ELECTRONIC DEVICES & CIRCUITS, Lecture notes of Electronics

Download ELECTRONIC DEVICES & CIRCUITS and more Lecture notes Electronics in PDF only on Docsity! UNIT-1 REVIEW OF SEMICONDUCTOR PHYSICS Atomic Structure According to the modern theory, All the materials are composed of very small particles called atoms. The atoms are the building bricks of all matter. Various scientists have given different theories regarding the structure of atom. However, for the purpose of understanding electronics, the study of Bohr’s atomic model is adequate. Bohr’s Atomic Model In 1913, Neils Bohr, Danish Physicist gave clear explanation of atomic structure. According to Bohr: (i) An atom consists of a positively charged nucleus around which negatively charged electrons revolve in different circular orbits. (ii) The electrons can revolve around the nucleus only in certain permitted orbits i.e. orbits of certain radii are allowed. The number of electrons in any orbit is given by 2n 2 where n is the number of the orbit. For example, First orbit contains 2 1 2 = 2 electrons Second orbit contains 2 2 2 = 8 electrons Third orbit contains 2 3 2 = 18 electrons (iii) The electrons in each permitted orbit have a certain fixed amount of energy. The larger the orbit (i.e. larger radius), the greater is the en- ergy of electrons. (iv) If an electron is given additional energy (e.g. heat, light etc.), it is lifted to the higher orbit. The atom is said to be in a state of excitation. This state does not last long, because the electron soon falls back to the original lower orbit. As it falls, it gives back the acquired energy in the form of heat, light or other radiations. Fig. shows the structure of silicon atom. It has 14 electrons. Two electrons revolve in the first orbit, 8 in the second orbit and 4 in the third orbit. The first, second, third orbits etc. are also known as K, L, M orbits respectively. These electrons can revolve only in permitted orbits (i.e. orbits of *radii r1, r2 and r3) and not in any arbitrary orbit. Thus, all radii between r1 and r2 or between r2 and r3 are forbidden. Each orbit has fixed amount of energy associated with it. 1 Energy Levels It has already been discussed that each orbit has fixed amount of energy associated with it. The electrons moving in a particular orbit possess the energy of that orbit. The larger the orbit, the greater is its energy. It becomes clear that outer orbit electrons possess more energy than the inner orbit electrons. A convenient way of representing the energy of different orbits is shown in Fig. 4.2 (ii). This is known as energy level diagram. The first orbit represents the first energy level, the second orbit indicates the second energy level and so on. The larger the orbit of an electron, the greater is its energy and higher is the energy level. Fig. 4.2 Fig. 4.1 2 PARAMATER CONDUCTORS SEMICONDUCTORS INSULATORS Conduction The conduction in conductors is due to the free electrons in metal bonding. The conduction in semiconductor is due to the movement of electron & holes. There are no free electrons or holes thus, there is no conduction. Band gap There is no or low energy gap between the conduction & valance band of a conductor. It does not need extra energy for the conduction state. The band gap of semiconductor is greater than the conductor but smaller than an insulator i.e. 1 eV. Their electrons need a little energy for conduction state. The band gap in insulator is huge (+5 eV), which need an enormous amount of energy like lightning to push electrons into the conduction band. Valence Electron in Outer Shell 1 Valence electron in outer shell. 4 Valence electron in outer shell. 8 Valence electron in outer shell. Conductivity High (10-7 mho/m) Medium (10-7 to 10- 13 mho/m) Very Low (10-3 mho/m) Almost negligible. Resistivity Low Moderate High Temperature coefficient of resistance Positive Negative Negative Charge carriers in conduction band Completely filled Partially filled Completely vacant Charge carriers in valence band Almost vacant Partially filled Completely filled Example Copper, Aluminium, graphite etc. Silicon, Germanium, arsenic etc. Paper, rubber, glass, plastic etc. Application The metals like iron & copper etc. that can conduct electricity are made into wires and cable for carrying electric current. Semiconductors are used every day electronic devices such as cellphone, computer, solar panel etc as switches, energy converter, amplifiers, etc. The insulators are used for protection against high voltages & prevention of electrical short between cables in circuits. 5 SEMICONDUCTORS • Semiconductors are the materials which have 4 electrons in its outer most orbit . OR • Semiconductor are materials whose electrical conductivity lies between conductor and an insulator. Some common semiconductors elemental  Si - Silicon (most common)  Ge - Germanium compound  GaAs - Gallium arsenide GaP - Gallium phosphide  AlAs - Aluminum arsenide AlP - Aluminum phosphide  InP - Indium Phosphide  There are also three-element (ternary) compounds (GaAsP) and four-elements (quaternary) compounds such as InGaAsP. Compounds are widely used in high-speed devices and devices requiring the emission or absorption of light. (i) Germanium. Germanium has become the model substance among the semiconductors; the main reason being that it can be purified relatively well and crystallised easily. Germanium is an earth element and was discovered in 1886. It is recovered from the ash of certain coals or from the flue dust of zinc smelters. Generally, recovered germanium is in the form of germanium dioxide powder which is then reduced to pure germanium. (i) (ii) The atomic number of germanium is 32. Therefore, it has 32 protons and 32 electrons. Two electrons are in the first orbit, eight electrons in the second, eighteen electrons in the third and four electrons in the outer or valence orbit [See Fig. 5.2 (i)]. It is clear that germanium atom has four valence electrons i.e., it is a tetravalent element. Fig. 5.2 (ii) shows how the various germanium atoms are held through co-valent bonds. As the atoms are arranged in an orderly pattern, therefore, germanium has crystalline structure. Fig. 5.2 6 (ii) Silicon. Silicon is an element in most of the common rocks. Actually, sand is silicon diox- ide. The silicon compounds are chemically reduced to silicon which is 100% pure for use as a semiconductor. The atomic number of silicon is 14. Therefore, it has 14 protons and 14 electrons. Two electrons are in the first orbit, eight electrons in the second orbit and four electrons in the third orbit [See Fig. 5.3 (i)]. It is clear that silicon atom has four valence electrons i.e. it is a tetravalent element. Fig. 5.3 (ii) shows how various silicon atoms are held through co-valent bonds. Like germanium, silicon atoms are also arranged in an orderly manner. Therefore, silicon has crystalline structure. Energy Band Description of Semiconductors It has already been discussed that a semiconductor is a substance whose resistivity lies between conductors and insulators. The resistivity is of the order of 10 4 to 0.5 ohm metre. However, a semi- conductor can be defined much more comprehensively on the basis of energy bands as under : A semiconductor is a substance which has almost filled valence band and nearly empty conduc- tion band with a very small energy gap (j 1 eV) separating the two. Figs. 5.4 and 5.5 show the energy band diagrams of germanium and silicon respectively. It may be seen that forbidden energy gap is very small; being 1.1 eV for silicon and 0.7 eV for germanium. Therefore, relatively small energy is needed by their valence electrons to cross over to the conduction band. Even at room temperature, some of the valence electrons may acquire sufficient energy to enter Fig. 5.5 Fig. 5.4 Fig. 5.3 7 same time, another current – the hole current – also flows in the semiconductor. When a covalent bond is broken due to thermal energy, the removal of one electron leaves a vacancy i.e. a missing electron in the covalent bond. This missing electron is called a *hole which acts as a positive charge. For one electron set free, one hole is created. Therefore, thermal energy creates hole-electron pairs; there being as many holes as the free electrons. The current conduction by holes can be explained as follows : The hole shows a missing electron. Suppose the valence electron at L (See Fig. 5.8) has become free electron due to thermal energy. This creates a hole in the co-valent bond at L. The hole is a strong centre of attraction **for the electron. A valence electron (say at M) from nearby co-valent bond comes to fill in the hole at L. This results in the creation of hole at M. Another valence electron (say at N) in turn may leave its bond to fill the hole at M, thus creating a hole at N. Thus the hole having a positive charge has moved from L to N i.e. towards the negative terminal of supply. This constitutes hole current. It may be noted that hole current is due to the movement of ***valence electrons from one co- valent bond to another bond. The reader may wonder why to call it a hole current when the conduc- tion is again by electrons (of course valence electrons !). The answer is that the basic reason for current flow is the presence of holes in the co-valent bonds. Therefore, it is more appropriate to consider the current as the movement of holes. Energy band description. The hole current can be beautifully explained in terms of energy bands. Suppose due to thermal energy, an electron leaves the valence band to enter into the conduc- tion band as shown in Fig. 5.9. This leaves a vacancy at L. Now the valence electron at M comes to fill the hole at L. The result is that hole disappears at L and appears at M. Next, the valence electron at N moves into the hole at M. Consequently, hole is created at N. It is clear that valence electrons move along the path PNML whereas holes move in the opposite direction i.e. along the path LMNP. Fig. 5.8 Fig. 5.9 10 Intrinsic Semiconductor A semiconductor in an extremely pure form is known as an intrinsic semiconductor. In an intrinsic semiconductor, even at room temperature, hole-electron pairs are created. When electric field is applied across an intrinsic semiconductor, the current conduction takes place by two processes, namely ; by free electrons and holes as shown in Fig. 5.10. The free electrons are pro- duced due to the breaking up of some covalent bonds by thermal energy. At the same time, holes are created in the covalent bonds. Under the influence of electric field, conduction through the semicon- ductor is by both free electrons and holes. Therefore, the total current inside the semiconductor is the sum of currents due to free electrons and holes. It may be noted that current in the ex- ternal wires is fully electronic i.e. by electrons. What about the holes ? Re- ferring to Fig. 5.10, holes being posi- tively charged move towards the negative terminal of supply. As the holes reach the negative terminal B, electrons enter the semiconductor crys- tal near the terminal and combine with holes, thus cancelling them. At the same time, the loosely held electrons near the positive terminal A are attracted away from their atoms into the positive terminal. This creates new holes near the positive terminal which again drift towards the negative terminal. 