Energy Levels and Atomic Structure. The Rutherford Model of the Atom

University of Technology 2016 2017 First Year, Lecture One Basic Construction of the Atom: Energy Levels and Atomic Structure The atom is a basic unit of material that consists of central nucleus surrounded by a cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charges called (protons) and uncharged particles called (neutrons) (except in the case of hydrogen, which is the no include neutrons). The electron is the basic unit of the negative charge in atom, where the charge of electron is 1.6 10-19 C (coulomb) and the mass of electron is 9.11 10-31 Kg. An atom containing an equal number of protons and electrons is electrically neutral. A group of atoms can remain bound to each other forming a molecule. The Rutherford Model of the Atom Rutherford, in 1911 proposed the following model of an atom: 1. The atom consist of a small nucleus in which the enter all the positive charge (protons) and almost the whole mass of the atom. 2. The nucleus occupies a very small space as compared to the size of the atom, he proposed that the atom was mostly empty space. 3. The electrons revolve around the nucleus in various orbits just as planets revolve around the sun. 4. The charge on the electron is negative and is equal in magnitude to the charge on the proton. Therefore the atom as a whole is electrically neutral. As a specific illustration of this atom model consider the hydrogen atom, Which is the simplest atom consisting of one electron revolving around the nucleus that consist of one proton. The electron will be effected by two forces: 1. Electrostatic Attraction Force (F e ): This force is resulting from the attraction between the electron and the proton, it can be calculated by Coulombs Law. See Fig.1-1. in newton (N). Where: : Negative electron charge = coulomb (C). : Positive nucleus charge = Ze, in coulomb (C). : Atomic number, Z =1 for hydrogen. : Permittivity of free space = farad / meter (F/m). : Radius of the orbit, in meter (m). 1

University of Technology 2016 2017 First Year, Lecture One 2. The centrifugal force (F c ): This force is resulting from the circular motion of the electron, it can be calculated by Newton s Second Law of motion. See Fig.1-1., in newton (N). Where: : The Electron mass = Kg. : The electron velocity in its circular path, in meter / second (m/s). : Acceleration toward the nucleus, in meter / second square (m/s 2 ). Fig. (1-1) Forces on the electron. In order the electron to be stable in orbit, the two forces must be equal in magnitude and opposite in direction, hence:, [1] The kinetic energy of the electron is The potential energy of the electron and the distance between the charges: in joule (J)., is the product of the electrostatic force in joule (J). Thus the total energy of the electron is This model could not explain the stability of the atom because according to classical electromagnetic theory the electron revolving around the nucleus must continuously radiate energy in the form of electromagnetic radiation and hence it should fall into the nucleus. If the electron is radiating energy, its total energy must decrease by the amount of this emitted energy as shown in Fig.1-2. 2

University of Technology 2016 2017 First Year, Lecture One Fig. (1-2) The electron moves in a spiral orbit toward the nucleus. The energy of the electron is often measured using an energy unit called electron volt (ev). This unit is defined as the energy that the electron has when it falls in a (1V) potential difference, therefore: The Bohr Model of the Atom He postulated the following three fundamental law: 1. The electron revolves around the nucleus in certain stable circular orbits without radiating energy. 2. The electron moves only in the orbits that which the angular momentum of electron are multiple integers of. [2] n = Principle quantum number, integer (1, 2, 3,, ). Since r should be denoted r n to indicate the n th orbit. Then from equ. (1) and (2) we obtain:, radius of the n th orbit of the electron in the hydrogen atom. Substituting (n=1) gives the smallest orbit (r 1 ) in the hydrogen atom, it equals: r 1 = 0.529 10-10 m, is called Bohr radius, it represent the radius of the electron orbit in the ground state (n=1) for hydrogen atom. While the other orbit radius determine by:, Also, substituting (r n ) in the equation of the total energy ( ), gives: This equation shows that there is a definite energy for each orbit, This 3

University of Technology 2016 2017 First Year, Lecture One equation shows that the total energy of the electron is always negative. This result is necessary for the electron in order to remain linked to the atom because if the total energy is greater than zero then the electron will have enough energy to separate from the nucleus. The lowest energy state E T1 is called the ground level and the other stationary states of the atom are called excited levels as shown in Fig. 1-3. The magnitude ( ) is constant and equal to (-13.6 ev) which represent the ground state energy for hydrogen atom. Fig. (1-3) The energy states in hydrogen atom. 3. The energy is radiated (emitted) from the atom only when the electron moves from a high level to a lower level as shown in Fig. 1-4, this energy equals the difference between the two levels energies and it will be released once in the form of photons where: h: Planck constant = f: The frequency of the radiation emitted or absorbed by the atom (Hz). E ni : Higher orbit energy (initial). E nf : Lower orbit energy (finial). n i : The number of the level from which the transition initiates. n f : The number of the level at which the transition terminates. Fig. (1-4) The emission of photon. 4

University of Technology 2016 2017 First Year, Lecture One, Balmar equation, Rydberg equation Where: f: Frequency of emitted photon (HZ). C: Speed of light. (C =3 10 8 m/s). : The wavelength (m). : The wave number (m -1 ). R: Rydberg constant. (R = 1.097 10 7 m -1 ). Example 1-1: Calculate the potential energy of an electron orbiting in ground state for a hydrogen atom with a kinetic energy of +13.6ev, hence calculate the total energy of this electron. Solution: 13.6 27.135-13.6ev. check the units? Example 1-2: Hydrogen atom electron is transferred from energy level 3 to energy level 1, calculate the wavelength of the radiated light and the radius of the orbit in level 3. Solution:. Example 1-3: Find for Balmer series whose n f = 2, using Rydberg formula. Solution: 5

University of Technology 2016 2017 First Year, Lecture One Exercises 1. In the Bohr model for a hydrogen atom, an electron revolving in the third excited state. If the electron drops into second state, calculate the following: a. The total energy of the electron in the third excited state. b. The transition energy. c. The kinetic energy of electron in first excited state. 2. Prove that the total energy of electron is measured in units of joule. 3. Hydrogen atom electron moves from third orbit to the first orbit, calculate the wave number of the radiated light. (Repeat using Rydberg equation). 4. Calculate the radius of orbit of an electron moving around a hydrogen atom with an energy of -13.6ev. If a photon of energy 10.2ev hits electron and losses all its energy to the electron, Calculate the new radius of orbit. What frequency of light will be emitted if the above electron changes it orbital radius to its old value. 5. An electron at rest hit by light of wavelength 4.96*10-7 m. Calculate the velocity of the electron. 6. An single electron orbits around a nucleus of charge (+Ze). It requires 47.2ev to excite the electron from third Bohr orbit to the second orbit, calculate: a. The value of Z. b. The wavelength of electromagnetic radiation required to transferred the electron from fourth to third orbit. c. The radius of the first excited orbit. 6

University of Technology 2016 2017 First Year, Lecture Two Waves and Particles Nature for light Modern laboratory experiments and facts physics process has duality that light has a completely different two properties where he owns particles and waves property. Through the phenomenon of the photoelectric effect, which Einstein developed prove particles properties of light, while through the phenomenon of diffraction effect, that the world has set De Broglie prove properties waves of light. photoelectric effect: When certain kinds of light are shone upon metals, electrons are liberated. Each of the liberated electrons leaves the metal with a kinetic energy anywhere from near zero up to a well-defined maximum as shown in Fig. 2-1. When the color, and hence the frequency, of the light is changed, the maximum energy of the liberated electrons is greater or smaller, depending upon the frequency change; it is a directly proportional relationship. If the frequency drops below a certain number, no electrons are liberated, no matter how intense the light. The maximum kinetic energy (K.E) observed from the metal as a result of incident light by 1 frequency (f) is: mv 2 2 max = hf Φ = h(f f o ), Einstein equation Where: Φ : Work function, f o : Threshold frequency. Fig. (2-1) Photoelectric effect phenomenon. These result in Fig. 2-2 indicate that the electrons receive an energy hf from the light (as a photons) and lose an amount of energy Φ in escaping from the surface of the metal. K.E hf C s K C u Photo-current Photo-current f f o f hf Fig. (2-2) The photoelectric effect experiment results. 10-9 s 1

