a plot will show a + and a subband for each allowed line. 3 Look at the plot, and make sure you understand its conceptual origin. (a) We said that if such a plot shows a gap or energy gap or bandgap between the lowest + subband and the highest subband, then the nanotube would exhibit semiconducting properties; electrical conduction through the tube can be controlled by controlling (electrically or chemically) the number of electrons having energies in the + ranges versus those in the ranges. (b) On the other hand, if such a plot shows no gap between the + and subbands, then we said the nanotube would be metallic. We went on to derive the following condition for a metallic tube: m n = ν = an integer (4) 3 where m and n describe how the tube has been formed by determining the circumferential vector c = ma 1 + na 2, and where the pair (m, n) is hence said to define the tube s chirality. Please look at the derivation, and make sure you can follow the steps and that you can understand the conclusion. 6. We ended our look at nanotubes by examining zigzag and armchair tubes. A zigzag tube occurs when graphene is rolled in the y-direction, whereas an armchair tube occurs when graphene is rolled in the x-direction. The names derive from the atomic patterns in the respective directions. 7. We found that zigzag tubes may or may not be metallic, but armchair tubes are always metallic. You should understand why this is the case, and be able to make the appropriate arguments yourself. Density of States 1. The density of states or DOS D(E) tells us the number of energy states (levels) per unit energy. Recall from the beginning of the course that D(E) is needed to determine current. 2 There can be some confusion in the language used to define subbands. For example, in a nanotube, every line of allowed k gives rise to + and a allowed solutions to the E-k relation along the allowed line, where E(k) = E 0 ± t c cos k y b cos k x a + 4 cos 2 k y b with E 0 0. Some authors (and your humble instructor) would count these as two subbands, whereas other authors (including the author of your textbook) might count them as one. It doesn t really matter, as long as we all agree on a convention and stick to it. I ll count them as two. 3 These correspond to the ± solutions in the E(k) expression of footnote 1.

4 Mani Vaidyanathan 4 2. We learned how to obtain D(E) from E-k information. To do this, we considered parabolic E-k relations: E = E c + 2 2m c k 2 (5) E = E v + 2 2m v k 2 (6) These equations are assumed to describe conduction and valence subbands of a material, where d 2 E/dk 2 > 0 in a so-called conduction subband, and d 2 E/dk 2 < 0 in a so-called valence subband. The parameters E c and E v refer to the minimum and maximum energy values of the respective bands. The parameters m c and m v are called band-structure effective masses, and are sometimes chosen to fit the above forms to actual E-k relations. 4 In this context, note that the forms (5) and (6) imply that m c > 0 and m v < To get the density of states, we talked about how to count states. Look at that material now. Make sure you are comfortable with the main results: (a) The components of the wave vector k are discretized, according to k x = 2π L x integer, k y = 2π L y integer, k z = 2π L z integer (7) where L x, L y, and L z are the material (solid) dimensions in the x, y, and z directions, respectively. Usually, these dimensions are large, in the sense that the solid is comprised of many unit cells in every direction; hence, for most purposes, as we previously discussed, k x, k y, and k z can be considered to be essentially continuous. However, for counting states, we need to admit that the values are strictly discrete. (b) We can convert any discrete sum over wave vectors k into a continuous integral. Specifically, for an arbitrary function g(k), we can write g(k) k g(k) dk x dk y dk z (2π/L x )(2π/L y )(2π/L z ) 4. We used the form (5) to investigate the density of states in one-, two-, and three-dimensional systems, with k representing the magnitude of the wave vector k. Look at the derivations, and make sure you can repeat the steps for an arbitrary E-k relation. 4 As I mentioned in class, E-k relations are also called dispersion relations. (8)

5 Mani Vaidyanathan 5 Here are the results: D(E) = Lm c π D(E) = Sm c 2π 2 [ 1 2m c (E E c ) ] 1/2 D(E) = Ωm c 2π 2 3 [2m c(e E c )] 1/2 where L L x, S L x L y, and Ω L x L y L z are the length, area, and volume of the material in the one-, two-, and three-dimensional cases, respectively. Current in Ballistic Nanowires and Nanotransistors 1. To finish off your introductory course in nanoelectronics, I tried to bring together all the things we ve learned in order to develop an ac (small-signal) equivalent circuit for a (fieldeffect) nanotransistor To do this, I began by reviewing three approaches to describing current in electronic devices. (a) One approach is to employ the rate-equation model from the beginning of the course. Doing this requires knowledge of the rate constants γ 1 / and γ 2 /, which characterize how easily electrons move between the channel and the respective contacts. Such constants have to be calculated from quantum mechanics. If the rates are known, then the rate-equation approach can be viewed as a simplification of a rigorous quantum-mechanical method for obtaining the current. Such an approach is necessary when the channel length L of the transistor is extremely small (L 10 nm). (b) A second approach is to use the famous drift-diffusion equation (DDE): J n (x) = qµ n E F (x)n(x) + qd n dn dx where x is position, J n (x) is the current density, n(x) is the electron concentration, E F (x) is the electric field, and µ n and D n are the electron mobility and diffusivity, respectively. This is a classical method that is reasonably valid for large channel lengths (L 100 nm). (c) A third approach is to use the so-called Boltzmann transport equation (BTE). It is known as a semiclassical approach because it mixes classical and nonclassical ideas, and it is suitable for channel lengths that are between the quantum-mechanical and classical limits (10 nm L 100 nm). We stated that in moving towards an equivalent circuit, we d use the semiclassical approach (c). Our approach is hence quite valid for modern nanotransistors with channel lengths between 10 and 100 nm. 5 A nanowire can be obtained from a nanotransistor simply by neglecting the gate terminal. (9) (10)

