1. Document that explains the chosen concept to the animator 1

2. DC JOSEPHSON EFFECT The phenomenon in which a DC current flows across the Josephson junction in the absence of an applied DC voltage across the junction is known as the DC Josephson Effect. Authors Anura.B.Kenkre Course Name: Superconductivity 2

3. Pre-requisites: Basic Quantum Mechanics(Phenomenon of tunneling.) Target Audience: TYBsc 3

4. Learning Objectives After interacting with this Learning Object, the learner will be able to: 1.Define a Josephson junction. 2.Explain the Josephson effect. 3.List the parameters on which the current flowing through the junction depends. 4.Predict the effect of the critical current on the voltage developed across the junction. 4

5. Definitions of the components/Keywords: 5 3 2 4 1 1.Two superconductors separated by an insulating barrier-- is called a Josephson junction. 2.The phenomenon in which a dc current can flow across a barrier without a driving voltage, is called the DC Josephson effect. 5

6. 6 IMPORTANT NOTE TO THE ANIMATOR: All the instructions/labels or anything WRITTEN in blue are CONTENT NOT TO BE DISPLAYED! All the instructions WRITTEN in black are CONTENT TO BE DISPLAYED! This is not applicable for images as there can be overlapping of these colours there. This should be followed for all the instructions,labels,etc… Kindly keep a note of this while displaying text in the animation.

7. Master layout or diagram Make a schematic diagram of the concept Explain to the animator about the beginning and ending of the process. Draw image big enough for explaining. In above image, identify and label different components of the process/phenomenon. (These are like characters in a film)‏ Illustrate the basic flow of action by using arrows. Use BOLD lines in the diagram, (minimum 2pts.)‏ In the slide after that, provide the definitions of ALL the labels used in the diagram 5 3 2 4 1 INSTRUCTIONS SLIDE 7

8. Master layout or diagram You may have multiple master layouts. –In this case, number the master layout. ( e.g. Master layout 1)‏ – Each Master layout should be followed by the stepwise description of the animation stages related to it. 5 3 2 4 1 INSTRUCTIONS SLIDE 8

9. Master Layout 1 5 3 2 4 1 9 Basic DC Josephson Effect I as a function of ∆φ Critical current versus voltage developed. Interactivity menu: DC JOSEPHSON EFFTECT

10. Master Layout 2 5 3 2 4 1 LEFT SIDE RIGHT SIDE Ψ(x) x 1 2 cooper pair 12 10 This is called a sine wave Josephson junction graph Marking to denote position of the cooper pair

11. 11 Master Layout 3 mV Current supply Slider bar voltmeter Use this font for the display. this is the LED display for the voltmeter. Use this font for the display. this is the LED display for the current supply. Josephson junction Redraw all the images given here.the ones given here are for reference.and do not retain the company names on them… 0.00.10.2 0.3 0.40.50.70.80.91.0 0.6 Current(I)(mA) Super conductor Super conductor

12. 12 How to draw a sine wave in our animation?? Time Amplitude 0 This height from zero on the x axis is called the amplitude of the sine wave. These are equal in height from the zero on the x axis. X axis Y axis This distance from zero till the end of the sine wave is called the Time period of the sine wave. End of sine wave This point is called half of the time period Note: 1. Distance between zero and half the time period has to be same as the distance between half the time period and end of sine wave. 2.Always take the amplitude and the time period as 2 units each. Master Layout 4

13. Animation design Please see the design template provided in the next slide. This is a sample template, and you are free to change as per your design requirements. Try and recreate the sections/subsections as shown in the template. 1 5 2 4 3 13

14. Want to know more… (Further Reading) ‏ Definitions Formula with derivation (if any) ‏ Graphs/Diagram (for reference) ‏ Test your understanding (questionnaire) ‏ Lets Learn! Concepts Assumptions (if any) ‏ Lets Sum up (summary) ‏ Instructions/ Working area Radio buttons (if any)/Drop down (if any) ‏ Play/pauseRestart What will you learn Credits Basic DC Josephson Effect I as a function of ∆φ Critical current versus voltage developed. Interactivity: 14 x Ψ(x) Wave function versus position graph for an electron Experimental Setup