University of Technology 2016 2017 First Year, Lecture Two Diffraction effect: Diffraction refers to various phenomena which occur when a wave encounters an obstacle or a slit. It is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle, see Fig. 2-3. sin θ = mλ, Bragg equation a Where: m = 1, 2, 3, Determination the position of diffraction cilices, λ: Wavelength of light, a: windows width, θ: Scattering angle. Fig. (2-3) Diffraction effect phenomenon. De Broglie Hypothesis Since the light is wave aspect in certain states and particle aspect in another states, therefore we can used these two aspect the wave and particle on material too. He set theory that determine the wavelength (λ) of the free electron has mass (m) and velocity (v). h λ =, De Broglie equation. mv = h p The electron which move has a wave the wavelength λ =, instrument from assume the mv electron moves at fixed orbit it expect produce waves when orbit around the nucleus at allowed orbit as shown in Fig. 2-4. 2πr = nλ 2πr = nh nh mvr = mv 2π Where: r: Radius of orbit. n: Integer number of waves. h 2 Fig. (2-4) Stable electron wave.

University of Technology 2016 2017 First Year, Lecture Two Waves and Particles Nature for Electron The electron similarly light which have so far treated as particles may behave like wave. Electrons arriving at screen produce a glow of light proportional to their number and energy. If the electron regarded as charged particle the energy imparted to the electron beam as a potential energy equal ev. If the electrons are to regarded as waves the gained energy define as the change in potential energy of the wave in passing through a potential difference of V volt =ev electron volt. If the electrons are particles we would expect then to be scattered randomly in all directions. The rings of intense light can only explained by assuming that the electron is a wave as shown in Fig. 2-5. Fig. (2-5) Thermionic emission Effect. Example 2-1: A tungsten surface having a working function of 4.52ev and wavelength is 0 2537 AP P, what is the maximum speed of the emitted electrons? Solution: f = c = 3 10 8 m/s = 1.182 λ 2537 10 10 m 1015 Hz 1 2 = hf Φ v max = 3.622 10 5 m/s 2 mv max Example 2-2: Find De Broglie wavelength for electron in a kinetic energy is 10 10P Pev. Solution: mv = 2 K. E. m = 5.399 10 23 kgm/s h λ = = 6.626 10 34 Js = mv 5.399 10 23 kgm/s 0.1227A0 3 3

University of Technology 2016 2017 First Year, Lecture Two Internal Structure of Solid Material A crystalline solid is distinguished by the fact that the atoms making up the crystal are arranged in a periodic fashion. Thus the crystal appears exactly the same at one point as it does at a series of other equivalent points, once the basic periodicity is discovered. However, not all solid are crystals Fig. 2-6, some have no periodic structure at all (amorphous solids), and other are composed of many small regions of single crystal material (polycrystalline solids). Fig. (2-6) Three general types of solids. Cubic Lattices The simplest three dimensional lattice is one in which the unit cell is a cubic volume, such as the three cells shown in Fig. 2-7. The simple cubic structure (sc) has an atom located at each corner of the unit cell. The body centered cubic (bcc) lattice has an additional atom at the center of the cube, and the face centered cubic (fcc) unit cell has atoms at the eight corners and centered on the six faces. Fig. (2-7) Unit cells for three types of cubic lattice structures. Example 2-3: Find the fraction of the fcc unit cell volume filled with hard spheres as in Fig. 2-8. Solution: Atoms per cell = 1(corners) + 3(faces) = 4 Nearest neighbor distance = 1 2 (a 2) Radius of each sphere = 1 4 (a 2) Volume of each sphere = 4 3 π[1 4 a 2 ]3 = πa3 2 Maximum fraction of cell filled = = 4 (πa3 2)/24 a 3 = 74 percent filled 24 no. of spheres vol.of each sphere total vol.of cell Fig. (2-8) 4

University of Technology 2016 2017 First Year, Lecture Two Energy Bands Structures in Solids The electrons of a single atom revolve around the nucleus in certain orbits, each orbit has a specific constant energy level and each level consists of several secondary level while in solids, which consist of a large number of close atoms, when these atoms unite to form the crystal any electron does not affected by the charges in its atom only but by the nuclei and the electrons in all other atoms forming the crystal too; therefore, the energy levels of the outer electrons change as a result of the interaction between atoms and instead of the specified energy levels of the single atom, the crystal contains a band of large number of very close energy levels that seems to be continuous, this band is called the valence band (V.B.). The electrons in the valence band are restricted to the atom and do not participate in the electrical conduction. When an electron in the valence band gains enough energy so that it can be separated from the atom, it jumps to the next band which is the conduction band (C.B.), see Fig 2-9. Fig. (2-9) Energy bands in carbon atoms. Insulators: The energy band structure at the normal lattice spacing is indicated schematically in fig. 2-10c. For a diamond (carbon) crystal the region containing no quantum states is several electron volts high (E.g. 6ev). The energy which can be supplied to an electron from an applied is too small carry the particle from the filled into the cant band. Since the electron cannot acquire sufficient applied energy. Semiconductor: The forbidden gap is small 1ev (Ge, germanium and Si, silicon) at 0 0 K. Energy cannot be acquired from an applied field, so the valence band remains fill, and the conduction band empty, and it is insulator at low temperature. At the temperature increase, some of these valence electrons obtain thermal energy greater than E g, and the electrons move into conduction band. These are free electrons can move under the effect of a small applied field, and result a current. The insulator has now become slightly conducting, it is a semiconductor, which has a free electrons in a conduction band and a holes in a valence band. 5

University of Technology 2016 2017 First Year, Lecture Two Metal: A solid which contains a partly filled band structure is called a metal. Under the influence of an applied electric field the electrons may acquire additional energy and move higher states. Since these mobile electrons constitute a current, this substance is a conductor and the partly filled region is the conduction band. One example of the band structure of a metal is given in fig.2-10a, which shows overlapping valence and conduction bands. Fig. (2-10) The energy gap in solids. Exercises 1. The wavelength of a ultraviolet rays is 3500A o falls on the surface of potassium, the maximum kinetic energy of electron optic is 1.6ev. Find work function for potassium. 2. An electron initially moves in first excited state, if it absorb energy with frequency 5 10 16 Hz. Find the momentum of this electron before and after interaction with photon energy. 3.If the threshold wavelength of a target material is 4.36 10-7 m. Calculate the work function of electron. 4. An electron initially moving in the third orbit around the nucleus of the hydrogen atom. Find the photo energy which emitted from electron transition to the first excited orbit. Calculate the De-Broglie wavelength of the electron in the orbit before transition, (using Rydberg formula). 5. If the ( 1 mv 2 2 max = hf Φ), Prove that by using the scale units of each quantities. 6

