Solids and Semiconductor PDF

Solids and Semi-conductor 127 Ge Ge P Ge 60 Ge E3 Solids. It is a state of matter which has a definite shape and a definite volume. The characteristic ID properties of the solid depends upon the nature of forces acting between their constituent particles (i.e. ions, atoms or molecules). Solids are d...

Solids and Semi-conductor 127 Ge Ge P Ge 60 Ge E3 Solids. It is a state of matter which has a definite shape and a definite volume. The characteristic ID properties of the solid depends upon the nature of forces acting between their constituent particles (i.e. ions, atoms or molecules). Solids are divided into two categories. Crystalline solids Amorphous or glassy solids These solids have no definite external geometrical form. form. U (1) These solids have definite external geometrical U D YG (2) Ions, atoms or molecules of these solid are arranged in a definite fashion in all it’s three dimensions. Ions, atoms or molecules of these solids are not arranged in a definite fashion. Example : Rubber, plastic, paraffin wax, cement etc. (4) They have well defined facets or faces. They do not possess definite facets or faces. (5) They are ordered at short range as well as at long range. These may be short range order, but there is no long range order. (6) They are anisotropic, i.e. the physical properties like elastic modulii, thermal conductivity, electrical conductivity, refractive index have different values in different direction. They are isotropic i.e. physical properties are similar in all direction. (7) They have sharp melting point. They may not have a sharp melting point. (8) Bond strengths are identical throughout the Bond strengths vary. ST (3) Examples : Quartz, calacite, mica, diamond etc. 128 Solids and Semi-conductor solid. These are considered as pseudo-solids or super cooled liquids. (10) An important property of crystals is their symmetry. Amorphous solids do not have any symmetry. 60 (9) These are considered as true solids. Terms Related With Crystal Structure. E3 (1) Crystal lattice It is a geometrical arrangement of points in space where if atoms or molecules of a solid are placed, we obtain an actual crystal structure of the solid. (2) Basis ID + The atoms or molecules attached with every lattice point in a crystal structure is called the basis of crystal structure. Thus, Crystal structure Lattic e U (3) Unit cell = Basis D YG Is defined as that volume of the solid from which the entire crystal structure can be constructed by the translational repetition in three dimensions. The length of three sides of a unit cell (3D) are called primitives or lattice constant they are denoted by a, b, c b   c  b a  2D unit cell 3D unit cell U a (4) Primitive cell ST A primitive cell is a minimum volume unit cell or the simple unit cell with particles only at the corners is a primitive unit cell and other types of unit cells are called non-primitive unit cells. There is only one lattice point per primitive cell. (5) Crystallographic axis The lines drawn parallel to the lines of intersection of the faces of the unit cell are called crystallographic axis. Note :  The location of each atom or molecule in a crystal lattice may be marked as a point which is called lattice point. All the crystals on the basis of the shape of their unit cells, have been divided into seven crystal systems as shown in the following table. Solids and Semi-conductor 129 S. No. System No. of lattice s Lattice constant s Angle between lattice constants Examples Cubic 3 a =b=c  =  =  = 90o Diamond, NaCl, Li, Ag, Cu, NH4Cl, Pb etc. (ii) Tetragonal 2 a =bc  =  =  = 90o White tin, NiSO4 etc. HgCl2, KNO3, gallium etc. 60 (i) (iii) Orthorhombi c 4 a bc  =  =  = 90 (iv) Monoclinic 2 a bc  =  = 90o and   90o KclO3, FeSO4 etc. 1 a bc       90 K2Cr2O7, CuSO4 etc. Triclinic o E3 (v) o (vi) Rhombohedral or Trigonal 1 a =b=c  =  =   90 (vii ) Hexagonal 1 a =bc  =  = 90o and  = 120o Calcite, As, Sb, Bi etc. ID Zn, Cd, Ni etc. U Cubic Lattices. (1) Different symmetry o (i) Centre of symmetry : An imaginary point within the crystal that any line