JEE-Physics Ans. (A,B,C,D) Solution P In this situation T1 T2 a a m1 m2 g T2 T1 r I a m2 m1 a m1 m2 r2 Example#19 A uniform rod AB of length is free to rotate about a horizontal axis passing through A. The rod is released from rest from horizontal position. If the rod gets broken at midpoint C when it becomes vertical, then just after breaking of the rod : A C B (A) Angular velocity of upper part starts to decrease while that of lower part remains constant. (B) Angular velocity of upper part starts to decrease while that of lower part starts to increase (C) Angular velocity of both the parts is identical (D) Angular velocity of lower part becomes equal to zero Solution . Ans. (A,C) v/2 On upper part torque of mg about A will decrease the angular velocity. v/2 Lower part of rod will not experience any couple hence its angular velocity can't change. v Initially both parts are having same angular velocities. Example# 20 to 22 A uniform hollow sphere is released from the top of a fixed inclined plane of inclination 37° and height 3m. It rolls without sliding. 37° fixed 2 0 . The acceleration of the centre of mass of the hollow sphere is NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 30 18 9 15 (A) 7 m/s2 (B) m/s2 (C) m/s2 (D) 7 m/s2 5 5 2 1 . The speed of the point of contact of the sphere with the inclined plane when the sphere reaches half–way of the incline is (A) 42 m/s (B) 21 m/s (C) 84 m/s (D) zero 2 2 . The time taken by the sphere to reach the bottom is 3 5 5 (D) None of these (A) 5 s (B) 3 s (C) 4 s 60 E
JEE-Physics Solution 20. Ans. (B) a g sin 103 / 5 18 m s2 K2 5 12 1 R2 3 21. Ans. (D) Speed of point of contact in pure rolling is always zero 22. Ans. (B) s ut 1 at2 3 1 18 t 2 t 5 s 2 sin 37 2 5 3 Example#23 to 25 A mouse, searching for food, jumped onto the rim of a stationary circular disk mounted on a vertical axle. The disk is free to rotate without friction. The velocity of the mouse was tangent to the edge of the disk before it landed. When the mouse landed, it gripped the surface, remained fixed on the outer edge of the disk at a distance R from the center, and set it into rotation. The sketch indicates the situation. v0 R 1 Before After The mass of the mouse is m = 0.10 kg, the radius of the disk is R = 0.20 m, and the rotational inertia of the disk is I = 0.0080 kg·m². The speed of the mouse, just before it landed on the disk is v = 1.5 o m/s. 2 3 . Magnitude of the angular velocity of the disk plus mouse, after it landed becomes (A) 0.25 rad/s (B) 2.5 rad/s (C) 0.375 rad/s (D) 3.75 rad/s 2 4 . Find the magnitude of the impulse received by the mouse as it landed on the disk. (A) 0.01 kg.m/s opposite to direction of motion (B) 0.01 kg.m/s in the direction of motion (C) 0.10 kg.m/s opposite to direction of motion (D) 0.10 kg.m/s in the direction of motion 2 5 . The mouse, still searching for food, crept to the center of the disk (where r = 0). Find angular velocity of the disk plus mouse, when the mouse was at the center of the disk. (A) 0.25 rad/s (B) 2.5 rad/s (C) 0.375 rad/s (D) 3.75 rad/s NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 Solution 23. Ans. (B) By conservation of angular momentum mv0R = (I + mR2) mv0R 0.1 1.5 0.2 2.5 rad/s I mR2 0.008 0.004 24. Ans. (C) Impulse received by mouse = change in momentum = 0.1 (2.5 × 0.2 – 1.5) = – 0.1 kg m/s 25. Ans. (D) By conservation of angular momentum : mv R = I = 0.1 1.5 0.2 = 3.75 rad/s 0 0.008 E 61
JEE-Physics Example#26 to 27 A hollow sphere is released from the top of a wedge, friction is sufficient for pure rolling of sphere on the wedge. There is no friction between the wedge and the ground. Radius of sphere is R. At the instant it leaves the wedge horizontally. m m 2 6 . Velocity of centre of mass of sphere w.r.t. ground is- (A) 5 (B) 2gh (C) 3 (D) 11 gh gh gh 7 7 7 2 7 . Angular velocity of sphere is- 12gh 27 gh 20gh 44gh (A) 7R 2 (B) 7 R 2 (C) 7R 2 (D) 7R 2 Solution 26. Ans. (C) m 2v v R v v m v R 27. Ans. (A) Enegy conservation mgh = 1 mv2 1 mv2 1 2 m R 2 4v2 v 3 gh 2 2 2 3 R2 7 2v 4 3 12gh R2 7 gh R 7R 2 Example#28 to 30 A disc of mass 2 M and radius R is placed on a fixed plank (rough) of length L. The coefficient of friction between the plank and disc is = 0.5. String (light) is connected to centre of disc and passing over a smooth light pulley and connected to a block of mass M as shown in the figure. Now the disc is given an angular velocity 0 in clockwise direction and is gently placed on the plank. Consider this instant as t=0. Based on above information, answer the following questions : 2M Fixed plank M NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 2 8 . Mark the correct statement w.r.t. motion of block and disc. (A) The block remains at rest for some time, t > 0. (B) The block starts accelerating just after placing of disc on plank. (C) The disc is performing pure rotational motion for some time t > 0 (D) Both (A) and (C) are correct. 2 9 . Time t upto which the block remains stationary is 0 (A) 0R (B) 40R (C) Zero (D) Question is irrelevant g g E 62
JEE-Physics 3 0 . Time time t at which the disc will cross the other end of the plank is- 01 8L (B) 0R 8L (C) 8L 40R (D) 0R 4L (A) g g g gg g g Solution 2M 28. Ans.(D) fr 29. Ans. (A) M Frictional force = Mg Friction will act up to the instant when the velocity of contact point becomes zero f = i + t t = i ....(i) fr R ....(ii) By solving (i) and (ii) time 0R I g 30. Ans. (B) T – f = 2Ma ...(ii) Mg – T = Ma ...(i) r By adding (i) and (ii) Mg – f = 3 Ma ...(iii) f R = I = 2MR2 a f = Ma ...(iv) r r 2R r From equation (iii) and (iv) a = g/4 L 1 at2 t 8L 2 g Total time = 0R 8L g g Example#31 3 1 . Four different situations of a moving disc are shown in column I and predictions about its final motion and forces acting on it are given in column - II. Column I Column II v v (P) finally disc will roll along the initial direction of velocity (v) 2R (A) 2v R (B) v (Q) finally, disc will roll in direction opposite to the initial direction of velocity (v) NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 2v v (R) finally, disc stops R (C) 4v v (S) Initially friction force acts in the direction opposite to that of initial velocity R (T) None of these (D) 63 E
JEE-Physics Solution Ans. (A)(S,P);(B)(P);(C)(S,R);(D)(S,Q) Final direction of pure rolling will be in direction of initial angular momentum about point of contact L = mv r + I cm cm For (A) : m vr MR2 v 3 m vR : hence in clockwise direction. 2 2R 4 For (B) : m vR mR2 2v 2mvR : hence in clockwise direction. 2 R For (C) : mvR mR2 2v 0 2 R mR2 4v For (D) : mvR – 2 R = –mRv : hence in anticlockwise direction. Direction of friction force : For (A) : Velocity of point of contact v/2 v Friction will be opposite to velocity For (B) : Velocity of point of contact 2v v Friction will be in the direction of velocity For (C) : Velocity of point of contact v 2v Friction will be opposite to the velocity For (D) : Velocity of point of contact v 4v Friction will be opposite to the velocity Example#32 Co l u m1n II (P) 3 mv Column I 2 0 (A) A ring of mass m is projected on rough horizontal plane with velocity v . 0 The magnitude of work done by frictional force to start rolling 0 (B) Kinetic energy of pivoted rod of mass m, velocity of centre of mass is v . (Q) 1 m v 2 0 8 0 0 (C) Kinetic energy of translation of a smooth rod of mass m, (R) 1 m v 2 where velocity of one end is v . 4 0 0 (S) 2 m v 2 NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 3 0 v0 m v=0 (D) Kinetic energy of a rod of mass m, as shown in figure. (T) 1 m v 2 9 0 45° v0 64 E