University of Technology 2016 2017 First Year, Lecture Three Mobility and Conductivity Mobility (µ) defined as the drift velocity per unit electric field. µ = v d E, in m2 v -1 s -1 The minus sign means the motion of electron is opposite direction of electric field. v d = eτe m µ = eτ m Where: v d : The drift velocity in m/s, e: The electron charge in c, τ: The relaxation time in s, E: The electric field intensity in v/m. When an electrical potential V is applied across a piece of material, a current of magnitude I flows, see Fig.3-1. The current is proportional to V, and can be described by Ohm's law: I = V/R R is the electrical resistance [ohms, Ω, V/A] and it is defined as: e L R = ρ L A ρ is resistivity in Ω.m. I = 1 V A ρ L J = δe δ = J E E I A Fig. (3-1) Piece metal connected to a voltage source. V J: The current density (A/m 2 ) through the cross section of the metal. δ: The conductivity in Ω -1 m -1. J = nev d δ = ne2 τ δ = ne µ m n: The number of free electrons per unit volume in electron/ m 3. This equation show that the conductivity depends on number of charges per unit volume, free electrons density and mobility of charges. Example 3-1: A silicon crystal having a cross section area of 0.001cm 2 and a length of 10-3 cm is connected at its ends with 10V battery at temperature 300 0 K, when the current is 100mA. Find the resistivity and the conductivity? Solution: ρ = V A = 10 I L 100 10 δ = 1 ρ 10 Ω-1 m -1. 0.001 10 4 3 10 3 10 2 = 0.1 Ωm. 1

University of Technology 2016 2017 First Year, Lecture Three Effects of Temperature and Doping on Mobility In lattice scattering a carrier moving through the lattice encounters atoms which are positions due to thermal vibration. At low temperature impurity scattering becomes the dominant mechanism. (lattice scattering is less important). A slowly moving carrier is likely to be scattered move strongly by an interaction with a charged greater momentum as shown in Fig. 3-2 and Fig. 3-3. Fig. (3-2) Approximate temperature dependence of mobility with both lattice and impurity scattering. 1 µ = 1 + 1 µ 1 µ 2 Impurity concentration Fig. (3-3) Variation of mobility with total doping impurity concentration for Ge at 300k. 2

University of Technology 2016 2017 First Year, Lecture Three Fermi-Dirac Level The Fermi function f(e) gives the probability that a give available electron energy state will be occupied at a given temperature. At room temperature, the thermal energy allow to a small amount of electrons to move from valence band to conduction band in semiconductor. The probability of filling the band by electrons depend on temperature is given by an equation called Fermi-Dirac function. f(e) = 1 1+e (E E F KT ) Where: f(e): The probability of finding the state E by electrons. E F : Energy of the Fermi level. K: Boltzmann constant = 8.62 10-5 ev/k =1.38 10-23 J/k. T: Absolute temperature in Kelvin. At absolute zero of temperature 0k, all electron states will be filled to an energy level E F. At higher temperature some electrons may be thermally excited into levels above E F. If E=E F, then the probability is 0.5 regardless to temperature T as shown in Fig. 3-4. Fig. (3-4) The Fermi-Dirac distribution function. Example 3-2: What the probability of finding electrons when the Fermi energy level equal the energy level. Solution: f(e) = 1 1+e (E E F KT ) = 1 1+e 0 = 0.5%. 3

and P mp P mp P and University of Technology 2016 2017 First Year, Lecture Three Exercises 1. A metal object with length 10m, resistance is 1kΩ and mobility is 4000cmP P/v.s. If the 28-3 concentration of free electrons equal 8.5 10P the carrying current is 2mA. Find the current density. 2. Find the drift velocity, if the voltage is 4V, length is 60cm, area is 2mP P, resistance is 5 kω 20-3 and the number of electrons is 4 10P P. 3. Prove the probability of finding electrons equal zero, if E > ERF Rat 0K temperature. 4. If E < ERFR T=0K, prove the probability of finding electrons equal 100%. 5. Calculate the temperature when probability is 0.99 and Fermi energy level below energy level by 0.3ev. 2 2 4

University of Technology 2016 2017 First Year, Lecture Four Semiconductors Fundamentals A semiconductor material is one whose electrical properties lie in between those of insulators and good conductors. Examples are: germanium, silicon and gallium arsenide. Semiconductor may be classified as under: Intrinsic Semiconductors A perfect semiconductor crystal containing no impurities or lattice defects is called an intrinsic semiconductor. In such a material there are no charge carriers at absolute zero but as the temperature raises electron-hole pairs are generated in pairs the concentration n of electron in the conduction band equals the concentration p of holes in the valance band as 2 shown in Fig. 4-1. Thus we have n=p=ni and np=nip P. Suppose the n is the number of electrons in the C.B, p is the number of holes in the V.B and ni is the intrinsic carrier concentration. Fig. (4-1) Generation of electron-hole pair in an intrinsic semiconductors. Since both electrons and holes are charge carriers in an intrinsic semiconductor, the total conductivity and the total current density in intrinsic semiconductor is: δi = neµrer +peµrhr δi = nie(µre R+µRhR) and Ji = nie(µre R+µRhR)E 1

University of Technology 2016 2017 First Year, Lecture Four The number of electrons per unit volume n(e) whose energies lies between E and E+dE is given by: n(e) = g C (E). F n (E), g C (E) = 8π 2 m 3 2 h 3 e E E c, for E E C, F n (E) = 1 1+ e (E E F KT ) where: g C (E) = Density of energy state in C.B. m e = Effective mass of electron, m e = Constant m. m = 9.11 10-31 Kg (electron rest mass), E c = Energy of the bottom of C.B. F n (E) = Fermi- Dirac probability function of finding electron in a particular level. K = Boltzman constant, T = Absolute temperature in Kelvin K Approximation solution, Assume E- E F >> KT F n (E) e (E E F KT ), KT = 0.026 ev at 300 k top of C.B n = n(e)de = g E c (E) F n (E)dE c E c n = 8π 2 E c h 3 3 m 2 e E E c e (E E F KT ) de n = 8π 2 h 3 m 32 e E c E E c e (E EF KT ) de n = N C e (E F E c KT ), N c = 2[ 2πm e KT ] 3 h 2 2 = 4.828 10 21 ( m e m )3 2.T 3 2 in m 3 Where N C = Is the effective density of state for electrons in the C.B. Similarly the number of holes in the valance band: p(e) = gv(e). F p (E) g V (E) = 8π 2 m 3 2 h 3 h E V E, for E E V, F p (E) = 1 F n (E) = 1 1 = 1+ e (E E F KT ) e ( E E F KT ) 1+ e (E E F KT ) Where: g v (E) = Density of energy state in V.B, m h = Effective mass of hole. E v = Energy of the bottom of V.B. Approximation solution, Assume E- E F >> KT (but ve), F p (E) e (E E F KT ) p = E v p(e)de E = v g v (E) F p (E)dE, p = 2 E v 8π 2 h 3 m h 3 2 E v E e (E E F KT ) de