6 Mani Vaidyanathan 6 3. The semiclassical approach requires us to compute the average electron velocity from the E-k curve of the channel. For an electron in a state k (assuming a one-dimensional system), we stated that the average velocity is v(k) = 1 de dk (11) 4. At this point, I digressed to sketch out the general visualization and approach. Look again at the diagrams and equations, and note the following: (a) In the semiclassical approach, the channel is subdivided into a sequence of small segments, each of which is at a local potential V (x). 6 (b) The local potential shifts the local E-k curve that applies to the local channel segment. For this to work, we stated that V (x) has to be slowly varying, which simply means that it doesn t have any sharp spikes or abrupt changes. (c) A sketch of the overall energy picture would hence show a sequence of shifted E-k curves, sitting side-by-side. Each curve has the (presumed parabolic) form E(k) = 2 2m c k 2 + E c (x) = 2 2m c k 2 + E c,eq qv (x) (12) where E c,eq is the band minimum at equilibrium, and E c (x) = E c,eq qv (x). (d) A plot of the band minimum E c (x) versus x is called a conduction-band diagram for the device. Note that it is essentially a plot of the potential variation V (x) through the device, since de c (x) = q dv (13) dx dx (e) If you study the overall picture, it s easy to see that it s quite involved. There are electrons all through the device at various positions x, and at each position, the electrons are (somehow) distributed among the available states k. To handle the situation, we hence define a distribution function f(x, k), which tells us the number of electrons in the state k at the point x. (f) If we can get our hands on f(x, k), then we can find the electron concentration n(x) and current J n (x) at x by appropriately summing f(x, k) over all electronic states k, as follows: n(x) = 1 f(x, k) (14) Ω J n (x) = k k qv(k) f(x, k) Ω where Ω is just the volume of the channel segment at x. 6 We worked in one dimension througout this discussion. (15)

7 Mani Vaidyanathan 7 (g) I then mentioned that getting f(x, k) is tricky, because we d have to solve the Boltzmann transport equation, which is an involved differential equation: 0 = v f x + qe F f k + C(f) (16) where C is known as the collision operator, and where f is subject to the boundary conditions that f = f 1 in the source and f = f 2 in the drain. Don t worry about the details of the equation, but just note that what it says is reasonable: the distribution of electrons (f) depends on their velocities (v), forces acting on them (E F ), and on the collisions they experience (C). The BTE was originally developed for classical gas particles; however, note that in the form (16), it contains the constant, a number intimately tied to nonclassical (quantum) mechanics. This is one reason that the BTE is known as a semiclassical equation. It s also worth noting that when the channel is large (L 100 nm), then the BTE (16) is equivalent to the DDE (10). 7 (h) Solving the BTE (16) is very tricky. Hence, I suggested that we use a simplified visualization and approach that would work when the electrons did not experience any collisions in moving between the source and the drain, i.e., under so-called ballistic conditions. This is a good first approximation for many types of nanotransistors. 5. The simplified visualization under ballistic conditions contained two key parts: (i) A single E-k diagram and a single channel potential V for the entire channel; in other words, we use only a single channel segment to represent the entire channel when the transport is ballistic, and spatial variation within the channel is neglected. (ii) A simple three-capacitor model for the electrostatics (instead of Poisson s equation), just like at the beginning of the course; remember that transport equations (like the BTE) always have to be solved self-consistently with the electrostatics. 6. With the aid of (11), and the simplified visualization, we then wrote the transport equations for charge and current. 8 I won t repeat the discussion here, I ll just quote the final results. Try to become comfortable with the picture we used, which hinges on how states are filled from the source and drain, and how to convert sums over the wave vector k into integrals, i.e., how to count states. Here are the results for the numbers 9 of right- and left-going electrons: n + = 1 2 n = 1 2 E c E c f 1 (E) D(E) de f 2 (E) D(E) de (17) 7 This can be shown using some mathematics, but we ll have to skip it in ECE 456 due to a lack of time. 8 As a warning, even though the equations look eerily similar to ones we had with the rate-equation approach, please note that they are not the same. I strongly suggest that you do not mix the approaches or even try to reconcile them. They look similar because they describe similar things, but they derive from quite different visualizations. 9 In the forms (17) that we derived, n + and n are strictly numbers, not numbers per unit volume, since we didn t divide by the channel s volume.