JEE-Physics Solution Ans. (A)(R), (B)(S), (C)(Q), (D)(P) For (A) : Final velocity v0 v v 2 m v 0 R mR2 R m vR So work done by friction = 1 m v 2 1 m v0 2 1 m v0 2 1 m v 2 2 0 2 2 2 4 0 2 v0 2v0 1 m 2 2v 2 2 2 3 3 For (B) : so KE = 0 m v 2 = 2v0 2 0 v0 /2 v0 1 1 / 2 2 2 8 For (C) : vcm so KE = m v 2 m v 2 cm 0 v0 For (D) : KE of rod = 1 IC2 1 m2 m2 2v0 1 m v 2 2 2 12 4 3 0 Example#33 A disc of radius R is rolling without slipping with an angular acceleration , on a horizontal plane. Four points are marked at the end of horizontal and vertical diameter of a circle of radius r (<R) on the disc. If horizontal and vertical direction are chosen as x and y axis as shown in the figure, then acceleration of points 1, 2, 3 and 4 are a1, a2 , a3 and a4 respectively, at the moment when angular velocity of the disc is . Match the following R1 y 4 r2 x3 Column-I Column-II (A) a1 (P) R rˆi r2 ˆj (Q) R rˆi r2 ˆj (B) a2 (R) R r2 ˆi rˆj (C) a3 (S) R r2 ˆi rˆj (D) a4 NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 (T) None of these Solution Ans. (A) Q; (B) R; (C) P; (D) S For (A): Acceleration of 1 w.r.t. centre of mass = rˆi 2rˆj rˆi 2rˆj Rˆi R r ˆi 2rˆj a1 ˆi rˆj For (B) : a2 rˆj 2rˆi Rˆi R 2r rˆi 2rˆj Rˆi R rˆi 2rˆj For (C) : a3 ˆi rˆj For (D) : a 4 rˆj 2rˆi Rˆi R 2r E 65
JEE-Physics Example#34 A solid uniform cylinder of mass m = 6 kg and radius r = 0.1 m is kept in balance on a slope of inclination = 37° with the help of a thread fastened to its jacket. The cylinder does not slip on the slope. The minimum required coefficient of friction to keep the cylinder in balance when the thread is held vertically is given as . Find the value of 4. F m r Solution Ans. 3 Fr – fr = 0 and mg sin – F sin – f = 0 F mg sin f F = f = mg 1 sin F cos + N – mg cos = 0 N = mg mg sin cos = mg cos ; f = N; mg sin = mg cos = tan = 0.75 1 sin 1 sin 1 sin 1 sin max Example#35 A uniform rod ABC of mass M and length is placed vertically on a rough horizontal surface. The coefficient of friction between the rod and surface is .A force F = 1.2 mg is applied on the rod at point B at a distance / 3 below centre of rod horizontally as shown in figure. If the initially acceleration of point A is k then find value of k. (Friction is sufficient to prevent slipping) A BF C Solution Ans. 6 Taking torque about C F m2 3 g So, a = 3 g 6 6 3 5 A 5 NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 66 E
JEE-Physics SEMI CONDUCTOR - ELECTRONICS ENERGY BANDS IN SOLIDS Based on Pauli's exclusion principle In an isolated atom electrons present in energy level but in solid, atoms are not isolated, there is interaction among each other, due to this energy level splitted into different energy levels. Quantity of these different energy levels depends on the quantity of interacting atoms. Splitting of sharp and closely compact energy levels result into energy bands. They are discrete in nature. Order of energy levels in a band is 1023 and their energy difference = 10–23 eV. Energy Band Range of energy possessed by an electron in a solid is known as energy band. Valence Band (VB) Range of energies possessed by valence electron is known as valence band. (a) Have bonded electrons. (b) No flow of current due to such electrons. (c) Always fulfill by electrons. Conduction Band (CB) Range of energies possessed by free electron is known as conduction band. (a) It has conducting electrons. (b) Current flows due to such electrons. (c) If conduction band is fully empty then current conduction is not possible. (d) Electrons may exist or not in it. Forbidden Energy gap (FEG) (Eg) E = (C B) – (V B) g min max Energy gap between conduction band and valence band, where no free electron can exist. E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 band energy (eV) Width of forbidden energy gap depends upon the nature of substance. conduction band Width is more, then valence electrons are strongly attached with nucleus forbidden energy gap Width of forbidden energy gap is represented in eV. valence band As temperature increases forbidden energy gap decreases (very slightly). CLASSIFICATION OF CONDUCTORS, INSULATORS AND SEMICONDUCTOR : - On the basis of the relative values of electrical conductivity and energy bands the solids are broadly classified into three categories (i) Conductors (ii) Semiconductors (iii) Insulator 1 E
JEE-Physics Comparison between conductor, semiconductor and insulator : Properties Conductor Semiconductor Insulator Resistivity 10–2 – 10–8 m 10–5 – 106 m 1011 – 1019 m Conductivity 102 – 108 mho/m 105 – 10–6 mho/m 10–11 – 10–19 mho/m Negative (Very slightly) Temp. Coefficient Positive Negative No current of resistance () Current Due to free Due to electrons electrons and holes Conduction Band Conduction Band Conduction Band Electron Energy Electron Energy ForbiddeEng 3eV Electron Energy No gap Overlapping Forbidden Eg 1ev Gap Valence Band region Gap Energy band diagram Valence Band Valence Band Insulator Conductor Semi conductor Forbidden energy gap 0eV 1eV 3eV Ge, Si, GaAs, Wood, plastic, Example : Pt, Al, Cu, Ag GaF2 Diamond, Mica CONCEPT OF \"HOLES\" IN SEMICONDUCTORS Due to external energy (temp. or radiation) when electron goes from valence band to conduction band (i.e. bonded electrons becomes free) a vacancy of free e– creats in valence band, hole free e– which has same charge as electron but positive. This positively Si Si charged vacancy is termed as hole and shown in figure. Si Si Si Si • It is deficiency of electron in VB. Si Si Si • It's acts as positive charge carrier. • It's effective mass is more than electron. Si Si Si • It's mobility is less than electron. Note : Hole acts as virtual charge carrier, although it has no physical significance. GOLDEN KEY POINTS E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 • Number of electrons reaching from VB to CB at temperature T kelvin n = A T 3 /2 e – E g AT 3 / 2 exp Eg 2kT – 2kT where T = absolute temperature k = Boltzmann constant = 1.38 10-23 J/K A = constant E = energy gap between CB and VB g • In silicon at room temperature out of 1012 Si atoms only one electron goes from VB to CB. • In germanium at room temperature out of 109 Ge atoms only one electron goes from VB to CB. 2E
