(a) Unde zeo-bias conditions, thee ae no lled states on one side of the junction which ae at the same enegy as the empty allowed states on the othe si

Similar documents
P-N Junction in equilibrium. Diode. Diode bias. Forward and reverse bias. Diode current-voltage characteristics. Models. Applications.

Experiment I Voltage Variation and Control

16.1 Permanent magnets

PES 3950/PHYS 6950: Homework Assignment 6

Lecture 2 Date:

SAMPLE PAPER I. Time Allowed : 3 hours Maximum Marks : 70

Quantum tunneling: α-decay

AP Physics C: Electricity and Magnetism 2003 Scoring Guidelines

Magnetic Dipoles Challenge Problem Solutions

Nuclear and Particle Physics - Lecture 20 The shell model

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

Numerical Integration

221B Lecture Notes Scattering Theory I

Module 9: Electromagnetic Waves-I Lecture 9: Electromagnetic Waves-I

ELECTROSTATICS::BHSEC MCQ 1. A. B. C. D.

[ ] [ ] 3.3 Given: turning corner radius, r ε = 0 mm lead angle, ψ r = 15 back rake angle, γ p = 5 side rake angle, γ f = 5

763620SS STATISTICAL PHYSICS Solutions 2 Autumn 2012

Motion along curved path *

Chapter 5 Force and Motion

Chapter 5 Force and Motion

2018 Physics. Advanced Higher. Finalised Marking Instructions

QUASI-STATIONARY ELECTRON STATES IN SPHERICAL ANTI-DOT WITH DONOR IMPURITY * 1. INTRODUCTION

55:041 Electronic Circuits

you of a spring. The potential energy for a spring is given by the parabola U( x)

PHYS 1444 Lecture #5

17.1 Electric Potential Energy. Equipotential Lines. PE = energy associated with an arrangement of objects that exert forces on each other

Physics Tutorial V1 2D Vectors

Physics 107 TUTORIAL ASSIGNMENT #8

The Millikan Experiment: Determining the Elementary Charge

Physics NYB problem set 5 solution

Fields and Waves I Spring 2005 Homework 8. Due: 3 May 2005

working pages for Paul Richards class notes; do not copy or circulate without permission from PGR 2004/11/3 10:50

11) A thin, uniform rod of mass M is supported by two vertical strings, as shown below.

The geometric construction of Ewald sphere and Bragg condition:

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum

Physics 4A Chapter 8: Dynamics II Motion in a Plane

Reading Assignment. Problem Description for Homework #9. Read Chapters 29 and 30.

Current, Resistance and

ev dm e evd 2 m e 1 2 ev2 B) e 2 0 dm e D) m e

Annihilation of Relativistic Positrons in Single Crystal with production of One Photon

6 Matrix Concentration Bounds

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

anubhavclasses.wordpress.com CBSE Solved Test Papers PHYSICS Class XII Chapter : Electrostatics

PHYS 1444 Section 501 Lecture #7

Substances that are liquids or solids under ordinary conditions may also exist as gases. These are often referred to as vapors.

Exam 3, vers Physics Spring, 2003

A NEW VARIABLE STIFFNESS SPRING USING A PRESTRESSED MECHANISM

EM Boundary Value Problems

Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 9 Solutions

STUDY ON 2-D SHOCK WAVE PRESSURE MODEL IN MICRO SCALE LASER SHOCK PEENING

Lecture 28: Convergence of Random Variables and Related Theorems

Lab #0. Tutorial Exercises on Work and Fields

Phys 1215, First Test. September 20, minutes Name:

Power efficiency and optimum load formulas on RF rectifiers featuring flow-angle equations

7.2. Coulomb s Law. The Electric Force

Introduction to Nuclear Forces

INTRODUCTION. 2. Vectors in Physics 1

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

EM-2. 1 Coulomb s law, electric field, potential field, superposition q. Electric field of a point charge (1)

On the Sun s Electric-Field

( n x ( ) Last Time Exam 3 results. Question. 3-D particle in box: summary. Modified Bohr model. 3-D Hydrogen atom. r n. = n 2 a o

Electric Field, Potential Energy, & Voltage

Algebra-based Physics II

Field emission of Electrons from Negatively Charged Cylindrical Particles with Nonlinear Screening in a Dusty Plasma