University of Technology 2016 2017 First Year, Lecture Four p = N V e (E v E F KT ), N V = 2[ 2πm h KT] 3 h 2 2 = 4.828 10 21 ( m h m )3 2.T 3 2 in m 3 Where N V = Is the effective density of state for holes in the V.B. We can then write two relations between the intrinsic carrier density and the intrinsic Fermi energy, namely: ni = N C e (E Fi E c KT ), ni = N V e (E v E Fi KT ) Eg (, Also ni = N C N V e 2KT ) Eg = E C E V, Energy gap Example 4-1: Calculate the total intrinsic conductivity at 300K of the compound semiconductor gallium arsenide (GaAs). For which the energy gap is 0.70 ev, Take NC = N V = 2.5 10 25 m -3 and the electron and hole mobility to be 5000 cm 2 /v.s and 1000 cm 2 /v.s respectively. Solution: Eg ( ni = N C N V e 2KT ) = [2.5 10 25 2.5 10 25 ] 1 2 e 0.7 = 3.561 2 0.026 1019 m 3. δi = 3.561 10 19 1.6 10 19 (5000 10 4 + 1000 10 4 ) = 3.419 Ω 1 m 1. Fermi Level in Intrinsic Semiconductor To determine the position of Fermi level relative to the valence and conduction bands (E Fi ): N C e (EFi E c KT E Fi = E c+ E v 2 ) = N V e (E v E Fi ), ln N C KT = E c+ E v 2E Fi N V KT + 1 KT 2 ln(n V ) or E N Fi = E c+ E v + 3 KT C 2 4 ln(m h m ) e, E Fi is in the middle of energy gap, as shown At T = 0 K, equation reduces to E Fi = E c+ E v 2 in Fig. 4-2. If m e > m h, ERFiR is slightly below centre of the gap. Fig. (4-2) Fermi-level location in an intrinsic semiconductors. 3

University of Technology 2016 2017 First Year, Lecture Four Extrinsic Semiconductor We have seen that the conductivity of intrinsic semiconductors is very small and hence they are not suitable for any useful work except as a heat or light sensitive resistances. The conductivity can, however, be enormously increased by addition of suitable impurity in a very small proportion. This process is called doping. A doped or impurity semiconductor is known as an extrinsic semiconductor. 1. N-type Semiconductors In order for our silicon crystal to conduct electricity, we need to introduce an impurity atom such as Arsenic, Antimony or Phosphorus into the crystalline structure making it extrinsic (impurities are added). These atoms have five outer electrons in their outermost orbital to share with neighbouring atoms and are commonly called "Pentavalent" impurities. This allows four of the five orbital electrons to bond with its neighbouring silicon atoms leaving one "free electron" to become mobile when an electrical voltage is applied (electron flow). As each impurity atom "donates" one electron, pentavalent atoms are generally known as "donors". Antimony (symbol Sb) or Phosphorus (symbol P), are frequently used as a pentavalent additive as they have 51 electrons arranged in five shells around their nucleus with the outermost orbital having five electrons, as shown in Fig. 4-3. The resulting semiconductor material has an excess of current-carrying electrons, each with a negative charge, and is therefore referred to as an "N-type" material with the electrons called "Majority Carriers" while the resulting holes are called "Minority Carriers". Then, n n.p n = ni 2, n n N D, and p n = ni2 ND Fig. (4-3) Structure and lattice of the donor impurity atom Antimony. Fermi Level in N-type Semiconductor n N D = N C e (E Fn E C ) KT ln( N D ) = (E Fn E C ) N C KT E Fn = E C + KT ln( N D ) N C 4

University of Technology 2016 2017 First Year, Lecture Four 2. P-Type Semiconductors If we go the other way, and introduce a "Trivalent" (3-electron) impurity into the crystalline structure, such as Aluminium, Boron or Indium, which have only three valence electrons available in their outermost orbital and the fourth closed bond cannot be formed. Therefore, a complete connection is not possible, giving the semiconductor material an abundance of positively charged carriers known as "holes" in the structure of the crystal where electrons are effectively missing. As there is now a hole in the silicon crystal, a neighbouring electron is attracted to it and will try to move into the hole to fill it. However, the electron filling the hole leaves another hole behind it as it moves. Trivalent impurities are generally known as "Acceptors" as they are continually "accepting" extra or free electrons. Boron (symbol B) is commonly used as a trivalent additive as it has only five electrons arranged in three shells around its nucleus with the outermost orbital having only three electrons, as shown in Fig. 4-4. The doping of Boron atoms causes conduction to consist mainly of positive charge carriers resulting in a "P-type" material with the positive holes being called "Majority Carriers" while the free electrons are called "Minority Carriers". Then a semiconductor material is classed as P-type when its acceptor density is greater than its donor density. Therefore, a P-type semiconductor has more holes than electrons. Then, n p.p p = ni 2, P p N A, and n n = ni2 N A Fig. (4-4) Structure and lattice of the acceptor impurity atom Boron. P N A = N V e (E V E Fp ) KT ln( N A ) = (E V E Fp ) N V KT E Fp = E V + KT ln( N V ) N A Fermi Level in P-type Semiconductor 5

University of Technology 2016 2017 First Year, Lecture Four Example 4-2: A p-n Junction employs the following doping levels: N A =10 16 cm -3 and N D =5 10 15 cm -3, ni =1.08 10 10 cm -3. Determine the hole and electrons on the two sides. Solution: Concentration of holes on P side, P P N A = 10 16 cm 3 Concentration of electron on P side, n P ni 2 N A, n P = (1.08 1010 cm 3 ) 2 10 16 cm 3 1.1 10 14 cm 3 Concentration of electron on n side, n n N D = 5 10 15 cm 3 Concentration of holes on n side, p n ni2 N D, p n = (1.08 1010 cm 3 ) 2 5 10 15 cm 3 2.4 10 4 cm 3 The Hall Effect The Hall effect describes the behavior of the free carriers in a semiconductor when applying an electric as well as a magnetic field. If a specimen carrying a current I is placed transverse magnetic field B, an electric field E is induced in the direction perpendicular to both I and B. This phenomenon known as the (Hall effect) as shown in Fig.4-5, is used to determine whether a semiconductor is n or p type and to find the carrier concentration and measuring the conductivity δ, the mobility µ can be calculated. Fig. (4-5) Hall setup and carrier motion for holes and electrons. To calculated the Hall field, we first calculate the Lorentz force (F L ) on the free carriers, at the equilibrium the Hall force (F H ) equal the Lorentz force. FH = F L, qe H = qvb, E H = VB, V = J E qn H = JB qn where: E H = Hall electric field, V = The drift speed, B = The magnetic field. R H = 1 Hall coefficient, q= -e for n-type semiconductor, nq semiconductor. 6 and q= +e for p-type

University of Technology 2016 2017 First Year, Lecture Four V H = E. w = JB.w = B.w.I = B.I V qn qnwt qnt H = R H. B.I Hall voltage t where: w = The width of piece, t = The thickness of piece. tanθ = µb, µ = R H δ, from this equation it is possible to measure the mobility if the conductivity and hall coefficient are known. Example 4-3: A sample of Si is doped with 10 17 phosphorus atom/cm 3. What would you expect to measure for its resistivity. What Hall voltage would you expect in a sample10-2 cm thickness if I = 1mA, B= 10-5 wb/cm 2 and µ = 700 cm 2 /v.s. Solution: δ = 1.6 * 10-19 *700 *10 17 = 11.2 (Ω.cm) -1 ρ = δ -1 = 0.0893 Ω.cm. R H = 1 = 62.5 cm 3 /c, V ne H = R H. B.I = 62.5 10 5 10 3 62.5 µv. t 10 2 Exercises 1. The concentration of acceptor silicon atom is 5 10 21 m -3 at 300 o K. How far from the edge of the valence band the fermi level if m h = 0.6m. 2. By adding phosphor with carrier 10 22 m -3 to pure silicon has electron concentration 1.45 10 16 m -3 at 300ºK. find the electron and hole densities at 300ºK and 500ºK, using E g =1.1ev. 3. Prove that N c = 4.828 10 21 ( m e m )3 2.T 3 2. 4. A sample of silicon have electron concentration 3.2 10 11 m -3 and L=86mm, t=10mm and w=35mm, when sample is tested by Hall effect, the value of resistance= 2.7KΩ and the Hall coefficient=3.6m 3 /c, is obtained. 1. Which kind of semiconductor that the sample? 2. Determine the value of mobility for the sample in (1). 5. Find hall coefficient, electron density and the angle between the field, for a silicon wire with thickness 2mm, while the applied magnetic field is 0.1T, carrying a current 10mA, hall voltage 1mv and the mobility= 0.36m 2 /v.s. 7