5.6 9 Extrinsic Semiconductor The intrinsic semiconductor has little current conduction capability at room temperature. To be useful in electronic devices, the pure semiconductor must be altered so as to significantly increase its conducting properties. This is achieved by adding a small amount of suitable impurity to a semicon- ductor. It is then called impurity or extrinsic semiconductor. The process of adding impurities to a semiconductor is known as doping. The amount and type of such impurities have to be closely controlled during the preparation of extrinsic semiconductor. Generally, for 10 8 atoms of semicon- ductor, one impurity atom is added. The purpose of adding impurity is to increase either the number of free electrons or holes in the semiconductor crystal. As we shall see, if a pentavalent impurity (having 5 valence electrons) is added to the semiconductor, a large number of free electrons are produced in the semiconductor. On the other hand, addition of trivalent impurity (having 3 valence electrons) creates a large number of holes in the semiconductor crystal. Depending upon the type of impurity added, extrinsic semicon- ductors are classified into: (i) n-type semiconductor (ii) p-type semiconductor 5.7 0 n-type Semiconductor When a small amount of pentavalent impurity is added to a pure semiconductor, it is known as n-type semiconductor. Fig. 5.10 11 The addition of pentavalent impurity pro- vides a large number of free electrons in the semiconductor crystal. Typical examples of pentavalent impurities are arsenic (At. No. 33) and antimony (At. No. 51). Such impurities which produce n-type semiconductor are known as donor impurities because they do- nate or provide free electrons to the semicon- ductor crystal. To explain the formation of n-type semi- conductor, consider a pure germanium crys- tal. We know that germanium atom has four valence electrons. When a small amount of pentavalent impurity like arsenic is added to germanium crystal, a large number of free electrons become available in the crystal. The reason is simple. Arsenic is pentavalent i.e. its atom has five valence electrons. An arsenic atom fits in the germanium crystal in such a way that its four valence electrons form covalent bonds with four germa- nium atoms. The fifth valence electron of arsenic atom finds no place in co-valent bonds and is thus free as shown in Fig. 5.11. Therefore, for each arsenic atom added, one free electron will be available in the germanium crystal. Though each arsenic atom provides one free electron, yet an extremely small amount of arsenic impurity provides enough atoms to supply millions of free electrons. Fig. 5.12 shows the energy band description of n-type semi-conductor. The addition of pentavalent impurity has produced a number of conduction band electrons i.e., free electrons. The four valence elec- trons of pentavalent atom form covalent bonds with four neighbouring germanium atoms. The fifth left over valence electron of the pentavalent atom can- not be accommodated in the valence band and trav- els to the conduction band. The following points may be noted carefully : (i) Many new free electrons are produced by the addition of pentavalent impurity. (ii) Thermal energy of room temperature still generates a few hole-electron pairs. However, the number of free electrons provided by the pentavalent impurity far exceeds the number of holes. It is due to this predominance of electrons over holes that it is called n-type semiconductor (n stands for negative). n-type conductivity. The current conduction in an n-type semiconductor is predominantly by free electrons i.e. negative charges and is called n-type or electron type conductivity. To understand n-type conductivity, refer to Fig. 5.13. When p.d. is applied across the n-type semiconductor, the free electrons (donated by impurity) in the crystal will be directed towards the positive terminal, constitut- ing electric current. As the current flow through the crystal is by free electrons which are carriers of negative charge, therefore, this type of conductivity is called negative or n-type conductivity. It may be noted that conduction is just as in ordinary metals like copper. Fig. 5.12 Fig. 5.11 12 Differences Between N-Type and P-type Semiconductors Parameter P-Type N-Type Impurity doped Trivalent impurity Pentavalent impurity Also known as Acceptor atom because of presence of additional hole. Donor atom due to the existence of additional electron. Doped group Group III elements. For eg - boron, gallium, indium, aluminium etc. Group V elements. for eg - arsenic, antimony, bismuth, phosphorus etc. Majority carriers Holes Electrons Minority carriers Electrons Holes Conductivity Due to presence of holes. Due to presence of electrons. Presence of fermi level Fermi level appears closer to the valence than the conduction band. Fermi level is present nearer to the conduction band than the valence band. Concentration of electrons Low Very high as compared to p type semiconductor Concentration of holes High Comparatively less than p type semiconductor. 15 Difference Between Intrinsic and Extrinsic semicoductors Parameter Intrinsic Semiconductor Extrinsic Semiconductor Form of semiconductor Pure form of semiconductor. Impure form of semiconductor. Conductivity It exhibits poor conductivity. It possesses comparatively better conductivity than intrinsic semiconductor. Band gap The band gap between conduction and valence band is small. The energy gap is higher than intrinsic semiconductor. Fermi level It is present in the middle of forbidden energy gap. The presence of Fermi level varies according to the type of extrinsic semiconductor. Dependency The conduction relies on temperature. The conduction depends on the concentration of doped impurity and temperature. Carrier concentration Equal amount of electron and holes are present in conduction and valence band. The majority presence of electrons and holes depends on the type of extrinsic semiconductor. Type It is not further classified. It is classified as p type and n type semiconductor. 16 Example Si, Ge etc. GaAs, GaP etc. 17 Electrical and Electronics Engineering 7-7 Semiconductor Physics and Diode conduct only at very high temperatures or if they are subjected to high voltage. Such a conduction is rare and ‘s called breakdown of an insulator. The other insulating materials are glass, wood, mica, paper etc. Energy Energy Energy t ‘Conduction band Conduction ‘Conduction band Atarge forbidden Eq 3p ‘Asma g foroidden gap £0 (0) Conductor {) Insulator (6) Semiconductor Fig. 7.4 Energy band diagrams 7.4.3. Semiconductors Now let us come to an important category of materials, which are neither insulators nor conductors. The forbidden gap in such materials is very narrow as shown in Fig. 7.4(c). Such materials are called semiconductors. The forbidden gap is about 1 eV. In such materials, the energy provided by the heat at room temperature is sufficient to lift the electrons from the valence band to the conduction band. Therefore at room temperature, semiconductors are capable of conduction. But at 0 °K or absolute zero (-273 °C), all the electrons of semiconductor materials find themselves locked in the valence band. Hence at © k, the semiconductor materials behave as perfect insulators. In case of semiconductors, forbidden gap energy depends on the temperature. For silicon and germanium, this energy is given by, Eg =121-3.6x104XxT eV (for Silicon) Eg = 0.785 - 2.23 x104xT eV (for Germanium) where T = Absolute temperature in °K Assuming room temperature to be 27 °C ie. 300 °K, the forbidden gap energy for Si and Ge can be calculated from the above equations. The forbidden gap for the germanium is 0.72 eV while for the silicon it is 112 eV at room temperature. The silicon and germanium are the two widely used semiconductor materials in electronic devices, as mentioned earlier. Koy Point : While calculating E,,, substitute T in °K. 20 Electrical and Electronics Engineering 7-8 Semiconductor Physics and Diode Why Silicon is most widely used ? Looking at the structure of silicon and germanium atom, it can be seen that valence shell of silicion is 3"! shell while valence shell of germanium is 4" shell. Hence valence electrons of germanium are at larger distance from nucleus than valence electrons of silicon. Hence valence electrons of germanium are more loosely bound to the nucleus tan those of silicon. Thus valence electrons of germanium can easily escape from the atom, due to very small additional energy imparted to them. So at high temperature, germanium becomes unstable than silicon and and hence silicon is widely used semiconductor material. mm Example 7.1: Calculate the value of forbidden gap for silicon and germanium at the temperature of 35 °C. Solution : Forbidden gap for silicon is given by, Eg = 121-3.6x104xT Now T = 35+273= 308% Eg = 121-3.6x10+ x 308 = 1.099 eV. While forbidden gap for germanium is given by, Eg = 0.785~2.23x104 xT = 0.785- 2.23 10 x 308 = 0.7163 ev 7.5 Intrinsic Semiconductors A sample of semiconductor in its purest form is called an intrinsic semiconductor. The impurity content in intrinsic semiconductor is very very small, of the order of one part in 100 million parts of semiconductor. For achieving such a pure form, the semiconductor materials are carefully refined. To understand the conduction in an intrinsic semiconductor let us study the crystalline structure of an intrinsic semiconductor. 7.5.1 Crystal Structure of Intrinsic Semiconductor Consider an atomic structure of an intrinsic semiconductor material like silicon. An outermost shell of an atom is capable of holding eight electrons. It is said to be completely filled and stable, if it contains eight electrons. But the outermost shell of an intrinsic semiconductor like silicon has only four electrons. Each of these four electrons form a bond with another valence electron of the neighbouring atoms. This is nothing but sharing of electrons. Such bonds are called covalent bonds. The atoms align themselves to form a three dimensional uniform pattern called a crystal. The crystal structure of germanium and silicon materials consists of repetitive occurrence in three dimensions of a unit cell. This unit cell is in the form of a tetrahedron with an atom at each vertex. But such a three dimensional structure is very difficult to 21 Electrical and Electronics Engineering 7-10 Semiconductor Physics and Diode Electron: Hole pa [eanaocbon ¢ (2) Breaking of covalent bond —_—_(b) Electromhole pair in a silicon crystal (c) Energy band diagram Fig. 7.6 Thermal generation The concentration of free electrons and holes is always equal in an intrinsic semiconductor. The hole also serves as a carrier of electricity similar to that of free electron. An electron is negatively charged particle. Thus a hole getting created due to electron drift is said to be positively charged. Key Point: Thus in an intrinsic semiconductors both holes as well as free electrons are the charge carriers. 