Nanoscience and Molecular Engineering (ChemE 498A) Semiconductor Nano Devices

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1 Homework 7 Dec 9, 1 General Questions: 1 What is the main difference between a metal and a semiconductor or insulator, in terms of band structure? In a metal, the Fermi level (energy that separates full and empty states) lies in the middle of a band, so there are accessible empty states an arbitrarily small energy above full states The band is partially filled, so that when an electric field is applied the electrons can shift in wave-vector and carry a current In a semiconductor or insulator, there is a finite energy gap between the full and empty electronic states The bands are either completely full or completely empty, and thus can carry no current (no states are available for the electric field to shift them into and unbalance the states at +k and k) It is necessary to dope or ecite the material for it to carry current What do we mean when we talk about a hole in a semiconductor? What are its properties? A hole is an empty state in an otherwise full (or nearly full) electron band Since the total momentum of a filled band is zero, the momentum of the hole is the inverse of that of the missing electron Likewise, the effective mass, energy and effective charge The net result is that the hole acts like a positive charge carrier that can move throughout the crystal 3 Why does a good solar cell design make a poor light emitter? In a solar cell, you want to quickly and efficiently collect the electron and hole induced by the absorption of a photon, separating them before they have a chance to recombine It is thus designed with high fields in the junction region For a light emitter, however, you need the electron and hole in the same place at the same time long enough for them to find each other an recombine (typically several nanoseconds) 4 How can band-gap engineering improve the efficiency of a solar cell? When a photon is absorbed, all the energy in a photon that lies above the band gap of a material typically goes into heat rather than electrical current If you can layer the material so that a large band gap material is on top, followed by layers of decreasing band gap, then less energy goes into heat and more into electrical energy as the high energy photons get absorbed in the top layer, etc You can also have a larger band gap material for the top layer, so the light gets absorbed in the active junction region, and not in the region without an electric field 5 What is the difference between spontaneous and stimulated emission of light? Spontaneous emission may occur anytime an electron in a higher energy electron state transitions into an empty state at lower energy Spontaneous emission has random 1

2 Homework 7 Dec 9, 1 phase and emission direction Stimulated emission is emission induced by the presence of a photon with the right energy, and the stimulated emission has the same phase and direction as the incident photon Spontaneous emission can be thought of as emission stimulated by vacuum fluctuations in the zero-point energy 6 Why does putting a quantum well in the middle of a pn-junction improve its ability to emit light by stimulated emission? The quantum well holds the electrons and holes in the same place long enough that they can find each other and recombine It also allows for an inversion to be set up, with more electrons in the conduction band than in the valence band in that small region of space, so that stimulated emission wins out over absorption when another photon comes by 7 Why does modulation doping improve the speed of a field effect transistor? By putting the dopants far from the interface, the Coulomb scattering of electrons off the ionized dopants is way down Also, the ionized dopants create an electric field that can help confine the electrons into a DEG, cutting out-of-plane scattering 8 Describe qualitatively how to start with a -dimensional electron gas (DEG), for eample at a GaAs/AlAs interface or Si/SiO interface, and pattern it to a 1D or D quantum structure Gate electrodes deposited on the surface, and or etching to pin the Fermi level in a position that empties the relevant states can be used to deplete the electrons from a region of space By varying the potential on gates, electrons may be ecluded form a region, forcing them to travel in a narrow region between the gates The advantage of this is that there are no interfacial scattering defects at the edges of the new 1D or D confined strutures 9 Why does confining conduction to a narrow 1D channel lead to quantized conductance? The requirement that the wavefunction of the electron go to zero at the edges of the channel quantizes the wavevector, k, and hence the state energy A fied number of electrons fits into that quantized state such that the current density that can respond to a given potential difference across the junction gives a quantized ratio 1 Why are some carbon nanotubes metallic and some semiconducting? The way in which the graphene sheet is rolled up to make a nanotube controls the nanotube metallicity If it rolls in such a way that the quantization implied by having an integral number of wavelengths fit around the tube gives a state that passes through the Dirac point where graphene s bands cross, then the tube is metallic If not, then the tube will have a gap