JEE-Physics EFFECT OF TEMPER ATURE ON SEMICONDUCTOR At absolute zero kelvin temperature Above absolute temperature At this temperautre covalent bonds are very With increase in temperature few valence strong and there are no free electrons and semiconductor behaves as perfect insulator. electrons jump into conduction band and hence it behaves as poor conductor. hole free e– Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si Si at 0 K at high temperature Valence band fully filled Valence band partially empty Conduction band fully empty Conduction band partially filled EFFECT OF IMPURITY IN SEMICONDUCTOR Doping is a method of addition of \"desirable\" impurity atoms to pure semiconductor to increase conductivity of semiconductor. or Doping is a process of deliberate addition of a desirable impurity atoms to a pure semiconductor to modify its properties in controlled manner. Added impurity atoms are called dopants. The impurity added may be 1 part per million (ppm). The dopant atom should take the position of semiconductor atom in the lattice. The presence of the dopant atom should not distort the crystal lattice. The size of the dopant atom should be almost the same as that of the crystal atom. The concentration of dopant atoms should not be large (not more than 1% of the crystal atom). It is to be noted that the doping of a semiconductor increases its electrical conductivity to a great extent. GOLDEN KEY POINTS • The concentration of dopant atoms be very low, doping ratio is vary from impure : pure :: 1 : 106 to 1 : 1010 In general it is 1 : 108 • There are two main method of doping. (i) Alloy method (ii) Diffusion method (The best) • The size of dopant atom (impurity) should be almost the same as that of crystal atom. So that crystalline E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 structure of solid remain unchanged. CLASSIFICATION OF SEMICONDUCTOR SEMICONDUCTOR Intrinsic semiconductor Extrinsic semiconductor (doped semicondutor) (pure form of Ge, Si) N-type P-type ne = nh = ni trivalent impurity pentavalent impurity E (P, As, Sb etc.) (Ga, B, In, Al) acceptor impurity (NA) donar impurity (ND) ne >> nh nh >> ne 3
JEE-Physics N type semiconductor When a pure semiconductor (Si or Ge) is doped by pentavalent impurity (P, As, Sb, Bi) then four electrons out of the five valence electrons of impurity take part, in covalent bonding, with four silicon atoms surrounding it and the fifth electron is set free. These impurity atoms which donate free e– for conduction are called as Donar impurity (N ). Due to donar impurity free e– increases very much so it is called as \"N\" type semiconductor. By D donating e– impurity atoms get positive charge and hence known as \"Immobile Donar positive Ion\". In N-type semiconductor free e– are called as \"majority\" charge carriers and \"holes\" are called as \"minority\" charge carriers. thermally generated e– Si As Si Si thermally free electrons generated hole minority Si Si Si Si donar e– positive hole As Si Si As donar ions N-type semiconductor N-type semiconducting crystal P type semiconductor When a pure semiconductor (Si or Ge) is doped by trivalent impurity (B, Al, In, Ga) then outer most three electrons of the valence band of impurity take part, in covalent bonding with four silicon atoms surrounding it and except one electron from semiconductor and make hole in semiconductor. These impurity atoms which accept bonded e– from valance band are called as Acceptor impurity (N ). Here holes increases very much so it A is called as \"P\" type semiconductor and impurity ions known as \"Immobile Acceptor negative Ion\". In P-type semiconductor free e– are called as minority charge carries and holes are called as majority charge carriers. Extra hole created by thermally holes E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 acceptor impurity atom generated e– minority e Si Al– Si Si thermally generated hole Si Si Si Si Al– Si Si Al– negative acceptor ions P-type semiconductor P-type semiconducting crystal 4E
Intrinsic Semiconductor N-type (Pentavalent impurity) JEE-Physics P-type(Trivalent impurity) CB CB CB 1. donor acceptor impurity impurity VB level VB level VB 2. free hole electron negative positive acceptor donar ion ion 3. Current due to Mainly due to electrons Mainly due to holes electron and hole n << n (N ~ n ) n >> n (N ~ n ) 4. n = n = n h eD e h eA h ehi I ~ I I ~ I 5. I = I + I e h eh Entirely neutral Entirely neutral 6. Entirely neutral Majority - Electrons Majority - Holes 7. Quantity of electrons Minority - Holes Minority - Electrons and holes are equal Mass action Law In semiconductors due to thermal effect, generation of free e– and hole takes place. Apart from the process of generation, recombination also occurs simultaneously, in which free e– further recombine with hole. At equilibrium rate of generation of charge carries is equal to rate of recombination of charge carrier. The recombination occurs due to e– colliding with a hole, larger value of n or n , higher is the probability of their e h recombination. Hence for a given semiconductor rate of recombination n × n eh so rate of recombination = R n × n R = recombination coefficient, eh The value of R remains constant for a solid, according to the law of thermodynamics until crystalline lattice structure remains same. E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 For intrinsic semiconductor n = n = n ehi so rate of recombination = R n 2 i R ne × nh = R ni2 ni2 = ne × nh Under thermal equilibrium, the product of the concentration 'n ' of free electrons and the concentration n of eh holes is a constant and it is independent of the amount of doping by acceptor and donor impurities. Thus from mass action law n e n h n12 E5
JEE-Physics Electron-hole Recombination : It is necessarly to complete a bond that electron is shared from neighbouring atoms or it may also be received from conduction bond. In the second case electron recombines with the hole of valnce bond. This process is known as electron-hole recombination. The breaking of bonds or generation of electron-hole pairs, and completion of bonds due to recombination is taking place continuously. At equilibrium, the rate of generation becomes equal to the rate of recombination, giving a fixed number of free electrons and holes. E x . 1 The energy of a photon of sodium light ( = 589 nm) equals the band gap of a semiconducting material. Find : (a) the minimum energy E required to create a hole-electron pair. E (b) the value of at a temperature of 300 K. kT Sol. (a) E hc (in eV) so E 12400 (E is in eV and is in Å) = 5890 Å e so E 12400 2.1eV (b) E 2.1 1.6 1019 J 81 5890 kT 1.38 1023 300 E x . 2 A P type semiconductor has acceptor level 57 meV above the valence band. What is maximum wavelength of light required to create a hole ? Sol. E hc hc 6.62 1034 3 108 E = 57 103 1.6 1019 = 217100 Å Ex .3 A silicon specimen is made into a p-type semiconductor by doping on an average one indium atom per 5 × 107silicon atoms. If the number density of atoms in the silicon specimen is 5 × 1028 atoms/m3; find the number of acceptor atoms in silicon per cubic centimeter. S o l . The doping of one indium atom in silicon semiconductor will produce one acceptor atom in p-type semiconductor. Since one indium atom has been dopped per 5 × 107 silicon atoms, so number density of acceptor atoms in 5 1028 1021 atom/m3 = = 1015 atoms/cm3 silicon 5 107 E x . 4 A pure Ge specimen is doped with A. The number density of acceptor atoms is approximately 1021 m–3. If E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 density of electron holes pair in an intrinsuc semiconductor is approximately 1019m–3, the number density of electrons in the specimen is : S o l . In pure semiconductor electron-hole pair n = 1019 m–3 i acceptor impurity N = 1021 m–3 A Holes concentration nh = 1021 m–3 electrons concentration = n = n 2 1019 2 = 1017 m–3 e i 1021 nh 6 E
