ECE 802 604 Nanoelectronics Prof Virginia Ayres Electrical

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ECE 802 -604: Nanoelectronics Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University

ECE 802 -604: Nanoelectronics Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu. edu

Lecture 14, 15 Oct 13 In Chapters 02 and 03 in Datta: How to

Lecture 14, 15 Oct 13 In Chapters 02 and 03 in Datta: How to correctly measure I = GV Brief discussion: Roukes article Two typo correction HW 02 Pr. 2. 3 Add scattering to Landauer-Buttiker Caveat: when Landauer-Buttiker doesn’t work When it does: Sections 2. 5 and 2. 6: motivation: why: 2. 5: Probes as scatterers especially at high bias/temps Buttiker approach for dealing with incoherent scatterers 2. 6 occupied states as scatterers 3. 1 scattering/S matrix VM Ayres, ECE 802 -604, F 13

Roukes article: VM Ayres, ECE 802 -604, F 13

Roukes article: VM Ayres, ECE 802 -604, F 13

Roukes article: VM Ayres, ECE 802 -604, F 13

Roukes article: VM Ayres, ECE 802 -604, F 13

Roukes article: TR possibilities: rebound direct rebound + seems to be direct X B

Roukes article: TR possibilities: rebound direct rebound + seems to be direct X B F = q(v x B) = -|e| (v X B) VM Ayres, ECE 802 -604, F 13

Roukes article: TL possibilities: direct rebound F = q(v x B) = -|e| (v

Roukes article: TL possibilities: direct rebound F = q(v x B) = -|e| (v X B) VM Ayres, ECE 802 -604, F 13

Roukes article: HW 01 VA Pr. 01: x = 0. 4, here it is

Roukes article: HW 01 VA Pr. 01: x = 0. 4, here it is x = 0. 3 HW 01 Datta E 1. 2 Find n and m HW 01 Datta E 1. 1 Find lf VM Ayres, ECE 802 -604, F 13

Two typos Pr. 2. 3: Roukes article: TL direct rebound Roukes article: TR rebound

Two typos Pr. 2. 3: Roukes article: TL direct rebound Roukes article: TR rebound direct VM Ayres, ECE 802 -604, F 13

Two typos Pr. 2. 3, marked in magenta: Datta Pr. 2. 3, p. 113:

Two typos Pr. 2. 3, marked in magenta: Datta Pr. 2. 3, p. 113: TR X B T 2 1 Datta Pr. 2. 3, p. 113: TL VM Ayres, ECE 802 -604, F 13

Lecture 14, 15 Oct 13 In Chapters 02 and 03 in Datta: How to

Lecture 14, 15 Oct 13 In Chapters 02 and 03 in Datta: How to correctly measure I = GV Brief discussion: Roukes article: HW 02 Pr. 2. 3: Add scattering to Landauer-Buttiker Caveat: when Landauer-Buttiker doesn’t work When it does: Sections 2. 5 and 2. 6: motivation: why: 2. 5: Probes as scatterers especially at high bias/temps Buttiker approach for dealing with incoherent scatterers 3. 1 scattering/S matrix VM Ayres, ECE 802 -604, F 13

Lec 13: In Section 2. 5: 2 -t example: with broadened Fermi f 0

Lec 13: In Section 2. 5: 2 -t example: with broadened Fermi f 0 Practical example: Roukes VM Ayres, ECE 802 -604, F 13

Lec 13: i as a function of how much energy E/what channel M the

Lec 13: i as a function of how much energy E/what channel M the e- is in If T = T’, can get to Landauer-Buttiker but no reason why T should = T’. Especially if energies from probes took e- far from equilibrium. VM Ayres, ECE 802 -604, F 13

Lec 13: Can expect T = T’ at equilibrium. Consider: if energies from probes

Lec 13: Can expect T = T’ at equilibrium. Consider: if energies from probes don’t take e- far from equilibrium: VM Ayres, ECE 802 -604, F 13

Lec 13: New useful G: VM Ayres, ECE 802 -604, F 13

Lec 13: New useful G: VM Ayres, ECE 802 -604, F 13

Lec 13: Basically I = G^V = G^ (m 1 -m 2) e that

Lec 13: Basically I = G^V = G^ (m 1 -m 2) e that works when probes hotted things up but not too far from equilibrium VM Ayres, ECE 802 -604, F 13

Lec 13: Example: does the figure shown appear to meet the linear (I =

Lec 13: Example: does the figure shown appear to meet the linear (I = G^ V) regime criteria? Criteria is: m 1 -m 2 << k. BT FWHM shown is k. BT Answer: No, they appear to be about the same (red and blue). However, part of FT(E) is low value. Comparing an ‘effective’ m 1 -m 2 (green) maybe it’s OK. VM Ayres, ECE 802 -604, F 13

Lec 13: If T(E) changes rapidly with energy, the “correlation energy” ec is said

Lec 13: If T(E) changes rapidly with energy, the “correlation energy” ec is said to be small. 0. 85 T(E) 0. 09 5 e. V 5. 001 e. V E A very minimal change in e- energy and you are getting a different and much worse transmission probability. VM Ayres, ECE 802 -604, F 13