7.5.3 Conduction by Electrons and Holes The electrons and holes generated due to thermal generation move randomly and hence cannot constitute any current. Now consider that battery is connected across the intrinsic semiconductor. Under the influence of applied voltage there is electron as well as hole motion in one particular direction, causing the flow of current. ‘The free electrons which are available in the conduction band are moved under the influence of applied voltage. The electrons as negatively charged get repelled from the negative terminal of battery and attracted towards the positive terminal. Thus there is an electric current due to the movement of electrons in conduction band. This is called electron current . There are electrons present in the valence band which are involved in forming the covalent bonds. Some holes are also present in the valence band due to escape of electrons from valence to conduction band. Under the influence of applied voltage, the electrons involved in covalent bonds break the covalent bonds and try to fill the holes present. The electron breaking the covalent bond jumps to the hole of neighbouring atom, leaving a hole behind. This is illustrated in the Fig. 7.7 (a), (b) and (c). The atom x has a hole due to escape of an electron to the conduction band. The electron fpgm atom y breaks its covalent bond and fill the hole of atom x. Now the hole is Electrical and Electronics Engineering 7-15 Semiconductor Physics and Diode E ~ Q) where v = Drift velocity in m/sec. E = Applied electric field in V/m where 1 is constant of proportionality and is called mobility of the electrons. This is applicable to the free electrons as well as the holes whichever are the majority carriers. So in general, = Mobility of a charged particle = z i! m/sec Units of w= TRS = = @) So it is measured in square metres per volt-second. Such steady movement of majority charge carriers with drift velocity constitutes a current. This current is called drift current. 7.8 General Expression for Conductivity Consider a tube of metal with large number of free elecrons as shown in the Fig. 7.12. Fig. 7.12 Tube of metal subjected to voltage Let, A = Cross-sectional area in m* L = Length inm V = Voltage applied in volts T = Time required by an electron to travel distance of 'L’ m L ° v = Drift velocity of electron = = eo) Y= mtectric field 25 v = pE where p = Mobility of electrons (3) Electrical and Electronics Engineering 7-16 ‘Semiconductor Physics and Diode Consider any cross-section as shown in the Fig. 7.13. Cross-section SE Soe ° Fig. 7.13 N electrons crossing in T Let N be the number of electrons passing through area A in time T. So number of electrons crossing the area A in unit time is If Charge on each electron x10 %C then the total charge crossing the cross-section area A in unit time is, dq = Charge on each electron x Ree But charge passing per unit time through a cross-section is the current. Charge passing 1 = Charge passing Unit Ne I= "FA 4) The current density J for the bar is current per unit cross-sectional area of the conducting material. J= + A/m? a 6) }-& but T= = vs from (1) But LA = Volume of the tube n = Concenration of free electrons Number of elecrons per unit volume = Nop ne oy /m 26 J = nev but v=pE Electrical and Electronics Engineering 7-17 Semiconductor Physics and Diode This is the general expression for current density in a given material. The current density is related to electric field E by the relation, where © = Conductivity of the material in (Qn) Key Point: The conductivity indicates the ease with which current can flow through the given material. Comparing (6) and (7) This is the general expression for the conductivity of the given material @) Key Point: The resistivity p is the reciprocal of the conductivity. 7.9 Conductivity of an Intrinsic Semiconductor In a semiconductor, there are two charged particles. One is negatively charged free electrons while the other is positively charged holes. These parties move in opposite direction, under the influence of an electric field but as both are of opposite sign, they constitute current in the same direction. For the semiconductor, n = Concentration of free electrons /m? P = Concentration of holes/m? ¥, = Mobility of electrons in m*/V-s Hp = Mobility holes in m?/V-s then the current density is given by, This can be obtained from the general expression for J derived in last section, ‘equation (6). ‘Hence the conductivity for a semiconductor is given by, 27 Electrical and Electronics Engineering 7 - 21 Semiconductor Physics and Diode Solution : Given values are 25 x 10° / cm? n= 25x10" _ 95 19! / mm? a = SE = 25 x 10? / mm Ha = 3800 cm?/V-s = 3800 x 10°* m? / V-s Bp = 1800 cm?/V-s = 1800 x 10°* m? / V-s o = tHe (3800 + 1800) x 1.8 x 107! x 1074 x 25 x 10% = 224 (Q-my\ 14 ay 7 Ea 7 OMA Q- em % 0 0.4464 Q- cm 7.10 Law of Mass Action If n is the concentration of free electrons and p is the concentration of holes then the law of mass action states that the product of concentrations of electrons and holes is always constant, at a fixed temperature. Mathematically it is expressed as, ~() where n, is intrinsic concentration. Important Observations 1. The law can be applied to both intrinsic and extrinsic semiconductors. 2. Asn, depends on temperature, the law is applicable at a fixed temperature. 3. In case of extrinsic semiconductors, n, is the intrinsic concentration of the basic semiconductor material used. Key Poirt : The low can be used to find the electron and hole densities in intrinsic semiconde :tors. 30 Electrical and Electronics Engineering 7-22 Semiconductor Physics and Diode 7.11 Extrinsic Semiconductors In order to change the properties of intrinsic semiconductors a small amount of some other material is added to it. The process of adding other material to the crystal of intrinsic semiconductors to improve its conductivity is called doping. The impurity added is called dopant. Doped semiconductor material is called extrinsic semiconductor. The doping increases the conductivity of the basic intrinsic semiconductors hence the extrinsic semiconductors are used in practice for manufacturing of various electronic devices such as diodes, transistors etc. Depending upon the type of impurit 1. type and 2. p-type the two types of extrinsic semiconductors are, 7.14.4 Types of Impurities The impurity material having five valence electrons is called pentavalent atom. When this is added to an intrinsic semicondcutor, it is called donor doping as each impurity atom donates one free electron to an intrinsic material. Such an impurity is called donor impurity. The examples of such impurity are arsenic, bismuth, phosphorous etc. This creates an extrinsic semiconductor with large number of free electrons, called n-type semiconductor. Another type of impurity used is trivalent atom which has only three valence electrons. Such an impurity is called acceptor impurity. When this is added to an intrinsic semiconductor, it creates more holes and ready to accept an electron hence the doping is called acceptor doping. The examples of such impurity are gallium, indium and boron. The resulting extrinsic semiconductor with large number of holes is called p-type semiconductor. 7.12 n-Type Semiconductor When a small amount of pentavalent impurity is added to a pure semiconductor, it is called n-type semiconductor. The pentavalent impurity has five valence electrons. These elements are such as arsenic, bismuth, phosphorous and antimony. Such an impurity is called donor impurity. Consider the formation of n-type material by adding arsenic (As) into silicon (Si). The arsenic atom has five valence electrons. An arsenic atom fits in the silicon crystal in such a way that its four valence electrons form covalent bonds with four adjacent silicon atoms. The fifth electron has no chance of forming a covalent bond. This spare electron enters the fig. 7.15 n-type material formation conduction band as a free electron. Electrical and Electronics Engineering 7-23 Semiconductor Physics and Diode Such n-type material formation is represented in the Fig. 7.15. This means that each arsenic atom added into silicon atom gives one free electron. The number of such free electrons can be controlled by the amount of impurity added to the silicon. Since the free electrons have negative charges, the material is known as n-type material and an impurity donates a free electron hence called donor impurity. Key Point: One donor impurity atom donates one free electron in n-type material. The free electrons are majority charge carriers. 7.12.1 Conduction in n-Type Semiconductor When the voltage is applied to the n-type semiconductor, the free electrons which are readily available due to added impurity, move in a direction of positive terminal of voltage applied. This constitutes a current. Thus the conduction is predominantly by free electrons. The holes are less in number hence electron current is dominant over the hole current. Hence in n-type semiconductors free electrons are called majority carriers while the holes which are small in number are called minority carriers. The conduction in n-type material is shown in the Fig. 7.16. Free elecvons large in number Holes small in number (nasty) (minony —— Battery Conventional current Fig. 7.16 Conduction in n-type material 7.43 p-Type Semiconductor When a small amount of trivalent impurity is added to a pure semiconductor, it is called p-type semiconductor. The trivalent impurity has three valence electrons. These elements are such as gallium, boron or indium. Such an impurity is called acceptor impurity. Consider the formation of p-type material by adding gallium (Ga) into silicon (Si). The gallium atom has three valence electrons. So gallium atom fits in the silicon <rystal in such a way that its three valence electrons form covalent bonds with the three adjacent 3Fig. 7.17 p-type material formation Electrical and Electronics Engineering 7-26 Semiconductor Physics and Diode 7.14.3 Law of Mass Action for Extrinsic Semiconductors It is seen that, mathematically the law is expressed as, where nj is intrinsic concentration. Important Observations ‘The law can be applied to both intrinsic and extrinsic semiconductors. 2. In case of extrinsic semiconductors, n, is the intrinsic concentration of the basic semiconductor material used. 3. For n-type material, n= n,, while p = p,, hence law can be stated as, n,P, = nf ~» (10) 4. For p-type material n = n,, while p = p, hence law can be stated as, yp, = 0? o (11) 5. The law is applicable irrespective of amount of doping. Asn, depends on temperature, the law is applicable at a fixed temperature. 7. The law can be used to find both majority and minority carrier concentrations in an extrinsic semiconductor. > 7.14.4 Carrier Concentrations in Extrinsic Semiconductors Let us obtain the concentrations of minority and majority carriers in n-type and p-type materials using Law of mass action. nn type material : For n-type material it is seon that, n,, = Np. At any fixed temperature, according to law of mass action, RaX Pa = nF where n,, is electrons ie. majority carrier concentration while p,, is hole ie. minority carrier concentration. Using n,, “Np, we can write minority carrier concentration as, No Pa = 9G nt Pa = NO = (12) Knowing n, and Np, the number of holes in n-type material i.e. minority carrier concentration can be obtained. 35 Electrical and Electronics Engineering 7-30 Semiconductor Physics and Diode = 1% 10% + (1 x10)? 