3 Homework 7 Dec 9, 1 Problems: 1 Effective Mass Show that the one dimensional effective mass is given by the inverse curvture of the band structure E(k), 1 m * 1 d E(k) h dk Start from the basic relationships: F dp h dk for a quantum system with momentum p ħk and show that this definition of the effective mass m* lets us also use the standard form F dp m *a where the acceleration a is the time rate of change of the group velocity v g 1 de(k) Now F is just the eternal force, and the h dk internal forces from the variation in the energy with position (or wavevector) is included in the effective mass Quantum mechanically, momentum is related to the wavevector as p hk, so Newton s relationship F dp d(hk) Meanwhile, we can also think of this relationship as the force changing the group velocity: F ma m dv g average energy and wavevector is given by v g 1 h F m dv g m h F m d E dk h dk d 1 m * 1 d E(k) h dk PN junction Fields de dk m d E h dk The group velocity of a wavepacket at a particular de Thus we can write dk de We can use calculus to rewrite de dk, so that dk d(hk) m d E Thus the two formulations are equivalent if h dk 1, or Assume that a p-n junction is characterized by two sharply defined charge regions of charge density ρ as shown, where N a and N d are the density of acceptors and donors, respectively, and the space charge depletion wihs are W a and In this central region, the free carrier densities are orders of magnitude smaller than in the bulk of the device since the electrons and holes recombine a) For <, you can write the Poisson equation (MKS units) as V qn a where dielectric constant (κ o) We are using MKS so the answers will be in Volts, and not stat-volts or esu Integrate this equation and a similar one for > to show that the barrier voltage is given by: ρ +qn d W a qn a Wd 3

4 Homework 7 Dec 9, 1 V b V 1 + V q N W ( a a + N d ) Note that from charge neutrality, N d W a N a, so that it may also be written as: V b qw a N a 1 + N a N d Note that V at -W a and Wd, since the potential is constant in the uncharged region, and we epect both potential and field to be continuous Define the zero of potential to be that at - <: Poisson equation: V qn a Integrate LHS: Integrate RHS: W a W a V d qn a qn a V V V W a V ( + W a), for qn a ( + W a) [the above gives the negative of the electric field] Integrating again for the potential: W a V d V() qn a + W a W a qn a W a >: Poisson Eqn: V RHS: qn d d qn d qn d ; LHS: ( ); qn d V d V V ; ( )d qn d V d V ( ) + V () Rearranging, we have V ( ) V () + qn d q N aw a + N ( d ) Using charge neutrality we have N aw a N d, for a net result of V barrier qw N a a 1+ N a N d b) Sketch the potential as a function of through the junction Justify your linear, quadratic, or whatever you find, dependence of the potential, as well as which side is positive or negative Show that the wih of depletion region i (j is the other region) is given by: W i V b qn i 1 + N i N j 4

5 Homework 7 Dec 9, 1 Potential (Volts) N a 3e-17 N d e o p n Field (Volts/micron) p n Position (microns) Position (microns) The potential is quadratic in position, with a different curvature on each side of the junction This is due to the constant charge density The electric field set up counteracts the diffusive chemical potential gradient It is in a direction to prevent majority carriers from crossing the junction Thus the potential for positive charges is low on the p-side and high on the n-side Typically, band diagrams are shown for negatively charged electrons, so they have the opposite sign Note that the side of the junction with lower doping levels has the larger potential drop, the wider depletion region and the higher field Qualitatively, the larger wih is because you need to ionize donors or acceptors deeper into the crystal to get the same total charge; the larger fields and potentials are because they are not as well screened by the lower charge density N d + N a To find the total wih of the junction, note that W a + W tot W a From above, we N d have V barrier qw N a a N a + N d qw N d d N a + N d qw i N j N i + N j Rearranging, we find: N d N a N i V b W i qn j 1+ N We can also do some messy algebra that I won t type up to solve for j N i W tot V b q N a + N d N a N d c) Show that in the case where N a «N d, the wih of the depletion region to a good approimation may be written as W tot μv b, where σ conductivity and μ mobility σ When a single type of carrier dominates, the ratio of the conductivity to the mobility is the charge carrier density nq: σ J E nqv d E nqμ ( J σe and v drift μe) If Na «Nd, then W tot V b qn a If we measure the conductivity on the more lightly doped, p-side of the junction, and use the hole mobility, then qna σ/μ, and W tot V b μ / σ For a given piece of silicon, you can measure σ, and μ is roughly constant for similar samples 5

6 Homework 7 Dec 9, 1 d) The junction capacitance is defined by C j dq dq In the practical case for which one side of the junction is very lightly doped while the other is heavily doped (N a «N d ), show the capacitance may be written: C j σ μv b Junction capacitance:c j dq dq The second term is d μv b σ 1 / μ σv b The total charge on this capacitor is: Q qn a WA σ WA, where A area of the junction Thus μ dq σ A Note that σ is NOT charge density here, but conductivity The capacitance per unit area μ is then: C j dq / A σ μ μ σv b σ μv b e) Si-doped GaAs (n-type) with carrier density ~ 1 16 cm -3 has room temperature mobility ~ 8 cm /V-sec, for σ ~ 1 mho/cm, 13 o, V b ~ 7 V Find the wih of the depletion region and the junction capacitance in the case N a «N d Convert standard numbers to MKS: hole mobility ~ 1 3 cm /V-sec, or about 1 m /V-sec, σ ~ 1 mho/cm 1 mho/m, 119o 1-1 MKS, Vb 7 V, then W tot V b μ / σ 4 microns and C 7e-4 F/m 7 nf/cm Note this value of σ/μ yields a doping level of ~ 1 15 cm -3 f) Sketch the bands for a double heterostructure pn-junction, with a central region of undoped GaAs (band gap 144 ev) of wih ¼ what you found in (e) with n-type AlGaAs (n ~ 1 16 cm -3 ) on one side (band gap 175 ev) and p-type AlGaAs (p ~ 1 15 cm -3 ) on the other Eplain your reasoning This is the eample that was discussed in class The wide band gap material creates barriers for electrons and holes to go fully through the junction, allowing them to accumulate under bias and build up an inversion 6