drawn D YG through it intersects the surface of the crystal at equal distances in both directions. (ii) Plane of symmetry : It is an imaginary plane which passes through the centre of a ST U crystal and divides it into two equal portions such that one part is exactly the mirror image of the other. Rectangular plane of symmetry A cubical crystal possesses Diagonal plane of symmetry Six diagonal plane of symmetry Three rectangular plane of symmetry (iii) Axis of symmetry : It is an imaginary straight line about which, if the crystal is rotated, it will present the same appearance more than once during the complete revolution. 130 Solids and Semi-conductor In general, if the same appearance of a crystal is repeated on rotating through an angle , around an imaginary axis, the axis is called an n-fold axis. 60 360 n o A cubical crystal possesses in all 13 axis of symmetry. Axis of four-fold symmetry (three) Because of six faces Axis of two-fold symmetry (six) Because of twelve faces E3 Axis of three-fold symmetry (four) Because of eight faces ID (iv) Elements of symmetry : The total number of planes, axes and centre of symmetry possessed by a crystal are termed as elements of symmetry. A cubic crystal possesses a total of 23 elements of symmetry. U Planes of symmetry  (3  6)  9 , Axes of symmetry  (3  4  6)  13 , Centre of symmetry = 1. D YG Total number of symmetry elements = 23 (2) Different lattice in cubic crystals There are three lattice in the cubic system. The simple cubic (sc) lattice. The body-centered cubic (bcc). SC BCC FCC U The face-centered cubic (fcc). (3) Atomic radius ST The half of the distance between two atoms in contact is defined as atomic radius. r D r r a r sc unit cell r = a/2 (4) Atoms per unit cell r a C r r A B bcc unit cell r  r r a a 3 a/4 C A B a bcc unit cell r  a/2 2 a a Solids and Semi-conductor 131 An atom located at the corner of a unit cell of a lattice is shared equally by eight other unit cells in the three dimensional lattice. Therefore, each unit cell has 1/8 th share of an atom at its each corner. Similarly, a face of the unit cell is common to the two unit cells in the lattice. Therefore, each unit cell has 1/2 share of an atom at its each face. The atom located at the 60 centre of the unit cell belongs completely to the unit cell. Let Nc, Nb and Nf be the number of atoms at the corners, centre and face of the unit cell (i) In sc lattice : N b  0, N f  0, N c  8 so N  1 (ii) In bcc lattice : N b  1, N f  0, N c  8 so N  2 (iii) In fcc lattice : N b  0, N f  6, N c  8 so N  4 2  Nc 8 ID (5) Co-ordination number Nf E3 respectively. Therefore the number of atoms per unit cell is given by N  N b  It is defined as the number of nearest neighbours that an atom has in a unit cell. It depends upon structure. U (i) Simple cubic structure : Each atom has two neighbours along X-axis, two along Y-axis and two along Z-axis so coordination number = 6. D YG (ii) Face-centred cubic structure: Every corner atom has four neighbours in each of the three planes XY, YZ, and ZX so coordination number = 12 (iii) Body-centred cubic structure: The atom of the body of the cell has eight neighbours at eight corner of the unit cell so coordination number = 8. (6) Atomic packing fraction (or packing factor or relative packing density) U The atomic packing fraction indicates how close the atoms are packed together in the given crystal structure or the ratio of the volume occupied by atoms in a unit cell in a crystal and the volume of unit cell is defined as APF. (i) For sc crystal : Volume occupied by the atom in the unit cell  4 3 a 3 r . Volume of the 3 6 ST unit cell  a 3 Thus P.F.  a 3 / 6 a 3  (ii) For bcc : P.F.   6  0.52 = 52% 3  68 % 8 (iii) For fcc : P.F.    74 % 3 2 (7) Density of unit cell Density of unit cell  Mass of the unit cell nA nA   Volume of the unit cell NV Na 3 where n = Number of atoms in unit cell (For sc lattice n = 1, for bcc lattice n = 2, for fcc lattice n = 4), A = atomic weight, N = Avogadro’s number, V = Volume of the unit cell. 132 Solids and Semi-conductor (8) Bond length The distance between