4 6.25 x 10) z = 5,901 x10" per m> -» Neglecting negative value p = 1x 10%+n =1.059x 10% per m3 ‘These are the actual concentrations of electrons holes in the sample. As p>n, holes are much more than electrons and sample will behave as p-type material i> Example 7.7 : If a donor impurity is added to the extent of one atom per 10° germanium atoms, calculate its resistivity at 300 °K. If its resistivity without addition of impurity at 300 °K is 44.64 Q-cm., comparing two values, comment on the result. Assume : jt, = 3800 on?/V-sec. Solution : Referring to the Table 7.2 of properties of germanium, germanium has 44x10 atoms/cm’. For 108 germanium atom there is 1 atom impurity added, as given. Thus, for 4.4% 10” germanium atoms, we have, 44x 107 -_=- 7 44x 10™ atoms of impurity /cm? This is nothing but concentration of donor atoms ie. Np Np = 44x10" per m3 = 4X0 ‘Now as donor impurity is added, n-type material will form, On = Mead = Np bad where n, = Np and p, = 3800 em2/V-sec = 3800x10+m2/V-sec ©, = 44x 10% x 3800x 104 x 1.6x 10-9 = 26.752 (Q-m)7 seth a . Resistivity = Py = G-= zgzqp = 00373 Q-m = 3.73 Q-em Comparing this with resistivity of intrinsic germanium it can be observed that resistivity reduces considerably due to addition of impurity. Hence conductivity of n-type material is much higher and hence it can carry more current as compared to the intrinsic semiconductor. By controlling amount of doping we can control the conductivity. 36 Electrical and Electronics Engineering 7-31 ‘Semiconductor Physics and Diode 7.15 Diffusion Current This is the current which is due to the transport of charges occurring because of nonuniform concentration of charged particles in a semiconductor. Consider a piece of semiconductor which is nonuniformly doped. Due to such nonuniform doping, one type of charge carriers occur at one end of a piece of semiconductor. The charge carriers are either electrons or holes, of one type depending upon the impurity used. They have the same polarity and hence experience a force of repulsion between them. The result is that there is a tendency of the charge carriers to move gradually i.e. to diffuse from the region of high carrier density to the low carrier density. This process is called diffusion. This movement of charge carriers under the process of diffusion constitutes a current called diffusion current. This is shown in the Fig. 7.19. (a) Nonuniform concentration (b) Process of diffusion Fig. 7.19 Diffusion current ‘The diffusion current continues till all the carriers are evenly distributed throughout the material. A diffusion current is possible only in case of nonuniformly doped semiconductors while drift current is possible in semiconductors as well as conductors. Key Point: The difusion current exists without external voltage applied while drift current can not exist without an external voltage applied. 7.15.1 Concentration Gradient Consider a p-type semiconductor bar which is nonuniformly doped. Along its length, in the direction of x as shown in Fig. 7.20 (a), there exists a nonuniform doping. As x increases, the doping concentration decreases. p Po x=0 x x20 * (a) Nonuniform concentration (b) The graph of concentration Ina semiconductor bar joles against x 1. 7.20 37 Be “Electrical and Electronics Engineering 7 - 34 ‘Semiconductor Physics and Diode Drift current density for electrons and holes can be expressed as, Jn = new, E and J,=pen,E (Refer section 7.8) Diffusion current density due to the electrons and holes can be expressed from equations (3) and (4) as, dn dp Jn = +eD, Gand Jp=-eD, -: Total current density due to the electrons can be expressed as, J, = nen, E+eD, 2 6 and Total current density due to the holes can be expressed as, dy Jp = pen, E-eD, © =O ‘And hence the total current density due to the electrons and holes (drift + diffusion) 7.15.4 Einstein's Relationship It is now known that drift current density is proportional to the mobility (1) while diffusion current density is proportional to the diffusion constant (D). There exists a fixed relation between these two constants which is called Einstein's relation. It states that, at a fixed temperature, the ratio of diffusion constant to the mobility is constant. This is Einstein’s relation. Mathematically it is expressed a3, D, _ D, Ge 7 ge KT = constant at fixed temperature, = @) P . is, where T is the temperature in °K And k is the Boltzmann’s constant = 86210 eV/°K 7.15.5 Voltage Equivalent of Temperature In the equation (8), the product KT is called voltage equivalent of temperature. The voltage equivalent of temperature is denoted by Vr. At room temperature i.e. at 27°C, 40 Electrical and Electronics Engineering 7-35 Semiconductor Physics and Diode = 273 +27 =300°K Vy = KT = 862x10% x 300 = 0.02586 V = 26 mV at 300°K Key Point: The value of Vz = 26 mV at 27 °C ie. 300 °K is very commonly used while solving the examples. Substituting this in equation (8) we get, Di _ Pp Fe 7 G7 Vr = 002586. at room temperature in P aut jp=9D, froom temperate. «0 In general we can express the relation between mobility and diffusion constant as, 39D _ at room temperature (11) 7.16 The p-n Junction Diode The p-n junction forms a popular semiconductor device called p-n junction diode. The Pn junction has two terminals called electrodes, one each from p-region and n-region. Due to the two electrodes it is called diode i.e. di + electrode. To connect the n and p regions to the external terminals, a metal is applied to the heavily doped n and p-type semiconductor regions. Such a contact between a metal and a heavily doped semiconductor is called ohmic contact. Such an ohmic contact has two important properties, 1. It conducts current equally in both the directions. 2. The drop across the contact is very small, which do not affect the performance of the device. ‘Thus ohmic contacts are used to connect n and p-type regions to the electrodes. The Fig. 7.22 (a) shows schematic arrangement of p-n junction diode while the Fig. 7.22 (b) shows the symbol of p-n junction diode. The p-region acts as anode while the reregion acts as cathode. The arrowhead in the symbol indicates the direction of the conventional current, which can flow when an external voltage is connected in a specific manner across the diode. Ohmic contacts, k J Angde Cathode Eh Electrode 4 Eo Electrode 2 ptype type (a) Two electrodes (b) Symbol of a diode Fig. 7.22 41 Electrical and Electronics Engineering 7 - 37 ‘Semiconductor Physics and Diode The large number of majority carriers constitute a current called forward current. Once the conduction electrons enter the p-region, they become valence electrons. Then they move from hole to hole towards the positive terminal of the battery. The movement of valence electrons is nothing but movement of holes in opposite direction to that of electrons, in the p-region. So current in the p-region is the movement of holes which are majority carriers. This is the hole current. While the current in the n-region is the movement of free electrons which are majority carriers. This is the electron current. Hence the overall forward current is due to the majority charge carriers. The action is shown in the Fig. 7.24. These majority carriers can then travel around the closed circuit and a relatively large current flows. The direction of flow of electrons is from negative to positive of the battery. While direction of the conventional current is from positive to negative of the battery as shown in the Fig. 7.24. Current. limiting resistor Fig. 7.24 Forward current in a diode Key Point : The direction of flow of electrons and conventional current is opposite to each other. 7.17.2 Effect on the Depletion Region Due to the forward bias voltage, more electrons flow into the depletion region, which reduces the number of positive ions. Similarly flow of holes reduces the number of negative ions. This reduces the width of the depletion region. This is shown in the Fig. 725. pletion region narrows, (2) Unbiased dose (0) Forward biased dlode Fig. 7.25 Key Point : Depletion region narrows due to forward bias voltage. 42 Electrical and Electronics Engineering 7 - 41 Semiconductor Physics and Diode of the stable atoms. So this is not due to the collision of carriers with atoms. Such a creation of free electrons is called zener effect which is different than the avalanche effect. These minority carriers constitute very large current and mechanism is called zener breakdown. Key Point: The normal p-n junction diode is practically not operated in reverse breakdown region though may be operated in reverse biased condition. The breakdown effects are not required to be considered for p-n junction diode. These effects are required to be considered for special diodes such as zener diode as such diodes are always operated in reverse breakdown condition. The Fig. 7.29(a) shows the avalanche effect while the Fig. 7.29 (b) shows the zener effect. Intense electric feld is responsible to generate ‘Due to high veloaly when onarge caters. Generated carer colides with atom, generate charge carriers Narrow depletion region Highty Generated Highly tt coped free electron (a) Avalanche breakdown {b) Zener effect, Fig. 7.29 Breakdown mechanisms. 7.19 The Current Components in a p-n Junction Diode It is indicated earlier that when a p-n junction diode is forward biased a large forward current flows, which is mainly due to majority carriers. The depletion region near the junction is very very small, under forward biased condition. In forward biased condition holes get diffused into n-side from p-side while electrons get diffused into p-side from msde. So on p-side, the current carried by electrons which is diffusion current due to minority carriers, decreases exponentially with respect to distance measured from the junction. This current due to electrons, on p-side which are minority carriers is denoted as Typ: Similarly holes from p-side diffuse into mvside carry current which decreases exponentially with respect to distance measured from the junction. This current due to holes on n-side, which are minority carriers is denoted as I,,. If distance is denoted by x then, 45 Electrical and Electronics Engineering 7-42 Semiconductor Physics and Diode ip (x) = Current due to electrons in p-side as a function of x Tp» (x) = Current due to holes in n-side as a function of x At the junction ie. at x = 0, electrons crossing from n-side to p-side constitute a current, 1,9(0) in the same direction as holes crossing the junction from p-side to n-side constitute a current, 1,, (0) - Hence the current at the junction is the total conventional current I flowing through the circuit. po (0)+ Tp () ~@) Now Iq (x) decreases on n-side as we move away from junction on n-side. Similarly 1,p(x) decreases on p-side as we move away from junction on p-side. But as the entire circuit is a series circuit, the total current must be maintained at I, independent of x. This indicates that on p-side there exists one more current component which is due to holes on p-side which are the majority carriers. It is denoted by I,» (x) and the addition of the two currents on p-side is total current I. Iyp(x) = Current due to holes in p-side. Similarly on n-side, there exists one more current component which is due to electrons on n-side, which are the majority carriers. It is denoted as 1,, (x) and the addition of the two currents on n-side is total current I. Iga (#) = Current due to electrons in n-side. On p-side, 1 = Typ (8) + Tp (X) s+ Q) On n-side, I &)+ Tyna 0) ~~) These current components are plotted as a function of distance in the Fig. 7:30. The current Typ decreases towards the junction, at the junction enters the mside and becomes I,, which further ‘Tetalcurent! decreases exponentially. Similarly the current Ij, decreases towards the i junction, at the junction enters the slocroneuront —y_ | ng roetivont side and becomes I,, which also ° Top Son oe Goren electron curent we x20 nae further decreases exponentially. 