JEE-Physics E x . 5 Pure Si at 300 K has equal electron (n ) and hole (n ) concentrations of 1.5 × 1016 m–3. Dopping by indium eh increases n to 3 × 1022 m–3. Calculate n in the doped Si. he Sol. For a doped semi-conductor in thermal equilebrium n n = n 2 (Law of mass action) eh i ne n 2 (1.5 1016 )2 = 7.5 × 109 m–3 i 3 1022 hh A RESISTIVITY AND CONDUCTIVITY OF SEMICONDUCTOR Conduction in conductor Relation between current (I) and drift velocity (v ) V d I = ne A v n = number of electron in unit volume E= d A= cross sectional area I V J = amp/m2 = ne v current density drift velocity of electron v = E Ad d J = ne E J = E Conductivityne Resistivity N - type Mobility vd P - type n >> n E n >> n eh Conduction in Semiconductor he J e ne ve Intrinsic semiconductor n = n J e nh vh eh J = ne [ ve + vh] 1 = en [ e + h ] 1 e n h 1 e n e h e GOLDEN KEY POINTS • Due to impurity the conductivity increases approximately 105 times E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 • sc = e + h = neee + n eh = e(n µ + nµ) h ee hh E x . 6 What will be conductance of pure silicon crystal at 300K Temp.. If electron hole pairs per cm3 is 1.072 x 1010 at this Temp, n = 1350 cm2 / volt sec & P = 480 cm2 / volt sec S o l . = niee + nieh = nie (e + h) = 3.14×10–6 mho/cm E x . 7 Pure Si at 300 K has equal electron n and hole n concentration of 1.5 × 1016/m3. Doping by indium increases eh n to 4.5 × 1022/m3. Calculate n in doped silicon. he n 2 (1.5 1016 )2 5 109 m 3 i (4.5 1022 ) Sol. n = e nh E 7
JEE-Physics E x . 8 A semiconductor has equal electron and hole concentration of 6 × 108/m3. On doping with certain impurity electron concentration increases to 9 × 1012/m3. (i) Identify the new semiconductor obtained after doping. (ii) Calculate the new hole concentration. S o l . n = 6 × 108/m3 and n = 9 × 1012 /m3 i e (i) n > n so it is N-type semiconductor ei 2 n 2 36 1016 i i 9 1012 (ii) n nenh n = = 4 × 104 /m3 h ne P - N JUNCTION Technique s for maki ng P-N junct ion (i) Alloy Method or Alloy Junction Here a small piece of III group impurity like indium is placed over n–Ge or n–Si and melted as shown in figure ultimetely P – N junction form. indium indium liquid indium buttom PN Jn N-type N-type N-type arsenic arsenic liquid arsenic buttom PN Jn P-type P-type P-type (ii) Diffusion Junction vacuum A heated P–type semiconductor is kept in pentavalent heat N-type Ge/Si P-N Jn impurity vapours which diffuse into P–type semiconductor as Al diffused shown and make P–N junction. vapours P-type layer (iii) Vapour deposited junction or epitaxial junction If we want to grow a layer of n–Si or p–Si then p–Si wafer is kept in an atmosphere of Silane (a silicon compound which to vacuum dissociates into Si at high temperatures) plus phosphorous pump vapours. On craking of silane at high temperature a fresh layer on n–Si grows on p–Si giving the \"P–N junction\". Since this junction growth is layer by so it is also referred as layer growth or epitaxial junction formation of P–N junction. –+ n Description of P-N Junction without applied voltage or bias p Given diagram shows a P–N junction immediately after it is formed. P region has mobile majority holes and immobile negatively charged impurity ions. E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 N region has mobile majority free electrons and immobile positively hole free electron charged impurity ions. Due to concentration difference diffusion of holes starts from P to N charge deplition layer side and diffusion of e– s starts N to P side. density width (10)-6 Due to this a layer of only positive (in N side) and negative + (in P–side) started to form which generate an electric field (N to P – distance side) which oppose diffusion process, during diffusion magnitude of electric electric field increases due to this diffusion it gradually decreased field distance intensity and ultimately stopes. Electric Potential The layer of immobile positive and negative ions, which have no V0 = Potential free electrons and holes called as depletion layer as shown in distance barrier diagram. 8E
JEE-Physics GOLDEN KEY POINTS • Width of depletion layer 10-6 m (a) As doping increases depletion layer decreases (b) As temperature is increased depletion layer also increases. (c) P-N junction unohmic, due to nonlinear relation between I and V. • Potential Barrier or contact potential G e 0.3 V Si 0.7 V • Electric field, produce due to potential barrier E V 0.5 E 105 V/m d 106 This field prevents the respective majority carrier from crossing barrier region DIFFUSION AND DRIFT CURRENT (1) Diffusion current – P to N side (2) Drift current – N to P side If there is no biasing diffusion current = drift current So total current is zero BEHAVIOUR OF P–N JUNCTION WITH AN EXTERNAL VOLTAGE APPLIED OR BIAS Forward Bias V If we apply a voltage \"V\" such that P–side is positive and N–side VB is negative as shown in diagram. The applied voltage is opposite to the junction barrier potential.Due to PN this effective potential barrier decreases, junction width also decreases, so more majority carriers will be allowed to flow across junction. It means the current flow in principally due to majority charge carriers and it is in the order of mA called as forward Bias. Reverse Bias V N If we apply a voltage \"V\" such that P–side is negative and VB N–side is positive as shown in diagram. P E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 The applied voltage is in same direction as the junction barrier potential. Due to this effective potential barrier increase junction, width also increases, so no majority carriers will be allowed to flow across junction. Only minority carriers will drifted. It means the current flow in principally due to minority charge carriers and is very small (in the order of µA). This bias is called as reversed Bias. GOLDEN KEY POINTS • In reverse bias, the current is very small and nearly constant with bias (termed as reverse saturation current). However interesting behaviour results in some special cases if the reverse bias is increased further beyond a certain limit, above particular high voltage breakdown of depletion layer started. • Breakdown of a diode is of following two types : (i) Zener breakdown (ii) Avalanche breakdown E9
