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מבוא למצב מוצק ולמוליכים למחצה מעבדה 4מח' מטרה: להסביר בקצרה את עיקרי החומר התאורטי בתחום הפיסיקה של המצב המוצק וכן הפיסיקה של מוליכים למחצה ,הנדרשים להבנת הניסויים: • • • • הולכה חשמלית במוצקים. אפקט הול. תא פוטוולטאי. פוטולומינסנציה. References: • Paul A. Tipler, Modern Physics, Chapter 9 • Charles Kittel, Introduction to Solid State Physics, Chapters 7,8 (Sixth edition) • מוליכים למחצה- גדי גולן,לב-אדיר בר • Prof. Dr. Beşire GÖNÜL presentation. • E.E. Technion s.c. course. Outlines: • Atoms and bonding • Energy bands and effective mass • Conduction in solids • Temperature dependant of solid conductance • p – n junction Atoms and bonding • The periodic table • Ionic bonding • Covalent bonding • Metallic bonding • van der Waals bonding Atoms and bonding • In order to understand the physics of semiconductor (s/c) devices, we should first learn how atoms bond together to form the solids. • Atom is composed of a nucleus which contains protons and neutrons; surrounding the nucleus are the electrons. • Atoms can combine with themselves or other atoms. The valence electrons, i.e. the outermost shell electrons govern the chemistry of atoms. • Atoms come together and form gases, liquids or solids depending on the strength of the attractive forces between them. • The atomic bonding can be classified as ionic, covalent, metallic, van der Waals,etc. • In all types of bonding the electrostatic force acts between charged particles. The periodic table 3 4 5 6 7 Valance electrons 4 out of 8 Electrons In outer shell First Shell He H Second Shell Li Be Na Mg B C Third Shell Al Si N O F Ne The periodic table • Ionic solids Alkali metals contains lithium (Li), sodium (Na), potassium (K),... and these combine easily with Halogens like fluorine (F), chlorine (Cl), bromine (Br),.. . and produce ionic solids of NaCl, KCl, KBr, etc. • Rare (noble) gases Elements of noble gases of helium(He), neon (Ne), argon (Ar),… have a full complement of valence electrons and so do not combine easily with other elements. • Elemental semiconductors Silicon(Si) and germanium (Ge) have 4 valance electrons. • Compound semiconductors 1) III-V compound s/c’s; GaP, InAs, AlGaAs etc. 2) II-VI compound s/c’s; ZnS, CdS etc. Ionic bonding The metallic elements have only up to the valence electrons in their outer shell will lose their electrons and become positive ions, whereas electronegative elements tend to acquire additional electrons to complete their octed and become negative ions, or anions. Na Cl Ionic bonding • • • • • Ionic bonding is due to the electrostatic force of attraction between positively and negatively charged ions. This process leads to electron transfer and formation of charged ions; a positively charged ion for the atom that has lost the electron and a negatively charged ion for the atom that has gained an electron. All ionic compounds are crystalline solids at room temperature. Ionic crystals are hard, high melting point, brittle and can be dissolved in ordinary liquids. NaCl and CsCl are typical examples of ionic bonding. Covalent bonding • The bonding is due to the sharing of electrons. • Covalently bonded solids are hard, high melting points, and insoluble in all ordinary solids. • Elemental semiconductors of Si, Ge and diamond are bonded by this mechanism and these are purely covalent. • Compound s/c’s exhibit a mixture of both ionic and covalent bonding. Comparison of Ionic and Covalent Bonding Metallic bonding • Valance electrons are relatively bound to the nucleus and therefore they move freely through the metal and they are spread out among the atoms in the form of a low-density electron cloud. • A metallic bond result from the sharing of a variable number of electrons by a variable number of atoms. A metal may be described as a cloud of free electrons. • Therefore, metals have high electrical and thermal conductivity. + + + + + + + + + Metallic bonding • All valence electrons in a metal combine to form a “sea” of electrons that move freely between the atom cores. The more electrons, the stronger the attraction. This means the melting and boiling points are higher, and the metal is stronger and harder. • The positively charged cores are held together by these negatively charged electrons. • The free electrons act as the bond (or as a “glue”) between the positively charged ions. • This type of bonding is nondirectional and is rather insensitive to structure. • As a result we have a high ductility of metals - the “bonds” do not “break” when atoms