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https://docs.google.com/document/d/1rvr2__r9CPDQDDvPJfnEhK3xK61OZb-N/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true UNIT 24 - SOLIDS AND SEMICONDUCTORS SOLID STATE ELECTRONICS (SEMICONDUCTORS) (A) Energy bands in solids: (i) In solids, the group of closely lying energy levels is known as energy band. (ii) In solids the energy bands are analogous to energy levels in an atom. (iii) In solids the atoms are arranged very close to each other. In these atoms there are discrete energy levels of electrons. For the formation of crystal these atoms come close together, then due to nucleus-nucleus, electron-electron and electron-nucleus interactions the discrete energy levels of atom distort and consequently each energy level spits into a large number of closely lying energy levels. (iv) The number of split energy levels is proportional to the number of atoms interacting with each other. If two atoms interact then each energy level splits into two out of which one will be somewhat above and another will be somewhat below the main energy level. In solids the number of atoms is very large (≈ 1023). Hence each energy level splits into large number of closely lying energy levels. Being very close to each other these energy levels assume the shape of a band. (v) In an energy band there are 1023 energy levels with energy difference of 10–23 ev. (vi) Curve between energy and distance i.e. U-r curve (a) When two atoms are interacting (b) When 1023 atoms are mutually interacting (vii) The are three types of energy bands in a solid viz. (a) Valence energy band (b) Conduction energy band (c) Forbidden energy gap. (viii) Difference between valence, forbidden and conduction energy bands. Valance Energy Band Forbidden Energy Band Conduction Energy Band In this band there are valence electrons. No electrons are found in this band In this band the electrons are rarely found This band may be partially or completely filled with electrons. This band is completely empty. This band is either empty or partially filled with electrons. In this band the electrons are not capable of gaining energy from external electric field. In this band the electrons can gain energy from electric field. The electrons in this band do not contribute to electric current. Electrons in this band contribute in this band contribute to electric current. In this band there are electrons of outermost orbit of atom which contribute in band formation. In this band there are electrons which are obtained on breaking the covalent bands. This is the band of maximum energy in which the electrons are always present. This is the band of minimum energy which is empty. This band can never be empty. This band can be empty. (ix) The conduction band is also known as first permitted energy band or first band. (x) Energy gap or Band gap (Eg): (a) The minimum energy which is necessary for shifting electrons from valence band to conduction band is defined as band gap (Eg) (b) The forbidden energy gap between the valence band and the conduction band is known as band gap (Eg). i.e. Eg = Ec – Ev (xi) As there are energy levels f electrons in an atom, similarly there are three specific energy bands for the electrons in the crystal formed by these atoms as shown in the figure (xii) Completely filled energy bands: The energy band, in which maximum possible number of electrons are present according to capacity is known as completely filled bank. (xiii) Partially filled energy bands: The energy band, in which number of electrons present is less than the capacity of the band, is known as partially filled energy band. (xiv) Electric conduction is possible only in those solids which have empty energy band or partially filled energy band. VARIOUS TYPES OF SOLIDS (i) On the basis of band structure of crystals, solids are divided in three categories. (a) Insulators (b) Semi-conductors (c) Conductors. (ii) Difference between Conductors, Semi-conductors and Insulators S.No. Property Conductors Semi-conductors Insulators 1. Electrical conductivity and its value Very high 10–7 mho/m Between those of conductors and insulators i.e. 10–7 mho/m to 10–13 mho/m Negligible 10–13 mho/m 2. Resistivity and its value Negligible Less than 10–5 Ω-m Between those of conductors and insulators i.e. 10–5 Ω-m to 105 Ω-m Very high more than 105 Ω-m 3. Band structure 4. Energy gap and its value Zero or very small More that in con-ductors but less than that in insu-lators e.g. in Ge, ΔEg =0.72 eV is Si, ΔEg =1.1 eV in Ga As ΔEg =1.3 eV Very large e.g. in diamond ΔEg = 7 eV 5. Current carriers and current flow Due to free electrons and very high Due to free electrons and holes more than that in insulators Due to free electrons but negligible. 6. Number of current carriers (electrons or holes) at ordinary temperature Very high very low negligible 7. Condition of valence band and conduction band at ordinary temperature The valence and conduction bands are completely filled or conduction band is some what empty (e.g. in Na) Valence band in somewhat empty and conduction band is somewhat filled Valence band is completely filled and conduction band is completely empty. 8. Behaviour at 0 K Behaves like a superconductor. Behaves like an insulator Behaves like an insulator 9. Temperature coefficient of resistance (α) Positive Negative Negative 10. Effects of temperature on conductivity Conductivity decreases Conductivity increases Conductivity increases 11. On increasing temperature the number of current carriers Decreases Increases Increases 12. On mixing impurities their resistance Increases Decreases Remains unchanged 13. Current flow in these takes place Easily Very slow Does not take place 14. Examples Cu, Ag, Au, Na, Pt, Hg etc. Ge, Si, Ga, As etc. Wood, plastic, mica, diamond, glass etc. (iii) Other properties of semiconductors: (a) Semi conducting elements are tetravalent i.e. there are four electrons in their outermost orbit. (b) Their lattice is face centered cubic (F.C.C.) (c) The number of electrons or cotters is given by i.e. on increasing temperature, the number of current carriers increases. (d) There are uncharged (iv) Holes or cotters: (a) The deficiency of electrons in covalent band formation in the valence band in defined as hole or cotter. (b) These are positively charged. The value of positive charge on them is equal to the electron charge. (c) Their effective mass is less than that of electrons. (d) In an external electric field, holes move in a direction opposite to that of electrons i.e. they move from positive to negative terminal. (e) They contribute to current flow. (f) Holes are produced when covalent bonds in valence band break. TYPES OF SEMICONDUCTORS AND DIFFERENCE BETWEEN THEM (i) The semiconductors are of two types. (a) Intrinsic or pure semiconductors (b) Extrinsic or dopes semiconductors (ii) Difference between intrinsic and extrinsic semiconductors: S.No. Intrinsic semiconductors Extrinsic semiconductors 1. Pure Ge or Si is known as intrinsic semiconductor The semiconductor, resulting from mixing impurity in it, is known as extrinsic semiconductors. 