Josephson JunctionsGraham Steinke15th of May 2008Graham Steinke Josephson JunctionsJosephson HistoryBrian Josephson theorizer of the aptly named Josephson JunctionpicturedThin SemiconductorSuperconductor 1 Superconductor 2Figure: A Josephson JunctionGraham Steinke Josephson JunctionsDC EffectI > 0I = Icsin(φ)Figure: A DC CircuitI is the constant current flowing out of the DC source while Icisthe critical current for the Junction. And I = Icsin(φ) is theJosephson current phase relation of the supercurrent, where φ isthe phase difference.Graham Steinke Josephson JunctionsRearing up!EnergyDC EffectFigure: The DC Effect schematicV =~2e˙φ (1)This is the Josephson Voltage-phase relation which josephsonfound in his studies.Graham Steinke Josephson JunctionsAnalogyCV RI > 0Figure: The Junction, Ordinary, Displacement currentsIn order to better simplify this situation must look at both ofKirchhoff’s Laws. 1stIin=PIoutand the 2ndisPVoltageDrops = 0Graham Steinke Josephson JunctionsThe current through the capacitor equals C˙V and the currentthrough the resistor equals V /R. The sum of all the currents alongwith the supercurrent Icsin(φ) minus the current I must equal zero.C˙V +VR+ Icsinφ − I = 0Or equivalentlyC˙V +VR+ Icsinφ = I . (2)If we replace V with~2e˙φ and˙V with~2e¨φ you get:Graham Steinke Josephson JunctionsC ~2e¨φ +~2eR˙φ + Icsin(φ) = I (3)AND THIS IS ANALOGOUS TO A DRIVEN DAMPEDPENDULUM!Now we need to take out all of the dimensions with:τ =2eIcR~t (4)Graham Steinke Josephson JunctionsDimensionless EquationAnd now we must redefine both˙φ and¨φ˙φ =dφdt=dφdτ·dτdt=dφdτ2eIcR~¨φ =d2φdt2=ddtdφdt=ddt2eIcR~dφdτ=2eIcR~ddtdφdτ=2eIcR~2eIcR~ddτdφdτ=2eIcR~2d2φdτ2Graham Steinke Josephson JunctionsSubstitutionsAnd now we will insert the new φ0and φ00into (3), divide by Icandcancel like terms to getβφ00+ φ0+ sinφ =IIc(5)where β =2eIcR2C~Graham Steinke Josephson JunctionsOverdamped LimitThough almost simple still not exactly what we were expectinghuh? Since this is analogous to the Pendulum lets look at theoverdamped limit where β 1 and if this is true we can disregardthe second order part of the equation and look at what is left.φ0=IIc− sin(φ) (6)Now lets analyze this equation a little more closely.Graham Steinke Josephson JunctionsAnalysisWe have two different cases to analyze the first being when I ≤ Ic.Figure: Flows on the circleGraham Steinke Josephson JunctionsGraphing itNow we can analyze (6) to actually get something out of it. Wewant to look at the graph of Voltage vs current to get a visualrepresentation of the DC effect. In order to do so we must take theexact average of various functions which in symbols looks like< f (x) > on [a, b] but is defined as< f >≈1b − aZbaf (x)dx (7)which we should all remember from calculus right?Graham Steinke Josephson Junctionssince V = (~/2e)φ then< V >=~2e<˙φ > .And<˙φ >=2eIcR~< φ0>∴< V >= IcR < φ0> (8)And as we saw from the previous slide I /Ic≤ 1 was a fixed pointsolution, and now we will look at the other solution which, as youmight have guessed, is not a fixed point solution.Graham Steinke Josephson JunctionsWhen I /Ic> 1 the system no longer has any equilibrium points itjust continually oscillates about the circle. And sense it continuallyoscillates it has a period governed by:T =2πsqrt(ω)2+ (α)2(9)this stems from the fact that the equation is now a non uniformoscillator. And in our case the ω is I /Icand the α is 1. Now wemay compute < φ0> by using (7) which will lead to< φ0>=1TZT0dφdτdτ =1TZ2π0dφ =2πT(10)Graham Steinke Josephson JunctionsNow we can combine (8) with (9) and with (10) to create:< V >= IcRsIIc2+ 12(11)which is true whenever I > Icor if I /Ic> 1. And the combinationof these two show what happens during the DC effect of aJosephson Junction. This can be summed up piece wise like:V =0 if I ≤ IcIcRq(IIc)2− 1 if I > Ic(12)Graham Steinke Josephson Junctions0 1 2 3 4 500.511.522.533.544.55VoltageCurrentVoltage vs. CurrentFigure: The DC EffectGraham Steinke Josephson
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