are rearranged – metals can experience a significant degree of plastic deformation. Van der Waals bonding • It is the weakest bonding mechanism. • It occurs between neutral atoms and molecules. • The explanation of these weak forces of attraction is that there are natural fluctuation in the electron density of all molecules and these cause small temporary dipoles within the molecules. It is these temporary dipoles that attract one molecule to another. They are as called van der Waals' forces. • Such a weak bonding results low melting and boiling points and little mechanical strength. Van der Waals bonding The dipoles can be formed as a result of unbalanced distribution of electrons in asymettrical molecules. This is caused by the instantaneous location of a few more electrons on one side of the nucleus than on the other. symmetric asymmetric Therefore atoms or molecules containing dipoles are attracted to each other by electrostatic forces. Classification of solids SOLID MATERIALS CRYSTALLINE Single Crystal POLYCRYSTALLINE AMORPHOUS (Non-crystalline) Crystalline Solid • Crystalline Solid is the solid form of a substance in which the atoms or molecules are arranged in a definite, repeating pattern in three dimension. Crystalline Solid • Single crystal has an atomic structure that repeats periodically across its whole volume. Even at infinite length scales, each atom is related to every other equivalent atom in the structure by translational symmetry Single Pyrite Crystal Amorphous Solid Single Crystal Polycrystalline Solid • • Polycrystal is a material made up of an aggregate of many small single crystals (also called crystallites or grains). The grains are usually 100 nm - 100 microns in diameter. Polycrystals with grains that are <10 nm in diameter are called nanocrystalline. Polycrystalline Pyrite form (Grain) Polycrystal Amorphous Solid • Amorphous (non-crystalline) Solid is composed of randomly orientated atoms, ions, or molecules that do not form defined patterns or lattice structures. מבני גבישים Simple Cubic Cubic Close Packed Body Centered Cubic Hexagonal Close Packed http://cst-www.nrl.navy.mil/lattice מודל הפסים במוצקים • מודל הפסים. • פס ערכיות ,פס הולכה ואנרגיית הפער. • איכלוס הפסים במבודד ,מוליך ומוליך למחצה. • זיהומים נוטלים וזיהומים תורמים. • רמת פרמי .EF • מסה אפקטיבית. מבנה רמות אנרגיה במוצקים מודל הפסים • מודל הפסים מתאר את מבנה רמות האנרגיה המותרות לאלקטרונים ,הנוצרות במוצק. • מהו התנאי לקיום פסים אחידים? • האטומים בגביש חייבים להיות מסודרים במבנה מרחבי מחזורי (האטומים יושבים על סריג מרחבי). • בשקפים הבאים נראה מדוע נוצרים פסי אנרגיה בגביש מסודר. נתחיל עם רמות האנרגיה של אטום בודד: כפי שלמדנו לאטום רמות אנרגיה בדידות (דיסקרטיות) ,וחלקן מאוכלסות באלקטרונים (האטום מיוצג ע"י בור פוטנציאל). אין שינוי ברמות האנרגיה כאשר יש שני אטומים רחוקים אחד מהשני. E E 0 0 Eex2 Eex1 EV Eex2 Eex1 EV E2 E2 E1 E1 כעת האטומים מתקרבים: רמות האנרגיה משתנות במקצת בגלל האינטראקציה בין האטומים. סביב כל רמת אנרגיה של האטום הבודד יש כעת שתי רמות אנרגיה. מספר המקומות הניתנים לאכלוס ע"י אלקטרונים נותר ללא שינוי. E E 0 0 Eex2 Eex1 EV Eex2 Eex1 EV E2 E2 E1 E1 מה קורה לרמות האנרגיה בגביש מוצק ? בכל סמ"ק של חומר מוצק יש כ 1022אטומים .לכן בגביש מסודר כל רמה מתפצלת ל 1022רמות צפופות מאוד .כך שכל רמת אנרגיה בדידה באטום המקורי הופכת להיות ,באופן אפקטיבי ,פס אנרגיה .בין כל שני פסי אנרגיה יש תחום אנרגיות שבו אסור לאלקטרונים להימצא – אנרגית הפער. פס הערכיות ,פס ההולכה ופער האנרגיה בינם ()Eg חשובים להולכה החשמלית. גביש 0 פס הולכה Conductance band אנרגיית הפער )Eg (gap E אטום בודד פס הולכה פס ערכיות Eex2 Eex1 EV E 0 X 1022 Eex2 Eex1 EV פס ערכיות Valance band E2 E2 E1 E1 :חישוב לדוגמא • Consider 1 cm3 of Silicon. How many atoms does this contain ? • Solution: The atomic mass of silicon is 28.1 g which contains Avagadro’s number of atoms. Avagadro’s number N is 6.02 x 1023 atoms/mol . The density of silicon: 2.3 x 103 kg/m3 so 1 cm3 of silicon weighs 2.3 gram and so contains 6.02 1023 2.3 4.93 1022 atoms 28.1 This means that in a piece of silicon just one cubic centimeter in volume , each electron energy-level has split up into 4.93 x 1022 smaller levels ! מבודדים • The magnitude of the band gap determines the differences between insulators, s/c‘s and metals. • The excitation mechanism of thermal is not a useful way to promote an electron to CB even the melting temperature is reached in an insulator. • Even very high electric fields is also unable to promote electrons across the band gap in an insulator. E CB (completely empty) Eg~several electron volts VB (completely full) Wide band gaps between VB and CB )מוליכים (מתכות E • פס ההולכה צמוד או .