2. Their conductivity is low (because only one electron in 109 contribute) Their conductivity is high 3. The number of free electrons (ni in conduction band is equal to the number of holes pi in valence band.) In these 4. These are not practically used These are practically used 5. In these the energy gap is very small In these the energy gap is more than that in pure semiconductors. 6. In these the Fermi energy level lies in the middle of valence band and conduction In these the Fermi level shifts towards valence or conduction energy bands. (iii) Properties of intrinsic semiconductors: (a) At absolute zero temperature (0 K) there are no free electrons in them. (b) At room temperature, the electron-hole pair in sufficient number are produced. (c) Electric conduction takes place via both electrons and holes. (d) The drift velocities of electrons and holes are different. (e) The drift velocity of electrons (Vdn) is greater than that of holes (Vdp). (f) The total current is (g) In connecting wires the current flows only via electrons. (h) The current density is given by Where Vdn = μn = Vdp = μp = drift velocity of electrons mobility of electrons drift velocity of holes mobility of holes (i) The electric conductivity is given by (j) Mobility of electron (k) Mobility of holes (l) At room temperature because where (iv) Extrinsic semiconductors: (a) Doping: The process of mixing impurities of other elements in pure semiconductors is known as doping. (b) Extrinsic semiconductors: the semiconductors, in which trivalent and pentavalent elements are mixed as impurities, are known as extrinsic semiconductors. (c) The extrinsic semiconductors are of two types (i) N-type semiconductors (ii) P-type semiconductors. (d) Difference between N-type and P-type semiconductors S.No. N-type semiconductors P-type semiconductors 1. In these the impurity of some pentavalent element like P, As, Sb, Bi, etc. is mixed In these, the impurity of some trivalent element like b, Al, In, Ga etc. is mixed 2. 3. In these the impurity atom donates one electrons, hence these are known as donor type semiconductors In these, the impurity atom can accept one electron, hence these are known as acceptor type semiconductors. 4. In these the electrons are majority current carriers and holes are minority current carriers. (i.e. the electron density is more than hole density nn >> np) In these the holes are majority current carriers and electrons are minority current carriers i.e. np >> nn 5. In these there is majority of negative particles (electrons) and hence are known as N-type semiconductors In these there is majority of positive particles (cotters) and hence are known as P-type semiconductors. 6. In these the donor energy level is close to the conduction band and far away from valence band. In these the acceptor energy level is close to the valence band and far away from conduction band. 7. Current density Jn = nq Vdn Jp = pq Vdp 8. Electric conductivity σn = nqμn ≈ nd qμn Where nd = number of donor atoms / cm3. σp = nqμp ≈ np qμp Where np = number of acceptor atoms / cm3. 9. The Fermi energy level lies close to conduction band (i.e. the Fermi energy level lies in between the donor energy level and conduction band) The Fermi energy level lies close to the valence band (i.e. the Fermi energy level lies in between the acceptor energy level and valence band) (v) Conductivity formulae: (a) (b) (c) (d) (vi) Resistivity formulae: (a) (b) (c) (d) (e) (f) (vii) Characterizes Si and Ge at 300 K Characteristics Ge Si Energy gap 0.7 (eV) 1.1 (eV) Electron mobility (μn) 0.39 (M2V–1S–1) 0.135 (M2V–1S–1) Cotter mobility (μp) 0.19 (M2V–1S–1) 0.048 (M2V–1S–1) Intrinsic current concentration ni = 2.4 × 1019 cm–3 ni = 1.5 × 1016 cm–3 Resistivity 0.46 Ω-m 2300 Ω-m Potential barrier 0.3 V 0.7 V SEMICONDUCTOR DIODE OR P-N JUNCTION, CONDUCTION IN P-N JUNCTION, DEPLETION LAYER AND BARRIER ENERGY P-N Junction (a) The device formed by joining atomically a wafer of P-type semiconductor to the wafer of N-type semiconductor is known as P-N junction. (b) There are three processes of making junctions (i) Diffusion (ii) Alloying (iii) Growth In majority of cases P-N junction is formed by diffusion process. The impurity concentration is maximum at surface and decreases gradually inside the semiconductor. (c) Conduction of current in P-N Junction: (i) In P-N junction the majority cotters in P-region and majority electrons in N-region start diffusing due to concentration gradient and thermal disturbance towards N-region and P-region respectively and combine respectively with electrons and cotters and become neutral. (ii) In this process of neutralization there occurs deficiency of free current carriers near the junction and layers of positive ions in N-region and negative ions in P-region are formed. These ions are immobile. Due to this an imaginary battery or internal electric field is formed at the junction which is directed from N to P. (iii) Depletion layer: (a) The region on both sides of P-N junction in which there is deficiency of free current carriers, is known as the depletion layer. (b) Its thickness is of the order of 1μm (= 10–6) (c) On two sides of it, there are ions of opposite nature. i.e. donor ion (+ve) on N-side and acceptor ions (–ve) on P-side. (d) This stops the free current carriers to crossover the junction and consequently a potential barrier is formed at the junction. (e) The potential difference between the ends of this layer is defined as the contact potential or potential barrier (VB). (f) The value of VB is from 0.1 to 0.7 volt which depends on the temperature of the junction. It also depends on the