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y x (b) r (0:067) me E j j } tan r (0:067) me E j j }! () Root is: E = 0:43 63 ev. r (0:067) me (j0:5 j j E j jj) } = Derive the tunneling probability (6.38) for

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y x (b) r (0:067) me E j j } tan r (0:067) me E j j }! () Root is: E = 0:43 63 ev. r (0:067) me (j0:5 j j E j jj) } = Derive the tunneling probability (6.38) for the double barrier junction depicted in Fig. 6. on p Research how tunneling is utilized in ash memories, and describe one such commercial ash memory product Research how eld emission is used in displays, and summarize the state of display technology based on eld emission Explain how a negative resistance device can be used to make an oscillator. Solution: An L circuit can exhibit a pure resonance, although the presence of a ordinary resistor will dissipate energy and damp the oscillation. A negative resistance can cancel the ordinary resistance, and lead to a sustained oscillation. Alternatively, a circuit consisting of an inductor, a capacitor, and a negative resistance can, in theory, exhibit oscillations that grow in time. 7 Problems hapter 7: oulomb Blockade and the Single Electron Transistor 7.. For a tunnel junction with = 0:5 af and R t = 00 k, what is the R time constant? What does this value mean for the tunnel junction circuit? Solution: = R t = 0: = 5:0 0 4 s () is the characteristic time between tunneling events. 7.. For a tunnel junction having = 0:5 af and R t = 00 k, what is the maximum temperature at which you would expect to nd oulomb blockade? Repeat if = : pf. Solution: We must have q e k BT; (3) 36 so that For = : pf, T :6 0 9 = k B (0:5 0 8 ) (: = 855K. (4) ) T :6 0 9 = k B (: 0 ) (: ) = 7: K. (5) 7.3. For the capacitor depicted in Fig. 7. on p. 5, for an electron to tunnel from the negative terminal to the positive terminal we determined the condition V (6) (see (7.8)). Show, including all details, that for an electron to tunnel from the positive terminal to the negative terminal we need V : (7) Solution: Let a single electron tunnel through the insulating layer from the positive terminal to the negative terminal, such that charge Q resides on the top plate, and Q + resides on the bottom plate. The energy stored in the eld of the capacitor is now such that the change in energy stored is E f = (Q ) ; (8) E = E i E f = Q (Q ) = (Q =) : (9) It must be energetically favorable for the tunneling event to occur, and so, if we require E 0, then we nd that for tunneling to occur. Thus, 7.4. The energy stored in a capacitor is (Q =) V 0; (30)! Q (3) (3) E = Q : (33) where Q is the charge on the capacitor plates (+Q on one plate, Q on the other plate). Using this equation, derive the condition on charge Q and voltage V for N electrons to tunnel across the junction at the same time, in the same direction. Solution: Before tunneling, we have +Q on one plate, and Q on the other plate. If N electrons tunnel from the negative to the positive plate, then after tunneling, on the positive plate, we have Q f = +Q + N, and on the negative plate, Q f = Q N. The change in energy is E i E f = Q (Q + N ) = N Q + N : (34) If we force E i E f 0; (35) 37 then or Q N (36) V N : (37) 7.5. There is always a capacitance between conductors separated by an insulating region. For the case of conductors associated with di erent circuits (i.e., circuits that should operate independently from one another), this is called parasitic capacitance, and, generally, / =d, where d is some measure of the distance between the conductors. For two independent circuits d is often fairly large, and thus very small parasitic capacitance are generally present. (a) Using the formula for the impedance of a capacitor, show that even for very small values of, at su ciently high frequencies the impedance of the capacitor can be small, leading to signi cant unintentional coupling between the circuits. (b) Describe why the small values of parasitic capacitance, which can easily be af or less, do not lead to oulomb blockade phenomena. Solution: (a). Z c = j! ; (38) and, therefore, even if is very small, if! is su ciently large jz c j can be small. (b). If is small due to large separation of the conductors, R t will be too large (i.e., tunneling will not occur), and no oulomb blockade phenomena will occur For the SET oscillator shown in Fig. 7.0 on p. 4, derive the oscillation period given by (7.3). Solution: Using (7.5), then so that V (t) = V Z t 0 I s dt = I s t; (39) T = I s T = e ; (40) T = e I s = jj I s : (4) 7.7. As another way of seeing that oulomb blockade is di cult to observe in the current-biased junction shown in Fig. 7.9 on p. 3, consider the equivalent circuit shown in Fig. 7. on p. 5. If R b is su ciently large so that it can be ignored, and if jz L j is su ciently small, construct an argument, from a total capacitance standpoint, for why the amplitude of the SET oscillations will tend towards zero. Solution: Referring to Fig. 7., if we ignore R b and Z L, then we have simply a current source in parallel with a capacitance and a tunnel junction. The tunnel junction itself consists of a capacitor in parallel with a tunnel resistance, and therefore the two capacitors combine in parallel (i.e., algebraically), resulting in, typically, a large capacitance. The amplitude of the oscillation is e =, and for large this amplitude is small For the quantum dot circuit depicted in Fig. 7.3 on p. 6, assume that initially there are n = 00 electrons on the dot. If a = b = : af, what is the condition on V s for an electron to tunnel onto the dot through junction b? Solution: From so that a a 38 n + ; (4) 00 + : (43) For a = : af, then : = 3:4 V (44) 7.9. For the quantum dot circuit depicted in Fig. 7.3 on p. 6, we found that for an electron to tunnel onto