4g. 7.30 Current components  Electrical and Electronics Engineering 7-43 Semiconductor Physics and Diode Key Point ; In forward bias condition, the current enters the p-side as a hole current and leaves the n-side as an electron current, of the same magnitude. So sum of the currents carried by electrons and holes at any point inside the diode is always constant equal to total forward current I. But the proportion due to holes and electrons in constituting the current varies with the distance, from the junction. 7.20 The Volt-Ampere (V-I) Characteristics of a Diode ‘The response of a diode when connected in an electrical circuit, can be judged from its characteristics known as Volt-Ampere commonly called V-I characteristics. The V-I characteristics in the forward biased and reverse biased condition is the graph of voltage across the diode against the diode current. 7.20.1 Forward Characteristics of p-n Junction Diode The response of p-n junction can be easily Curent tie y oe a ph indicated with the help of characteristics called V-I characteristics of p-n junction. It is the graph of voltage applied across the p-n junction and the current flowing through the p-n junction. The Fig. 7.31 shows the forward biased Fig, 7.31 Forward biased diode diode. The applied voltage is V while the voltage across the diode is V, The current flowing in the circuit is the forward current ,. The graph of forward current I, against the forward voltage V, across the diode is called forward characteristics of a diode. ‘The forward characteristics of a diode is shown in the Fig. 7.32. i Basically forward characteristics can lama, be divided into two regions : 1, Region O to P : As long as V, is less than cut-in voltage (V,) , the current flowing is very small. Practically this current is assumed to be zero. 2. Region P to Q and onwards : A: V; increases towards V, the width of depletion region goes on reducing. When V, exceeds V, ie. cut-in voltage, the depletion region becomes very thin and current J, increases suddenly. This wotage (4) increase in the current is exponential as shown in the Fig. 7.32 by the region Fig. 7.32 Forward characteristics of a diode pig q. 47 Electrical and Electronics Engineering 7-46 Semiconductor Physics and Diode 2. Reverse dynamic resistance : This is the reverse resistance under the a.c. conditions, denoted as r,. It is the ratio of incremental change in the reverse voltage applied to the corresponding change in the reverse current. > = __ Change in reverse voltage id eee Tx Change in reverse current P| The dynamic resistance is most important in practice whether the junction is forward or reverse biased. 7.20.3 Complete V-I Characteristics of a Diode The complete V-I characteristics of a diode is the combination of its forward as well as reverse characteristics. This is shown in the Fig. 7.35. ‘Normal ‘operating region My Fig. 7.35 Complete V-1 characteristics of a diode In forward characteristics, it is seen that initially forward current is small as long as the bias voltage is less that the barrier potential. At a certain voltage close to barrier potential, current increases rapidly. The voltage at which diode current starts increasing rapidly is called as cut-in voltage. It is denoted by V,. Below this voltage, current is less than 1% of maximum rated value of diode current. The cutin voltage for germanium is about 0.2 V while for silicon it is 0.6 V. It is important to note that the breakdown voltage is much higher and practically diodes are not operated in the breakdown condition. The voltage at which breakdown occurs is called reverse breakdown voltage denoted a8 Vpn. Key Point : Reverse current before the breakdown is very very small and can be practically need Electrical and Electronics Engineering 7-48 Semiconductor Physics and Diode The factor 1 is called emission coefficient or ideality factor. This factor takes into account the effect of recombination taking place in the depletion region. The effect is dominant in silicon diodes and hence for silicon n = 2. The range of factor is from 1 to 2 ‘The voltage equivalent of temperature indicates dependence of diode current on temperature. The voltage equivalent of temperature V; for a given diode at temperature T is calculated as, where k = Boltzmann's constant = 8.62 x10"5 eV/°K T = Temperature in °K. = 27 + 273 = 300 °K and the value of V; is 26 mV, At room temperature of 27 °C ie. as seen earlier. The value of V; also can be expressed as, T T T Vrs les ™ oa 8) 8.62x 10°5 Key Point : The diode current equation is applicable for all the conditions of diode ie. unbiased, forward biased and reverse biased. When unbiased, V = 0 hence we get, I = L[e’-1]=0A Thus there is no current through diode when unbiased. Key Point: For forward biased, V must be taken positive and we get current I positive ‘which is forward current. For reverse biased, V must be taken negative and we get negative current I which indicates that it is reverse current. If both sides of diode current equation is divided by cross-sectional area A of the junction, t = Bem =] ie, J= sofe%vr -1]A/me oo @& where J = Forward current density Jq = Reverse saturation current density 51  Electrical and Electronics Engineering 7-51 . Semiconductor Physics and Diode ‘> Example 7.9 : The voltage across a silicon diode at room temperature of 300 °K is 0.71 V when 2.5 mA current flows through it. If the voltage increases to 0.8 V, calculate the new diode current. Solution : The current equation of a diode is T = Ip (eV/r - 1) ’ At300°K, Vp = 26 mV = 26x10 -V V = O71Vforl=25mA and n= 2 for silicon 25x10 = Ip [ eO7/2x26x10) _ yy * Ty = 293x109 A Now V = 08 V I = 293x107 [ e8/2x26x10) _ 3) = 0.0141 A = 14.11 mA tm) Example 7.10: A germanium diode has a reverse saturation current of 3 yA. Calculate the forward bias voltage at the room temperature of 27 °C and 1% of the rated current is flowing through the forward biased diode. The diode forward rated current is 1 A. Solution : The given values are, ly = 3 wA, T = 27°C = 27 +273 = 300°K ,n=1 Now Tea = 1A for diode and T= 1% of Igugg at 27°C 1 1 = qqgx()= 001 A Vz = KT = 8.62x107 x 300 = 0.026 V According to diode equation, 1 = [evr - 0.01 = 3x10fe¥/!*0.% — 3333.333 = eV /00%6 7 eV/ 00% = 3334.3333 In fe¥!9-™9] = In [ 3334.3983 ] taking natural log v Toe = 8.112 V = 02109 V 52 ” Electronic Devices and Circuits vacancy. So because of the thermal agitation of the crystal lattice, an electron of another ion may come very close to the ion which has lost the electron. The ion which has lost the electron will immediately steal an electron from the closest ion, to fill its vacancy. The holes move from the first ion to the second ion. When no electric field is applied, the motion of free electron is random in nature. But when clectric field is applied, all the free electrons are lined up and they move towards the positive electrode. The life period of a free electron may be Iu-sec to I millisecond after which it is absorbed by another ion. 2.9 MASS ACTION LAW In an intrinsic Semiconductor number of free electrons n = n, = No. of holes p = p, Since the crystal is electrically neutral, n,p, = ne, Regardless of individual magnitudes of n and p, the product is always constant, np =n? ny= AT2 ¢ 2KT (2.13) This is called Mass Action Law. 2.10 LAW OF ELECTRICAL NEUTRALITY Let Np is equal to the concentration of donor atams in a doped semiconductor. So when these donor atoms donate an electron, it becomes positively charged ion, since it has lost an electron. So positive charge density contributed by them is Np. If ‘p’ is the hole density then total positive charge density is Np + p. Similarly if N, is the concentration of acceptor ions, (say Boron which is trivalent, ion, accepts an electron, so that 4 electrons in the outer shell are shared by the Ge atoms ), it becomes negatively charged. So the acceptor ions contribute charge = (Na +n ). Since the Semiconductor is electrically neutral, when no voltage is applied, the magnitude of positive charge density must equal that of negative charge density. Total positive charge, Np + p= Total negative charge (Na +n) Npotp=Natn This is known as Law of Electrical Neutrality. Consider n-type material with acceptor ion density Na = 0. Since it is n-type, number of electrons is >> number of holes. So ‘Pe can be neglected in comparison with n. Ny ~= Np. (Since every Donor Atom contributes one free electron. ) In n-type 1e material, the free electron concentration is approximately equal to the density of donor atoms. In n-type semiconductor the electron density n, = Np. Subscript n indicates that it is n-type semiconductor. But Th * Pp = (2.14) 55 Np Tn Electronic Devices and Circuits A~ 9.64 «1014 Eg = 0.25 ev n2= 6.25 x 10°fem? Natn=Nptp Total negative charge = Total positive charge or p—n=Na-Np=(3-2) * 10/4 = 10!4 or p=nt 10!" Then n(n t 10'4)=6.25 x 1076 or n= $.8 x 10! electrons. /em? and p=n+10!4 = 1,06 x 10'* holes/cm? As p>n, this is p-type semiconductor. Problem 2.17 Find the concentration of holes and electrons in a p type germanium at 300°K, if the conductivity is 100 Q- cm. p, Mobility of holes in Germanium = 1800 em?/V - sec. Solution It is p-type p >> n. Sp~ PE Bp 100 ee 17 3 Pp= an 1.610 x 1800 3.47 x 10°" holes/em’ P nx p= n,= 2.5 x 10°/em? _ 2.5x10!3f T7 = 1.8 * 10? electrons/em? 3.47x10 Problem 2.18 (a) Find the concentration of holes and electrons in p-type Germanium at 300°K, if o = 100 v/em. (b) Repeat part (a ) for n-type Si, if o = 0.1 v/em. Solution As it is p-type semiconductor, p >> n. Oo = petty o 100 Sun 7 16x10 x1g00 73:47 * 10"holes/em? Hp = 3.47 « 10! holes/em? nxp= n + _EGo nj= ATZ2e 2kT =2.5 x 10% /em3 56 Junction Diode Characteristics 3 p=3.47 « 10'"/em? 2 13 n n= 25x10") 1.8 « 10? electron/m? PP 3.47x10!7 (b) oO= ne ly: 0.1 = — i a3 1300x1.6x10-% ~ 481 * 10m =4.81 «10'om3 n= 1.5 « 10": 2 p = te = (Seto! f = 4.68 « 10!" hole/m? no 4.g8ixtol4 Problem 2.19 A sample of Ge is doped to the extent of 10'4 donor atoms/cm? and 7 x 10" acceptor atoms/ cm?. At room temperature , the resistivity of pure Ge is 60 Q-cm. If the applied electric field is 2 V/cm, find total conduction current density. Solution For intrinsic Semiconductor, n= oj Me ( Hy + Hy) ! te = Go vem. bp for Ge is 1800 cm?/V-sec ; Hy = 3800 cm?