two nearest atoms in a unit cell of a crystal is defined as bond length. (i) In a sc lattice : Bond length = a (ii) In a bcc lattice : Bond length  a 60 Bond length  3a (iii) In a fcc lattice : 2 2 Note :  Hexagonal closed packed (HCP) lattice –  Coordination number = 12  P.F.   2 6  Number of atoms per unit cell = 2 E3  abc c b a Types of Binding and Crystals. ID  Magnesium is a special example of HCP lattice structure. Binding Cause of binding D YG S. No. U (1) Binding : Bindings among the atoms or molecules are mainly of following types. Ionic Electrostatic force between positive and negative ions (ii) Covalent Sharing of electrons of opposite spins between two neutral atoms (iii) Hydrogen Vander Waal ST (iv) (v) Metallic Very high Electrical conductivit y Examples Very low NaCl, etc. High Semi conductor Ge, Si, diamond etc. Mutual electrostatic interaction between molecules of surface of different electron densities Low Insulator Non polar molecules or Vander Waal forces or dipole-dipole interaction Low Normally insulator Mutual interaction between electrons and ion lattice — High U (i) M.P. CsCl, LiF H2O HF, NH3etc. Cl2, I2, CO2 etc. (2) Three kinds of crystals (i) Single crystal : The crystals in which the periodicity of the pattern extends throughout the piece of the crystal are known as single crystals. Single crystals have anisotropic behaviour i.e. their physical properties (like mechanical strength, refractive index, thermal and electrical conductivity) are different along different directions. The small sized single crystals are called mono-crystals. Solids and Semi-conductor 133 (ii) Poly-crystals : A poly-crystal is the aggregate of the monocrystals whose well developed faces are joined together so that it has isotropic properties. Ceramics are the important illustrations of the poly-crystalline solids. (iii) Liquid crystals : The organic crystalline solid which on heating, to a certain temperature 60 range becomes fluid like but its molecules remain oriented in a particular directions, showing that they retain their anisotropic properties, is called liquid crystal. These crystals are used in a liquid crystal displays (L.C.D.) which are commonly used in electronic watches, clocks and microcalculators etc. E3 Energy Bands. In isolated atom the valence electrons can exist only in one of the allowed orbitals each of a Energy bands are of following types (1) Valence band ID sharply defined energy called energy levels. But when two atoms are brought nearer to each other, there are alterations in energy levels and they spread in the form of bands. U The energy band formed by a series of energy levels containing valence electrons is known as valence band. At 0 K, the electrons fills the energy levels in valence band starting from lowest one. D YG (i) This band is always fulfill by electron. (ii) This is the band of maximum energy. (iii) Electrons are not capable of gaining energy from external electric field. (iv) No flow of current due to such electrons. (v) The highest energy level which can be occupied by an electron in valence band at 0 K is called fermi level. (2) Conduction band U The higher energy level band is called the conduction band. (i) It is also called empty band of minimum energy. ST (ii) This band is partially filled by the electrons. (iii) In this band the electrons can gain energy from external electric field. (iv) The electrons in the conduction band are called the free electrons. They are able to move any where within the volume of the solid. (v) Current flows due to such electrons. (3) Forbidden energy gap (Eg) Energy gap between conduction band and valence band E g  (C.B.) min  (V.B.) max (i) No free electron present in forbidden energy gap. (ii) Width of forbidden energy gap upon the nature of substance. C.B. max. min. Eg V.B. (iii) As temperature increases (), forbidden energy gap decreases () very slightly. max. min. 134 Solids and Semi-conductor Types of Solids. On the basis of band structure of crystals, solids are divided in three categories. Properties (1) Electrical conductivity (2) Resistivity