General Railgun Function

University of Illinois at Chicago Department of Physics. Electricity & Magnetism Qualifying Examination

b Ψ Ψ Principles of Organic Chemistry lecture 22, page 1

RESONANCE SERIES RESONANT CIRCUITS. 5/2007 Enzo Paterno 1

AP Physics Electric Potential Energy

Physics 2B Chapter 22 Notes - Magnetic Field Spring 2018

Phys 222 Sp 2009 Exam 1, Wed 18 Feb, 8-9:30pm Closed Book, Calculators allowed Each question is worth one point, answer all questions

13. Adiabatic Invariants and Action-Angle Variables Michael Fowler


1.2 Differential cross section

Supporting information

Appendix B The Relativistic Transformation of Forces

15 B1 1. Figure 1. At what speed would the car have to travel for resonant oscillations to occur? Comment on your answer.

5.111 Lecture Summary #6 Monday, September 15, 2014

1) Emits radiation at the maximum intensity possible for every wavelength. 2) Completely absorbs all incident radiation (hence the term black ).

Electrostatics. 1. Show does the force between two point charges change if the dielectric constant of the medium in which they are kept increase?

Single Particle State AB AB

= e2. = 2e2. = 3e2. V = Ze2. where Z is the atomic numnber. Thus, we take as the Hamiltonian for a hydrogenic. H = p2 r. (19.4)

Magnetic Field. Conference 6. Physics 102 General Physics II

F g. = G mm. m 1. = 7.0 kg m 2. = 5.5 kg r = 0.60 m G = N m 2 kg 2 = = N

d 2 x 0a d d =0. Relative to an arbitrary (accelerating frame) specified by x a = x a (x 0b ), the latter becomes: d 2 x a d 2 + a dx b dx c

EXAM NMR (8N090) November , am

Tutorial Exercises: Central Forces

PHYS Summer Professor Caillault Homework Solutions. Chapter 5

SPH4U Unit 6.3 Gravitational Potential Energy Page 1 of 9

Multipole Radiation. February 29, The electromagnetic field of an isolated, oscillating source

B. Spherical Wave Propagation

Pulse Neutron Neutron (PNN) tool logging for porosity Some theoretical aspects

When two numbers are written as the product of their prime factors, they are in factored form.

ASTR415: Problem Set #6

Boundary Layers and Singular Perturbation Lectures 16 and 17 Boundary Layers and Singular Perturbation. x% 0 Ω0æ + Kx% 1 Ω0æ + ` : 0. (9.

EKT 345 MICROWAVE ENGINEERING CHAPTER 2: PLANAR TRANSMISSION LINES

LECTURER: PM DR MAZLAN ABDUL WAHID PM Dr Mazlan Abdul Wahid

EFFECTS OF FRINGING FIELDS ON SINGLE PARTICLE DYNAMICS. M. Bassetti and C. Biscari INFN-LNF, CP 13, Frascati (RM), Italy

Determining solar characteristics using planetary data

Homework 7 Solutions

Transcription:

1 Esaki Diode hen the concentation of impuity atoms in a pn-diode is vey high, the depletion laye width is educed to about 1 nm. Classically, a caie must have an enegy at least equal to the potential-baie height in ode to coss the junction. Howeve, quantum mechanics indicates that thee is a nonzeo pobability that a paticle might penetate though a baie as thin as that indicated above. This phenomenon is called tunneling, and because of this, these high-impuity density p-n devices ae called tunnel diodes, o Esaki diodes. The condition that the baie be less than 1 nm thick is a necessay but not a sucient condition fo tunneling. It is also equied that occupied enegy states exist on the side fom which the electons tunnel and that allowed empty states exist on the othe side. The enegy-band diagam fo a heavily - doped diode unde open cicuit and evese bias conditions is shown below. Figue 1: Esaki diode 1