/V-sec. =n; so 1 elt, +un) 604 6x107' en +1800) = 1.86 x 10° electron/em> pxm=n=(186*103 7% q) Natn=Np>p Na =7 = 10/om> Np = 10!4/em? (p-n)=Na-Np=-3% 108 0 (2 Solving 1) and (2) simultaneously to get p and n, p= 0.88 x 1013 n= 3.88 x 10! J = (nly + pp ). €& = { (3.88 )( 3800 ) + (0.88 X 1800) } « 10). es = §2.3 mA/om? n= 57 76 Electronic Devices and Circuits 4n y= 53 2m)? (1.6 « 10°97? = 6,82 « 1077 m = Mass of Flectron in Kgs h = Planck’s Constant is Joule-secs. The equation for { E ) is called the Fermi Dirac Probability Function. It specifies the fraction of all states at energy E (eV) occupied under conditions of thermal equilibrium. From Quantum Statistics, it is found that, 1 RE) = ER (2.16) l+e kT where k = Boltzmann Constant, eV/°K T = Temp °K Er = Fermi Level or Characteristic Energy The momentum of the electron can be uncertain. Heisenberg postulated that there is always uncertainty in the position and momentum of a particle, and the product of these two uncertainities is of the order of magnitude of Planck’s constant ‘h’. If A, is the Uncertainty in the Momentum ofa particle, A, is the uncertainty in the position of a particle A, x Ax x h. 2.11.1 EFFECTIVE Mass When an external field is applied to a crystal, the free electron or hole in the crystal responds, as if its mass is different from the true mass. This mass is called the Effective Mass of the electron or the hole. By considering this effective mass, it will be possible to remove the quantum features of the problem. This allows us to use Newton’s law of motion to determine the effect of external forces on the electrons and holes within the crystal. 2.11.2 Fermi LeveL Named after Fermi, it is the Energy State, with 50% probability of being filled if no forbidden hand exists. In other words, it is the mass energy level of the electrons, at 0°K. IFE=Ep RE)=% From Eq.{ 2.17 ). If a graph is plotted between ( E- Ef) and f E), it is shown in Fig. 2.7 At T = 0°K, if E > Ef then, KE) = 0. That is, there is no probability of finding an electron having energy > Ep at T= 0° K. Since fermi level is the max. energy possessed by the electrons at 0°k. AE) varies with temperature as shown in Fig. 2.7. 60 Junction Diode Characteristics 79 2.11.6 FERM! LEVEL IN INTRINSIC SEMICONDUCTOR @ 2 Conduction Band n=p=n; a n= No e-fec-ENkT E,, p=Ny eo Es-Ey KT n=p Ey or Ne e(Ec“Be VET _ Nye {E, Ey KT EY Electrons in the valence bond occupy energy levels up to ‘Ep’. Ep is defined that way. Then the additional energy that has to bo supplied so that free electron will Valence Band move from valence band to the conduction band is Ec N, x(E,-Ey) (EctEr) Qo a-5 1-0 —£_¢ KT KT ——> f(E) Ny -28, +E ¢+Ey Fig. 2.8 Energy band diagram. =e KT Taking logarithms on both sides, No _ Ec+Ey-2Ep In Ny KT _EctEy KT, Ne Bee Ny > 2am, KT)? Nc=2 rr 2.1 ale) am) 2nmy.KT \2 Nye (2.19) where m,and m, are effective masses of holes and electrons. If we assume that m, = mp, { though not valid ), The graphical representation is as shown in Fig. 2.8. Fermi Level in Intrinsic Semiconductor lies in the middle of Energy gap Eg. 61 30 Electronic Devices and Circuits Problem 2.23 In p-type Ge at room temperature of 300 °K, for what doping concentration will the fermi level coincide with the edge of the valence bond ? Assume [lp = 0.4 m. Solution Ep = Ey when Na=Ny Ep=Ey+kTl My =Eyt+ NL PO EV m-Ng / mp \2 . Ny = 4.82 x 10°5| mp}? x TF? = 4.82 x 10'(0.4)9?(300)°? vm?) = 6.33 x 1018, “ Doping concentration N q = 6.33 x 10!* atoms/em?. Problem 2.24 If the effective mass of an electron is equal to twice the effective mass of a hole, find the distance in electron volts (ev) of fermi level in as intrunsic semiconductor from the centre of the forbidden bond at room temperature. Solution For Intrunsic Semiconductor, _|(EctBy)_KT {Ne Ep= 2 Jj 2 \Ny If Mp = My then Nc=Nvy. Hence Er will be at the centre of the forbidden band. But ifm, # m,. Ef will be away from the centre of the forbidden band by KT Ne _y kT me 2 Ny ‘>in, wow a( 20a c= m2 2 oT)” rom, nwa me ] / 3 = 7 * 0.026 in (2) =13.5m.eV 62 Junction Diode Characteristics 83 But p= Ny xe (Er Ev/KT N Ne = Ee Ey KT Taking Logarithms, Na _ _(&-£y) Inyy = ORT N KTxIn-4 25g, Ny ~ Pv Ee or Ep=Ey+ktxin Nv. seenvte (2.23 ) Ny Na=Ny Fermi Level is close to Valance Band Ey in p-type semiconductor. Problem 2.25 In n type silicon, the donor concentration is 1 atom per 2 * 10° silicon atoms. Assuming that the effective mass of the electron equals true mass, find the value of temperature at which, the fermi level will coincide with the edge of the conduction band. Concentration of Silicon = 5 10” atom/ cm3. Solution Donor atom concentration = 1 atom per2 x 10° Si atom. Silicon atom concentration 5 x 10” atoms/em? 22 5x10 Np= = 2.5 * 10em?. 2x10 For n-type, Semiconductor, Ne "Bp = Be-KT In(—& Ep = Ec In ( Np ) If Ep were to coincide with Ec, then Nc = Np’ Np = 2.5 « 104 fom}, h= Plank’s Constant; K =Boltzman Constant m, the effective mass of electrons to be taken as = mg nf 23.14 «9.11079! x Kx T lw 3 = 4,28 x 10'5 72 65 86 Electronic Devices and Circuits Problem 2.27 In p-type Ge at room temperature of 300 °K, for what doping concentration will the fermi level coincide with the edge of the valence bond ? Assume B= 0.4 m. Solution EB. = EY when Ny =Ny Ny E,=E,+kTIn. Na 3 Ny = 4.82 x 108( mE? x T2? = 4.82 x 10'(0.4)>7(300)22 m = 6.33 «108, Doping concentration N a 633% 10'8 atoms/em?, Problem 2.28 If the effective mass of an electron is equal to thrice the effective mass of a hole, find the distance in‘electron volts (ev) of fermi level in as intrunsic semiconductor from the centre of the forbidden bond at room temperature. Solution For Intrunsic Semiconductor, AR] If m =m pn then No= Ny Hence E, will be atthe centre of the fo: bidden band. But ifm, mE, will be away from the centre of the forbidden band by N ¥ m in fen int 2 Ny 27° Mp, = v2 2am, KT , No=2 — A / z " 3/2 2nm,kT 2| 2 v n 3 =4* 0.026 In (3) =21.4meV 66 Junction Diode Characteristics 87 2.12.2 Drier CURRENT IN P-TYPE SEMICONDUCTOR The mechanism is the same to as explained above. The holes in the acceptor type semiconductor moves towards the negative electrode and enter p-Type into it, pulling out one electron from the negative electrode from the acceptor atoms ( Fig. 2.11 ), the hole has moved | away, i.e. it has acquired an electron. So electrical neutrality or of its original condition is disturbed. This results ina Fig. 2.2 Drift current in p-type semiconductor electrons from the acceptor atom being pulled away. These free electrons enter the positive electrode. The acceptor atoms having lost one electron steal another electron from the adjoining atom resulting in a new hole. The new holes created thus drift towards negative electrode. 2.12.3 Dirrusion CURRENT This current results due to difference in the concentration gradients of charge carriers. That is, free electrons and holes are not uniformly distributed all over the semiconductor. In one particular area, the number of free electrons may be more, and in some other adjoining region, their number may be less. So the electrons where the concentration gradient is more move from that region to the place where the electrons are lesser in number. This is true with holes also. Let the concentration of some carriers be as shown in the Fig 2.12. The concentration of carriers is not uniform and varies as shown along the semiconductor length. Area A, is a measure of the number of carriers between x, and x2. Area Az is a measure of the number of carriers between x2 and x3. Area Ay is greater than Area A). Therefore number of carriers in area A, is greater than the number of carriers in Ay. Therefore they will move from A; to A>. If these — Conductance of the Carriers w B C —+> Distance of ( along the Semiconductor ) Fig. 2.12 Diffusion current. 67 Junction Diode Characteristics n° Consider a semiconductor of area A, length dx (x+#dx-—x =dx ) as shown in Fig. 2.13. Let the average hole concentration be ‘p’. Let E, is a factor of x. that is hole current due to concentration is varying with distance along the semiconductor. Let I, is the current entering the volume at x at time t, and (I, + dI,) is the current leaving the volume at (x + dx ) at the same instant of time ‘t’. P So when only I, colombs is entering. (1, + dl, ) colombs are leaving. Therefore effectively there is a decrease of ( Ip + dl, — Ip ) = dlp colombs per second within the volume. Or in other words, since more hole current is leaving than what is entering, we can say that more holes are leaving than the no. of holes entering the semiconductor at ‘x’. If dl, is rate of change of total charge that is * + dx nxq Fig 2.13 Charge flow in semiconductor di, =d t dl 7. gives the decrease in the number of holes per second with in the volume A x dx. Decrease in holes per unit volume ( hole concentration ) per second due to I, is dl, 1 Py Axdx 4 But = Current Density a+, oo q° dx" But because of thermal agitation, more number of holes will be created. If po is the thermal equilibrium concentration of holes,( the steady state value reached after recombination ), then, the increase per second, per unit volume due to thermal generation is, _2 go Tp ‘Therefore, increase per second per unit volume due to thermal generation, Po Bo, But because of recombination of holes and electrons there will be decrease in hole concentration. The decrease = 70 92 Electronic Devices and Circuits Charge can be neither created nor destroyed. Because of thermal generation, there is increase in the number of holes. Because of recombination, there is decrease in the number of holes. Because of concentration gradient there is decrease in the number of holes. So the net increase in hole concentration is the algebraic sum of all the above. ep _ P,P ot t P aes . . : . d Partial derivatives are used since p and Jp are functions of both time t and distance x. i gives the variation of concentration of carriers with respect to time ‘t’. If we consider unit volume of a semiconductor ( 7-type ) having a hole density p,, some holes are lost due to recombination. If py, is equilibrium density, ( i.e., density in the equilibrium condition when number of electron = holes ). Pa = Pno The recombination rate is given as . The expression for the time rate of change in. carriers density is called the Continuity Equation. aes dp Recombination rate R = at Life time of holes in n-type semiconductors _AP _ Pn=Pno DR dp/dt or \ dp _ | Pn—Pno dt T where p, is the original concentration of holes in n-type semiconductors and yo is the concentration after holes and electron recombination takes place at the given temperature. In other word Ppo is the thermal equilibrium minority density. Similarly for a p-type semiconductors, the life time of electrons = Np Mo | dx tn “dx/dt° dt 2.15 THE HALL EFFECT If a metal or semiconductor carrying a current I is placed in a perpendicular magnetic field B, an electric field E is induced in the direction perpendicular to both I and B. This Phenomenon is known as the Hall Effect. It is used to determine whether a semiconductor is p-type or n-type. By measuring conductivity o, the mobility yu can be calculated using Hall Effect. In the Fig. 2.14 current | is in the positive X-direction and B is in the positive Z-direction. So a force will be exerted in the negative Y-direction. If the semiconductor is 7-fype, so that current is carried by electrons, these electrons will be forced downward toward side |. So side 1 becomes negatively charged with respect to side 2. Hence a potential V}; called the Hall Voltage appears between the surface | and 2. 