8 Mani Vaidyanathan 8 with f 1 and f 2 being the source and drain Fermi functions, given by f 1 (E) = f 2 (E) = e (E µ+qv S)/k B T e (E µ+qv D)/k B T (18) where µ is the equilibrium Fermi level, V S and V D are the source and drain voltages, k B is Boltzmann s constant, and T is the temperature. And here are the results for the (conventional) current in the right- and left-going directions: I + = q h I = q h E c E c f 1 (E) de f 2 (E) de (19) with the conventional drain current (defined as positive into the drain) then being given by I D = I I + (20) 7. As I ve already mentioned, transport equations must always be combined with electrostatic equations, and we used the same three-capacitor model for the electrostatics as with the rate-equation approach at the beginning of the course. The channel potential is then [ ] CS V S + C G V G + C D V D U = qv = q C E + q2 (n n 0 ) C E (21) where C E C S + C G + C D is the total electrostatic capacitance, with C S, C G, and C D being the electrostatic capacitances between the channel and the source, gate, and drain, respectively; these would have to be found from electrostatic considerations of the transistor s structure. I mentioned that in a field-effect nanotransistor, we often write C S C JS, C D C JD, and C G LC ox, where C JS and C JD are junction electrostatic capacitances associated with the source- and drain-channel junctions, and C ox is called the gate-oxide capacitance per unit length L. 8. Keep in mind the effect of the channel potential, which is to shift the entire E-k curve by shifting the band minimum: E c = E c,eq + U = E c,eq qv (22) where (as mentioned earlier) E c,eq is the band minimum at equilibrium. Thus, the lower limits of integration in (17) and (19) depend on V, and D(E) depends on V through E c in the first relation in (9).

9 Mani Vaidyanathan 9 9. So, how do we exploit these types of equations in order to get an equivalent circuit for the transistor in this case, an ac or small-signal equivalent circuit. 10 (a) The starting point is to draw the three-capacitor electrostatic network used to model the effects of the electrostatic equation (21). We obviously need at least this much. (b) To this, we add two transport or quantum capacitances, defined as the per-unitlength capacitances found from appropriately differentiating (17), as follows: C SQ = q n + L (V V S ) C DQ = q n L (V V D ) (23) Don t worry here about the math that might be required to actually find these from (17); rather, understand what these capacitances model, i.e., that for small ac signals, they model the transport equations for charge. (c) We also need to model the transport equations for current, which we do with a couple of transport or quantum transconductances, defined by appropriately differentiating (19), as follows: g SQ = g DQ = I + (V S V ) I (V D V ) (24) Once again, don t worry about the math that might be required to actually find 11 these from (19). Focus on the fact that for small ac signals, these transconductances model the transport equations for current. (d) The overall ac circuit is then found by connecting the electrostatic network with the quantum capacitances and transconductances. The connections may not be obvious, only because we did not discuss them at length. However, the point is that everything has to be hooked together such that, for ac signals, the electrostatics and transport agree on the charge and potential at the channel node, and on the proper currents for given ac terminal voltages. If you compare, you ll see that the circuit is the same (except for a difference in nomenclature) as the one on the last page of the course outline. Hence, we have come the full distance from Schrödinger s equation all the way up to something useful to electrical engineers, namely, a circuit diagram. 10 I m talking about something like the well-known hybrid-π model for bipolar transistors. 11 Similarly, don t worry about why I used V V S,D in the differentials of (23) versus V S,D V in those of (24).

10 Mani Vaidyanathan 10 Conclusions Let go of the details for a moment. Here s what you learned: 1. A subband is just a subset of the E-k relation for a material, where the subset s relevance typically has its origins in some physical constraint. As an example, we looked at subbands in carbon nanotubes, i.e., subbands that arise when graphene is rolled into the form of a tube, and we saw how such subbands could determine the tube s properties (metallic or semiconducting). 2. The density of states or DOS D(E), which is needed to determine current, can be found by appropriately counting states in E-k information. 3. Using a semiclassical approach, which is different from the rate-equation approach we studied at the beginning of the course, it is possible to write equations describing the operation of a nanotransistor. These equations can then be used to construct an ac equivalent circuit for a nanotransistor, something an electrical engineer might use to better understand the transistor s capabilities in analog-circuit applications. At this stage, it s more important than ever to view the material we ve covered in two levels of abstraction; keep the high-level results separate from the details, and delve into the details only when needed. Together, a view of the big picture, combined with an ability to handle the details if required, means you ve understood the course, i.e., you have an honest, introductory understanding of the emerging and challenging field of nanoelectronics.