nature of semiconductor and the doping concentration. For germanium and silicon its values are 0.3 V and 0.7 V respectively. (g) P-N Junction diode or semiconductor diode: (i) Symbolic representation of diode: (ii) The direction of current flow is represented by the arrow head. (iii) In equilibrium state current does not flow in the junction diode. (iv) In can be presumed to be equivalent to a condenser in which the depletion layer acts as a dielectric. (v) Potential distance curve at P-N Junction (vi) Charge density curve at P-N Junction (vii) Curve between electric field and distance near P-N junction BIASING OF JUNCTION DIODE (i) No current flows in the junction diode without an external battery. It is connected to a battery in two different ways. Hence two different bias are possible in junction diode. (a) Forward bias (b) Reverse bias (ii) Difference between forward bias and reverse bias: S.No. Forward bias Reverse bias 1. The P-region of junction diode is connected to positive terminal of battery and N-region is connected to negative terminal of battery. The P-region is connected to negative terminal and N-region is connected to positive terminal of the battery. 2. In this the width of depletion layer decreases In this the width of depletion layer increases 3. Current flows in it due to majority electrons and majority holes and hence high current (mA) flows in it. The direction of current in it is from P to N. Current flows in it due to minority electrons and minority holes and hence negligible current (in μA) flows in it. Direction of current is from N to P 4. The junction resistance is low The junction resistance is high 5. Curve between forward voltage and forward current Curve between reverse voltage and reverse current Illustration 1: When the reverse potential in a semiconductor diode are 10V and 20V, then the corresponding reverse currents are 25μ A and 50μA respectively. The reverse resistance of junction diode will be: (A) 40 (B) 4 × 105 (C) 40K (D) 4 × 10–5 Sol. (B) rr = = = 4 × 105 Illustration 2: The depletion layer in a silicon diode is 1μm wide and its knee potential is 0.6V, then the electric field in the depletion layer will be: (A) 0.6 V/m (B) 6 × 104 V/m (C) 6 × 105 V/m (D) Zero Sol: (C) E = Illustration 3: A semiconductor P-N junction is to be forward biased with a battery of e.m.f. 1.5 Volt. If a potential difference of 0.5V appears on the junction which does not depend on current and on passing 10mA current through the junction there occurs huge Joule loss, then to use the junction at 5mA current, the resistance required to be connected in its series will be: (A) 3 K (B) 300 (C) 300 K (D) 200 Sol: (D) VR = (1.5 − 0.5) = 1V = IR ∴ R = CHARACTERISTICS OF JUNCTION DIODE (i) The characteristic curves of junction diode are of two types (a) Static characteristic curves (b) Dynamic characteristic curves (ii) The static and the dynamic characteristics are also of two types (A) (a) Static forward characteristics curves (b) Static reverse characteristic curves (B) (a) Dynamic forward characteristic curves (b) Dynamic reverse characteristic curves (iii) Static forward characteristics (a) In the absence of load resistance, the curves drawn between the forward voltage (Vf) and forward current (Ιf) are known as the static forward characteristics of junction diode. (b) (c) On increasing the Vf the value of Ιf increases exponentially (d) Circuit diagram: (iv) Static reverse characteristics: (a) In the absence of load resistance, the curves drawn between the reverse voltage (Vr) and reverse current (Ιr) are known as the static reverse characteristics of junctions diode. (b) (c) (d) After the breakdown point at B, the reverse current (Ιr) does not depend on the reverse voltage (Vr) in the BC portion of curve. CONSTANTS OF JUNCTION DIODE (A) (i) Static forward and reverse resistances (ii) Dynamic forward and reverse resistances (B) Static forward resistance (Rf): (i) The ratio of the forward voltage (Vf) and forward current (Ιf) at any point on the static forward characteristic is defined as static forward resistance of junction diode. (ii) Its value is of the order of 102 Ω. (C) Static reverse resistance (Rr): (i) The ratio of reverse voltage (Vr) and reverse current (Ιr) at any point on static reverse characteristic is defined as the static reverse resistance of junction diode. (ii) Its value of is of the order of 106 (iii) (D) Dynamic forward resistance (Vr): (i) The ratio of small change in forward voltage to the corresponding small change in forwards current on static forward characteristic is defined as the dynamic forward resistance of junction diode (rf) (ii) (iii) (E) Dynamic reverse resistance (rr): (i) The ratio of the small change in reverse voltage to the corresponding small change in reverse current on the static reverse characteristics is defined as the dynamic reverse resistance of junction diode. (ii) (iii) Illustration 4: The value of current in the adjoining diagram will be: (A) 0 amp (B) 10−2 amp (C) 102 amp (D) 10−3 amp Sol. (B) If = = 10−2A Illustration 5: The value of current in the following diagram will be: (A) 0.10A (B) 10−2 A (C) 1 A (D) 0 A Sol. (D) In reverse bies Ir = 0 Illustration 6: The saturation current of a P-N junction germanium at 27oC is 10−5 amp. The potential required to be applied in order to obtain a current of 250mA in forward bias will be: (A) 0.57 V (B) 0.48 V (C) 0.26 V (D) 0.63 V Sol. (C) If = IS or = or ∴ V = = 0.26 V Illustration 7:The junction diode in the following circuit requires a minimum current of 1mA to be above the knee point (0.7V) of its I-V characteristic curve. The voltage across the diode is independent of current above the knee point, If VB = 5V, then the maximum value of R so that the voltage is above the knee point will be: (A) 4.3k (B) 860k (C) 4.3 (D) 860 Sol% (B) VB = Vknee + IR or 5 = 0.7 + 10–3 R or R = 4.3K ZENER BREAKDOWN, AVALANCHE BREAKDOWN AND ZENER DIODE: S.No. Avalanche breakdown Zener breakdown 1. The doping in the formation of P-N Junction is low The doping in the formation of P-N junction is high 2. The covalent bonds break as a result of collision of electrons and holes with the valence electrons In this the covalent bonds break spontaneously. 