Conductors Insulators Semiconductors 102 to 108 Ʊ/m 10– 8 Ʊ/m 10– 5 to 100 Ʊ/m 10–2 to 10–8 -m 108 -m (negligible) Band structure C.B. C.B. Energy gap Zero or very small Current carries Free electrons (6) Condition of V.B. and C.B. at ordinary temperature V.B. and C.B. are completely filled or C.B. is some what empty D YG (7) Eg (less) V.B. V.B. Very large; for diamond it is 6 eV –– U (5) Temperature co- C.B. Eg (maximum) ID V.B. (4) 105 to 100 -m E3 (3) 60 S.No. For Ge Eg = 0.7 eV for Si Eg = 1.1 eV Free electrons holes and V.B. – filled completely V.B. – empty somewhat C.B. – unfilled completely C.B. – filled somewhat Positive Zero Negative Decreases — Increases Increases — Decreases efficient of resistance () (9) Effect of temperature on conductivity U (8) Effect of temperature on resistance Examples ST (11) (12) Electron density Cu, Ag, Au, Na, Pt, Hg Wood, etc. mica, diamond, glass etc. 1029/m3 plastic, — Ge, Si, Ga, As etc. Ge ~ 1019 /m3 Si ~ 1016 /m3 Holes in semiconductors At absolute zero temperature (0 K) conduction band of semiconductor is completely empty and the semiconductor behaves as an insulator. Solids and Semi-conductor 135 When temperature increases the valence electrons acquires thermal energy to jump to the conduction band (Due to the braking of covalent bond). If they jumps to C.B. they leaves behind 60 the deficiency of electrons in the valence band. This deficiency of electron is known as hole or cotter. A hole is considered as a seat of positive charge, having magnitude of charge equal to that of an electron. (1) Holes acts as virtual charge, although there is no physical charge on it. (2) Effective mass of hole is more than electron. E3 (3) Mobility of hole is less than electron. Types of Semiconductors. ID (1) Intrinsic semiconductor A pure semiconductor is called intrinsic semiconductor. It has thermally generated current carriers (i) They have four electrons in the outermost orbit of atom and atoms are held together by covalent bond (ii) Free electrons and holes both are charge carriers and n e (in C.B.)  n h (in V.B.) U (iii) The drift velocity of electrons (v e ) is greater than that of holes (v h ) D YG (iv) For them fermi energy level lies at the centre of the C.B. and V.B. (v) In pure semiconductor, impurity must be less than 1 in 10 8 parts of semiconductor. (vi) In intrinsic semiconductor n e(o )  n h(o )  ni  AT 3 / 2 e  Eg / 2 KT ; where n e(o )  Electron density in conduction band, n h(o )  Hole density in V.B., n i  Density of intrinsic carriers. (vii) Because of less number of charge carriers at room temperature, intrinsic semiconductors have low conductivity so they have no practical use. Net current and conductivity U When some potential difference is applied across a piece of intrinsic semiconductor current flows in it due to both electron and holes i.e. i = ie + ih  i  n e eAv e i  eA [ne ve  nhvh ] Hence conductivity of semiconductor   e[n e  e  n h  h ] e– hole Electric field ST i where ve = drift velocity of electron, vh = drift velocity of holes, v v E = Applied electric field  e  e  mobility of e– and  n  h  mobility of holes V E E Note :  (ni)Ge ~ 2.4  10 19 / m 3 and (ni)Si ~ 1.5  10 16 / m 3  At room temperature  Ge   Si  e  h  Conductivity of semiconductor increases with temperature because number density of charge carriers increases.  In a doped semiconductor, the number density of electrons and holes is not equal. But it can be established that n e n h  n i2 ; where ne, nh are the number density of electrons and holes respectively and ni is the number density of intrinsic curries 136 Solids and Semi-conductor (i.e. electrons or holes) in a pure semiconductor. This product is independent of donor and acceptor impurity doping. (2) Extrinsic semiconductor Intrinsic semiconducto r (i) It is also called impure semiconductor. (ii) The process of adding impurity is called Doping. Impurity + 60 Extrinsic semiconducto r (iii) Impurities are of two types : Trivalent impurity The elements whose atom has five valance impurities e.g. As, P, Sb etc. These are also called donor impurities. These