Chapter 4.6, Problem 1E

Interpretation Introduction

Interpretation:

To sketch the super-current ${\text{I}}_{\text{c}}\text{ sin }\varphi \text{ (t)}$ as a function of $\text{t}$ assuming that $\frac{\text{I}}{{\text{I}}_{\text{c}}}$ is slightly greater than $\text{1}$ and $\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ >> 1}$, for a Josephson junction in the over-damped limit $\text{β = 0}$. To sketch the instantaneous voltage $\text{V(t)}$ for both the cases

Concept Introduction:

The Josephson relation in the over-damped limit $\text{β = 0}$ is given by $\dot{\varphi }\text{ = }\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ - sin }\varphi$.

Josephson voltage-phase relation is given by $\text{V = }\frac{\hslash }{\text{2e}}\dot{\varphi }$

Answer to Problem 1E

Solution:

a) The super-current ${\text{I}}_{\text{c}}\text{ sin }\varphi \text{ (t)}$ as a function of $\text{t}$ assuming that $\frac{\text{I}}{{\text{I}}_{\text{c}}}$ is slightly greater than $\text{1}$ and $\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ >> 1}$ for a Josephson junction in the over-damped limit $\text{β = 0}$ is sketched.

b) The instantaneous voltage $\text{V(t)}$ for both the cases is sketched.

Explanation of Solution

Josephson junctions are superconducting devices that are capable of generating voltage oscillations of extraordinarily high frequency. It consists of two closely spaced superconductors separated by a weak connection.

The Josephson relation in the over-damped limit $\text{β = 0}$ is given by

$\dot{\varphi }\text{ = }\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ - sin }\varphi$

Here, ${\text{I}}_{\text{c}}$ is the critical current.

Assuming that $\frac{\text{I}}{{\text{I}}_{\text{c}}}$ is slightly greater than $\text{1}$

Consider $\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ = 1}\text{.1}$

$\dot{\varphi }\text{ = 1}\text{.1 - sin }\varphi$

The graph of $\varphi$ vs. $\text{t}$ is shown below

Assuming that $\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ >> 1}$.

Consider $\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ = 10}$,

${\varphi }^{\prime }\text{ = 10 - sin }\varphi$

The graph of $\varphi$ vs. $\text{t}$ is shown below.

b) Josephson voltage-phase relation is given by

$\text{V = }\frac{\hslash }{\text{2e}}\dot{\varphi }$

Substituting $\dot{\varphi }\text{ = }\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ - sin }\varphi$, we get

$\text{V = }\frac{\hslash }{\text{2e}}\left(\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ - sin }\varphi \right)$

Consider $\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ = 1}\text{.1}$

The graph of voltage vs. $\varphi$ is shown below.

$\text{V = }\frac{\hslash }{\text{2e}}\left(\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ - sin }\varphi \right)$

Consider $\frac{\text{I}}{{\text{I}}_{\text{c}}}\text{ = 10}$

The graph of voltage vs. $\varphi$ is shown below.