2 -DEG VM Ayres, ECE 802 -604, F 13

2 -DEG VM Ayres, ECE 802 -604, F 13

Example: in HW 01 Pr. 1. 1 you solved for tf for a 2

Example: in HW 01 Pr. 1. 1 you solved for tf for a 2 -DEG in Ga. As @ 1 K using the graph shown in Figure 1. 3. 2. Estimate the corresponding correlation energy ec VM Ayres, ECE 802 -604, F 13

Estimate the corresponding correlation energy ec Answer: VM Ayres, ECE 802 -604, F 13

Estimate the corresponding correlation energy ec Answer: VM Ayres, ECE 802 -604, F 13

Lecture 14, 15 Oct 13 In Chapters 02 and 03 in Datta: How to

Lecture 14, 15 Oct 13 In Chapters 02 and 03 in Datta: How to correctly measure I = GV Brief discussion: Roukes article Two typo correction HW 02 Pr. 2. 3 Add scattering to Landauer-Buttiker Caveat: when Landauer-Buttiker doesn’t work When it does: Sections 2. 5 and 2. 6: motivation: why: 2. 5: Probes as scatterers especially at high bias/temps Buttiker approach for dealing with incoherent scatterers 3. 1 scattering/S matrix VM Ayres, ECE 802 -604, F 13

Lec 10: Scattering: Landauer formula for R for 1 coherent scatterer X: Reflection =

Lec 10: Scattering: Landauer formula for R for 1 coherent scatterer X: Reflection = resistance VM Ayres, ECE 802 -604, F 13

Lec 10: Coherent scattering means that phases of both transmitted and reflected e -

Lec 10: Coherent scattering means that phases of both transmitted and reflected e - waves are related to the incoming e- wave in a known manner E > barrier height V 0 E < barrier height V 0 VM Ayres, ECE 802 -604, F 13

Lec 10: Transmission probability for 2 scatterers: T => T 12: That’s interesting. That

Lec 10: Transmission probability for 2 scatterers: T => T 12: That’s interesting. That Ratio is additive: Assuming that the scatterers are identical: VM Ayres, ECE 802 -604, F 13

Therefore: Resistance for two coherent scatterers is: VM Ayres, ECE 802 -604, F 13

Therefore: Resistance for two coherent scatterers is: VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

Resistance is due to partially coherent /partially incoherent transmission VM Ayres, ECE 802 -604,

Resistance is due to partially coherent /partially incoherent transmission VM Ayres, ECE 802 -604, F 13

1 Deg, M = 1 m. R m. L X O m X Example:

1 Deg, M = 1 m. R m. L X O m X Example: probe VM Ayres, ECE 802 -604, F 13

1 Deg, M = 1 m. R m. L X O m X probe

1 Deg, M = 1 m. R m. L X O m X probe m VM Ayres, ECE 802 -604, F 13

Model the phase destroying impurity as two channels attached to an energy reservoir m

Model the phase destroying impurity as two channels attached to an energy reservoir m VM Ayres, ECE 802 -604, F 13

Influence of the incoherent impurity can be described using a Landauer approach as: VM

Influence of the incoherent impurity can be described using a Landauer approach as: VM Ayres, ECE 802 -604, F 13

m. L m. R m VM Ayres, ECE 802 -604, F 13

m. L m. R m VM Ayres, ECE 802 -604, F 13

m. L m. R m VM Ayres, ECE 802 -604, F 13

m. L m. R m VM Ayres, ECE 802 -604, F 13

m. L m. R m VM Ayres, ECE 802 -604, F 13

m. L m. R m VM Ayres, ECE 802 -604, F 13

m. L m. R m VM Ayres, ECE 802 -604, F 13

m. L m. R m VM Ayres, ECE 802 -604, F 13

Landauer-Buttiker treats all “probes” equally: what is going into “probe” 3: m. L m.

Landauer-Buttiker treats all “probes” equally: what is going into “probe” 3: m. L m. R m VM Ayres, ECE 802 -604, F 13

Landauer-Buttiker treats all “probes” equally: what is going into “probe” 4: m. L m.

Landauer-Buttiker treats all “probes” equally: what is going into “probe” 4: m. L m. R m VM Ayres, ECE 802 -604, F 13

Outline of the solution: Goal: V = IR, solve for R What is V:

Outline of the solution: Goal: V = IR, solve for R What is V: m. A – m. B What is I: I = I 1 = I 2 VM Ayres, ECE 802 -604, F 13

1 Deg, M = 1 m. R m. L X O m m. A

1 Deg, M = 1 m. R m. L X O m m. A X probe m. B m VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

Condition: net current I 3 + I 4 = 0 VM Ayres, ECE 802

Condition: net current I 3 + I 4 = 0 VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

Condition: net current I 3 + I 4 = 0 VM Ayres, ECE 802

Condition: net current I 3 + I 4 = 0 VM Ayres, ECE 802 -604, F 13

Condition: net current I 3 + I 4 = 0 VM Ayres, ECE 802

Condition: net current I 3 + I 4 = 0 VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

VM Ayres, ECE 802 -604, F 13

m. L m. R VM Ayres, ECE 802 -604, F 13

m. L m. R VM Ayres, ECE 802 -604, F 13