(a) Unde zeo-bias conditions, thee ae no lled states on one side of the junction which ae at the same enegy as the empty allowed states on the othe side. Hence, thee can be no ow of chage in eithe diection acoss the junction, and the cuent is zeo, an obviously coect conclusion fo an open - cicuited diode. (b) If a evese bias is a applied, the height of the baie is inceased above the open-cicuit value E. Hence, the n-side levels must shift downwad with espect to the p-side levels. In this case, thee ae some enegy states in the valence band on the p-side that lie at the same level as allowed empty states in the conduction band of the n-side. Hence, these electons might tunnel fom the p to the n - side, giving ise to the evese diode cuent. As the magnitude of the evese bias inceases, the heavily - shaded ae gows in size, causing the evese cuent to incease, as shown by section (1) in the Figue below. Figue 2: IV chaacteistics. Conside now the fowad bias case when the potential baie is deceased below E. The n-side level shifts upwad with espect to those on the p-side, and the enegy-band pictue fo this situation is indicated in Figue 3. 2

Figue 3: Fowad Bias 3

It is clea that thee will be occupied states in the conduction band on the n-side which ae at the same enegy as the allowed empty sates (holes) in the valence band on the p-side. Hence electons will tunnel fom the n to the p mateial giving ise to the fowad cuent of section (2). As the fowad bias inceases futhe, the condition in (b) is eached, giving ise to maximum cuent value. If moe fowad bias is applied, the situation in (c) is obtained and the tunneling cuent deceases, giving to section (3). At even lage bias, thee will not be empty allowed states on one side of the junction at the same enegy as the occupied states on the othe side and the tunneling cuent dops to zeo. In addition to the quantum-mechanical cuent descibed above, the egula pn-junction cuent is being collected at lage voltages, giving ise to section (4). Mathematical analysis of the tunneling cuent is based on the esults descibed in the pevious sections. In othe wods, one can use the KB appoximation to calculate the tunneling coecient (o pobability) and then use the esult into the Landaue expession fo the cuent. To calculate the tunneling coecient, we conside the case when E? = and E z 6=, and we appoximate the baie height bye G : Figue 4: Tunneling baie V (Z) =E G 1, Z (1) The attenuation of the baie is thus Z Z 2m 2 = 2 ~ [V (Z), E Z]dZ 2 4

Z s Z 2m = 2 Z Z = 2 Z Z = 2 s 2mE G Z Z = (,2 ) = (,2 ) =, 4 3 E G 1, Z, E Z dz 1, Z, E Z dz s 1, Z, E Z d s 1, Z, E Z d 1, Z, E Z " 1, Z, E Z, Z (, ) Z,, Z 1, E Z # (2) whee 1, Z, E Z = 1,, E Z, E Z =, E Z,, E Z (3) Hence = 4 3 4 3 1, E Z 1, 3 2 E Z (4) povided that E Z =E G 1. Substituting the last expession into the KB esult fo the tansmission coecient gives whee we have dened T (E Z ) = e,2 4 = e, 2 3 ~ {z 2 E } e Z T = T e E Z=E (5) 5

s E = 2 2m s Eg 2 1 = 2m 2 1 ~ = 2 2 2m The expession fo T (E Z ) suggests that the lage the enegy E Z, the lage the pobability fo tansmission, which should be expected. The next task is to calculate the cuent unde fowad bias conditions, fo which we use the peviously deived expession (6) Figue 5: Fowad bias conditions. J = J L!R, J R!L =, em whee Z 1 Z 1 de Z T (E Z ) de t [f L (E Z + E t ), f R (E Z + E t + ev a )] (7) Using the appoximation (valid at T! ); E = E Z + E t = E Z1 + E t, E cn (8) f L (E Z + E t ), f R (E Z + E t + ev a )= 6

@f = f(e Z + E t ), f(e Z + E t ), ev a @E E Z+E t = =,ev a (E Z + E t, E F ) (9) we get fo the fowad cuent J =, em (,ev a)t whee = e2 m V a T = e2 m V a T Z E max t =, e2 m V a T E Z 1 Z 1 de t de Z e EZ=E (E Z + E t, E F ) (,E )e,et=e de t e (E F%, E t )=E E max t h i e,emax t =E, 1 (1) Theefoe E max t = E VP +E cn, ev a (11) i.e., e,emax t =E, 1 = e,(ev P+Ecn,eVa)=E, 1. 1, E VP +E cn, ev. a, 1 E J =, e2 m V a T E = e2 m V a T = e2 m T E t V a, E VP +E cn, ev a E VP +E cn {z } E t 1, ev a E t E, ev a! (12) (13) The maximum tansvese enegy is obtained using,e cno + ev a + E max? = E VP (14) E? max = E VP +E cn, ev a (15) 7