15. Explain the process 1 5 3 2 4 In this step, use an example to explain the concept. It can be an analogy, a scenario, or an action which explains this concept/process/topic Try to use examples from day-to-day life to make it more clear You have to describe what steps the animator should take to make your concept come alive as a series of moving images. Keep the examples simple to understand, and also to illustrate/animate. 15

16. Stepwise description of process The goal of the document is to provide instructions to an animator who is not a expert. You have to describe what steps the animator should take to make your concept come alive as a moving visualization. Use one slide per step. This will ensure clarity of the explanation. Add a image of the step in the box, and the details in the table below the box. You can use any images for reference, but mention about it's copyright status The animator will have to re-draw / re-create the drawings Add more slides as per the requirement of the animation 1 5 3 2 4 16

17. Step 1: 1 5 3 2 4 Text to be displayed Description of the action/ interactivity Choice of interactivity 1.Let the user select a radio button from the interactivity menu. 2.If the user selects Basic DC Josephson effect then go to slide 18. 3.If the user selects I as a function of ∆φ then go to slide 26. 4.If the user selects Critical current versus voltage developed then go to slide 31. 17 Refer to master layout 1.

18. Step 2: 1 5 3 2 4 Text to be displayed Description of the action/ interactivity Josephson Junction RIGHT SIDE 1.First let the figure on the left side fade in. 2.Then let the graph on the right side fade in. 3.Display the objects in the order of the numbering..for example:The rectangular box should appear first since it is labeled 1,followed by the label ‘superconductor’ labeled 2 and so on… 4.Show this in a time length of 5seconds.(this time length is for basic reference..) 5.Go to step 2. Ψ(x) Super conductor Super conductor Josephson Junction Insulating Barrier LEFT SIDE 18 Wave function versus position graph for an electron 1 2 3 4 5 7 6 x

19. Step 3: 1 5 3 2 4 Text to be displayed Description of the action/ interactivity Cooper pair entering the insulating barrier. cooper pair 1.Let the black labels from images in previous slide disappear and begin with instruction 2 given below. 2.First let the cooper pair appear as shown in the figure on the left side of the screen. 3.Cooper pair: the two dark blue balls have to be joined and surrounded by the light blue ring when they appear. 4.Then show them moving from 1 to 2 along the red arrow at a time length of 4 seconds.(this time length is for basic reference..) (.Do not show the red arrow on screen. 5.When the electron pair moves from 1 to 2 on the left side of the screen, the graph has to appear,from 1 to 2 on the right side of the screen simultaneously. 6.For drawing the graph,refer to slide 12.It is a static graph. 7.Redraw the graph and sharpen it so that the sine wave looks more sharp and neat.This image is only for reference. 8.This effect is shown in the images above. Ψ(x) LEFT SIDE RIGHT SIDE x 12 19 Wave function versus position graph for an electron 12 graph

20. Step 4: 1 5 3 2 4 Text to be displayed Description of the action/ interactivity Cooper pair in the insulating barrier 1.Show the electron pair moving from 2 to 3 along the red arrow at a time length of 3seconds.(this time length is for basic reference..).Do not show the red arrow on screen. 2.When the electron pair moves from 2 to 3 on the left side of the screen, the graph has to appear, from 2 to 3 on the right side of the screen simultaneously. 3.For drawing the graph,refer to slide 12.It is a static graph. 4.Redraw the graph and sharpen it so that the sine wave looks more sharp and neat.This image is only for reference. LEFT SIDE RIGHT SIDE Electron pair 23 20 Wave function versus position graph for an electron 23 Ψ(x) x