71 Junction Diode Characteristics 93 Fig 2.14 Hall effect. In the equilibrium condition, the force due to electric field intensity ‘E’, because of Hall effect should be just balanced by the magnetic force or ee = Bev v = Drift Velocity of carriers in m / sec B = Magnetic Field Intensity in Tesla ( wb/m?) or C= By rennet (a) But e= Visd where Vii = Hall Voltage d = Thickness of semiconductors. J =nev or J=pv p = charge density. J = Current Density (Amp / m?) I or d= ba @ = width of the semiconductor; ad = cross sectional area J = current . [ J = Current Density = bd e= Vid or Vu = ed 72 96 Electronic Devices and Circuits 2.16 SEMICONDUCTOR DIODE CHARACTERISTICS If a junction is formed using p-type and n-type semiconductors, a diode is realised and it has the properties of a rectifier. In this chapter, the volt ampere characteristics of the diodes, electron- hole currents as a function of distance from the junction and junction capacitances will be studied. 2.16.1 THeory OF p-n JUNCTION Take an intrinsic Silicon or Germanium crystal. If donor ( -type ) impurities are diffused from one point and acceptors impurities from the other, a p-n junction is formed. The donor atoms will donate electrons. So they loose electrons and become positively charged. Similarly, acceptor atoms accept an electron, and become negatively charged. Therefore in the p-n junction on the p- side, holes and negative ions are shown and on the n-side free electrons and positive ions are shown. To start with, there are only p-type carriers to the left of the junction and only n-type carriers to the right of the junction. But because of the concentration gradient across the junction, holes are in large number on the left side and they diffuse from left side to right side. Similarly electrons will diffuse to the right side because of concentration gradient. Holes 8 O99 @ 00, 0 ° oO 8 <— Free Electrons (b) ©) -_Wv dx Electrostatic Potential Barrier fo: Holes qd) : Potential Barrier for Electrons (e) - Fig 2.15. Potential distribution in p-n junction diode. 75 Junction Diede Characteristics 97 Because of the displacement of these charges, electric field will appear across the junction. Since p-side looses holes, negative field exists near the junction towards left. Since n-side looses electrons, positive electric field exists on the n side. But at a particular stage the negative field on p-side becomes large enough to prevent the flow of electrons from 7-side. Positive charge on n-side becomes large enough to prevent the movement of holes from the p-side. The charge distribution is as shown in Fig. 2.15 (b). The charge density far away from the junction is zero, since before all the holes from p-side move to n-side, the barrier potential is developed. Acceptor atoms near the junction have lost the holes. But for this they would have been electrically neutral. Now these holes have combined with free electrons and disappeared leaving the acceptor atom negative. Donor atoms on n-side have lost free electrons. These free electrons have combined with holes and disappeared. So the region near the junction is depleted of mobile charges. This is called depletion region, space charge region or transition region. ‘The thickness of this region will be of the order of few microns 1 micron = 10 m= 10m. The electric field intensity near the junction is shown in Fig. 2.15 (¢ ). This curve is the integral of the density function p. The electro static potential variation in the depletion region is ‘dv shown in Fig. 2.15 (d).€= ~ iF This variation constitutes a potential energy barrier against further diffusion of holes across the barrier. When the diode is open circuited, that is not connected in any circuit, the hole current must be zero. Because of the concentration gradient, holes from the p-side move towards n-side. So, all the holes from p-side should move towards n-side. This should result in large hole current flowing even when diode is not connected in the circuit. But this will not happen. So to counteract the diffusion current, concentration gradient should be nullified by drift current due to potential barrier. Because of the movement of holes from p-side to n-side, that region ( p-region ) becomes negative. A potential gradient is set up across the junction such that drift current flows in opposite direction to the diffusion current. So the net hole current is zero when the diode is open circuited. The potential which exists to cause drift is called contact potential or diffusion potential. Its magnitude is a few tenths of a volt ( 0.01V ). 2.16.2 p-m JUNCTION AS A DIODE The p-n Junction shown here forms a semiconductor device called DIODE. Its symbol is A —-piK. A is anode. K is the cathode. It has two leads or electrodes and hence the name Diode. If the anode is connected to positive voltage terminal of a battery with respect to cathode, it is called Forward Bias, ( Fig. 2.16 (a) ). If the anode is connected to negative voltage terminal of a battery with respect to cathode, it is called Reverse Bias, ( Fig. 2.16 (b) ). Amik A K A ee n T > Pts, dy 1 —I\ += VE +1 (a) Forward bias (b ) Reverse bias (¢) p-n junction forward bias Fig 2.16 76 77 98 Electronic Devices and Circuits 2.16.3 Onumic CONTACT In the above circuits, external battery is connected to the diode. But directly external supply cannot be given to a semiconductor. So metal contacts are to be provided for p-region and n-region. A Metal-Semiconductor Junction is introduced on both sides of p-n junction. So these must be contact potentials across the metal-semiconductor junctions. But this is minimized by fabrication techniques and the contact resistance is almost zero. Such a contact is called ohmic contact. So the entire voltage appears across the junction of the diode. K — Pe ohmic contacts 4, TE Fig 2.17 Ohmic contacts. 2.17. THE P-N JUNCTION DIODE IN REVERSE BIAS Because of the battery connected as shown, holes in ( Fig. 2.16 (b)) p-npe and electrons in n- type will move away from the junction. As the holes near the junction in p-region they will move away from the junction and negative charge spreads towards the left of the junction. Positive charge density spreads towards right. But this process cannot continue indefinitely, because to have continuous flow of holes from right to left, the holes must come from the n-side. But n-side has few holes. So very less current results. But some electron hole pairs are generated because of thermal agitation. The newly generated holes on the 7-side will move towards junction. Electrons created on the p-side will move towards the junction. So there results some small current called Reverse Saturation Current. \t is denoted by 1,. 1, will increase with the temperature. So the reverse resistance or back resistance decreases with temperature. 1, is of the order of a few HA. The reverse resistance of a diode will be of the order of MQ For ideal diode, reverse resistance is 0. ihe same thing can be explained in a different way. When the diode is open circuited, there exists a barrier potential. If the diode is reverse biased, the barrier potential height increases by a magnitude depending upon the reverse bias voltage. So the flow of holes from p-side to n-side and electrons from n-side to p-side is restricted. But this barrier doesn’t apply to the minority carriers on the p-sides and n-sides. The flow of the minority carriers across the junction results in some current. 2.18 THE P-N JUNCTION DIODE IN FORWARD BIAS When a diode is forward biased, the potential barrier that exists when the diode is open circuited, is reduced. Majority carriers from p-side and n-side flow across the junction. So a large current results. For ideal diode, the forward resistance R, = 0. The forward current Ips will be of the order of mA ( milli-amperes ). ‘ Junction Diede Characteristics 101 Ec and Egy — Ep = % (Eg) - Ey = (2) -., ae (2) Adding (1) and (2), (Fy — Eyp) + ( Fog — Ep) = Eg - Ey - Ey or (E, + Ey) = Eg—( Eo ~ Ep) — (Ex Ey) (E, +E, )=Eo [Xe] (Egy - Ey) = KT In, —& ‘ "Np Np = Donor Atom Concentration N,/m?. Na = Acceptor Atom Concentration N,/m?. t Nv (Ep ~ Ey) = KT ne) NoNy E,=KT nf m2 ) ar [n( }-m(}-o(85) Ne-Ny Np Na N,.N, oy = KS KE Ae Eo Kr n *Ne Ny KT In mo} (2.32) The energy is expressed in electron volts eV. K is Botlzman’s Constant in eV /°K = 8.62 x 10 eV /°K Therefore, E, is in eV and Vj is the contact difference potential in volts V, is numerically equal to E,. In the case of n-type semiconductors, n, = Np. ( Subscript ‘n’ indicates electron concentration in m-type semiconductor ) n= A, * Py = Np * Pa Np n, = Intrinsic Concentration n, = Electron Concentration in p-type semiconductor n, = Electron Concentration in n-t,pe semiconductor Pp = Hole Concentration in p-type semiconductor Pp, = Hole Concentration in n-type semiconductor 2 2 = Fi = Fi P,= and Np= "Np Ph 2 n2 n= SL and Nye POONA ry n2=n, x Pp, 80 102 Electronic Devices and Circuits Substituting all these value in E) = KT In roy \ Pn oKT In -xr (2) ° Nno ( Pon E)=KT In|) |= KT In po} \Pno Taking reasonable values of ny = 10! / om3 Ngo = = 104/ room} KT = 0.026 eV wo" Ey = 0.026 x Inve =0.718 eV. 2.20 THE CURRENT COMPONENTS IN A p-n JUNCTION DIODE x=0 Distance at Fig 2.21 Current components in a p-n junction. Junction Diode Characteristics 103 When a forward bias is applied to the diode, holes are injected into the n-side and electrons to the p-side. The number of this injected carriers decreases exponentially with distance from the junction. Since the diffusion current of minority carriers is proportional to the number of carriers, the minority carriers current decreases exponentially, with distance. There are two minority currents, one due to electrons in the p-region Inp, and due to holes in the n-region Ipn. As these currents vary with distance, they are represented as Jpn(x). Electrons crossing from # to p will constitute current in the same direction as holes crossing from p ton. Therefore, the total current at the junction where x = 0 is I = Ipn(0) + Inp(0) The total current remains the same. The decrease in Jpn is compensated by increase in Inp on the p-side. Now deep into the p-region ( where x is large ) the current is because of the electric field (since bias is applied ) and it is drift current Jpp of holes. As the holes approach the junction, some of them recombine with electrons crossing the junction from 7 to p. So Ipp decreases near the junction and is just equal in magnitude to the diffusion current Inp. What remains of Ipp at the junction enters the #-side and becomes hole diffusion current Jpn in the n-region. Since holes are minority carriers in the »-side, [pn is small and as hole concentration decreases in the n-region, Jpn also exponentially decreases with distance. In a forward biased p-n junction diode, at the edge of the diode on p-side, the current is hole current ( majority carriers are holes ). This current decreases at the junction as the junction approaches and at a point away from the junction. on the n-side, hole current is practically zero. But at the other edge of the diode, on the n-side. the current is electron current since electrons are the majority carriers. Thus in a p-n junction diode, the current enters as hole current and leaves as electron current. 