3. Higher reverse potential is required for breakdown. Low reverse potential is required for breakdown 4. In this the thermally generated electrons due to electric field ionize other atoms and release electrons. In this the covalent bonds near the junction break due to high reverse potential ~20 V and consequently electrons become free. (ii) Zener diode: (a) The junction diode made of Si or Ge, whose reverse resistance is very high, is known as Zener diode. (b) It works at Zener voltage (Vz) i.e. the voltage at which breakdown starts. Zener voltage (Vz): The voltage at which breakdown starts in Zener diode and consequently the reverse current in the circuit abruptly increases, is defined as Zener voltage. (c) It is used in power supplies as a voltage regulator. (d) Symbolic representation of Zener diode. SALIENT FEATURES RELATING TO JUNCTION DIODE (i) In junction diode the current flow is unidirectional as in vacuum diode. (ii) Current flows in the semiconductor diode when it is forward biased. (iii) Its P-part behaves like a plate and N-apart behaves like a cathode. (iv) Relation between forward current and saturation current Where Ιf and Ιs are forward and saturation currents respectively, K = Boltzmann constant, T = absolute temperature V = potential difference (a) In forward bias (b) In reverse bias (v) The velocity gained by the charge carriers in an electric field of unit intensity, is defined as their mobility (vi) Forward and reverse characteristic curves of Si and Ge diodes: Knee Point: That point on the forward characteristics of junction diode after which the curve becomes linear, is known as the knee point. In the diagram it is represented by the point A. Knee voltage: The potential at knee point A is known as the knee potential or forward potential at which the forward current abruptly increases is known as the knee potential. (a) This potential does not depend on the current. (b) For Si its value is 0.7 V. (vii) Greater the value of ΔEg, stronger will be the binding of valence electrons to the nucleus. USES OF JUNCTION DIODE (i) Rectifier (ii) Off switch (iii) Condenser DIFFERENCE BETWEEN VACUUM TUBE DEVICES AND SEMICONDUCTOR DEVICES S.No. Vacuum tube devices semiconductor devices 1. These are voltage controlled devices. These are current driven devices 2. These work at high voltage These work at low voltage. 3. For these a filament battery is required. For these no filament batteries required 4. These are not temperature sensitive devices These are temperature sensitive devices 5. Their life is less and are more expensive Their life is longer and are cheap 6. Their size is big Their size is small. 7. Their efficiency is more Their efficiency is less 8. Power consumption is maximum Power consumption is minimum 9. These can not be used integrated circuit (ΙC's) These can be used as integrated circuits 10. Electric conduction is only via electrons. Electric conduction takes place both by electrons and cotters. VARIOUS TYPES OF P-N JUNCTIONS S.No. P-N Device Biasing Principle Uses Explanation 1. Light Emitting Diode (LED) Forward Production of light from electric current Burglar alarms, calculators, pilot lamps, telephone, digital watch and in switch boards In Ga, As, Electromagnetic radiations are emitted on account of transitions of electron from conduction band to valance band. 2. Photodiode Reverse Electric conduction from light In sound films, computers, tape, in reading computer cards and in light driven switches. The covalent bonds in semiconductors break due to electromagnetic radiations and more electrons become free and conductivity increases. 3. Zener diode Reverse Current is controlled In voltage regulation Voltage across it remains constant 4. Solar cell No biasing Production of potential difference by sun light For generating electrical energy in cooking food etc. Due to nuclear fusion process sun is constantly emitting light and heat energy. The upper surface of P-N junction is thin in this diode. Other salient features (a) The value of electric field across the P-N junction is 105 V/m (b) (c) The values of contact potential for Si and Ge are 0.7 V and 0.3 V respectively. SEMICONDUCTOR DIODE AS RECTIFIER (i) Rectification: The process in which an alternating current is converted into direct current, is defined as rectification. (ii) Rectifier: The device employing diode, used to convert an alternating current into direct current, is known as rectifier. (iii) The rectifiers are of two types: (a) Half wave rectifier (b) Full wave rectifier Half wave rectifier: The rectifier, in which only alternate half cycles of applied alternating signal are converted into direct current, is known as half wave rectifier. Full wave rectifier: The rectifier is which the whole cycle of applied alternating signal is converted into direct current, is known as full wave rectifier. (iv) Difference between half wave rectifier and full wave rectifier No. Half-Wave Rectifier Full Wave Rectifier centre taped Full wave Bridge Rectifier 1. 2. In this, one diode or one semiconductor diode is used In this, two diodes or one double diode or two junction diodes are used In this four junction diodes from the bridge circuit. 3. Ordinary transformer is used Centre tap transformer is used Transformer is not required 4. It converts half cycle of applied A.C. signal into D.C. signal It converts the whole cycle of applied A.C. signal into D.C. signal It converts the whole cycle of applied A.C. signal into D.C. signal 5. Input and output curves Input and output curves Input and output curves 6. The value of 7. 8. The value of ripple factor is % The value of r in it is 48.2% The value of r in it is 48.2% 9. Efficiency (η) (a) (b) When rp = RL then η = 20.3% (c) When rp << RL then η = 40.6% (a) (b) When rp = RL then η = 40.6% (c) When then η = 81.2% Its efficiency is 81.2% 10. Peak inverse voltage PΙV = E0 PΙV = 2E0 PΙV = 2E0 11. Form factor F = 1.11 F = 1.11 12. The ripple frequency is equal to the frequency of applied e.m.f. The ripple frequency is twice that of the applied e.m.f. The ripple frequency is twice that of the applied e.m.f. 13. Curve between the output voltage from filter circuit and time Curve Curve 14. The value of D.C. component in output voltage is less than the A.C. The value of D.C. component in output voltage is more than that of A.C. The value of D.,C. component in output voltage is more than that of A.C. 15. The value of peak inverse voltage (PΙV) is E0 The value of PΙV is E0 The value of PΙV is E0 16. The value of peak load current is