JEE-Physics Comparison between Forward Bias and Reverse Bias Reverse Bias Forward Bias P positive P negative N negative N positive PN PN V V +– –+ 1. Potential Barrier reduces 1. Potential Barrier increases. 2. Width of depletion layer decreases 2. 3. P-N jn. provide very small resistance 3. Width of depletion layer increases. 4 Forward current flows in the circuit 5 . Order of forward current is milli ampere. 5. P-N jn. provide high resistance 4 . Very small current flows. Order of current is micro ampere for Ge or Neno ampere for Si. 6 . Current flows mainly due to majority carriers. 6 . Current flows mainly due to minority carriers. 7. Forward characteristic curves. 7. Reverse characteristic curve Vr(volt) if Reverse saturation (mA) current 0 knee break down voltage voltage Vf(volt) Ir (A) 8. Forward resistance 8. Reverse resistance E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 Rf Vf 100 Rr Vr 106 If Ir 9 . Order of knee or cut in voltage 9. Breakdown voltage Ge Ge 0.3 V Si 25 V 35 V Si 0.7 V Special point : Generally Rr = 103 : 1 for Ge R r = 104 : 1 for Si Rf Rf 10 E
JEE-Physics Ex .9 The resistance of p-n junction diode decreases when forward biased and increases when reverse biased. Why? S o l . When p-n junction is forward biased, the width of depletion layer decreases and the barrier potential is opposed by the forward bias. In other words, potential barrier decreases. Hence the diffusion of holes and electrons through the junction increases. Due to this, the diode current increases and hence resistance decreases. When p-n junction is reverse biased, the barrier potential is supported and the width of depletion layer increases. As a result of this, the diode current becomes almost zero as there is no diffusion of majority carriers (electrons and holes) through the junction. Hence the resistance of the junction diode increases when reverse biased. Ex.10 What is an ideal diode ? Draw the output waveform across the load resistor R, if the input waveform is as shown in the figure. +6V V -6V S o l . An ideal diode has zero resistance when forward biased and infiniter resistance when it is reversed biased. Output wave form is shown in fig. +6V 0V Ex.11A potential barrier of 0.5 V exists across a p-n junction (i) If the depletion region is 5 × 10–7 m wide. What is the intensity of the electric field in this region ? (ii) An electron with speed 5×105 m/s approaches the p-n junction from the n-side with what speed will it enter the p-side. Sol.: (i) Width of depletion layer L = 5 × 10–7 m E= V V 0.5V L E L 5 10 7 = 106 volt/m PN (ii) Work energy theorm 1 M v 2 eV 1 M v 2 2 i 2 f E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 vf M v 2 2eV = 2.7 × 105 m/s i M Ex.12 Figure shows a diode connected to an external resistance and an e.m.f. Assuming that the barrier potential developed in diode is 0.5 V1 obtain the value of current in the circuit in milliampere. Sol. E = 45 V R = 100 100 1 voltage drop across p-n junction = 0.5 V effective voltage in the circuit V = 4.5 – 0.5 = 4.0 V 4.5V current in the circuit I V 4.0 0.04 A = 0.04 × 1000 mA = 40mA R 100 Ex.13Differentiate zener and avalanche breakdown. Sol. The difference between these two are as follows E 11
JEE-Physics Zener Break down Avalanche Break down Where covalent bonds of depletion layer, its Here covalent bonds of depletion layers are bro self break, due to high electric field of very ken by collision of \"Minorities\" which aquire high Reverse bias voltage. high kinetic energy from high electric field of very-very high reverse bias voltage. This phenomena predominant This phenomena predominant (i) At lower voltage after \"break down\" (i) At high voltage after breakdown (ii) In P – N having \"High doping\" (ii) In P – N having \"Low doping\" (iii) P – N Jn. having thin depletion layer (iii) P – N Jn. having thick depletion layer Here P – N not demage paramanently Here P – N damage peramanentaly due to \"In D.C voltage stablizer zener phenomenan \"Heating effect\" due to abruptly increament of is used\". minorities during repeatative collisoins. CHARACTERISTIC CURVE OF P-N JUNCTION DIODE PN symbol PN +(0-1)–V (0-10)V If(mA)Forward A Ir(A) bias curve + mA Vr (Volt) – E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 + D D O Vf(Volt) Rh R Reverse – Rh R bias curve Forward bias Reverse bias In forward bias when voltage is increased from 0V is steps and corresponding value of current is measured, the curve comes as OB of figure. We may note that current increase very sharply after a certain voltage knee voltage. At this voltage, barrier potential is completely eliminated and diode offers a low resistance. In reverse bias a microammeter has been used as current is very very small. When reverse voltage is increased from 0V and corresponding values of current measured the plot comes as OCD. We may note that reverse current is almost constant hence called reverse saturation current. It implies that diode resistance is very high. As reverse voltage reaches value V , called breakdown voltage, current increases very sharply. B For Ideal Diode in forward bias R f = 0 in reverse bias R r = 'ON' switch 'OFF' switch RECTIFIER It is device which is used for converting alternating current into direct current. E 12
JEE-Physics Half wave rectifier A.C. Supply S1 D A S1 D Secondry Voltage S2 RL A.C. Supply S2 B A Input RL B Output For positive half cycle For negative half cycle During the first half (positive) of the input signal, let S is at positive and S is at negative potential. So, the PN 12 junction diode D is forward biased. The current flows through the load resistance R and output voltage is L obtained. During the second half (negative) of the input signal, S1 and S2 would be negative and positive respectively. The PN junction diode will be reversed biased. In this case, practically no current would flow through the load resistance. So, there will be no output voltage. Thus, corresponding to an alternating input signal, we get a unidirectional pulsating output as shown. Peak inverse voltage (PIV) In half wave rectifier PIV = maximum voltage across secondary coil of transformer (V ) s = Peak value of output (V ) m Full wave rectifier When the diode rectifies the whole of the AC wave, it is called full wave rectifier. Figure shows the experiemental arrangement for using diode as full wave rectifier. The alternating signal is fed to the primary a transformer. The output signal appears across the load resistance RL. D1 D1 Secondry Voltage S1 A A. C. Supply S1 Input B BA A. C. Supply D1 D2 D1 D2 D1 D2 D1 RL RL S2 S2 Output For positive half cycle D2 For negative half cycle D2 E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 During the positive half of the input signal : Let S positive and S negative. 12 In this case diode D is forward biased and D is reverse biased. So only D conducts and hence the flow of 12 1 current in the load resistance R is from A to B. L During the negative half of the input signal : Now S1 is negative and S2 is positive. So D1 is reverse-biased and D2 is forward biased. So only D2 conducts and hence the current flows through the load resistance R from A to B. L It is clear that whether the input signal is positive or negative, the current always flows through the load resistance in the same direction and full wave rectification is obtained. E 13