impurities are also called donor impurities because they donates extra free electron. The elements whose each atom has three valance electrons are called trivalent impurities e.g. In, Ga, Al, B, etc. These impurities are also called acceptor impurities as they accept electron. ID E3 Pentavalent impurity (iv) The number of atoms of impurity element is about 1 in 10 8 atoms of the semiconductor. (v) n e  n h D YG U (vi) In these fermi level shifts towards valence or conduction energy bands. (vii) Their conductivity is high and they are practically used. (3) Types of extrinsic semiconductor N-type semiconductor (i) Intrinsic S.C. U Ge ST Ge Pentavale nt impurity + P P-type semiconductor Intrinsic S.C. Vacancy Ge Free electron Ge Ge Ge N-types S.C. (ii) Majority charge carriers – electrons Minority charge carriers – holes B Ge Ge P-types S.C. Majority charge carriers – holes Minority charge carriers – electrons (iii) ne >> nh; ie >> ih nh >> ne; ih >> ie (iv) Conductivity   ne e e Conductivity   nh h e (iv) N-type semiconductor is neutral (not negatively charged) Trivalent impurity + electrically (v) Impurity is called Donar impurity because one impurity atom generate one e–. P-type semiconductor is also neutral (not positively charged) electrically Impurity is called Acceptor impurity. Solids and Semi-conductor 137 (vi) Donor energy level lies just below the conduction band. C.B. Acceptor energy level lies just above the valence band. C.B. Acceptor energy level 60 Donor energy level V.B. E3 V.B. P-N Junction Diode. When a P-type semiconductor is suitably joined to an N-type semiconductor, then resulting P-N Junction ID arrangement is called P-N junction or P-N junction diode P N Anode Cathod e D YG (1) Depletion region N U P On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons from N-region diffuse through the junction into P-region and the hole from P region diffuse into N-region. Due to diffusion, neutrality of both N and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions appears near the junction in N-crystal. This layer is called depletion layer U (i) The thickness of depletion layer is 1 micron = 10–6 m. 1 Dopping ST (ii) Width of depletion layer  – + VB (iii) Depletion is directly proportional to temperature. (iv) The P-N junction diode is equivalent to capacitor in which the depletion layer acts as a dielectric. P N Depletion layer (2) Potential barrier The potential difference created across the P-N junction due to the diffusion of electron and holes is called potential barrier. For Ge VB  0.3 V and for silicon VB  0.7 V On the average the potential barrier in P-N junction is ~ 0.5 V and the width of depletion region ~ 10–6. 138 Solids and Semi-conductor So the barrier electric field E  V 0.5   5  10 5 V / m d 10 6 Some important graphs Potential Charge density N P Electri c field N 60 P N P +ve –ve distance Distance E3 Distance (3) Diffusion and drift current ID Because of concentration difference holes/electron try to diffuse from their side to other side. Only these holes/electrons crosses the junction, having high kinetic energy. This diffusion results is an electric current from the P-side to the N-side known as diffusion current (idf) idf  idr so inet  0 D YG Note :  In steady state U As electron hole pair (because of thermal collisions) are continuously created in the depletion region. These is a regular flow of electrons towards the N-side and of holes towards the P-side. This makes a current from the N-side to the P-side. This current is called the drift current (idr).  