The value of PLC is The value of PLC is (v) Similarities between half wave and full wave rectifiers (a) The alternating input signal to be rectified is connected to the primary of transformer. (b) The output voltage is obtained across the ends of load resistance. (c) The output voltage is unidirectional but it is not constant rather pulsating. (d) The output voltage is the mixture of alternating and direct voltages. (e) In output, the direct components are more than the alternating components. (f) Diode conducts only when the plate is positive with respect to the cathode. semiconductor diode conducts only when it is forward biased. (g) When the plate of diode is negative with respect to the cathode than it does not conduct. (vi) Definitions: (a) Efficiency of rectifier (i) (ii) (b) Ripple Factor (r): (i) (ii) (c) Form factor (i) (ii) (iii) For full wave rectifier (iv) For half wave rectifier TRANSISTOR (i) That current driven device, which is formed by three doped semiconductor regions, is known as transistor. (ii) That current driven device, in which the emitter current controls the collector current, is known as transistor. (iii) There are three semiconductor regions in a transistor viz Emitter (E), Base (B) and collector (C). (iv) Function of emitter: To send electrons or cotters into the base Function of base: To send electrons or cotters received from the emitter into the collector region. Function of collector: To collect electrons or cotters from the base region. (v) The distance between E and B in a transistor is less than that between B and C and the collector is marked with a dot (.) (vi) Transistors are of two types: (i) PNP transistor (ii) NPN transistor PNP-transistor: (i) Symbolic representation: (ii) In this conventional current flows from emitter (E) to base (B) hence the arrow head on emitter is from E to B. (iii) Sketch diagram (iv) Working of PNP transistor (a) The emitter-base junction is forward biased while base-collector junction is reverse biased. (b) A large number of holes enter from emitter to base and at the same time a very small number of electrons enter from the base to the emitter. (c) The electrons in the emitter region recombine with an equal number holes and neutralise them. (d) The loss of total number of holes in the emitter is compensated by the flow of an equal number of electrons from the emitter to the positive terminal of battery. (e) These electrons are released by breaking of covalent bonds among the crystal atoms in the emitter and an equal number holes is again created. (f) Thus in PNP transistor emitter current is mainly due to the flow of holes, but in eternal circuit it is due to flow of electron from emitter to the positive terminal of the battery. (g) The base is very thin and is lightly doped. Therefore only a few holes (~ 1%) combine with electrons in base. Hence the base current ΙB is very small. (h) Nearly 99% of the holes coming from the emitter are collected by the collector. (i) For each hole reaching the collector, an electron is released from the negative terminal of collector base battery to neutralise the hole. (j) The relation between three currents is as under (k) The input impedance is low and output impedance is high. The output voltage required to be applied is more than the input voltage. (l) The functions of E, B and C are to send cotters into base region, to send these cotters into collector region and to collect the cotters received from base region respectively. NPN-transistor: (i) Symbolic representation (ii) In this conventional current flows from base towards emitter, hence the arrow head on emitter is directed from B to E. (iii) Sketch diagram (iv) Working of NPN transistor (a) The emitter-base junction is forward biased whereas the collector-base junction is reverse biased. (b) The majority electrons in the emitter are pushed into the base. (c) The base is thin and is lightly doped. Therefore a very small fraction (say 1%) of incoming electrons combine with the holes. Hence base current is very small. (d) The majority of electrons are rushing towards the collector under the electrostatic influence of C-B battery. (e) The electrons collected by the collector move towards the positive terminal of C-B battery. (f) The deficiency of these electron is compensated by the electrons released from the negative terminal of E-B battery. (g) Thus in NPN transistors current is carried by electron both in the external circuit as well as inside the transistor. (h) The relation between these current is given by (i) The input impedance is low and output impedance is high. The output voltage required to be applied is more than the input voltage. Illustration 8: For a common emitter connection the values of constant collector and base current are 5mA and 50 μA respectively. The current gain will be: (A) 10 (B) 20 (C) 40 (D) 100 Sol. (D) β = = CHARACTERISTICS OF TRANSISTOR The study of variation in current with respect to voltage in a transistor is called its characteristic. For each configuration of transistor, there are two types of characteristics: (i) Input characteristics (ii) Output characteristics (A) Common base configuration (i) Circuit diagram (ii) Input characteristics (a) Input characteristics are obtained by plotting the emitter current ΙE versus emitter-base voltage VEB at constant collector base potential VCB. (b) ΙE is almost independent of VCB. (c) Due to very low input impedance, ΙE increases rapidly with small increase in VEB. (d) ΙE is finite at finite value of VCB even when VEB is zero. To reduce ΙE to zero, the emitter must be reverse biased. (iii) Output characteristics: (a) The output characteristics are obtained by plotting the collector current (ΙC) versus collector-base voltage (VCB) at constant emitter current (ΙE). (b) ΙC varies with VCB only at very low voltage (< 1 V). Transistor is not operated in this region. (c) As VCB increases beyond 1 volt, ΙC becomes independent of VCB but depends only upon the emitter current ΙE. (d) Due to high output impedance, a very large change in VCB produces a very small change in ΙC. (e) For the region to the left of VCB = 0 and for ΙE > 0, both emitter and collector are forward biased and it is called saturation region. (f) For ΙE < 0, both emitter and collector are reverse biased and the region is called the cut-off region. (g) For central region VCB > 0, the curves