חופף לפס הערכיות E CB VB Overlapping VB and CB CB • הפסים לא מלאים ולכן מתאפשרת תנועת .המטענים VB Touching VB and CB • No gap between valance band and conduction band מבודד לעומת מוליך פס הולכה פס הולכה פס הולכה פס ערכיות פס ערכיות פס ערכיות מבודד מוליך (לדוגמא )αSn cubic מוליך (לדוגמא )HgTe מוליכים למחצה • • • At 0K valance band full and conduction band emptylike Insulator. When enough energy is supplied to the e- sitting at the top of the valance band, e- can make a transition to the bottom of the conduction band. When electron makes such a transition it leaves behind a missing Empty conduction band electron state. • • • This missing electron state is called as a hole. Hole behaves as a positive charge carrier. Magnitude of its charge is the same with that of the electron but with an opposite sign. Forbidden energy gap [Eg] energy e+- e+- e+- e+Full valance band ? כיצד ניתן לעורר את האלקטרונים- מוליכים למחצה Answer : • Thermal energy • Electrical field • Electromagnetic radiation Partly filled CB Eg Partly filled VB Energy band diagram of a s/c at a finite temperature. To have a partly field band configuration in a s/c , one must use one of these excitation mechanisms. 1-Thermal Energy Thermal energy = k x T = 1.38 x 10-23 J/K x 300 K =25 meV Excitation rate = constant x exp(-Eg / kT) Although the thermal energy at room temperature, RT, is very small, i.e. 25 meV, a few electrons can be promoted to the CB. Electrons can be promoted to the CB by means of thermal energy. This is due to the exponential increase of excitation rate with increasing temperature. Excitation rate is a strong function of temperature. 2- Electric field • For low fields, this mechanism doesn’t promote electrons to the CB in common s/c’s such as Si and GaAs. • An electric field of 1018 V/m can provide an energy of the order of 1 eV. This field is enormous. So , the use of the electric field as an excitation mechanism is not useful way to promote electrons in s/c’s. 3- Electromagnetic Radiation c 1.24 34 8 E h h (6.62 x10 J s) x(3x10 m / s) / (m) E (eV ) (in m) h = 6.62 x 10-34 J-s c = 3 x 108 m/s 1 eV=1.6x10-19 J for Silicon Eg 1.1eV Near infrared 1.24 ( m) 1.1 m 1.1 To promote electrons from VB to CB Silicon , the wavelength of the photons must 1.1 μm or less 3- Electromagnetic Radiation Conduction Band • The converse transition can also happen. e- • An electron in CB recombines with a hole in VB and generate a photon. photon + Valance Band • The energy of the photon will be in the order of Eg. • If this happens in a direct band-gap s/c, it forms the basis of LED’s and LASERS. Intrinsic semiconductor • The conductivity of a pure (intrinsic) s/c is low due to the low number of free carriers. • The number of carriers are generated by thermally or electromagnetic radiation for a pure s/c. • For an intrinsic semiconductor n = p = ni n = concentration of electrons per unit volume p = concentration of holes per unit volume ni = the intrinsic carrier concentration of the semiconductor under consideration. Intrinsic semiconductor The intrinsic carrier concentration ni depends on; • the semiconductor material, and • the temperature. For silicon at 300 K, ni has a value of 1.4 x 1010 cm-3. Clearly , equation (n = p = ni) can be written as n.p = ni2 Donors and Acceptors What is doping and dopants impurities ? • To increase the conductivity, one can dope pure s/c with atoms from column lll or V of periodic table. This process is called as doping and the added atoms are called as dopants impurities. n-type p-type Addition of different atoms modify the conductivity of the intrinsic semiconductor. p-type doped semiconductor Si + Column lll impurity atoms Electron Have four valance e-’s Boron (B) has three valance e-’ s Hole Boron bonding in Silicon Boron sits on a lattice side Si Si Bond with missing electron B Si Si p >> n Normal bond with two electrons p-type doped semiconductor • Boron(column III) atoms have three valance electrons, there is a deficiency of electron or missing electron to complete the outer shell. • This means that each added or doped boron atom introduces a single hole in the crystal. There are two ways of producing hole 1) Promote e-’s from VB to CB, 2) Add column lll impurities to the s/c. p-type Energy Diagram CB Ec = CB edge energy level acceptor (Column lll) atoms Eg EA= Acceptor energ level Ev = VB edge energy level VB Electron Hole p-type doped semiconductor 1. 