When no external source is connected, diode is called unbiased. (4) Biasing Means the way of connecting emf source to P-N junction diode Reverse biasing (i) Positive terminal of the battery is connected to the P-crystal and negative E terminal of the battery – + is connected to NEb crystal (i) Positive terminal of the battery is connected to the N-crystal and negative E terminal of the battery is connected to P- ST U Forward biasing P N + – + crystal P – Eb N – + (ii) Width of depletion layer decreases (ii) Width of depletion layer increases (iii) RForward  10 - 25 (iii) RReverse  105 (iv) Forward bias opposes the potential (iv) Reverse bias supports the potential barrier and for V > VB a forward current is set barrier and no current flows across the Solids and Semi-conductor 139 up across the junction. junction due to the diffusion of the majority carriers. (A very small reverse currents may exist in the circuit due to the drifting of minority carriers across the junction) which the current starts to increase. For Ge it is 0.3 V and for Si it is 0.7 V. which break down of semiconductor occurs. For Ge it is 25 V and for Si it is 35 V. (vi) (vi) Reverse voltage E3 df – diffusion dr – drift N P Break down voltage Idf Idr Inet Reverse current ID Knee voltage Forward voltage 60 (v) Break down voltage : Reverse voltage at Forward current in mA (v) Cut-in (Knee) voltage : The voltage at Idf Idr Inet Reverse Breakdown and Special Purpose Diodes. U (1) Zener breakdown D YG When reverse bias is increased the electric field at the junction also increases. At some stage the electric field becomes so high that it breaks the covalent bonds creating electron, hole pairs. Thus a large number of carriers are generated. This causes a large current to flow. This mechanism is known as Zener breakdown. (2) Avalanche breakdown U At high reverse voltage, due to high electric field, the miniority charge carriers, while crossing the junction acquires very high velocities. These by collision breaks down the covalent bonds, generating more carriers. A chain reaction is established, giving rise to high current. This mechanism is called avalanche breakdown. ST (3) Special purpose diodes Zener diode Light emitting Photo diode Solar cells diode (LED) It is based on the photovoltaic effect. N It is a highly doped p- Specially designed In diodes One of the semiconductor region n junction which is not damaged by high diodes, which give out light radiations when electron and hole pairs are created by is made so thin that the light incident on reverse current. The breakdown voltage is made very sharp. In forward biases. LED’S are made of GaAsp, Gap etc. junction photoelectric effect. That is the covalent bonds are it reaches the p-n junction and gets absorbed. It converts these 140 Solids and Semi-conductor broken by the EM radiations absorbed by the electron in the V.B. These are used for detecting light signals. P-N Junction Diode as a Rectifier. + Input ac signal + U During positive half cycle Diode forward biased Output signal obtained During negative half cycle Diode reverse biased Output signal not obtained ST O/P (dc) D2 + + – D YG + D1 RL ID + – Output dc signal output dc RL U Input ac signal Full wave rectifier E3 Half wave rectifier Input ac solar energy into electrical energy. 60 the forward bias, the zener diode acts as ordinary diode. It can be used as voltage regulator – Output dc signal D1 D2 – D1 D2 Fluctuating dc During positive half cycle Diode : D1 forward biased D2 reverse biased Output signal obtained due to D1 only During negative half cycle Diode : D1 reverse biased D2 forward biased Filter Output signal obtained due to D2 only Note :  Fluctuating dc constant dc. Transistor. A junction transistor is formed by sandwiching a thin layer of P-type semiconductor between two N-type semiconductors or by sandwiching a thin layer of n-type semiconductor between two P-type semiconductor. Solids and Semi-conductor 141 P N C P E B P C B E C E C E – Emitter (emits majority charge carriers) C – Collects majority charge carriers B – Base (provide proper interaction between E and C) B E3 B Note :  In N 60 N E normal operation base-emitter is forward biased and collector base junction is reverse biased. ID (1) Working of Transistor : In both transistor emitter - base junction is forward biased and collector – base junction is reverse biased. N P PNP – transistor U NPN – transistor N Ie D YG Ib Ib + – A mA VEB mA VCB Ie Ic N U ST VCB N Ib – Ib – mA + A mA + VEB VCB Ie Ic Ib Ib VEB P Ie VEB VCB 5% emitter electron combine with the holes in the base region resulting in small base current. Remaining 95% electrons enter the collector region. 