are parallel and it is called active region. In this region emitter is forward biased and collector is reverse biased. (B) Common emitter configuration: In this configuration emitter is common to input and output circuits. (i) Circuit diagram (ii) Input characteristics (a) Input characteristics are obtained by plotting the base current (ΙB) versus base emitter voltage (VBE) for constant collector-emitter voltage (VCE). (b) ΙB increases with increase in VBE, but less rapidly as compared to common base configuration, indicating that input resistance of common emitter configuration is greater than that of common base configuration. (c) These characteristics resemble with those of a forward biased junction diode indicating that the base-emitter section of a transistor is essentially a junction diode. (iii) Output characteristics: (a) The output characteristics are obtained by plotting collector current ΙC versus collector-emitter voltage (VCE) at constant value of base current (ΙB). (b) ΙC increases with increase of VCE upto 1 volt and beyond 1 volt it becomes almost constant. (c) The value of VCE upto which ΙC increases is called the knee voltage. The transistor always operates above knee voltage. (d) Above knee voltage, ΙC is almost constant. (e) The region for VCE < 1 volt is called saturation region as both emitter and collector are forward biased. (f) In the region ΙB ≤ 0, both emitter and collector are reverse biased and it is called the cut-ff region. (g) The central region, where the curves are uniformly spaced and sloped, is called the active region. In this region the emitter is forward biased and the collector is reverse biased. (C) Common collector configuration: In this configuration collector is common to input and output circuits. (i) Circuit diagram (ii) Input characteristics: (a) The input characteristics are obtained by plotting the base current ΙB versus base-collector voltage (VBC) for constant emitter-collector voltage (VBC). (b) ΙB decreases with increase of VBC. These characteristics are quite different from those of common base and common emitter configurations. (c) As VBC increases, VBE decreases for constant value of VEC, thereby reducing ΙB. (iii) Output characteristics: (a) Output characteristics are obtained by plotting emitter current ΙE versus collector-emitter voltage (VCE) for constant base current (ΙB). (b) The curves are similar to those obtained in output characteristics in common emitter configuration indicating that ( ΙB is very small) (c) In this configuration the output can be obtained in either direction and hence it is used for matching the impedance for two way amplifier and switching circuits. Illustration 9:In a transistor amplifier, β = 62, RL = 5000 Ω and internal resistance of the transistor is 500 Ω. The voltage amplification of the amplifier will be: (A) 500 (B) 620 (C) 780 (D) 950 Sol. (B) Voltage amplification = Illustration 10:In the above problem, the power amplification will be: (A) 25580 (B) 33760 (C) 38440 (D) None of these Sol. (C) Power amplification = TRANSISTOR AS AN AMPLIFIER (i) Amplification: The phenomenon, in which the amplitude of input signal (Voltage, current or power) is increased, is defined as amplification. (ii) Amplifier: The device (electronic circuits), used to increase the amplitude of input signal is known as amplifier. (A) Common base amplifier: (i) NPN Transistor: (a) Circuit diagram (b) When no A.C. signal is applied, then only collector current ΙC flowing through the load resistance RL will produce a voltage drop ΙCRL across the load resistance.  net collector voltage VC = VCB – ΙCRL . . . (1) (c) Now input A.C. signal, to be amplified, is applied to the input circuit. During positive half of the input signal, the forward bias is reduced. This reduced emitter current (ΙE) and consequently ΙC is also reduced. According to Eq. (1) VC increases. (d) During negative half cycle of the input signal, the forward bias is increased So, ΙE and hence ΙC increases. According to Eq. (1), VC decreases. (e) The input and the output signals are in same phase. (f) Current amplification factor (α): (i) DC current gain (αDC): The ratio of the collector current to the emitter current at constant collector voltage is defined as D.C. current gain αDC is always less than one. (ii) AC current gain (αAC): The ratio of change in collector current to the change in emitter current at constant collector voltage is defined as AC current gain. (g) Voltage gain: (i) The ratio of the output voltage across the load resistance to the input signal voltage is defined as voltage gain. Voltage gain = (Ri is the resistance of the input circuit) (ii) Since α is approximately equal to one and RL >> Ri, hence very high voltage gain can be obtained. (h) Power gain: (i) The ratio of the output power to the input power is defined as power gain. = Current gain × voltage gain (ii) Since RL > > Ri, hence power gain is quite large. Transistor as an oscillator (i) The simplest electrical oscillating system consists of an inductance L and a capacitor C connected in parallel. (ii) Once an electrical energy is given to the circuit, this energy oscillates between capacitance (in the form of electrical energy) and inductance (in the form of magnetic energy) with a frequency (iii) The amplitude of oscillations is damped due the presence of inherent resistance in the circuit (iv) In order to obtain oscillations of constant amplitude, an arrangement of regenerative or positive feedback from an output circuit to the input circuit is made so that the circuit losses may be compensated. (v) Circuit diagram of NPN transistor in common emitter configuration as an oscillator. (vi) The tank circuit is used in emitter-base circuit of the transistor. The E-B circuit is forward biased whereas the C-B circuit is reverse biased. (vii) The coil L' in the emitter-collector circuit is inductively coupled with coil L. (viii) When the key K is closed, ΙC begins to increase. The magnetic flux linked with coil L' and hence with L also begins to increase. This supports the forward bias of B-E circuit. As a result of this ΙE increases. Consequently ΙC also continues increase till saturation. (ix) When ΙC attains saturation value, mutual inductance has no role to play. (x) When the capacitor begins to discharge through inductance L, the ΙE and hence ΙC begins to decrease. consequently, the