21. Step 5: 1 5 3 2 4 Audio Narration ‏ Description of the action/ interactivity RIGHT SIDE 1.Show the electron pair moving from 3 to 4 along the red arrow at a time length of 4seconds.(this time length is for basic reference..). Do not show the red arrow on screen. 2.When the electron pair moves from 3 to 4 on the left side of the screen, the graph has to appear, from 3 to 4 on the right side of the screen simultaneously. 3.For drawing the graph,refer to slide 12.It is a static graph. 4.Redraw the graph and sharpen it so that the sine wave looks more sharp and neat.This image is only for reference. Cooper pair leaving the insulating barrier 3 4 Electron pair LEFT SIDE 21 Wave function versus position graph for an electron Ψ(x) x 3 4

22. Step 6: 1 5 3 2 4 Audio Narration ‏ Description of the action/ interactivity Formative assesment LEFT SIDE RIGHT SIDE I V 1 4 1 I-V Characteristics 4 Answer this: (click on the right option) What happens when the electron encounters a barrier due to the insulating layer? 1.It stops wherever it is 2.It tunnels through the barrier 3.It retraces its path 1. Make the earlier lines vanish 2. Create a text box with a clickable button in a overlay screen and make it appear as shown. 3.Allow the user to click on any one of the radio buttons shown. 4.Go to step 6A. 22

23. Step 6A: 1 5 3 2 4 Audio Narration ‏ Description of the action/ interactivity feedback LEFT SIDE RIGHT SIDE I V 1 4 1 I-V Characteristics 4 That’s very good!!! The electron tunnels through the barrier and due to this tunneling, there exists a dc current through the junction and current flows through the circuit. 1.The correct answer is 2.If the user clicks on it then make the earlier box vanish. 2. Create a text box with a clickable button in a overlay screen and make it appear as shown above. 3. Then show the animation as explained in step7. 23

24. Step 6B: 1 5 3 2 4 Audio Narration ‏ Description of the action/ interactivity feedback LEFT SIDE RIGHT SIDE I V 1 4 1 I-V Characteristics 4 Have a look at the animation and try again… Try again?? CLICK 1.If the user clicks on 1 or 3 then display the above text box with a clickable button as an overlay screen. 2. If the user clicks on the red button then go to step 2. 24

25. Step 7: 1 5 3 2 4 Audio Narration ‏ Description of the action/ interactivity Dc current flowing through the junction. 1.Show the cooper pairs moving simultaneously from 1 to 4 along the red arrow for a time length of 4 seconds.Do not show the red arrow on screen. 2.These electron pairs should not touch each other neither should the black ring around them touch each other. 3.Show many cooper pairs moving from 1 to 4(only 2 are shown in the image above). 4.When the cooper pairs moves from 1 to 4 on the left side of the screen, show the pointer moving simultaneously as shown.. 5.Repeat the animation till the user selects some other radio button. 4 1 25 pointer

26. 1 5 3 2 4 Instructions for the animator Instruction to the learner Results and Output Boundary limits Interactivity type (IO1/IO2..) ‏ Step No:8 Input boxes. Enter values between 0 and 360, for φ 1 and φ 2 and observe the output. Allow values to be entered between 0 and 360 for φ 1 and φ 2 with a difference of 10 units. Eg:10,20,30,etc Allow the user to enter values for φ1 and φ2. When the user clicks on ‘show’,calculate I and |(φ2-φ1)| using the formula I=Ic sin(φ2- φ1) and |(φ2-φ1)| respectively. Here,|(φ2-φ1)| means mod of the value of (φ2-φ1)..It means that the sign of the answer should always be positive. Display these values in the respective data boxes. If the user clicks on ‘show’ before entering both the values of φ1 and φ2 the display ‘Enter values for φ1 and φ2 and then click on Show’. Default value for Ic=0.5mA. Plot a graph (as shown above)of I versus |(φ2-φ1) |using this value of I and |(φ2-φ1)|. Display the text in black above the graph. Include a proceed button. When the user clicks on ‘proceed’ go to slide 28. Enter a value for φ1: Enter a value for φ2: I |Φ 2 – φ 1 | Value I(current)value Φ2-φ1Φ2-φ1 Plot each value of I versus |(φ2-φ1)| as the user enters values. Go on joining the points on the graph so that a graph as given above is formed at some point. If the user clicks on the proceed button then go to slide 28. proceed For DC Josephson Effect:Voltage=0 +Ic -Ic 26 Show