University of Technology 2016 2017 First Year, Lecture Five The P-N Junction When we join (or fuse) P and N-type together these two materials behave in a very different way producing what is generally known as a PN Junction. When the N and P-type semiconductor materials are first joined together a very large density gradient exists between both sides of the junction so some of the free electrons from the donor impurity atoms begin to migrate across this newly formed junction to fill up the holes in the P-type material producing negative ions. However, because the electrons have moved across the junction from the N-type silicon to the P-type silicon, they leave behind positively charged donor ions (N D ) on the negative side and now the holes from the acceptor impurity migrate across the junction in the opposite direction into the region were there are large numbers of free electrons. As a result, the charge density of the P-type along the junction is filled with negatively charged acceptor ions (N A ), and the charge density of the N-type along the junction becomes positive. This charge transfer of electrons and holes across the junction is known as diffusion. This process continues back and forth until the number of electrons which have crossed the junction have a large enough electrical charge to repel or prevent any more carriers from crossing the junction. The regions on both sides of the junction become depleted of any free carriers in comparison to the N and P type materials away from the junction. Eventually a state of equilibrium (electrically neutral situation) will occur producing a "potential barrier" zone around the area of the junction as the donor atoms repel the holes and the acceptor atoms repel the electrons as shown in Fig. 5-1. Since no free charge carriers can rest in a position where there is a potential barrier the regions on both sides of the junction become depleted of any more free carriers in comparison to the N and P type materials away from the junction. This area around the junction is now called the Depletion Region. Fig. (5-1) Structure of P-N junction. V o = KT e ln N AN D ni 2 1

University of Technology 2016 2017 First Year, Lecture Five Drift and Diffusion Currents The net current that flows through a (PN junction diode) semiconductor material has two components drift current, which is the movement caused by electric fields as shown in Fig. 5-2 and diffusion current, which is the flow caused by variations in the concentration as shown in Fig. 5-3, that is, concentration gradients. Drift Current J e drift = enµ e E J h drift = epµ h E J T drift = enµ e E + epµ h E Fig. (5-2) Drift current flow. Diffusion Current J e diffusion = ed e dn dx J h diffusion = ed h dp dx J T diffusion = ed e dn dx ed h dp dx Fig. (5-3) Diffusion current flow. Where: D e : electron Diffusion coefficient in cm 2 s -1. D h : hole Diffusion coefficient in cm 2 s -1. Einstein relation, D e = D h = KT = V µ e µ h e T, where = KT 0.026V at 300K e V T : Thermal velocity In general, total current can flow by drift and diffusion separately. Total current density: J e T = J e drift + J e diffusion = enµ e E + ed e dn dx J h T = J h drift + J h diffusion = epµ h E ed h dp dx J total = J e T + J h T 2

University of Technology 2016 2017 First Year, Lecture Five Built-in Voltage ( V o ) The existence of an electric field within the depletion region suggests that the junction may exhibit a built-in potential see Fig. 5-4. Since the electric field E = dv dx. dp epµ h E = ed h, Pµ dv dx h = D dp dx h dx v n dv = KT p n dp, V e p p P o = KT ln P p e v p V o = KT e ln N AN D ni 2. P n Fig. (5-4) Chart showing the Built-in voltage. Example 5-1: A silicon junction employs N A = 2 10 16 cm -3 and N D = 4 10 16 cm -3, The intrinsic carrier concentration is (1.08 10 10 m -3 ), determine the Built-in Voltage at room temperature (T = 300K). Solution: V o = KT ln N AN D V e ni 2 o = 26 mv ln (4 1016 )(2 10 16 ) = 768 mv. (1.08 10 10 ) 2 Width of the Depletion Region The equilibrium width of the transition region is determined by how many in mobile charges must be uncovered to produce the built in voltage. Thus, the width will depend on the doping concentration on each side of the junction as shown in Fig. 5-5. The charge per unit area in the depletion region on the p-side must be equal in magnitude to that on the n-side. X p = [ 2Є oє r V o e N D N A ( N A+ N D ) ]1 2, X n = [ 2Є oє r V o e N A N D ( N A+ N D ) ]1 2 The total width of the depletion region (w) is given by: W = X n + X p = [ 2Є oє r v o e If N A >> N D, w = 2Є oє r V o And if N D >> N A, w = 2Є oє r V o en D en A Where: Є r = relative permittivity ( N A+ N D ) (N A N D ) ]1 2 3 Fig. (5-5) Width of depletion region.

University of Technology 2016 2017 First Year, Lecture Five P-N Junction Diode The hole diffusion current at the edge of the junction {(x = 0) in the n-region} can be shown to be given by: J p(0) = e p V D n D h V [e T 1], and Lh = (D L h τ h ) 1 2. h Where: L h = hole diffusion length, τ h = life time before recombination for holes, and V D = The applied voltage on the diode. Similarly, the current density of electron at the edge of the transition large in the p-region: J n(0) = e n V D p De V [e T 1], and Le = (D L e τ e ) 1 2. e The total current across the junction is [ J n(0) + J p(0) ] which must be constant throughout the diode. Hence. The total current is given by: V D V I D = A J = A (J n(0) + J p(0) ) I D = I s [e T 1] and Is = A ( e p n D h + e n p De), L h L e Where: I D = Diode current, I s =reverse saturation current, A = cross-section area of the diode. Example 5-1: A silicon p-n Junction has cross section area A=1mm 2 at 300K. The intrinsic carrier concentration is 8.4 10 20 m -3, the doping concentration in the n-side is 10 22 m -3 and in the p-side is 10 24 m -3, The relative permittivity of silicon is 12. Hole and electron mobility s are 0.05 m 2 v -1 s -1 and 0.13 m 2 v -1 s -1 respectively and τ e = τ h =10µs, Calculate: 1. The contact potential or Built in voltage V o. 2. The result conductivity from majority carrier in each region δ n, δ p. 3. The diffusion constant for each carrier D e,d h. 4. Calculate reverse saturation current. Solution: 1. V o = KT e ln N AN D = 26 ln 1024 10 22 = 0.248 volt. ni 2 (8.4 10 20 ) 2 2. δ n = e µ e N D = 1.6 10 19 0.13 10 22 = 208Ω 1 m 1. δ P = e µ h N A = 1.6 10 19 0.05 10 24 = 8000Ω 1 m 1. 3. D e = V T µ n = 26 0.13 = 0.0033 m 2 s 1. D h = V T µ h = 26 0.05 = 0.0013 m 2 s 1. 4. L e = (D e τ e ) 1 2 = (0.00338 10 10 6 ) 1 2 = 0.00018m. L h = (D h τ h ) 1 2 = (0.0013 10 10 6 ) 1 2 = 0.000114m. 4