2. The impurity atoms from column lll occupy at an energy level within Eg . These levels can be Shallow levels which is close to the band edge, Deep levels which lies almost at the mid of the band gap. If the EA level is shallow i.e. close to the VB edge, each added boron atom accepts an e- from VB and have a full configuration of e-’s at the outer shell. These atoms are called as acceptor atoms since they accept an e- from VB to complete its bonding. So each acceptor atom gives rise a hole in VB. The current is mostly due to holes since the number of holes are made greater than e-’s. n-type doped semiconductor Si + Column V impurity atoms Have four valance e-’s Arsenic (As) has five valance e-’ s Electron Si Weakly bound electron Conduction band Ec Si Ed Eg As Si Band gap is 1.1 eV for silicon Valance band Ev Electron Si n >> p Normal bond with two electrons n-type, p- type :מוליך למחצה Ge Ge n-type Ge E פס הולכה + Ge Ge − As Ge Excess +charge Eg פער האנרגיה Ge Excess electron from arsenic atom n-type germanium. Ge Ed≈0.012eV Ge Ge p-type 0 פס ערכיות E − Ge Ge Ge + B Ge Excess −charge פס הולכה Eg Ge Positive hole, as one electron was removed from a bond to complete the tetrahedral bonds of the boron atom. 0 Ea≈0.01eV פער האנרגיה פס ערכיות מוליך למחצה :אינטרינסי ואקסטרינסי פס הולכה פס הולכה פס הולכה Ed Ea פס ערכיות פס ערכיות פס ערכיות מוליך למחצה אינטרינסי מוליך למחצה אקסטרינסי מוליך למחצה אקסטרינסי n- type p- type Fermi level , EF • This is a reference energy level at which the probability of occupation by an electron is ½. • Since Ef is a reference level therefore it can appear anywhere in the energy level diagram of a S/C . • Fermi energy level is not fixed. • Occupation probability of an electron and hole can be determined by Fermi-Dirac distribution function, FFD ; FFD EF = Fermi energy level kB = Boltzman constant T = Temperature 1 E EF 1 exp( ) k BT Fermi level , EF FFD 1 E EF 1 exp( ) k BT • E is the energy level under investigation. • FFD determines the probability of the energy level E being occupied by electron. if E EF f FD • 1 1 1 exp 0 2 1 f FD determines the probability of not finding an electron at an energy level E; the probability of finding a hole . Carrier concentration equations The number density, i.e., the number of electrons available for conduction in CB is 3/ 2 2 m kT EC EF n 2 exp ( ) h kT E EF E Ei n NC exp ( C ) n ni exp( F ) kT kT * n 2 The number density, i.e., the number of holes available for conduction in VB is 3/ 2 2 m*p kT EF EV p 2 exp ( ) h2 kT E EV E EF p NV exp ( F ) p ni exp( i ) kT kT ()11 ריכוז נושאי מטען – מל"מ אינטרינסי )3/4 ( Eg 2kT ni pi = 2(2kT h ) ( me mh ) e * * 2 32 -M*eמסה אפקטיבית אלקטרונים. -M*hמסה אפקטיבית חורים. - Egהיא אנרגית הפער בין פס הערכיות לפס ההולכה. 2(2πmkT/h2)3/2 ≈ 1019 per cm3 T = 300K • לדוגמא ריכוזי האלקטרונים בפס הערכיות של מל"מ אינטרינסיים ב : T=300K ni (Si) ≈ 1.2 x 1010 cm-3 ni (Ge) ≈ 2.4 x 1013 cm-3 ni (GaAs) ≈ 2.2 x 106 cm-3 • ריכוזים אלו קטנים באופן משמעותי מצפיפות האטומים בגביש שהיא כ .5 x 1022 cm-3 ריכוז נושאי המטען כתלות בטמפרטורה במל"מ אקסטרינסי מקובל לחלק לארבעה תחומים: • תחום הקיפאון ()Freeze-out • תחום היינון החלקי של אטומי הזיהום ()Partial ionization • התחום האקסטרינסי ()Saturation • התחום האינטרינסי במל"מ אינטרינסי -נמצאים תמיד בתחום האינטרינסי ריכוז נושאי המטען כתלות בטמפרטורה תחום היינון החלקי של אטומי הזיהום ()Partial ionization kT << Ed, Ea << Eg • תחום זה מתחיל קצת מעל האפס המוחלט. • האנרגיה התרמית מספיקה כדי להעביר חלק מנושאי המטען מרמות האנרגיה של הזיהומים. )kT n = (2N d )1 2 (2me kT h 2 )3 2 e( Ed n-type p = (2N a ) (2m kT h ) e p-type ) ( E a kT 2 32 h Na, Ndריכוז האטומים הזיהומים הנותנים והנוטלים הכללי. Edנמדד מתחתית פס ההולכה. Eaנמדד מתקרת פס הערכיות. 12 ריכוז נושאי המטען כתלות בטמפרטורה התחום האקסטרינסי ()Saturation Ed, Ea < kT << Eg • כמעט כל נושאי המטען שמקורם בזיהומים מיוננים. • ריכוז נושאי מטען האינטרינסיים עודנו זניח. p=0 n = Nd n-type n=0 p = Na p-type Na, Ndריכוז האטומים הזיהומים הנותנים והנוטלים הכללי. ריכוז נושאי המטען כמעט קבוע בטמפרטורה. ריכוז נושאי המטען כתלות בטמפרטורה התחום האינטרינסי Ea, Ed < kT < Eg • האנרגיה התרמית מספיקה להעלות אלקטרונים מפס הערכיות לפס ההולכה. • מספר נושאי המטען האינטרינסיים גדול ממספר נושאי המטען שמקורם בסימום. )n= Nd + N(T )p = N (T n-type )p= Na + N(T )n = N (T p-type Na, Ndריכוז האטומים הזיהומים הנותנים והנוטלים הכללי. )3/4 ( E g 2kT N (T ) ni pi = 2(2kT h ) ( me mh ) e * * 2 32 ריכוז נושאי המטען כתלות בטמפרטורה (.)Si The Concept of Effective Mass • If the same magnitude of electric field is applied to both electrons in vacuum and inside the crystal, the electrons will accelerate at a different rate from each other due to the existence of different