2.21 LAW OF THE JUNCTION Poo = Thermal Equilibrium Hole Concentration on p-side Pag = Thermal equilibrium bole concentration on n side Pro = Pho eed (1) where Vo is the Electrostatics Barrier Potential that exists on both sides of the junction. But the thermal equilibrium hole concentration on the p-side Ppo = Py CO) MY (2) where p,(0) Hole concentration on n-side near the junction v = Applied forward bias voltage. This relationship is called Boltzman’s Relationship. Equating (1) and (2). p, (0) @(Yor-V) = py x eYo/ Vs Mo or Py(9) = Pag * e” Pa(©) = Pao * oY! Yt 82 106 Electronic Devices and Circuits If V is much larger than V;, 1 can be neglected. So I increases exponentially with forward bias voltage V. In the case of reverse bias, if the reverse voltage —V >> V.. then ev" can be neglected and so reverse current is—I, and remains constant independent of V. So the characteristics are as shown in Fig. 2.22 and not like theoretical characteristics. The difference is that the practical characteristics are plotted at different scales. If plotted to the same scale, (reverse and forward) they may be similar to the theoretical curves. Another point is, in deriving the equations the breakdown mechanism is not considered. As V increases Avalanche multiplication sets in. So the actual current is more than the theoretical current. Iqma) V, (Volts ) LUA) Fig 2.22. V-1 Characteristics of p-n junction diode. Cur IN VOLTAGE V, In the case of Silicon and Germanium, diodes there is a Cut In or Threshold or Off Set or Break Point Voltage, below which the current is negligible. It’s magnitude is 0.2V for Germanium and 0.6V for Silicon ( Fig. 2.23 ). Germanium XN Silicon — 0.2V 0.6V V, (Volts ) 85 Fig 2.23 Forward characteristics of a diode. Junction Diode Characteristics 107 2.23.1 DIODE RESISTANCE Vv The static resistance (R) of a diode is defined as the ratio of T of the diode. Static resistance varies widely with V and I. The dynamic resistance or incremental resistance is defined as the dv reciprocal of the slope of the Volt-Ampere Characteristic a: This is also not a constant but depends upon V and I. 2.24 TEMPERATURE DEPENDANCE OF P-N JUNCTION DIODE CHARACTERISTICS The expression for reverse saturation current Ip = Dp Da Ip = Ae cle + 2s n? 2aT3 n? = AgT? eFao'®r D, decreases with temperature. Ij a T? or Ip = KT™ e-VGonvT where V,, is the energy gap in volts. (Eg in eV) ForGermanium, 1 =1,m=2 For Silicon, n=2,m~ 1.5 Ip = KT™e-VGonvT Taking In, ( Natural Logarithms ) on both sides, ~YGo InIy = In K) +m x In (T) ~~ Vr Differentiating with respect to Temperature, Ady - ov -(- Ment) ly “aT T nV; T i; ilo 2 my, Yao “aT T nT, m 7 value is negligible 7 1 Alo Veo Ip “aT = nTVy Experimentally it is found that reverse saturation current increases ~ 7% /°C for both Silicon and Germanium or for every 10°C rise in temperature, 1, gets doubled. The reverse saturation current increases if expanded during the increasing portion 86 108 Electronic Devices and Circuits 25V+——— V, 10 +— 1A —10 pA Ik, 500 pA Fig 2.24 Reverse characteristics of a p-n junction diode. Reverse Saturation Current increases 7% /°C rise in temperature for both Silicon and Germanium. For a rise of 1°C in temperature, the new value of I, is, Iy = ff ah 9 \ 100)°° 1,07 Ip. For another degree rise in temperature, the increase is 7% of (1.07 Ip). Therefore for 10 °C rise in temperature, the increase is (1.07)!9 = 2. Thus for every 10 °C rise in temp I, for Silicon and Germanium gets doubled. 2.25 SPACE CHARGE OR TRANSITION CAPACITANCE Cy When a reverse bias is applied to a p-n junction diode, electrons from the p-side will move to the n-side and vice versa, When electrons cross the junction into the n-region, and hole away from the junction, negative charge is developed on the p-side and similarly positive charge on the 7- side. Before reverse bias is applied, because of concentration gradient, there is some space charge region. Its thickness increases with reverse bias. So space charge Q increases as reverse bias voltage increases. _Q But c= Vv Therefore, Incremental Capacitance, idQ c,=|(&I Tt lav where |dQ| is the magnitude of charge increase due to voltage dV. It is to be noted that there is negative charge on the p-side and positive charge on n-side. But we must consider only its magnitude. Cc r= 2 urrent [=~ 87 . Junction Diode Characteristics 11 y2e dw wee aw) (7Yy * fexNaxW € 1 = Xa exN, W € = exN,xW exA Cy = wl stn (2.35) This expression is similar to that of the Parallel Plate Capacitor. 2.26 DIFFUSION CAPACITANCE, C, When a p-n junction diode is forward biased, the junction capacitance will be much larger than the transition capacitance C,.. When the diode is forward biased, the barrier potential is reduced. Holes from p-side are injected into the n-side and electrons into the p-side. Holes which are the minority carriers in the n-side are injected into the n-side from the p-region. The concentration of holes in the n-side decrease exponentially from the junction. So we can say that a positive charge is injected into the n-side from p-side. This injected charge is proportional to the applied forward bias voltage “V’. So the rate of change of injected charge ‘Q’ with voltage ‘V’ is called the Diffusion Capacitance C,. Because of Cp total capacitance will be much larger than Cy in the case of forward bias, ( Cp is few mF (2mF.) ) Cy value will be a few pico farads. DERIVATION FoR C), Assume that, the p-side is heavily doped compared to n-side. When the diode is forward biased, the holes that are injected into the n-side are much larger than the electrons injected into the p-side. So we can say that the total diode current is mainly due to holes only. So the excess charge due to minority carriers will exist only on n-side. The total charge Q is equal to the area under the curve multiplied by the charge of electrons and the cross sectional area A of the diode. P,(0) is the Concentration of holes/em’. Area is in cm2, x in cm, p,(0) p,(0) = Hole concetration in n-region at x = 0 Carrier + . Pp = Thermal Equilibrium Concentration of Holes. Concentration| no Fig 2.26 Carrier concentration variation. 90 112 Electronic Devices and Circuits Q = JAeP, (Oe */P dx 0 W ~x/ Ly Ae Py(ay fe"? ax 0 Q fl = AeP,(0) [—Lp [0-1] ] AcL.,P,(0) dQ dp, (0) Cp= wW =AeL, x Wy We know that ex A x Dpxp, (0) xe"? Ipn(x) = + Lp AeD,p, (0) Ly p,(0) = LyxI/AeD, dp, (0) _ _t» — dl dv AeD, * dv or Ipn(0)= T= Ly - Zep; “8 evsssseeel) where g is the conductance of the diode. Substitute equation (2) in (1). vo Ly Cy= Ae x L, x ‘AeD, xg ia = xg Dy The lifetime for holes Tot is given by the eq. 2 L, D, t= p where D,, = Diffusion coefficient for holes. L, = Diffusion length for holes. Dy = cm/sec Cp=txg. 91 Junction Diode Characteristics 113 But diode resistance nVy ] where 1 = 1 for Germanium y= 2 for Silicon I a Vr _ it nVy Cp is proportional to I. In the above analysis we have assumed that the current is due to holes only. So it can be represented as Cpp. If the current due to electron is also to be considered then we get corresponding value of C,,. The total diffusion capacitance = Cpp + Cp, Its value will be around 20uF. " Cyp=txg or rxCyp=t rx Cp is called the time constant of the given diode. It is of importance in circuit applications. Its value ranges from nano-secs to hundreds of micro-seconds. Charge control description of a diode : Q=AxexL, x p,(0) Is AeD,p, (0) L Q La xexp,(0) Lp T=Qx DJL, 2Dp = But L, /Dp = t Q I== - Tt 2.27 DIODE SWITCHING TIMES When the bias of a diode is changed from forward to reverse or viceversa. the current takes definite time to reach a steady state value. 2.27.1 FORWARD RECOVERY TIME ( Trp y Suppose a voltage of 5V is being applied to the diode. Time taken by the diode to reach from 10% to the 90% of the applied voltage is called as the forward recovery time ty. But usually this is very small and so is not of much importance. This is shown in Fig, 2.26. 2.27.2 Diode REVERSE RECOVERY TIME (t,,) When a diode is forward biased, holes are injected into the ‘n’ side. The variation of concentration of holes and electrons on n-side and p-side is as shown in Fig. 2.27. P_,, is the thermal equilibrium concentration of holes on ”-side. P,, is the total concentration of holes on ‘n’-side. 92 Junction Diode Characteristics 119 2.28.1 AVALANCHE BREAKDOWN When there is no bias applied to the diode, there are certain number of thermally generated carriers. When bias is applied, electrons and holes acquire sufficient energy from the applied potential to produce new carriers by removing valence electrons from their bonds. These thermally generated carriers acquire additional energy from the applied bias. They strike the lattice and impart some energy to the valence electrons. So the valence electrons will break away from their parent atom and become free carriers. These newly generated additional carriers acquire more energy from the potential (since bias is applied). So they again strike the lattice and create more number of free electrons and holes. This process goes on as long as bias is increased and the number of free carriers gets multiplied. This is known as avalanche multiplication, Since the number of carriers is large, the current flowing through the diode which is proportional to free carriers also increases and when this current is large, avalanche breakdown will occur. 2.28.2, ZENER BREAKDOWN Now if the electric field is very strong to disrupt or break the covalent bonds, there will be sudden increase in the number of free carriers and hence large current and consequent breakdown. Even if thermally generated carriers do not have sufficient energy to break the covalent bonds, the electric field is very high, then covalent bonds are directly broken. This is Zener Breakdown, A junction having narrow depletion layer and hence high field intensity will have zener breakdown effect. (= 10°V/m). Ifthe doping concentration is high, the depletion region is narrow and will have high field intensity, to cause Zener breakdown. 2.28.3 THERMAL BREAKDOWN If a diode is biased and the bias voltage is well within the breakdown voltage at room temperature, there will be certain amount of current which is less than the breakdown current. Now keeping the bias voltage as it is, if the temperature is increased, due to the thermal energy, more number of carriers will be produced and finally breakdown will occur. This is Thermal Breakdown. In zener breakdown, the covalent bonds are ruptured. But the covalent bonds of all the atoms will not be ruptured. Only those atoms, which have weak covalent bonds such as an atom at the surface which is not surrounded on all sides by atoms will be broken. But if the field strength is not greater than the critical field, when the applied voltage is removed, normal covalent bond structure will be more or less restored. This is Avalanche Breakdown. But if the field strength is very high, so that the covalent bonds of all the atoms are broken, then normal structure will not be achieved, and there will be large number of free electrons. This is Zener Breakdown. In Avalanche Breakdown, only the excess electron, loosely bound to the parent atom will become free electron because of the transfer of energy from the electrons possessing higher energy. 2.29 ZENER DIODE This is a p-n junction device, in which zener breakdown mechanism dominates. Zener diode is always used in Reverse Bias. Its constructional features are: 1. Doping concetration is heavy on p and n regions of the diode, compared to normal p-n junction diode. 2. Due to heavy doping, depletion region width is narrow. 95 96 97