the dot through junction b, and then o of the dot through junction a, we need (45) if initially there were no electrons on the dot (n = 0), where a = b =. It was then stated that for the opposite situation, where an electron tunnels onto the dot through junction a, and then o of the dot through junction b, we would need V s : (46) Prove this result, showing all details. Solution: Assume that initially there are n electrons on the island, and that the initial voltages across the junctions a and b are Va i and Vb i, respectively. The initial charges on the junctions are Qi a and Q i b. Let one electron tunnel onto the island through junction a. The voltage drops across junctions a and b become Va f = (V s b (n + ) ) = Va i V f b ; (47) = (V s a + (n + ) ) = V i b + ; (48) such that V s = V a + V b is maintained, and the resulting charge stored by junction b is Q f b = bv b = b V i b + b = Q i b + b : (49) The change in charge Q b = Q i b Q b must come from the power supply (the change in charge on junction a is associated with the tunneling event), and is associated with the supply doing work W = V s b : (50) Upon the electron tunneling onto the island through junction a, the change in the total energy (the change in the stored energy minus the work done) is E t = E se W (5) = a b Vs + (n ) a b Vs + ((n + ) ) b + V s ; and requiring that this energy change be positive, i.e., that the tunneling event is energetically favorable, n + V s b 0; (5) such that V s qe If we let a = b = and n = 0, then we have b n + ; (53) V s : (54) 39 7.0. onsider the double-junction oulomb island system depicted in Fig. 7.3 on p. 6, and refer to Fig. 7.7 on p. 3. Assume that an electron tunnels onto the island subsequent to applying a voltage V = e =. Draw the resulting energy band diagram, showing the readjusted energy bands, and the re-establishment of oulomb blockade. 7.. For the SET shown in Fig. 7. on p. 6, assume that a = b = 0 af, g = :4 0 6 F, and V g = 0: V. If initially there are 75 electrons on the island, then what is the condition on V s for an electron to tunnel across junction b and onto the island? Solution: From so that we have n + + ( a V s + g V g ) 0; (55) (n + =) g V g a : (56) (75 + =) :4 0 6 (0:) = :4 V. (57) 7.. For the SET we considered electrons tunneling onto the island through junction b, then o of the island through junction a, resulting in positive current ow (top-to-bottom). Explicit details were provided for the generation of oulomb diamonds for I 0, and the results merely stated for I 0. Fill in the details of the derivation predicting oulomb diamonds for I Draw the energy band diagram for a SET (a) under zero bias (V s = V g = 0), (b) when 0 and V g = 0, and (c), when 0 and V g = j j = ( g ) onsider an electron having kinetic energy 5 ev. (a) alculate the de Broglie wavelength of the electron. (b) If the electron is con ned to a quantum dot of size L L L, discuss how big the dot should be for the electron s energy levels to be well-quantized. (c) To observe oulomb blockade in a quantum dot circuit, is it necessary to have energy levels on the dot quantized? Why or why not? Solution: (a) E KE = r m ev EKE! v = (58) m e = h p = h m e v = = 0:548 nm. h q = E m KE e m e h p me E KE j j (59) The electron will act like a classical particle when L. (b) L. (c) It is not necessary to have energy levels on the dot quantized. The developed formulas did not assume energy quantization Assume that a charge impurity q = resides in an insulating region. Determine the force on an electron 0 nm away from the impurity. Repeat for the case when the electron is 0 m away from the impurity. Solution: The force on an electron by a charge is F = E = br 4 0 r! jfj = qe 4 0 (0 0 9 ) = 4:67 0 N; (60) 40 and at 0 m, jfj = q e 4 0 (0 0 6 ) = 4: N. (6) 7.6. Universal conductance uctuations (UF) occur in oulomb blockade devices due to the interference of electrons transversing a material by a number of paths. Using other references, write one-half to one page on UFs, describing the role of magnetic elds and applied biases Summarize some of the technological hurdles that must be overcome for molecular electronic devices to be commercially viable There is currently a lot of interest in spintronics, which rely on the spin of an electron to carry information (see Section 0.4). In one-half to one page, summarize how spin can be used to provide transistor action. 8 Problems hapter 8: Particle Statistics and Density of States 8.. Energy levels for a particle in a three-dimensional cubic space of side L with hard walls (boundary conditions (4.50)) were found to be (4.54), E n = } m el n x + n y + n z ; (6) n x;y;z = ; ; 3; :::, which leads to the density of states (8.6). Using periodic boundary conditions (4.55), energy levels were found to be (4.59) E n = } m el n x + n y + n z ; (63) n x;y;z = 0; ; ; :::. Following a derivation similar to the one shown for (8.6), show that the same density of states arises from (8.65). Solution: In the hard wall case we only count the rst octant of the aforementioned sphere, since sign changes don t lead to additional states. For the case of periodic Bs, sign is important, and so we are interested in the total number of states N T having energy less than some value (but with E E ). This is approximately the volume of the sphere N T = 4 3 n3 = 4 3 E E 3= (64) (a factor of =8 was present for the hard-walled case). The total number of states having energy in the range (E; E E) is N T = 4 3 = 4 3 ' 4 3 = 4 3 E 3= E 3= E 3= E 3= E 3= E 3= (E) 3= (E E) 3= (65)! 3= (E) 3= E 3= E E 3 E E 3 E = E= (E) ; E where we used ( x) p ' px for x. The density of states (DOS), N (E), is de ned as the number of states per unit volume per unit energy around an energy E. The total number of states in E 3= 4

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