potentials inside the crystal. Comparing Free e- in vacuum In an electric field mo =9.1 x 10-31 kg Free electron mass • The electron inside the crystal has to try to make its own way. • So the electrons inside the crystal will have a different mass than that of the electron in vacuum. • This altered mass is called as an effectivemass. An e- in a crystal In an electric field In a crystal m = ? m* effective mass What is the expression for m* • Particles of electrons and holes behave as a wave under certain conditions. So one has to consider the de Broglie wavelength to link partical behaviour with wave behaviour. • Partical such as electrons and waves can be diffracted from the crystal just as X-rays . (Bragg diffraction) • Certain electron momentum is not allowed by the crystal lattice. This is the origin of the energy band gaps. n 2d sin n = the order of the diffraction λ = the wavelength of the X-ray d = the distance between planes θ = the incident angle of the X-ray beam n = 2d (1) The waves are standing waves 2 = k is the propogation constant The momentum is By means of equations (1) and (2) certain e- momenta are not allowed by the crystal. The velocity of the electron at these momentum values is zero. Energy P = k (2) The energy of the free electron can be related to its momentum E= P 2 P= 2m0 h free e- mass , m0 2 1 2 k2 h h E 2m 2 2m (2 ) 2 h = 2 E 2k 2 2m The energy of the free e- is related to the k k momentum E versus k diagram is a parabola. Energy is continuous with k, i,e, all energy (momentum) values are allowed. E versus k diagram or Energy versus momentum diagrams Find effective mass , m* We will take the derivative of energy with respect to k ; 2 dE k dk m - m* is determined by the curvature of the E-k curve 2 d2E 2 m dk Change m* m* instead of - m* is inversely proportional to the curvature m 2 2 d E dk 2 This formula is the effective mass of an electron inside the crystal. Positive and negative effective mass m* Direct-band gap s/c’s (e.g. GaAs, InP, AlGaAs) E 2 d 2 E dk 2 CB e- • The sign of the effective mass is determined directly from the sign of the curvature of the E-k curve. • The curvature of a graph at a minimum point is a positive quantity and the curvature of a graph at a maximum point is a negative quantity. • Particles(electrons) sitting near the minimum have a positive effective mass. • Particles(holes) sitting near the valence maximum have a negative effective mass. • A negative effective mass implies that a particle will go ‘the wrong way’ when an extrernal force is applied. k + VB band Direct an indirect-band gap materials Direct-band gap s/c’s (e.g. GaAs, InP, AlGaAs) E • For a direct-band gap material, the minimum of the conduction band and maximum of the valance band lies at the same momentum, k, values. • When an electron sitting at the bottom of the CB recombines with a hole sitting at the top of the VB, there will be no change in momentum values. • Energy is conserved by means of emitting a photon, such transitions are called as radiative transitions. CB ek + VB Indirect-band gap materials Indirect-band gap s/c’s (e.g. Si and Ge) • E • CB For an indirect-band gap material; the minimum of the CB and maximum of the VB lie at different k-values. When an e- and hole recombine in an indirect-band gap s/c, phonons must be involved to conserve momentum. ePhonon Eg k + VB Atoms vibrate about their mean position at a finite temperature.These vibrations produce vibrational waves inside the crystal. Phonons are the quanta of these vibrational waves. Phonons travel with a velocity of sound . Their wavelength is determined by the crystal lattice constant. Phonons can only exist inside the crystal. Indirect-band gap materials • The transition that involves phonons without producing photons are called nonradiative (radiationless) transitions. • These transitions are observed in an indirect band gap s/c and result in inefficient photon producing. • Momentum conservation requires phonon and photon emit together Much lower probability. • So in order to have efficient LED’s and LASER’s, one should choose materials having direct band gaps such as compound s/c’s of GaAs, AlGaAs, etc… Photon vs. Phonon • For GaAs, calculate a typical (band gap) photon energy and momentum , and compare this with a typical phonon energy and momentum that might be expected with this material. photon phonon E(photon) = Eg(GaAs) = 1.43 ev = hc / λ E(photon) = h s /λ = hvs / a0 λ (phonon) ~a0 = lattice constant =5.65x10-10 m c= 3x108 m/sec P=h/λ = hv E(phonon) = h h=6.63x10-34 J-sec