University of Technology 2016 2017 First Year, Lecture Five P n = ni2 N D = (8.4 1020 ) 2 10 22 = 7.056 10 19 m 3, n P = ni2 N A = (8.4 1020 ) 2 10 24 = 7.056 10 17 m 3 I s = (1 10 6 ) 1.6 10 19 7.056 1019 0.0013 0.000114 + 7.056 1017 0.00338 = 1.3 10 4 A. 0.00018 Basic Construction of Diode A diode is made from a small piece of semiconductor material, usually silicon, in which half is doped as a p region and half is doped as an n region with a pn junction and depletion region in between. The p region is called the anode and is connected to a conductive terminal. The n region is called the cathode and is connected to a second conductive terminal. The basic diode structure and schematic symbol are shown in Fig. 5-6. A pn junction diode is a one way device offering low resistance when forward biased and behaving almost as an insulator when reverse biased. Hence such diodes are mostly used as rectifiers for converting alternating current into direct current. (a) Fig. (5-6) (b) Effect of Applied Voltage If V Bias = 0 V (No Applied Bias) In the absence of an applied bias voltage, the net flow of charge in any one direction for a semiconductor diode is zero, see Fig. 5-7. Fig. (5-7) 5

University of Technology 2016 2017 First Year, Lecture Five If V Bias < 0 V (Reverse Bias) The potential barrier to carrier diffusion across the junction increases and current that exists under reverse-bias conditions is called the reverse saturation current and is represented by I s, see Fig. 5-8. Fig. 5-8 If V Bias > 0 V (Forward Bias) The potential barrier to carrier diffusion across the junction decreases and current increases exponentially, see Fig. 5-9. Fig. (5-9) 6

University of Technology 2016 2017 First Year, Lecture Five The V-I Characteristic Carve Combine the curves for both forward bias and reverse bias, and you have the complete V-I characteristic curve for a diode, as shown in Fig. 5 10. Fig. (5-10) Characteristic carve of diode. 1. DC or Static Resistance Resistance Levels It is already defined, that the resistance of the diode under d.c. conditions is called d.c. resistance. When the voltage applied is constant, the operating point also remains constant. The ratio of the constant d.c. diode voltage to constant d.c. diode current, at the operating point is d.c. resistance of the diode as shown in Fig. 5-11 and applying the following equation. R D = V D I D 2. Ac or Dynamic Resistance Fig. (5-11) When an a.c. voltage is applied to the diode then the operating point of the diode keeps on changing its position instantaneously, according to the changes in the input voltage. This defines a particular change in diode current corresponding to the change in the diode voltage. This is shown in the Fig. 5-12. The change in a.c. voltage applied to diode V D produces a change in the current I D. The ratio of these changes define the a.c. resistance of the diode. 7

University of Technology 2016 2017 First Year, Lecture Five r d = V d I d Fig. (5-12) According to mathematics, the forward dynamic resistance represents the reciprocal of the slope of the diode characteristic curve. By making use of diode equation, the forward dynamic resistance expression can be derived as follows, see Fig. 5-13. V D V I D = I s [e T 1], di D = I D, by neglected I dv D V S T dv D = V T di D I D At room temperature, r d = 0.026V I D Fig. (5-13) 3. Average AC Resistance The average AC resistance of the diode is the resistance determined by joining a straight line through the intersection point, obtained corresponding to the maximum and minimum values of the input voltage such as indicated in Fig.5-14. r av = V d I P.to P. d 8 Fig. (5-14)

University of Technology 2016 2017 First Year, Lecture Five Exercises 1. A silicon p-n junction has the hole and electron resistivity are 0.001Ω.m, 0.004Ω.m respectively in T=250 ºK. The electric field is 3 10 22 v/m, the intrinsic carrier density is 6.2 10 10 m -3, the electron and hole diffusion length are 1.8 10-4 m, 1.1 10-4 m respectively, the contact voltage is 0.35V, and ι e = ι h =8µs. Find Reverse saturation current density. 2. Given a Si diode with the following physical ch/s the doping concentration in the p-type is 2.25 10 17 cm -3 and in the n-type is 2.25 10 14 cm -3, junction area is 2 10-3 cm 2, life time of hole in n-region=10µs, life time of electron in p-region =50µs, hole and electron mobility s are 600cm 2 v -1 s -1 and 2100cm 2 v -1 s -1 respectively, the intrinsic carrier concentration is 1.4 10 10 cm -3 and the voltage is 0.66v operation at room temperature. Calculate J p(0) and J n(0) then calculate the total diode current. 3. A silicon p-n junction has energy of C.B. and V.B. are 0.9ev, 0.1ev respectively at 200ºK. the hole and electron mobility, resistivity are 0.16m 2 /v.s, 0.36 m 2 /v.s, 0.0025Ω.m and 0.005Ω.m. the electric field is 2 10 28 v/m, m * e =0.8m 0, m * h=0.5m o, and ι e = ι h =10µs, find: 1.The Built in voltage. 2. The drift currents. 3.Electron and hole diffusion length. 9

University of Technology 2016 2017 First Year, Lecture Six 1. Ideal Model: (see Fig. 6-1); Forward-bias, Diode Equivalent Circuits (Models) Reverse-bias, Fig. (6-1) An ideal diode has the following properties: When V D = 0, I D > 0 Diode behaves like a switch: When VD < 0, I D = 0 closed in forward bias mode, open in reverse bias mode. 2. Practical Model: (see Fig. 6-2); Forward-bias, Reverse-bias, Fig. (6-2) An simplified diode has the following properties: When V D = V T, I D > 0 Diode behaves like a voltage source in series with a switch. When VD < V T, I D = 0 3. Complex Model: (see Fig. 6-3); Forward-bias, Reverse-bias, Fig. (6-3) 1

University of Technology 2016 2017 First Year, Lecture Six Example 6-1: Determine whether each silicon diode in Fig. 6 4 is forward-biased or reverse-biased. (a) Fig. (6-4) (b) Solution: Appling a golden rule, a diode is forward biased when V Anode V Catode = positive voltage V T, yields: In (a): In (b): (+5V) (+8V) = -3V. Therefore, the diode is reversed-biased. (0V) (-10V) = +10V. Therefore, the diode is forward-biased. Example 6-2: Find the current passing through diodes D 1 and D 2 for the network of Fig. 6-5. D1 R D2 Solution: Applying KVL yields: E 1 V D1 V R V D2 E 2 = 0 I = E 1 V D1 V D2 E 2, Fig. (6-5) R 25 0.7 0.3 8 I = = 16mA = I 1K D1 = I D2. E1 Si 25V 1K Ge 8V E2 Example 6-3: Calculate V o, I, I D1 and I D2 for the parallel silicon diode network in Fig. 6-6. R Solution: 1. V o = 0.7V. 2. Applying KVL yields: Fig. (6-6) E IR V D1 = 0 I = 28.18mA. 3. I D1 = I D2 = I 2 = 14.09mA. E=10V 0.33K I I D1 D1 I D2 D2 + V o - 2

University of Technology 2016 2017 First Year, Lecture Six Diode Applications Diode Switching Circuits The Switching of a semiconductor diode is of great importance in computer circuit design. In such applications, pulse type or square voltage waveforms are used to switch ON and OFF the p-n junction diodes. The p-n junction diodes are used to construct the digital logic gates which represent the building blocks in computers. The diodes here are used as switches as Fig. 6-7, so their operation is rapidly change between their high resistance and low resistance states due to the instantaneous voltage change between a low value often 0V to a high value often 5V respectively. It is worth noting that the for the diode to be in its ON state it should be forward biased which in turn implies that its anode voltage with respect to ground should be more positive than that of cathode otherwise the diode is in its OFF state. One of the very important application of diode switching circuit is logic gates. Fig. (6-7) Diode Logic Gates In logic gates, logical functions are performed by parallel or series connected switches. Diode logic is implemented by diodes which exhibit low impedance when forward biased and a very high impedance when reverse biased. There are two kinds of diode logic gates - OR and AND. It is not possible to construct NOT (Invert) diode gates because the NOT or Invert function requires an active component such as a transistor. Diode OR Gate: If all inputs are at 0V, current flowing through R will pull the output voltage down until the diodes clamp the output. Since these diodes are treated as ideal, the output is clamped to 0 volts, which is logic level 0. If any input is 1V, current flowing through the now forwardbiased diode will pull the output voltage up, providing a positive voltage at the output, a logic 1. Any positive voltage will represent a logic 1 state; the summing of currents through multiple diodes does not change the logic level. The other diodes are reverse biased and conduct no current (see Fig. 6-8). Fig. (6-8) 3