magnetic flux linked with L' and hence with L decreases. The forward bias of E-B circuit is opposed thereby further reducing ΙE and ΙC. This process continues till ΙC becomes zero. (xi) At this stage too, the mutual inductance has once again no role to play. ASSIGNMENT 1. Zener breakdown will occur if : (A) Impurity level is low (B) Impurity level is high (C) Impurity is less in n-side (D) Impurity is less in p-side Solution : (D) 2. The correct symbol for zener diode is : (A) (B) (C) (D) Solution : (A) 3. The reason of current flow in PN junction in forward bias is : (A) Drifting of charge carriers (B) Minority charge carries (C) Diffusion of charge carries (D) all of the above Solution : (C) 4. In a half wave rectifier circuit, the input signal frequency is 50 Hz, the output frequency will be : (A) 25 Hz (B) 50 Hz (C) 200 Hz (D) 100 Hz Solution : (D) 5. The reverse bias in a junction diode is changed from 5V to 15V then the value of current changes from 38μA to 88 μA. The resistance of junction diode will be: (A) 4 × 105 (B) 3 × 105 (C) 2 × 105 (D) 106 Sol.(C) rr = = = 2 × 105 6. The inverse saturation current in a P-N junction diode at 27oC is 10−5 ampere. The value of forward current at 0.2 volt will be: (A) 2037.6 × 10−3A (B) 203.76 × 10−3A (C) 20.376 × 10−3A (D) 2.0376 × 10−3A Sol.(C) If = = = 10–5 × [2038.6 − 1] = 20.376 × 10−3A 7. The potential barrier in the depletion layer is due to (A) Ions (B) Holes (C) Electrons (D) Forbidden band Solution : (A) We konw that the recombination of free and mobile electrons and holes produces the narrow region at the junction called depletion layer. It is so named because this region is devoid of free and mobile charges carriers take electrons and holes and there being present only positive ions which are not free to move. 8. In a forward biased p-n junction, the potential barrier (A) Decreases (B) Increases (C) Remains constant (D) Becomes zero Solution : (A) We know that for a forward-biased p-n junction, the forward current rises exponentially with the applied forward voltage and get a certain potential, it is ready to break the potential barrier. Therefore in forward biased p-n junction, the potential barrier will decrease. 9. The logic behind 'NOR' gate is that it gives (A) High output when both the inputs are low (B) Low output when both the inputs are low (C) High output when both the inputs are high (D) None of these Solution : (A) We know that the 'NOR' gate is the opposite of 'OR' gate. Therefore its gives a high output only when both its inputs are low. 10. In n-type semi conductors, majority charge carriers are (A) Holes (B) Protons (C) Neutrons (D) Electrons Solution : (D) 11. In the above problem, if VB = 5V, what should be the value of R to establish a current of 5mA in the current of −5mA in the circuit? (A) 4.3k (B) 860k (C) 4.3 (D) 860 Sol. (D) VB = Vknee + I'R' or 5 = 0.7 + 5 × 10−3 R' or R' = 860 or 5 = 0.7 + 10–3 R or R = 4.3K 12. In Q. 70, what is the power dissipated in the resistance R when a current of 5mA flows in the circuit at VB = 6V- (A) 3.5W (B) 3.5 mW (C) 26.5 W (D) 26.5mW Sol. (D) 6 = 0.7 + 5 × 10−3 R' or R' = 1060 P = I2R'' = (5 × 10−3)2 × 1060 = 26.5 mW 13. In the above problem, the power dissipated in the diode will be: (A) 3.5W (B) 3.5 mW (C) 26.5 W (D) 26.5mW Sol. (B) Power dissipated in diode = VI = 0.74 × 5 × 10−3 = 3.5 mW 14. In Q. 70, if R = 1 KΩ the minimum voltage VB required to keep the diode above the knee point will be: (A) 170 V (B) 17 V (C) 1.7 V (D) 0.17 V Sol.% (C) VB = Vknee + IR = 0.7 + 103 × 10−3 = 1.7V output resistance 15. In a common emitter circuit if VCC is changed by 0.2 V, collector current changes by 4 x 10–3 mA. The output resistance will be: (A) 10 kΩ (B) 30 kΩ (C) 50 kΩ (D) 70 kΩ Sol.(C) R = 16. A transistor has a base current of 1 mA and emitter current 100 mA. The collector current will be: (A) 100 mA (B) 1 mA (C) 99 mA (D) None of these Sol. (C) IC = Ie − Ib = (100 − 1) = 99mA 17. In the above problem, the current transfer ratio will be: (A) 0.90 (B) 0.99 (C) 1.1 (D) None of these Sol. (B) α = 18. In Q.17 the current amplification factor (β) will be: (A) 89 (B) 95 (C) 99 (D) 101 Sol. (C) β= 19. A transistor has α = 0.95. If the emitter current is 10 mA, then collector current will be: (A) 9.5 mA (B) 10 mA (C) 0.95 mA (D) None of these Sol. (A) Collector current Ic = αIe = 0.95 × 10 = 9.5 mA 20. In the above problem, the base current will be: (A) 0.1 mA (B) 0.2 mA (C) 0.3 mA (D) 0.5 mA Sol.(D) Ib = Ie − Ic = (10 − 9.5) = 0.5mA 21. In Q. 90 the current amplification factor will be: (A) 11 (B) 19 (C) 35 (D) 79 Sol.(B) β = 22. Which one of the following gates can be served as a building block for any digital circuit? (A) OR (B) AND (C) NOT (D) NAND Solution : (D) We know that the basic gates (i.e. OR, AND and NOT) can be obtained by the repeated use of the NAND gates. Therefore in the digital circuits NAND gates can be served as a building block. 23. The cause of the potential barrier in a p-n junction diode is (A) Depletion of positive charges near the junction (B) Depletion of negative charges near the junction (C) Concentration of positive charges near the junction (D) Concentration of positive and negative charges near the junction Solution : (D) We know that the concentration of holes and electrons are greater in p and n regions respectively with some concentration difference. This concentration difference establishes density gradient and it results into diffusion. Therefore a potential barrier is produced due to the concentration of positive and negative charges near the junction. 