Vs= 5x103 m/sec ( velocity of sound) λ (photon)= 1.24 / 1.43 = 0.88 μm E(phonon) = hvs / a0 = 0.037 eV P(photon) = h / λ = 7.53 x 10-28 kg-m/sec P(phonon)= h / λ = h / a0 = 1.17x10-24 kg-m/sec Photon vs. Phonon • • • • Photon energy = 1.43 eV Phonon energy = 37 meV Photon momentum = 7.53 x 10-28 kg-m/sec Phonon momentum = 1.17 x 10-24 kg-m/sec Photons carry large energies but negligible amount of momentum. On the other hand, phonons carry very little energy but significant amount of momentum. Electric conduction • • • • • • Carrier drift Carrier mobility Mobility variation with temperature A derivation of Ohm’s law Drift current equations Semiconductor band diagrams with an electric field present • Carrier diffusion • The Einstein relation • Total current density Drift and Diffusion • As recalls, current is the rate of flow of charge. • So current depend on the number of charge carriers and their flowing capabilities. • There are two current mechanisms which cause charges to move. • The two mechanisms are drift and diffusion. Carrier Drift • Electron and holes will move under the influence of an applied electric field since the field exert a force on charge carriers (electrons and holes). F qE • These movements result a current of ; Id I d nqVd A Id : drift current Vd : drift velocity of charge carrier n : number of charge carriers per unit volume q : charge of the electron A : area of the conductor / semiconductor Carrier Mobility , Vd E E: : 2 cm V Sec applied field mobility of charge carrier is a proportionality factor Vd E So is a measure how easily charge carriers move under the influence of an applied field or determines how mobile the charge carriers are. Carrier Mobility , Macroscopic understanding Vd E In a perfect Crystal 0 It is a superconductor Microscopic understanding? (what the carriers themselves are doing?) q * m me* mh* in general m ; n type * e m ; p type * h Microscopic understanding of mobility? How long does a carrier move in time before collision ? The average time taken between collisions is called as relaxation time, (or mean free time) How far does a carrier move in space (distance) before a collision? The average distance taken between collisions is called as mean free path, l. Carrier Mobility , • A perfect crystal has a perfect periodicity and therefore the potential seen by a carrier in a perfect crystal is completely periodic. • So the crystal has no resistance to current flow and behaves as a superconductor. The perfect periodic potential does not impede the movement of the charge carriers. However, in a real device or specimen, the presence of impurities, interstitials, subtitionals, temperature , etc. creates a resistance to current flow. • The presence of all these upsets the periodicity of the potential seen by a charge carrier. The mobility two components in s.c. The mobility has two component Lattice interaction component L Impurity interaction component I Mobility variation with temperature in s.c. T T Low temperature High temperature 1 T 1 L 1 ln( ) I This equation is called as Mattheisen’s rule. I L ln( T ) Peak depends on the density of impurities Variation of mobility with temperature s.c. L At high temperature (as the lattice warms up) L component becomes significant. decreases when temperature increases. L C1 T 3 2 It is called as a T 3 2 C1 is a constant. T 1.5 power law. Carriers are more likely scattered by the lattice atoms. Variation of mobility with temperature s.c. At low temperatures I I component is significant. decreases when temperature decreases. I C2 T 3 2 C2 is a constant. Carriers are more likely scattered by ionized impurities. Thermal velocity • Assume crystal is at thermodynamic equilibrium (i.e. there is no applied field). What will be the energy of the electron at a finite temperature? • The electron will have a thermal energy of kT/2 per degree of freedom. So , in 3D, electron will have a thermal energy of 3kT 1 * 2 3kT E m Vth Vth 2 2 2 Vth : thermal velocity of electron Vth T 1 2 Vth m * 1 2 3kT m Random motion no current • Since there is no applied field, the movement of the charge carriers will be completely random. This randomness result no net current flow. As a result of thermal energy there are almost an equal number of carriers moving right as left, in as out or up as down. Vdrift Vs. Vth • Calculate the velocity of an electron in a piece of n-type silicon due to its thermal energy at RT and due to the application of an electric field of 1000 V/m across the piece of silicon. Vth ? Vd ? V th RT 300 K E 1000 V / m me* 1.18 m0 0.15 m 2 /(V s ) 3kT 5 m / sec 10 x 1.08 V th m Vd E Vd 150 m / sec Calculation Drift velocity=Acceleration x Mean free time Vd F * m Force is due to the applied field, F=qE Vd F qE * m m* Vd E q m Calculation- Drift Velocities • Calculate the mean free time and mean free path for electrons in a piece of n-type silicon and for holes in a piece of p-type silicon. ? l ? me* 1.18 mo e 0.15 m 2 /(V s ) e e me q 1012 sec vthelec 1.08 x105 m / s mh 0.59mo h 0.0458 m 2 /(V s ) h h mh q 1.54 x10 13 sec vthhole 1.052 x105 m / s le vthelec e (1.08 x105 m / s )(10 12 s ) 10 7 m lh vthhole h (1.052 x105 m / s )(1.54 x10 13 sec) 2.34 x10 8 m Saturated Drift Velocities • The equation of Vd .E does not imply that Vd increases linearly with applied field E. • Vd increases linearly for low values of E and then it saturates at some value of Vd which is close Vth at higher values of E. • Any further increase in E after saturation point does not increase Vd instead warms up the crystal. A Derivation of Ohm’s Law Vd E I d nqVd A Id Jd A q m J x nqVd nq E nq m 2 J x Ex nq 2 Jx m 1 Ex m 1 ( m) Drift Current Equations For undoped or intrinsic semiconductor ; n=p=ni For electron For hole J p pqE p J n nqEn drift current for electrons number of free electrons per unit volume mobility of electron drift current for holes number of free holes per unit volume mobility of holes Drift Current Equations Total current density Ji Je Jh J i nqE n pqE P since n p ni J i ni q ( n p ) E For a pure intrinsic semiconductor Drift Current Equations J total ? for doped or extrinsic semiconductor n-type semiconductor; n p JT nqn E N Dqn E where ND is the shallow donor concentration p-type semiconductor; p n J T pq p E N Aq p E where NA is the shallow acceptor concentration Carrier Diffusion Current mechanisms Drift Diffusion photons P nkT dP dn kT dx dx dn 1 dP dx kT dx Contact with a metal Einstein Relation Einstein relation relates the two independent current mechanicms of mobility with diffusion; Dn kT n q and Dp kT p q for electrons and holes Constant value at a fixed temperature 2 cm sec volt 2 cm V sec kT 25 mV q kT J / K K volt q C at room temperature Total Current Density When both electric field (gradient of electric potential) and concentration gradient present, the total current density ; dn J n q n nE qDn dx dp J p q p pE qD p dx J total J n J p Semiconductor Band Diagrams with Electric Field Present At equilibrium ( with no external field ) EC All these energies are horizontal Eİ Pure/undoped semiconductor EV How these energies will change with an applied field ? + qV e- EC Ef Eİ n – type Electric field Electron movement Hole flow EV hole Semiconductor Band Diagrams with Electric Field Present • With an applied bias the band energies slope down for the given semiconductor. Electrons flow from left to right and holes flow from right to left to have their minimum energies for a p-type semiconductor biased as below. _ + e- EC qV Eİ p – type Ef Electric field Electron movement Hole flow EV hole Under drift conditions; •Under drift conditions; holes float and electrons sink. Since there is an applied voltage, currents are flowing and this current is called as drift current. •There is a certain slope in energy diagrams and the depth of the slope is given by qV, where V is the battery voltage. תלות בטמפרטורה של הולכה חשמלית במוליכים המוליכות σמוגדרת ע"י : J=σE σ=ne2τ/m -nצפיפות נושאי המטען. - eמטען האלקטרון. - mהמסה האפקטיבית של נושא המטען. - τהוא זמן הרלקסציה. m ,eלא תלויים בטמפרטורה ולכן התלות נובעת רק מהשינוי בריכוז נושאי המטען ובזמן הרלקסציה. במוליכים ריכוז נושאי המטען גבוה מאוד ואינה משתנה עם הטמפרטורה, ולכן התלות בטמפ' נובעת מהתלות של τבטמפרטורה. במתכות – גבישים מסודרים – ההתנגשויות בעיקר עם פונונים .ההתנגדות עולה עם הטמפרטורה באופן מתון – כמו 1/Tαכאשר αמסדר גודל של .1 בסגסוגות – ריבוי פגמים בגביש – ההתנגשויות בעיקר עם פגמים .ההתנגדות כמעט לא תלויה בטמפרטורה .מוליכות גרועה מזו של מתכות מסודרות. תלות בטמפרטורה של הולכה חשמלית במוליכים למחצה חלוקה ל 4תחומי טמפרטורה: • תחום הקיפאון ( σ=0 T=0 )Freeze outהמל"מ מבודד. • תחום היינון החלקי (kT << Ed, Ea << Eg )Partial ionization )σ exp(-Ed/kT )σ exp(-Ea/kT עבור מוליך למחצה מסוג n עבור מוליך למחצה מסוג p • התחום האקסטרינסי (Ed, Ea < kT << Eg )Saturation ריכוז נושאי המטען כמעט קבוע בטמפרטורה ,ולכן השינוי בהולכה קטן ונובע מהתלות של הנידות בטמפרטורה. • התחום האינטרינסי Ea, Ed < kT < Eg )σ exp(-Eg/2kT במל"מ צפיפות נושאי המטען החופשיים nו pהוא בדרך כלל הגורם דומיננטי בתלות המוליכות בטמפרטורה .