7 Homework 7 Dec 9, 1 3 D Subbands in a MOSFET (adapted from Kittel ISSP Problem 17) The contact plane between two semiconductors with different band gaps, or a semiconductor and an insulator, gives rise to an offset in the bands at the interface When an electric field is applied, electrons can be trapped by a triangular well at the interface The potential energy for an electron in the conduction band may be approimated as V (z > ) eez and V (z < ), where the interface is at z The wavefunction must go to zero for z <, and for z > may be separated as ψ (, y, z) u(z)e ik e ik y y, where u(z) satisfies the differential equation h d u + V (z)u u m dz The eact solution to the triangle well is an Airy function, but the variational trial function u(z) Azep(-az) gives a reasonable approimation to the ground state energy* a) Using the trial wavefunction, show that the energy is h a m + 3eE a The trial wave function in the z-direction is u Ae a First, normalize the wave function so that 1 u *u d A e d A (!)(a) 3 A 1 4a 3, or A a3/ The epectation value for the energy for this trial wave function is: u * Hu d A e a ħ m + ee e a d The electric field term is: A ee 3 e d 4a 3 ee( 3!(a) 4 ) 4a3 ee 16a 4 3 ee a The kinetic energy term requires derivatives of u: d d e a (1 a)e a d ; d e a (a a)e a The integral becomes: A ħ ( m a a )e a d 4a3 ħ m [ a( a) a ( a ) 3 ] ħ a m Therefore: ħ a m + 3 ee for this trial wave-function a b) Show the energy has a minimum when a 3eEm h To find the value of a that minimizes the energy, check when the derivative is zero: d da ħ a m 3 ee a ; a3 3eEm ħ To check for minimum or maimum, d da ħ m + 3 ee a 3 > Positive curvature means this is a minimum energy for this form of the wave function 7

8 Homework 7 Dec 9, 1 c) Show that min 189 h m 3eE / 3 Substituting the value of a found in part (b) to the energy epression in part (a) yields: ħ a m + 3 ee a ħ 3eEm m ħ 3eE / 3 ħ m / eE + 3 ħ ee 3eEm / 3 ħ m 3 3eE 189 / 3 ħ m This wave function u needs to be multiplied by a -dimensional Bloch wave in the y and z directions for the total wave function The energy is then tot 189 3eE / 3 ħ + ħ k m * m *, where k is the wave-vector of that -D Bloch wave and m* is the effective mass in that -D periodic potential When making a quantum well with electrons at the conduction band minimum, that effective mass is essentially the same as the 3-D value You should use m* for the confinement energy, as well, since the interface state was built out of CBM states In the eact solution, the factor of 189 is replaced by 178 As the electric field increases, the etent of the wavefunction in the z direction (a) decreases The function u(z) defines a surface conduction channel on the low energy side of the junction The various eigenvalues of u(z) define what are known as electric subbands The eigenfunctions cannot carry current in the z direction, but do carry a surface channel current in the y plane The electric field can turn the channel on or off by moving the quantized level through the Fermi level, resulting in a field effect transistor or allowing the DEG to be patterned into 1D or D d) Eplain how changing the applied electric field can move the ground state through the Fermi level and turn the interface layer on or off for conduction At low fields, the ground state will be low, and a state might be occupied Increasing the electric field can then raise the ground state to a point where it is no longer occupied There will then no longer be electrons in the channel, and it can no longer carry current 4 Carbon Nanostructures Read any one of the thousands of papers on nanotubes, graphene or buckyballs Choose one figure from these papers (that includes data or the results of a theoretical calculation) Eplain the figure what happened to generate the result (ie, what was the technique, sample geometry, etc), what the result is, and what the authors say it means Include a copy of the figure and the citatioin of the reference from which you took it HINT: go to UW libraries home page, and select Articles and Research Databases The two I find most useful are Web of Science and Engineering Village Enter a few terms into the topic line (eg nanotube and transport and blockade) In Engineering Village you can ask it to restrict the result to General or Review papers 8