JEE-Physics D2 Input D4 RL Bridge Rectifier D1,D4 D2,D3 D1,D4 D2,D3 D1,D4 D1 Output D3 During positive half cycle During negative half cycle D and D are forward biased on switch D and D are forward biased on switch 1 4 2 3 D and D are reverse biased off switch D and D are reverse biased offswitch 23 14 In bridge rectifier peak inverse voltage PIV = V = V sm Form Factor Irms E rms F = Idc or E dc for full wave rectifier F = for half wave rectifier F = 22 2 Ripple and ripple factor In the output of rectifier some A.C. components are present. They are called ripple & there measurement is given by a factor known as ripple factor. For a good rectifier ripple factor must be very low. Total output current I = I2ac + I2dc Where I = rms value of AC component present in output rms ac Iac r = I2 1 = F2 1 Ripple factor r = Idc rms I2dc Rectifier efficiency = Pdc = I2dc R L Half wave rectifier Pac I2 (R F + R L ) rms Full wave rectifier or bridge wave rectifier 0.406 0.812 E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 = 1 Rf 1 Rf RL RL If Rf < < 1 , then = 40.6% If Rf < < 1 , then 81.2% RL RL Special Note If R = R Special Note If R = R fL fL 20.3% 40.6% Note : In brige full wave rectifier R is two times of resistance of P-N jn. diode in FB. f 14 E
JEE-Physics Ripple Frequency (i) For half wave rectifier input output input frequency = 50Hz ripple frequency = 50Hz (ii) for full wave rectifier input output ripple frequency = 100Hz input frequency = 50Hz output Pulse Pulse frequency = 50 Hz (i) For half wave rectifier input Pulse frequency = 100 Hz (ii) For full wave rectifier input output Pulse frequency = 100 Hz Pulse frequency = 100 Hz Comparison Between Average Rectifiers E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 Number of Diodes Half-wave Centre-tap Full-wave Transformer necessary 1 2 Bridge Peak secondary voltage No Ye s 4 Peak Inverse Voltage V V No (when peak of output = V ) V s s m s V = V V = 2V Peak load Current, I sm s m V = V m s m Vin Vin RMS Current, Irms rd R L rd R L Vin 2rd R L Im Im 2 2 Im 2 DC current, I Im 2 Im 2 Im dc Ripple factor, r 1.21 0.482 0.482 Rectification efficiency (max) 40.6% 81.2% 81.2% Ripple frequency (when input = 50 Hz) 50 Hz 100 Hz 100 Hz E 15
JEE-Physics Ex . 14 A sinusoidal voltage of amplitude 25 volts and frequency 50 Hz is applied to a half wave rectifier using PN diode. No filter is used and the load resistor is 1000 . The forward resistance Rf ideal diode is 10 . Calculate (i) Peak, average and rms values of load currrent. (ii) d.c. power output (iii) a.c. power input (iv) % Rectifier efficiency (v) Ripple factor S o l . (i) m Vm 25 = 24.75 mA Rf RL (10 1000) (ii) (iii) dc m 24.75 7.88mA 3.14 rms m 24.75 12.38mA 2 2 Pdc = Idc2 × RL = (7.88 × 10–3)2 × 103 62 mW Pac = Irms2(Rf + RL) = (12.38 × 10–3)2 × (10 + 1000) 155 mW Pdc (iv) Rectifier efficiency = Pac × 100 62 = × 100 = 40 % 155 (v) Ripple factor = Irms 2 1/2 = 12.38 2 = 1.21 Iac 7.88 1 1 Ex.15 The halfwave rectifier supplies power to a 1 k load. The input supply voltage is 220 V neglecting forward resistance of the diode, calculate (i) Vdc (ii) dc and (iii) Ripple voltage (rms value) S o l . (i) Vdc Vm 2 Vrms 2 220 99 volt 3.14 (ii) dc Vdc 99 99 mA RL 1000 (iii) r = (Vr )rms Vdc or (Vr)rms = r × vdc = 1.21 × 99 = 119.79 volt. Ex.16 A fullwave rectifier supplies a load of 1 K. The a.c. voltage applied to the diodes is 220 volt rms. If diode resistance is neglected, calculate. (i) Average d.c. voltage (ii) Average d.c. current (iii) Ripple voltage (rms) E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 Sol. (i) Average d.c . voltag e Vdc = 2 Vm = 0.636 Vm where Vm = maximum across each half of the secondary winding. If V be the rms voltage across each half of the secondary winding then V Vm 2 Vdc = 0.636 × V 2 = 0.9 V = 0.9 × 220 = 198 volt. Vdc 198 (ii) For fullwave rectifier Idc = RL = 1000 = 198 mA Vr (rms) Vr(rms) = r × Vdc Vr(rms) = 0.482 × 198 = 95.436 volt (iii) r = Vdc 16 E
JEE-Physics Ex.17 A fullwave P.N. diode rectifier used load ressitor of 1500 . No filter is used. Assume each diode to have idealized charcteristic with Rf = 10 and Rr = . Since wave voltage applied to each diode has amplitude of 30 volts and frequency 50Hz. Calculate. (i) Peak, d.c. rms load current (ii) d.c. power input (iii) A.C. power input (iv) Rectifier efficiency Sol. (i) Peak current Im = Vm Rf RL 30volts Im = 10 1500 = 19.9 mA d.c. load current Idc = 2Im = 0.636 Im = 0.636 × 19.9 mA = 12.66 mA. rms = m 19.9 14 mA 2 2 (ii) D.C. Power output Pdc = Id2c × RL = (12.66 × 10–3)2 × 1500 Watt = 240.41 mW (iii) A.C. power input Pin = Irms2 (Rf + R2) = (14 × 10–3)2 (10 + 1500) watt = 295.96 mW FILTER CIRCUIT To reduce A.C. Components Capacitor Filter DC AC + DC AC C RL output of rectifier output L - C Filter DC + low AC DC E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 AC + DC AC C RL output of rectifier output - Filter (Best Filter) output DC + low AC approx pure D.C DC AC + DC AC C1 AC C2 RL output of rectifier E 17
JEE-Physics ZENER DIODE A specifically doped crystal diode which can work in break down region is known as Zener diode. It is always connected in reverse biased condition manner. + Used as a voltage regulator RS RL regulated output Symbol of + Fluctuating Zener diode input In forward biased it works as a simple diode. Voltage regulating circuit of Zener diode SOME OTHER SPECIAL DIODES Photodiode A junction diode made from “light or photo sensitive semiconductor” is called a “photo diode” its symbol . When light of energy \"h'' falls on the photodiode (Here h > energy gap) more electrons move from valence band, to conduction band, due to this current in circuit of photodiode in \"Reverse bias\", increases. As light intensity is increased, the current goes on increases so photo diode is used, \"to detect light intensity\" for example it is used in \"Vedio camera\". Light emitting diode (L.E.D) When a junction diode is “forward biased” energy is released at junction in the form of light due to recombination of electrons and holes. In case of Si or Ge diodes, the energy released is in infra-red region. In the junction diode made of GaAs, InP etc energy is released in visible region such a junction diode is called \"light emitting diode\" (LED) Its symbol Solar cell Solar cell is a device for converting solar energy into electrical. A junction diode in which one of the P or N sections is made very thin (So that the light energy falling on diode is not greatly asorbed before reaching the junction) can be used to convert light energy into electric energy such diode called as solar cell. Its symbol E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 (i) It is operated into photo voltaic mode i.e., generation of voltage due to the bombardment of optical photon. (ii) No external bias is applied. (iii) Active junction area is kept large, because we are intrested in more power. Materials most commonly used for solar cell is Si, As, Cds, CdTe, CdSe, etc. Variable capacitor (Varactor) P – N junction diode can be used as a \"Capacitor\" here depletion layer acts as \"dielectric material\" and remaining \"P\" and \"N\" part acts as metallic plates. its symbol E Diode laser It is intersting form of LED in which special construction helps to produce stimulated radiation as in laser. 18