University of Technology 2016 2017 First Year, Lecture Six Diode AND Gate: If all inputs are at 1V, current flowing through R will pull the output positive till the diodes clamp the output to 1 volts, the logical 1 output level. If any input switches to 0 V, current flowing through the diode will pull the output voltage down to 0 volts. The other diodes would be reverse biased and conduct no current (see Fig. 6-9). Fig. (6-9) Example 6-4: Determine which diodes are forward biased and which are reverse biased in the circuits shown in Fig. 6-10. Assuming a 0.7-V drop across each forward-biased diode, determine the output voltage V o. Solution: Fig. (6-10) In (a): D 1 : +8 - (-12) = +20V, D 2 : -8 - (-12) = +4V, D 3 : -12-(-12) = 0V. Therefore, D 1 is forward biased and D 2 and D V o = +8-0.7 = +7.3V. 4 3 is reverse biased.

University of Technology 2016 2017 First Year, Lecture Six In (b): D 1 : +25 - (+25) = 0V, D 2 : +25 0 = +25V, D 3 : +25 - (+5) = +20V, D 4 : +25 - (-10) = +35V. Therefore, D 4 is forward biased and D 1 and D 2 and D 3 V o = -10 + 0.7 = -9.3V. are reverse biased. In (c): D 1 : +18 - (-18) = +36V, D 2 : +18 - (-3) = +21V, D 3: +18- (+18) = 0V, D4: +18- (-18) = +36V, D5: +18- (+18) = 0V. Therefore, D 1 and D4 are forward biased and D 2 and D 3 and D 5 V o = -18+0.7 = -17.3V. are reverse biased. Exercises 1. Find the value of I D and V o in the circuits shown in Fig. 6-11. Fig. (6-11) 5

University of Technology 2016 2017 First Year, Lecture Six 2. Determine Vo and I for each circuit in Fig. 6-12. Assume that each of the diodes in these circuits has a forward voltage drop of 0.7 V. Fig. (6-12) 6

University of Technology 2016 2017 First Year, Lecture Seven Diode Rectifier Circuits Basic Definition: All active electronic devices require a source of constant dc that can be supplied by a dc power supply, rectifier one of important part of it (see Fig. 7-1). A diode circuit that converts an ac voltage to a pulsating dc voltage and permits current to flow in one direction only is called "rectifier" and the ac-to-dc conversion process is termed "rectification". Fig. (7-1) Half-Wave Rectifier (HWR) Fig. (7-2) For the half-wave rectifier circuit of Fig. 3-2: The average (dc) value of a half-wave rectifier sine-wave voltage (V dc ) is 1

University of Technology 2016 2017 First Year, Lecture Seven T V dc = 1 v T o(wt). dwt = 1 V 0 2π 0 Po π sin wt. dwt = V Po π = 0.318V Po For Ideal Model: V dc = 0.318V P(Sec) For Practical Model: V dc = 0.318(V P(Sec) V T ) [3-1a] [3-1b] The root mean square (rms) value of the load voltage V o(rms) is V o(rms) = 1 T v T o 2 (wt). dwt 0 π = 1 V 2π Po 2 sin 2 wt. dwt 0 = V Po 2 = 0.5V Po For Ideal Model: V o(rms) = 0.5 V P(Sec) For Practical Model: V o(rms) = 0.5(V P(Sec) V T ) [3-2a] [3-2b] The rms value of the ac component or (the ripple voltage) V r(rms) of the rectified signal is V r (rms) 2 = V o(rms) V 2 dc = (0.5V Po ) 2 (0.318V Po ) 2 = 0.385V Po For Ideal Model: V r(rms) = 0.385 V P(Sec) For Practical Model: V r(rms) = 0.385(V P(Sec) V T ) [3-3a] [3-3b] The percent ripple (r) in the rectified waveform or (ripple factor) is r = V r(rms) V dc 100% = 0.385V Po 0.318V Po 100% = 121% Efficiency (η) = P dc P ac 100% = V 2 dc R L V2 o(rms) R L 100% = 40.5% The peak inverse voltage (PIV) of the diode is PIV = V P(Sec) [3-4] The frequency of the output rectified signal (f o ) is f o = f i [3-5] 2

University of Technology 2016 2017 First Year, Lecture Seven Full-Wave Rectifiers (FWRs) 1. A Bridge Full-Wave Rectifier: Fig. (7-3) For the bridge full-wave rectifier circuit of Fig. 3-3: π 0 V dc = 1 π V Po sin wt. dwt = 2V Po π = 0.636V Po For Ideal Model: V dc = 0.636V P(Sec), For Practical Model: V dc = 0.636(V P(Sec) 2V T ) [3-6a] [3-6b] π 0 V o(rms) = 1 V π Po 2 sin 2 wt. dwt = V Po 2 = 0.707V Po For Ideal Model: V o(rms) = 0.707 V P(Sec), For Practical Model: V o(rms) = 0.707(V P(Sec) 2V T ) [3-7a] [3-7b] V r (rms) 2 = V o(rms) V 2 dc = (0.707V Po ) 2 (0.636V Po ) 2 = 0.308V Po For Ideal Model: V r(rms) = 0.308V P(Sec) For Practical Model: V r(rms) = 0.308(V P(Sec) 2V T ) [3-8a] [3-8b] r = V r(rms) V dc 100% = 0.308V Po 0.636V Po 100% = 48.4% V 2 dc R L (η) = P dc 100% = P ac V2 100% = 81% o(rms) R L PIV = V P(Sec) for Ideal Model [3-9a] PIV = V P(Sec) V T for Practical Model [3-9b] f o = 2f i [3-10] 3

University of Technology 2016 2017 First Year, Lecture Seven 2. A Center-Tapped (CT) Full-Wave Rectifier: Fig. (7-4) For the center-tapped full-wave rectifier circuit of Fig. 3-4: π 0 V dc = 1 π V Po sin wt. dwt = 2V Po π = 0.636V Po For Ideal Model: V dc = 0.636V P(Sec), For Practical Model: V dc = 0.636(V P(Sec) V T ) [3-11a] [3-11b] π 0 V o(rms) = 1 V π Po 2 sin 2 wt. dwt = V Po 2 = 0.707V Po For Ideal Model: V o(rms) = 0.707V P(Sec), For Practical Model: V o(rms) = 0.707(V P(Sec) V T ) [3-12a] [3-12b] V r (rms) 2 = V o(rms) V 2 dc = (0.707V Po ) 2 (0.636V Po ) 2 = 0.308V Po For Ideal Model: V r(rms) = 0.308V P(Sec) For Practical Model: V r(rms) = 0.308(V P(Sec) V T ) [3-13a] [3-13b] r = V r(rms) V dc 100% = 0.308V Po 0.636V Po 100% = 48.4% V dc 2 R L V2 o(rms) R L (η) = P dc 100% = P ac 100% = 81% PIV = 2V P(Sec) for Ideal Model [3-14] PIV = 2V P(Sec) V T for Practical Model [3-15] f o = 2f i 4