רק בתחום האקסטרינסי ,כאשר ריכוז נושאי המטען קבוע, ניתן להבחין בשינוי בזמן הרלקסציה. Conductivity in solids Organic (plastic) Semiconductors הולכה חשמלית במוצקים -סיכום ההולכה החשמלית במוצקים תלויה בגורמים הבאים: .1כמות נושאי המטען שיכולים לנוע כאשר מופעל מתח חשמלי על המוצק. .2התנגשויות של נושאי המטען עם תנודות השריג ,הפונונים. .3התנגשויות של נושאי המטען עם פגמים במבנה השריגי (נקעים דחיקים וכו'). • תלות ההולכה בטמפרטורה תלויה בהשתנות הגורמים הנ"ל כפונקציה של הטמפרטורה. • בחומרים מסוימים ישנו גורם אחד דומיננטי והוא יאפיין את תלות הולכה בטמפרטורה .לדוגמה :מתכת – פונונים ,מל"מ – ריכוז נושאי מטען, סגסוגת – התנגשות בפגמים. p – n junction Hole Movement n-type +++++ +++++ +++++ +++++ +++++ +++++ +++++ +++++ --------------------------------- p-type Electron Movement ++++ ++++ ++++ Fixed positive space-charge Metallurgical junction ---------- Fixed negative space-charge Ohmic end-contact p – n junction p-type n-type EC EC Ef Eİ Eİ Ef EV EV EC Eİ p-type n-type EC Ef Eİ Ef EV EV There is a big discontinuity in the fermi level accross the p-n junction. p – n junction Lots of electrons on the left hand side of the junction want to diffuse to the right and lots of holes on the right hand side of the junction want to move to the left. The donors and acceptors fixed,don’t move (unless you heat up semiconductors, so they can diffuse) because they are elements (such as arsenic and boron) which are incorporated to lattice. However, the electrons and holes that come from them are free to move. Idealized p-n junction Holes diffuse to the left of the metalurgical junction and combine with the electrons on that side. They leave behind negatively charged acceptor centres. Similarly, electrons diffusing to the right will leave behind positively charged donor centres. This diffusion process can not go on forever. Because, the increasing amount of fixed charge wants to electrostatically attract the carriers that are trying to diffuse away(donor centres want to keep the electrons and acceptor centres want to keep the holes). Equlibrium is reached. This fixed charges produce an electric field which slows down the diffusion process. This fixed charge region is known as depletion region or space charge region which is the region the free carriers have left. It is called as depletion region since it is depleted of free carriers. p – n junction The drift and diffusion currents are flowing all the time. But, in thermal equilibrium, the net current flow is zero since the currents oppose each other. Under non-equilibrium condition, one of the current flow mechanism is going to dominate over the other, resulting a net current flow. The electrons that want to diffuse from the ntype layer to the p-layer have potential barier. DR Neutral p-region ---------- +++ +++ +++ Current Mechanisms, Neutral n-region Diffusion of the carriers cause an electric in DR. Field Direction Electron Drift Hole energuy Electrıon energy p – n junction in thermal equilibrium Drift current is due to the presence of electric Electron Diffusion field in DR. EC Ef Diffusion current is due to the majority carriers. EV Drift current is due to the minority carriers. Hole Diffussion Hole Drift Appliying bias to p-n junction + - p n forward bias - + p n reverse bias How current flows through the p-n junction when a bias (voltage) is applied. The current flows all the time whenever a voltage source is connected to the diode. But the current flows rapidly in forward bias, however a very small constant current flows in reverse bias case. Appliying bias to p-n junction I(current) Reverse Bias Vb I0 Forward Bias V(voltage) Vb ; Breakdown voltage I0 ; Reverse saturation current There is no turn-on voltage because current flows in any case. However , the turn-on voltage can be defined as the forward bias required to produce a given amount of forward current. If 1 m A is required for the circuit to work, 0.7 volt can be called as turn-on voltage. Appliying bias to p-n junction Zero Bias p Forward Bias + - -- ++ n -- ++ p Ec Ev qVbi E v n qVbi VF Vbi p --- ++ ++ n Ec Ev Potential Energy Ec - + - + Reverse Bias + q Vbi Vr Vbi VR Vbi VF Ideal diode equation J Total Dn n po D p pno qV q exp L p kT Ln qV 1 J o exp kT 1 multiplying by area ; qV I I o exp kT 1 Ideal diode equation This equation is valid for both forward and reverse biases; just change the sign of V. Ideal diode equation reverse bias • Change V with –V for reverse bias. When qV > a few kT; exponential term goes to zero as qV I I o exp kT 1 I Io Reverse saturation current Current Forward Bias VB I0 Voltage VB ; Breakdown voltage Reverse Bias I0 ; Reverse saturation current