Chapter 4.6, Problem 4E

Interpretation Introduction

Interpretation:

a) To determine the relation between dc bias current, current through array, and current through load resistance.

b) To show that $I_a=I_C\sin {\Phi }_1+\frac{v_1}{r}$ and $I_a=I_C\sin {\Phi }_2+\frac{v_2}{r}$.

c) To relate $v_k$ and ${\Phi }_k$.

d) To show $I_b=I_C\sin {\Phi }_k+\frac{\hslash }{2er}{\dot{\Phi }}_k+\frac{\hslash }{2eR}\left({\dot{\Phi }}_1+{\dot{\Phi }}_2\right)\mathrm{......................}\text{For k=1,2}\text{.}$

e) To write the equation derived in part d) in standard form of ${\dot{\Phi }}_k=\Omega +a\sin {\Phi }_k+K{\displaystyle \sum _{j=1}^2\sin {\Phi }_j}$

Concept Introduction:

A Josephson Junction consists of two superconductors which are coupled by a week connection of insulators, a weakened superconductor, or a semiconductor.

Josephson Junctions are superconducting devices, which can produce voltage oscillations of very high frequency in the range of ${10}^{10}-{10}^{11}$ Hz.

If a Josephson junction is connected to a dc current source so that a non-zero current $I$ can be driven through the junction, it is possible to show that if this current is less than critical current $I_C$, there will be no voltage generated across the junction. But phases of both the superconductors will have constant phase difference $\Phi ={\Phi }_2-{\Phi }_1$. $\Phi$ satisfies the Josephson current phase relation

$I=I_C\sin \Phi$

When the current $I$ exceeds the critical current $I_C$, a constant phase difference cannot be maintained, and a potential difference is developed across the junction. The phase difference begins to slip, and the rate of slippage is given by the equation

$V=\frac{\hslash }{2e}\dot{\Phi }$