University of Technology 2016 2017 First Year, Lecture Seven Summary Different parameters for the HWR and FWR circuits are listed in Table 7-1. Parameter HWR FWR Bridge V Po V P(Sec) (Ideal) V P(Sec) -V T (practical) V P(Sec) (Ideal) V P(Sec) -2V T practical) V dc 0.318V Po 0.636V Po CT V P(Sec) (Ideal) V P(Sec) -V T (practical) V o(rms) 0.5V Po 0.707V Po V r 0.385V Po r 121% η 40.5% PIV V P(Sec) (Ideal) V P(Sec) (practical) V P(Sec) (Ideal) V P(Sec) -V T (practical) 0.308V Po 48.4% 81% f o f i 2f i 2V P(Sec) (Ideal) 2V P(Sec) -V T (practical) Example 7-1: For the full wave bridge rectifier circuit shown in Fig. 7-5: 1. Find the peak inverse voltage on each silicon diode. 2. Calculate the output DC voltage and current. 3. Sketch the output voltage. 4. If diode D 4 is burned, sketch the output voltage in this case. Fig.(7-5) Solution: 1. V P(Pri) = 2V i(rms) = 2 220 = 311V. V P(Sec) = N 2 V N P(Pri) = 1 311 = 20. 7V. 1 15 PIV = V P(Sec) V T = 20. 7 0. 7V = 20V. 5

University of Technology 2016 2017 First Year, Lecture Seven 2. V Po = V P(Sec) 2V T = 20. 7 1. 4V = 19. 3V. V dc = 0. 636V Po = 0. 636 19. 3V = 12. 27V. 3. I dc = V dc R L = 12.27V 2.2k = 5. 57mA. 4. In this case, the circuit will behave as a half wave rectifier. Exercises 1. For the center tapped Full Wave Rectifier circuit in Fig. 7-6. Find the peak and average power delivered to R L. the percent ripple factor, and the minimum PIV rating required for each diode. draw the output voltage. Fig. (7-6) 6

University of Technology 2016 2017 First Year, Lecture Eight Diode Clipping Circuits Basic Definition: There are a variety of diode circuits called clippers (limiters or selectors) that have the ability to "clip" off a portion of the input signal above (positive) or below (negative) certain level without distorting the remaining part of the alternating waveform. Depending on the orientation of the diode, the positive or negative region of the input signal is "clipped" off. There are two general categories of clippers: series and parallel. The series configuration is dined as one where the diode is in series with the load. While the parallel variety has the diode in a branch parallel to the load (see Fig. 8-1). Series (Negative) Clipper Series (Positive) Clipper Parallel (Negative) Clipper Fig. (8-1) Parallel (Positive) Clipper Example 4-1: Sketch the shape of the output voltage waveform for this clipper circuits. 1. Series (Negative) Clipper, see Fig. 8-2. Fig. (8-2) 1

University of Technology 2016 2017 First Year, Lecture Eight Solution: Fig. (8-2) (cont.) 2. Double Diode Series Clipper, see Fig. 8-3. Solution: Fig. (8-3) 2

University of Technology 2016 2017 First Year, Lecture Eight Fig. (8-3) (cont.) 3. Parallel (Negative) Clipper, see Fig. 8-4. Assuming an ideal diode with no forward voltage drop, see Fig. 8-4. Fig. (8-4) 3

University of Technology 2016 2017 First Year, Lecture Eight Solution: Fig. (8-4)(cont.) 4. Double Diode Parallel Clipper, see Fig. 8-5. Solution: Fig. 8-5 4

University of Technology 2016 2017 First Year, Lecture Eight Fig. (8-5) (cont.) 5. Special Type Clipper (Comparator), see Fig. 8-6. Fig. 8-6 Solution: Fig. (8-6) (cont.) 5

University of Technology 2016 2017 First Year, Lecture Eight Summary: A variety of series and parallel clippers with the resulting output for the sinusoidal input are provided in Fig. 8-7. Fig. (8-7) 6

University of Technology 2016 2017 First Year, Lecture Eight Exercises: 1. Design biased series clippers (with silicon diodes) to perform the functions indicated in the output waveforms of Fig. 8-8. (a) Fig. (8-8) (b) 2. Design biased parallel clippers (with silicon diodes) to perform the functions indicated in the transfer characteristics of Fig. 8-9. (a) (b) Fig. (8-9) 3. Sketch the output voltage (v o ) and the transfer characteristics (v o against v i ) for each circuit of Fig. 8-10 for the input (v i ) shown. Fig. (8-10) 7

University of Technology 2016 2017 First Year, Lecture Nine Diode Clamping Circuits Basic Definition: A clamper is a circuit designed to shift ac waveform either above or below a given reference voltage without changing the appearance of the applied signal. The circuit must have a capacitor, a diode, and a resistive element, but it can also employ an independent dc supply to introduce an additional shift. The magnitude of R and C must be chosen such that the time constant τ = RC is large enough to ensure that the voltage across the capacitor does not discharge significantly during the interval (T/2) the diode is non conducting. Throughout the analysis we will assume that for all practical purposes the capacitor will fully charge or discharge in five time constants. Therefore, the condition required for the capacitor to hold its voltage during the discharge period between pulses of the input signal is 5τ = 5RC T 2 = 1 2f Example 5-1: What is the output voltage that you would expect to observe across RL in the clamping circuit of Fig. 9-1. Solution: Fig. (9-1) Start with the part of the input signal which makes the diode ON (forward biased), draw the equivalent circuit, calculate V C and V o. The circuit of Fig. 5-1, the diode is forward bias during the positive half period of the input signal. For the input section KVL will result in +24 - V C 0.7 = 0 V C = 23.3V, and V o = 0.7V. Consider the other part of the input signal (which makes the diode off (reverse biased)), draw the equivalent circuit and calculate V o using V C from step 1. The output voltage (Vo) can be determine by KVL in the output section, -24 23.3 Vo = 0 V o = -47.3V. Draw the output for two steps, see Fig. 9-2. 1

University of Technology 2016 2017 First Year, Lecture Nine 5τ = 5RC = 5(10 10 3 )(0.5 10-6 ) = 25ms. T = 1/f = 1/1k =1ms, T/2 = 0.5ms. 5τ T/2, 25ms/0.5ms = 50 times. So that, it is certainly a good approximation that the Capacitor will hold its voltage (23.3V) during the Discharge period between pulses of the input signal. Output waveform across RL. Fig. 9-2 Example 9-2: Determine (v o ) for the following network with the input shown. Fig. (9-3) Solution: For the input section KVL will result in -20 + V C + 0.7 + 10 = 0 V C = 9.3V, and V o = -10.7V. The output voltage (Vo) can be determine by KVL in the output section, +20 + 9.3 Vo = 0 V o = +29.3V. 5τ = 5RC = 5(40 10 3 )(0.2 10-6 ) = 40ms. T = 1/f = 1/1k =1ms, T/2 = 0.5ms. 5τ T/2, 40ms/0.5ms = 80 times. 2 Fig. (9-4)

University of Technology 2016 2017 First Year, Lecture Nine Summary: A number of clamping circuits and their effect on the square-wave input signal are shown in Fig. 9-5. Negative Clampers Positive Clampers Fig. 9-5 3

University of Technology 2016 2017 First Year, Lecture Nine Exercise: 1. Design a clamper circuit (with silicon diode) that will produce output in fig. 9-6. Fig. 9-6 2. Sketch the output (v o ) for the circuit of Fig. 9-7 for the input (v i ) shown. Assume ideal diodes. Fig. 9-7 3. Design a clamper circuit (with Germanium diode) that will produce output v o = 5+10Sinωt V when the input is v i = +5 +10Sinωt V. Draw the circuit diagram and the input and output signals. 4