JEE-Physics E x . 1 8 A zener diode of voltage VZ (=6V) is used to maintain a constant voltage across a load resistance RL (=1000) by using a series resistance RS (=100). If the e.m.f. of source is E (= 9 V), calculate the value of current through series resistance, Zener diode and load resistance. What is the power being dissipated in Zener diode. Sol. Here, E = 9V ; VZ = 6V ; RL = 1000 and RS = 100, Potential drop across series resistor VR = E – VZ = 9 – 6 = 3 V Current through series resistance RS is VR 3 0.03A R 100 Current through load resistance RL is L VZ 6 0.006A RL 1000 Current through Zener diode is IZ = I - IL = 0.03 - 0.006 = 0.024 amp. Power dissipated in Zener diode is PZ = VZ IZ = 6 x 0.024 = 0.144 Watt Ex.19 A Zener diode is specified having a breakdown voltage of 9.1 V with a maximum power dissipation of 364 mW. What is the maximum current that the diode can handle. Sol. Maximum current that the given diode can handle is 364 103 A i.e., 40 mA. 9.1 TRANSISTORE:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 Inventor William Bradford Shockley, John Bardeen and Walter Houser Brattain. Transistor is a three terminal device which transfers a singal from low resistance circuit to high resistance circuit. It is formed when a thin layer of one type of extrinsic semiconductor (P or N type) is sandwitched between two thick layers of other type of extrinsic semiconductor. Each transistor have three terminals which are :- (i) Emitter (ii) Base (iii) Collector Emitter : It is the left most part of the transistor. It emit the majority carrier towards base. It is highly doped and medium in size. Base : It is the middle part of transistor which is sandwitched by emitter (E) and collector (C). It is lightly doped and very thin in size. Collector : It is right part of the transistor which collect the majority carriers emitted by emitter. It has large size and moderate doping. There are two semiconducting PN-junctions in a transistor (i) The junction between emitter and base is known as emitter-base junction (J ). EB (ii) The junction between base and collecter is known as base-collector junction (J ). CB E 19
JEE-Physics TRANSISTOR ARE OF TWO TYPES N-P-N Transistor If a thin layer of P-type semiconductor is sandwitched between two thick layers of N-type semiconductor is known as NPN transistor. E NPN NPN C EC BB P-N-P Transistor If a thin layer of N-type of semiconductor is sandwitched between two thick layer of P-type semiconductor is known as PNP transistor. E P NP PNP C EC BB GOLDEN KEY POINTS • Transistor have two P-N Junction J and J , therefore it can be biased in four following ways as given below: EB CB E NPN C Emitter-Base B Region of working Forward biased Active Reverse biased Collector-Base Inverse Active Reverse biased Reverse biased Cut off Forward biased Forward biased Saturation Reverse biased • Comparsion between E, B and C Forward biased High dopping Low dopping Emitter Medium size Medium dopping Smallest size Base Largest size Collector • The collector region is made physically larger than the emitter. Because collector has to dissipiate much E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 greater power. • Transistor all mostly work in active region in electronic devices & transistor work as amplifier in Active region only. • Transistor i.e. It is a short form of two words \"Transfer resistors\". Signal is introduced at low resistance circuit and out put is taken at high resistance circuit. • Base is lightly doped. Otherwise the most of the charge carrier from the emitter recombine in base region and not reaches at collector. • Transistor is a current operated device i.e. the action of transistor is controlled by the motion of charge carriers. i.e. current 20 E
JEE-Physics WORKING OF NPN TRANSISTOR The emitter Base junction is forward bias and collector base junction is reversed biased of n-p-n transistor in circuit (A) and symbolic representation is shown in Figure. emitter-base collectoremitter-base IE IC junction junction N PN IE e IC h mA E BC mA VEE IB VCC VEE IB VCC When emitter base junction is forward bias, electrons (majority carriers) in emitter are repelled toward base. The barrier of emitter base junction is reduced and the electron enter the base, about 5% of these electron recombine with hole in base region result in small current (I ). b The remaining electron ( 95%) enter the collector region because they are attracted towards the positive terminal of battery. For each electron entering the positive terminal of the battery is connected with collector base junction an electron from negative terminal of the battery connected with emitter base junction enters the region. The emitter current (I ) is more than the collector (I ). ec The base current is the difference between I and I and proportional to the number of electron hole recombination ec in the base. I = I +I e bc WORKING OF PNP TRANSISTOR When emitter-base junction is forward biased holes (majority carriers) in the emitter are repelled towards the base and diffuse through the emitter base junction. The barrier potential of emitter-base junction decreases and hole enter the n-region (i.e. base). A small number of holes ( 5%) combine with electron of base-region resulting small current (I ). The remaining hole ( 95%) enter into the collector region because they are attracted towards b negative terminal of the battery connected with the collector-base junction. These hole constitute the collector current (I ). c E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 emitter-base collectoremitter-base IC IC junction junction VEE IB + VCC P NP IE mA E BC mA VEE IB VCC As one hole reaches the collector, it is neutralized by the battery. As soon as one electron and a hole is neutralized in collector a covalent bond is broken in emitter region. The electron hole pair is produced. The released electron enter the positive terminal of bettary and hole more towards the collector. E 21
