Josephson junction

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Recently I am confused about this thing: according to the I-V curve, the resistance is 0 at zero bias, however, the conductance is finite by Andreev reflection? Is there a contradictory? Thanks!
 
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Hi,

You should think this phenomena not in the classical picture but as a quantum phenomena where this effect is due to a Macroscopic Wave Function, just as an example you can think as an electron "turns" around the nuclei, it does not loss any energy.
Finally remember that in the Josephson equations the current is more an oscillation where current goes back and forth keeping the mean current equal to zero.
Cheers
 
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JJ zero bias conductance

The current through the JJ is due to tunneling of Cooper pairs. Unlike quasiparticle tunneling, pair tunneling does not involve excitations and can occur even without bias across the junction. Thus one
could connect a current source to a junction and, for currents less than a certain critical value, no
voltage would be developed if the current were carried across the insulator by Cooper pairs. The tunnel current is mediated by the phase difference of the macroscopic wave function across the junction and expressed in the Josephson relation for the current density: J = (Jc)sin(phi), where Jc is the critical current density and phi is the phase difference.

Andreev reflection is only observed in junctions with an highly transparent tunnel barrier (oxide barriers disrupt the coherent quasiparticle scattering channel) and is not observed at zero bias.
 

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