27. This page is for animators reference: Example: Suppose the user has entered the following values: φ2=180 and φ1=90. Then (φ2-φ1)=180-90=90 And I=Ic sin (φ2-φ1) I=0.5sin(90) I=0.5*1=0.5. In this case, you have to show only the blue spot since only first set of values are given by the user. As the user enters more values, you will finally get the sinusoidal shape as shown. In short, plot the points as the user enters values and go on joining these values by a line. 90 0.5 I (Φ 2 -φ 1 ) 27

28. Step 8.1: 1 5 3 2 4 Audio Narration ‏ Description of the action/ interactivity Assessment 1.Create a text box with a clickable button and make it appear as shown above. 2.Allow the user to click on any one of the radio buttons shown. 3.Go to step 8.2 Now since you know in what way the current depends on the phase difference, can you guess which of the following equations best describes the dependence of I with respect to ∆φ??? 1) I=Ic sin(φ2-φ1) 2) I=Ic cos(φ2-φ1) 3) I=(φ2-φ1) 28

29. Step 8.2: 1 5 3 2 4 Audio Narration ‏ Description of the action/ interactivity feedback 1.The correct answer is 1.If the user clicks on it then make the text in the above box appear. That’s very good!!!! You have got the correct answer! 29

30. Step 8.3: 1 5 3 2 4 Audio Narration ‏ Description of the action/ interactivity feedback Enter more values for φ 1 and φ 2 and try it again…. Try??? CLICK 1.If the user clicks on 2 or 3 then display the above text box with a clickable button. 2. If the user clicks on the red button then go to slide 26. 30

31. 1 5 3 2 4 Step No: 9 0.00.10.2 0.3 0.40.50.70.80.91.0 0.6 Current(I)(mA) Super conductor Super conductor (Ic)0.5 0.0 *For animator: markings on scales have to be equidistant! Current source mV Instructions are on the next slide…

32. 32 Instructions for the animator Instruction to the learner Results and Output Boundary limits Interactivity type (IO1/IO2.. ) ‏ Slider bar Vary the value of current and observe what happens to the value of voltage developed across the junction. I can take values from 0 to 1 with a difference of 0.1. Eg:0.1, 0.2,0.3.. etc The user will vary the values for I. Find the value of V corresponding to this value of I from the table given on the next slide. Plot the Values of I versus V dynamically(as shown above) according to the selected value of I and its corresponding value of V. Plot the graph only upto the point to which the user has dragged the slider bar. As the user varies the slider bar display the values from the column for current given on slide 33 in the display of current supply. As the user varies the slider bar display the values from the column for voltage given on slide 33 in the display of voltmeter. Till I=0.5mA,the voltage value will be zero. Above I=0.5mA,The value of voltage varies linearly with the applied current. Instructions:

33. Current(I)(mA)Voltage(V) (mV) 0.0 0.10.0 0.20.0 0.30.0 0.40.0 0.51.0 0.62.0 0.73.0 0.84.0 0.95.0 1.06.0 33

34. Interactivity and Boundary limits In this section, you will add the ‘Interactivity’ options to the animation. Use the template in the next slide to give the details. Insert the image of the step/s (explained earlier in the Section 3) in the box, and provide the details in the table below. The details of Interactivity could be: Types: Drop down, Slider bar, Data inputs etc. Options: Select one, Multiple selections etc Boundary Limits: Values of the parameters, which won’t show results after a particular point Results: Explain the effect of the interaction in this column Add more slides if necessary 1 2 5 3 4 34