5% emitter holes combine with the electrons in the base region resulting in small base current. Remaining 95% holes enter the collector region. Ie > Ic , and Ic = Ib + Ic Ie > Ic , and Ic = Ib + Ic Note :  In a transistor circuit the reverse bias is high as compared to the forward bias. So that it may exert a large attractive force on the charge carriers to enter the collector region. 142 Solids and Semi-conductor (2) Characteristics of transistors : A transistor can be connected in a circuit in the following three different configurations. (i) Common base (CB) (ii) Common emitter (CE)(iii) Common collector (CC) 60 (i) CB characteristics : The graphs between voltages and currents when base of a transistor is common to input and output circuits are known as CB characteristic of a transistor. VCB = – 10 V Ic E C Ie (mA) + B – RL Ib VCB = output – VEB = input VCB = 0 VCB = – 20 V + VCB (in volt) ID VEB (in volt) Ie = 15 mA Ie = 10 mA Ie = 5 mA Ie = 0 mA Ic (mA) E3 PNP Ie (ii) CE characteristics : The graphs between voltages an d currents when emitter of a PNP – + C Ic D YG Ie U transistor is common to input and output circuits are known as CE characteristics of a transistor. B Ib ( A) E Ie RL VCE = – 0.2 V Ib = 30 mA Ib = 20 mA Ib = 10 mA Ib = 0 mA Ic (mA) VCE = output – VEB = input VCE = – 0.1 V + VEB (in volt) VCE (in volt) Knee voltage U (3) Transistor as an amplifier : A device which increases the amplitude of the input signal ST is called amplifier. Amplifie r Input signal Output amplified signal The transistor can be used as an amplifier in the following three configuration (i) CB amplifier (ii) CE amplifier (iii) CC amplifier PNP Ib VEB – VC RL VCB – + CB amplifier Ic C B C B + Ie Output signal E Inpu ~ t sign al PNP Ic Ie E Inpu ~ t sign al Ie RL – VEB – + Output signal VCE + CE amplifier Solids and Semi-conductor 143 60 (4) Parameters of CE/CB amplifiers Transistor as C.E. amplifier Transistor as C.B. amplifier (i) Current gain () Small change in collector current (ic ) ; Small change in collector current (ie )  i  (a)  ac   c   ib  VB (constant) Collector current( ic ) Emitter current (ie ) (b)  dc  valve of dc lies between 0.95 to 0.99 (ii) Voltage gain valve of ac lies between 15 and 20 (ii) Voltage gain Change in output voltage( Vo ) Change in input voltage (Vi ) U Av  ic ib ID (b)  dc (or  )  VCE = constant E3 (a)  ac  (i) Current gain () Av  Vo   ac  Resistance gain Vi D YG  Av = ac  Resistance gain (iii) Power gain  Change in output power( Po ) Change in input power( Pc ) (iii) Power gain  2  Power gain   ac  Resistance gain Note :  Trans conductance (gm) : The ratio of the change in collector current to the change in emitter base voltage is called ic gm  trans conductance. i.e.. VEB U ST (5) Relation between  and  : Also g m    1  Po 2   ac  Resistance gain Pi or   AV ; RL = Load resistance RL  1  (6) Comparison between CB, CE and CC amplifier S.No. Characteristic (i) Input resistance (Ri) (ii) Output resistance Amplifier CB CE CC  50 to 200  low  1 to 2 k medium  150 – 800 k high  50 k medium  k low 20 – 200 high 20 – 200 high  1 – 2 k high (Ro) (iii) Current gain 0.8 – 0.9 low 144 Solids and Semi-conductor (iv) Voltage gain Medium High Low (v) Power gain Medium High Low (vi) Phase Zero 180o Zero current Power difference (vii) and Used as amplifier for Example: 1 What is the coordination number of sodium ions in the case of sodium chloride structure[CBSE 1988 (a) 6 (b) 8 (d) 12 In NaCl crystal Na+ ion is surrounded by 6 Cl  ion, therefore coordination number of Na+ is 6. Example: 2 (c) 4 ID Solution : (a) Voltage E3 Example s 60 between input output voltages A Ge specimen is doped with Al. The concentration of acceptor atoms is ~1021 atoms/m3. U Given that the intrinsic concentration of electron hole pairs is ~ 10 19 / m 3 , the concentration of electrons in the specimen is (b) 10 15 / m 3 D YG (a) 10 17 / m 3 (c) 10 4 / m 3 [AIIMS 2004] (d) 10 2 / m 3 Solution : (a) ni2  n h n e  (10 19 )2  10 21  n e  n e  10 17 / m 3. Example: 3 A silicon specimen is made into a P-type