24. A transistor having α = 0.99, is used in a common-base amplifier. If the load resistance is 4.5 KΩ and the dynamic resistance of the emitter junction is 50 KΩ, the voltage gain of the amplifier will be: (A) 79.1 (B) 89.1 (C) 99.1 (D) None of these Sol. (B) 25. In the above problem, the power gain will be: (A) 88.2 (B) 98.2 (C) 78.2 (D) None of these Sol. (A) Ap = Current gain × voltage gain 0.99 × 89.1 = 88.2 26. The binary equivalent of 25 is: (A) 111001 (B) 11001 (C) 10001 (D) 10011 Sol. (B) 25 = 16 + 8 + 1 = 1 × 24 + 1 × 23 + 0 × 22 + 0 × 21 + 1 × 20 ∴ Binary equivalent = 11001 27. The decimal equivalent of 1111: (A) 25 (B) 35 (C) 15 (D) 5 Sol. (C) 1111 = 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20 = 8 + 4 + 2 + 1 = 15 28. In binary system, 11000101 represents the following number on decimal system: (A) 4 (B) 401 (C) 197 (D) 204 Sol. (C) 11000101 = 1 × 27 + 1 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 1 × 20 = 128 + 64 +0 + 0 + 0 + 4 +1 = 197 29. In a transistor circuit, when the base current is increased by 50 μA, keeping collector voltage fixed at 2 volt, the collector current increase by 1.0 mA. The current amplification factor of the transistor will be: (A) 10 (B) 20 (C) 30 (D) 40 Sol. (B) β = 30. The value of α for a transistor is 0.9. What would be the change in the collector current corresponding to a change of 4 mA in the base current in a common emitter arrangement ? (A) 36 mA (B) 72 mA (C) 18 mA (D) None of these Sol. (A) ΔIc = β ΔIb = = 25 mA For Test 1. In a common base circuit at VC = 3V, a change in emitter current from 12.0 mA to 18.5 mA produces a change in collector current from 11.8 to 17.4 mA. The current gain will be: (A) 0.9521 (B) 0.8615 (C) 0.7351 (D) None of these Sol. (B) α = 2. For a common base amplifier, the values of resistance gain and voltage gain are 3000 and 2800 respectively. The current gain will be: (A) 0.93 (B) 0.83 (C) 0.73 (D) 0.63 Sol. (A) AI = 3. In the above problem the power gain of the amplifier will be: (A) 351.2 (B) 1575.30 (C) 2173.6 (D) 2613.3 Sol. (D) AP = 4. In the above problem, the collector current will be: (A) 0.44 A (B) 1.44 A (C) 0.44 mA (D) 1.44 mA Sol. (D) Ic = Ie − Ib = 1.5 × 10−3 − 60 × 10−6 = 1.44 mA 5. In the above problem, the collector current will be: (A) 6.676 mA (B) 5.382 mA (C) 4.987 mA (D) None of these Sol. (A) Ic = Ie − Ib = 6.8 × 10−3 − 0.124 × 10−3 = 6.676 mA 6. In a given transistor, the emitter current is changed by 2.1 mA. This results in a change of 2 mA in the collector current and a change of 1.05 V in the emitter-base voltage. The input resistance is: (A) 5000 Ω (B) 3000 Ω (C) 1000 Ω (D) 500 Ω Sol. (D) Input resistance = 7. In the above problem, the A.C. gain of the transistor A.C. gain of the transistor (αAC) will be: (A) 0.90 (B) 0.95 (C) 0.99 (D) None of these Sol. (B) αAC = 8. In Q. 132, if the transistor is used in common emitter configuration, then the value of βAC will be: (A) 9 (B) 19 (C) 29 (D) 39 Sol. (B) βAC = 9. The DC current gain of a transistor in common base configuration is 0.98. Its current gain in common emitter configuration will be: (A) 99 (B) 69 (C) 49 (D) 39 Sol. (C) β = 10. When the emitter current in a transistor is changed by 1 mA, then its collector current changes by 0.99 mA. The value of its common base current gain will be: (A) 1.99 (B) 9.99 (C) 0.999 (D) 0.99 Sol. (D) α = 11. In the above problem, the value of common emitter current gain will be: (A) 0.99 (B) 9.9 (C) 99 (D) 999 Sol. (C) β = 12. In the above problem, the value of base current will be: (A) 0.0132 mA (B) 0.132 mA (C) 0.00132 mA (D) 1.32 mA Sol. (A) IB = IE= 1.1 − 1.0868 = 0.0132 mA 13. The current gain in a common base circuit is 0.98. When the emitter current is changed by 5 mA, then the change in collector current will be: (A) 0.196 mA (B) 2.45 mA (C) 4.90 mA (D) 5.10 mA Sol. (C) or 14. A doped semiconductor has impurity levels 32 meV below the conduction band. The semiconductor is: (A) N-type (B) P-type (C) N-P junction (D) None of the above Sol. (A) The impurity level shifts towards conduction band in N-type semiconductor. 15. A potential barrier of 0.5 V exists across a p - n junction. If the width of depletion layer is 10−6 m, the intensity of electric field in this region will be: (A) 1 x 106 V/m (B) 5 x 105 V/m (C) 4 x 104 V/m (D) None of these Sol. (B) E = V/m 16. When a p-n junction in a diode is reverse biased, the thickness of the depletion layer (A) increases (B) decreases (C) become zero (D) remains constant Solution : (A) In reverse bias, as electrons and holes move away from the junction, the thickness of the depletion layer increases. 17. The two diodes A and B are biased as shown, then (A) The diodes A and B are reverse biased (B) The diodes A is forward biased and B is reverse biased (C) The diodes B is forward biased and diode A is reverse biased (D) The diodes A and B are forward biased. Solution : (D) 18. The potential in the depletion layer is due to (A) electrons (B) holes (C) ions (D) forbidden band Solution : (C) 19. The width of forbidden gap in silicon crystal is 1.1 eV. When the crystal is converted in to a n-type semiconductor the distance of Fermi level from conduction band is (A) greater than 0.55 eV (B) equal to 0.55 eV (C) lesser than 0.55 eV (D) equal to 1.1 eV Solution : (C) 20. Consider the following statements A and B and identify the correct choice of the given answers. (1) The width of the depletion layer in a p-n junction diode increases in forward bias. (2) In an intrinsic semiconductor the Fermi energy level is exaclty in the middle of the forbidden gap. (A) A is true and B is False (B) Both A and B are false (C) A is false and B is true (D) Both A and B is true Solution : (C) When a p-n junction diode is forward biased the thickness of depletion layer decreases and thereby the current across the junction increases. In an intrinsic semi-conductor, Fermi energy level will be in the middle of the forbidden gap. 21. In the study of transistor as amplifier, if and where IC, IB, IE are the collector, base and emitted currents, then (A) (B) (C) (D) Solution : (C) 22. A gate has the following truth table A B R 1 1 1 1 0 0 0 1 0 0 0 0 The gate is: (A) AND (B) OR (C) NOR (D) NAND Solution : (A) 23. Out of following the forward based diode is (A) (B) (C) (D) Solution : (C) 24. The approximate ratio of resistance in the forward and reverse bias of the PN junction diode is (A) 102 : 1 (B) 10–2 : 1 (C) 1 : 10–4 (D) 1 : 104 Solution : (D) 25. The resistance of a reverse biased pn junction diode is about : (A) 1 ohm (B) 102 ohm (C) 103 ohm (D) 106 ohm Solution : (D) 26. On reducing forward voltage, the potential barrier (A) increases (B) decreases (C) unchanged (D) none of these Solution : (A) 27. In the depletion layer, there are (A) Only electrons (B) only holes (C) electrons and holes (D) None of the above Solution : (D) 28. Zener diode is used : (A) As a amplifier (B) As a rectifier (C) As a oscillator (D) As a voltage regulator Solution : (D) 29. Which of the following is a correct relation for a transistor (A) (B) (C) (D) Solution : (B) IE = IB + IC Differentiating, we get 30. Which one is in forward bias : (A) (B) (C) (D) None of these Solution : (B)