35. INSTRUCTIONS SLIDE Self- Assessment Questionnaire for Learners Please provide a set of questions that a user can answer based on the LO. They can be of the following types: –These questions should be 5 in number and can be of objective type (like MCQ, Match the columns, Yes or No, Sequencing, Odd One Out). –The questions can also be open-ended. The user would be asked to think about the question. The author is requested to provide hints if possible, but a full answer is not necessary. –One can include questions, for which the user will need to interact with the LO (with certain parameters) in order to answer it. 35

36. INSTRUCTIONS SLIDE Please make sure that the questions can be answered by interacting with the LO. It is better to avoid questions based purely on recall. Questionnaire for users to test their understanding 36

37. 37 Questionnaire 1. Which of the following best represents a Josephson Junction? Answers: a)Two superconductors placed together b)a superconductor joined to a normal metal c)Two superconductors separated by a thin insulating barrierd) ‏ two metals separated by a thin insulating barrier. Correct answers:1)c Feedback: If user clicks correct answer then display “Correct! Make sure you can explain the reasoning!” If user clicks incorrect answer then display “Have a look at the simulation and Try again! ”

38. 38 Questionnaire 2. What is the function of the insulating barrier? Answers: a)to allow the electrons to flow without any resistance b)To introduce a phase difference between the two parts of the wave function on the opposite sides of the junctionc)to stop the the electrons from tunnelling through the barrierd) ‏ To make the phase difference between the two parts of the wave function zero. Correct answers:2)b Feedback: If user clicks correct answer then display “Correct! Make sure you can explain the reasoning!” If user clicks incorrect answer then display “Have a look at the simulation and Try again! ”

39. 39 Questionnaire 3. In the formula for the current flowing through the junction, what does Io stand for? Answers: a)maximum current that the junction can carry without a potential difference b) maximum current that the junction can carry with a potential difference c)minimum current that the junction can carry with a potential differenced) ‏ minimum current that the junction can carry without a potential difference Correct answers:3)a Feedback: If user clicks correct answer then display “Correct! Make sure you can explain the reasoning!” If user clicks incorrect answer then display “Have a look at the simulation and Try again! ”

40. 40 Questionnaire 4. A DC current flows through the junction on account of which parameters? Answers: a)phase difference between the wave functions only b)maximum current through the junctiononlyc)both a and bd) ‏ neither a or b Correct answers:4) ‏ c Feedback: If user clicks correct answer then display “Correct! Make sure you can explain the reasoning!” If user clicks incorrect answer then display “Have a look at the simulation and Try again! ”

41. 41 Questionnaire 5. How do the electrons travel through the junction? Answers:a) as cooper pairsb)singlec)as a bunchd)in pairs of three. Correct answers:5)a Feedback: If user clicks correct answer then display “Correct! Make sure you can explain the reasoning!” If user clicks incorrect answer then display “Have a look at the simulation and Try again! ”

42. Links for further reading Books: 1)Introduction to Solid state physics-Charles Kittel(chapter 12) 2)Solid state physics-MA Wahab.(chapter 17) 3)Solid state physics-Ashcroft/Mermin. (Chapter 34) 42

43. INSTRUCTIONS SLIDE Please provide points to remember to understand the concept/ key terms of the animation The summary will help the user in the quick review of the concept. Summary 43

44. Summary 1.The phenomenon in which a dc current can flow across a barrier without a driving voltage, is called the DC Josephson effect. 2.The Dc Josephson current is expressed in the form of the following equation: I=Icsin(φ 2 -φ 1 ), where (φ 2 -φ 1 ) is the phase difference and Ic is the critical current. 3.Critical current is the maximum zero-voltage current that can be passed by the junction. 4.If a current having value more than the critical current is passed through the junction, then the metal returns to its normal conducting state and is no longer superconducting. 44