semi-conductor by doping, on an average, one Indium atom per 5  10 7 silicon atoms. If the number density of atoms in the silicon specimen is 5  10 28 atoms/m3, then the number of acceptor atoms in silicon will be (a) 2.5  10 30 atoms/cm3 (b) 1.0  10 13 atoms/cm3 (d) 2.5  10 36 atoms/cm3 Number density of atoms in silicon specimen = 5  1028 atom/m3 = 5  1022 atom/cm3 U Solution : (c) (c) 1.0  10 15 atoms/cm3 Since one atom of indium is doped in 5  107 Si atom. So number of indium atoms doped per ST cm-3 of silicon. n Example: 4 5  10 22 5  10 7  1  10 15 atom / cm 3. A P-type semiconductor has acceptor levels 57 meV above the valence band. The maximum wavelength of light required to create a hole is (Planck’s constant h = 6. 6  10 34 J-s) (a) 57 Å hc   (b) 57  10 3 Å (c) 217100 Å (d) 11.61  10 33 Å 6. 6  10 34  3  10 8 hc  = 217100 Å. E 57  10  3  1.6  10 19 Solution : (c) E Example: 5 A potential barrier of 0.50V exists across a P-N junction. If the depletion region is  5. 0  10 7 m wide, the intensity of the electric field in this region is Solids and Semi-conductor 145 (a) 1.0  10 6 V / m Example: 6 E (d) 2.0  10 6 V / m (c) 2.0  10 5 V / m V 0.50 = 1  106 V/m.  d 5  10 7 A 2V battery is connected across the points A and B as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery when its positive terminal is connected to A is 60 Solution : (a) (b) 1.0  10 5 V / m 10  (a) 0.2 A 10  E3 (b) 0.4 A (c) Zero A B (d) 0.1 A i So Example: 7 Current in the circuit will be 5 40 5 (b) 50 5 (c) 10 5 (d) 20 A D YG (a) Solution : (b) V 2   0.2 A. R 10 ID Since diode in upper branch is forward biased and in lower branch is reversed biased. So V current through circuit i  ; here rd = diode resistance in forward biasing = 0 R  rd U Solution : (a) 20 30 A i A 5V 20 A The diode in lower branch is forward biased and diode in upper branch is reverse biased i 5 5  A 20  30 50 U  Example: 8 Find the magnitude of current in the following circuit ST (a) 0 (b) 1 amp –4V 3 –1 V (c) 0.1 amp (d) 0.2 amp Solution : (a) Diode is reverse biased. Therefore no current will flow through the circuit. Example: 9 The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the R resistor R, connected in series with the diode for obtaining maximum 0.5 V current (a) 1.5  1.5 V 146 Solids and Semi-conductor (b) 5  (c) 6.67  (d) 200  The current through circuit i  P 100  10 3   0.2 A V 0.5  voltage drop across resistance = 1.5 – 0.5 = 1 V  R  For a transistor amplifier in common emitter configuration for load impedance of 1 k (hfe = 50 and hoe = 25) the current gain is (a) – 5.2 (b) – 15.7 (c) – 24.8 (d) – 48.78 E3 Example: 10 1  5 0.2 60 Solution : (b) h fe  50 Solution : (d) In common emitter configuration current gain A i  Example: 11 In the following common emitter configuration an NPN transistor with current gain  = 100 is used. The output voltage of the amplifier will be ID (a) 10 mV 1K (b) 0.1 V D YG (d) 10 V Voltage gain  Output voltage Input voltage = – 48.78. 10K Vout  Vout = Vin  Voltage gain  Vout = Vin  Current gain  Resistance gain = Vin    Example: 12 1  25  10 6  10 3 U 1mV (c) 1.0 V Solution : (c) 1  h oe R L  RL 10 = 10  3  100   1V. R BE 1 While a collector to emitter voltage is constant in a transistor, the collector current U changes by 8.2 mA when the emitter current changes by 8.3 mA. The value of forward current ratio hfe is [KCET 2002] (a) 82 Example: 13 (c) 8.2 (d) 8.3  i  8.2   82 h fe   c   8. 3  8.2 i  b  Vce ST Solution : (a) (b) 83 The transfer ratio of a transistor is 50. The input resistance of the transistor when used in the common-emitter configuration is 1 K. The peak value for an ac input voltage of 0.01 V peak is [CBSE PMT 1998] (a) 100 A (b) 0.01 mA (c) 0.25 mA (d) 500  A Vi 0.01  50   500  10 6 A  500  A Ri 1000 Solution : (d) ic  ib    Example: 14 In a common base amplifier circuit, calculate the change in base current if that in the emitter current is 2 mA and  = 0.98 (a) 0.04 mA (b) 1.96 mA (c) 0.98 mA (d) 2 mA Solids and Semi-conductor 147 Solution : (a) ic   ie  0.98  2  196 mA ST U D YG U ID E3 60  ib  ie  ic  2  1.96  0.04 mA.