JEE-Physics Basic Transistor Circuit Configurations :- To study about the characterstics of transistor we have to make a circuit [In which four terminals are required. But the transistor have three terminals, so one of the terminal of transistor is made common in input and output both. Thus, we have three possible configuration of transistor circuit. (i) Common base configuration (ii) Common emitter configuration (iii) Common collector configuration In these three common emitter is widely used and common collector is rarely used. Common emitter characterstics of a transistor Circuit Diagram : R1 R2 ( ) ( ) Circuit diagram for characteristic curve of n-p-n transistor in CE mode Input characterstics input characteristic curves The variation of base current (I ) (input) with base emitter voltage (V ) 100 0V 5V b EB 75 = = at constant-emitter voltage (V ) is called input characterstic. 50 CE 25 V e ce (i) Keep the collector-emitter voltage (V ) constant (say V = 1V) 0 V CE CE c (ii) Now change emitter base voltage by R and note the corresponding I b(A) 1 value of base current (I ). 0.2 0.4 0.6 0.8 0.8 b (iii) Plot the graph between V and I . EB b (iv) A set of such curves can be plotted at different (V = 2V) Vbe(volt) CE Output characterstics The variation of collector current I (output) with collector-emitter 10 Ib = 100A c 8 Ib = 75A 6 Ib = 50A voltage (V ) at constant base current (I ) is called output characterstic. CE b 4 Ib = 25A 2 Ib =0 (i) Keep the base current (Ib) constant (say Ib = 10A) (ii) Now change the collector-emitter voltage (V ) using variable resistance R and 1 234 5 VCE(volt) CE 2 IC(mA) note the corresponding values of collector current (I ). E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 c (iii) Plot the graph between (V versus I ) CE c (iv) A set of such curves can be plotted at different fixed values of base current (say 0, 20 A, 30 A etc.) TRANSISTOR AS AN AMPLIFIER The process of increasing the amplitude of input signal without distorting its wave shape and without changing its frequency is known as amplification. A device which increases the amplitude of the input signal is called amplifier. A transistor can be used as an amplifier in active state. A basic circuit of a common emitter transistor amplifier is shown. 22 E
IC – JEE-Physics intput C Vo amplifier signal B IB RL 3. Common Collector (CC) + amplified E VCE VCC output signal C C – Vi IE + IB IC +– VBB common emitter amplifier NPN transistor Comparative study of transistor configurations 1. Common Base ( C B ) 2. Common Emitter (CE) C B C E E C B C B E CB B CE E CC C B E C IE IC C E IC IE IB B B IC C BB IB IE IB E E C Input Resistance Low (100 ) High (750 ) Very High 750 k Output resistance Current Gain Very High High Low (A or ) (A or ) (A or ) I I I IC IC IE = IE = IB = IB Voltage Gain AV Vo = IC R L A V Vo IC R L AV Vo = IE R L Vi IE R i Vi = IB R i Vi IBRi A v = RL A v = RL A v = RL Ri Ri Ri E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 less than 1 Power Gain Ap Po Ap Po Ap Po Pi Pi Pi Ap 2 RL A p 2 RL Ap 2 RL Ri Ri Ri Phase difference same phase opposite phase same phase (between output and input) Application For High Fequency For Audible fequency For Impedance Matching E 23
JEE-Physics Relation Between and -- I = I + I I = I + I 1+ E B C E B C 1- divide by I divide by I 1 C B 1- IE IB 1 IC IC IE 1+ IC 11 IB IB +1 1- GOLDEN KEY POINTS • In transistor charge carriers move from emitter to collector. Emitter send the charge carriers and collector collect them this happen only when emitter-base junction is forward bias and collector-base junction is reverse bias (base of amplifier) • In transistor reverse bias is high as compared to forward bias so that the charge carriers move from emitter to base exert a large attractive force to enter in collector region so base current is very less. • CE configuration is widely used becasue it have large voltage and power gain as compared to other amplifiers. • In amplifier negative feed back is used to stabilized the gain. • CC is used for impdence matching for connecting two transistors in cascade. Q . 2 0 A transistor is a current operated device. Explain why ? E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 A n s . The action of a transistor is controlled by the charge carriers (electrons or holes). That is why a transistor is a current operated device. Q . 2 1 In a transistor, reverse bias is quite high as compared to the forward bias. Why ? A n s . In a transistor, charge carriers (electrons or holes) move from emitter to collector through the base. The reverse bias on collector is made quite high so that it may exert a large attractive force on the charge carriers to enter the collector region. These moving carriers in the collector constitute a collector current. Q . 2 2 A transistor is a temperature sensitive device. Explain. A n s . In a transistor, conduction is due to the movement of current carriers electrons and holes. When temperature of the transistor increases, many covalent bonds may break up, resulting in the formation of more electrons and holes. Thus, the current will increase in the transistor. This current gives rise to the production of more heat energy. the excess heat causes complete breakdown of the transistor. Q . 2 3 The use of a transistor in common-emitter configuration is preferred over the common-base configuration. Explain why ? A n s . The current gain and hence voltage gain in the common-emitter configuration is much more than i of common- base configuration. Hence the former is preferred over the later. 24 E
JEE-Physics Q . 2 4 Why do we prefer transistor over the vacuum tubes in the portable radio receivers ? A n s . This is because of two reasons : (i) Transistor is compact and small in size than the vacuum tube. (ii) Transistor can operate even at low voltage which can be supplied with two or three dry cells. Q . 2 5 Why a transistor cannot be used as a rectifier ? A n s . If transistor is to be used as a rectifier the either emitter-base or base-collector has to used as diode. For equated working of the said set of diodes, the number density of charge carriers in emitter and base or base and collector must be approximately same. As base is lightly doped and comparatively thin, so emitter cannot work as a rec t i f i er. Ex.26 In a transistor, the value of is 50. Calculate the value of . S o l . = 50 50 – 50 = 50 = 0.98 1 50 = 1 51 Ex.27 Calculate the collector and emitter current for which Ib = 20 A, = 100 S o l . = 100, I = 20 A b I = I = 100 × 20 × 10–6 = 2000 A c b I = I + I = 20 + 2000 = 2020 A = 2.02 × 10–3 A = 2.02 mA ebc Ex.28 For a common emitter amplifier, current gain = 50. If the emitter current is 6.6 mA, calculate the collector and base current. Also calculate current gain, When emitter is working as common base amplifier. S o l . = 50 ; I = 6.6 mA e Ic I = Ib = 50I ...(i) Ib b c I = I + I using equation (i) we get 6.6 = 50 I + I = 51I bb b ecb 6.6 or Ib 51 = 0.129 mA HenceI = 50 × 6.6 = 6.47 mA and = 50 0.98 c 1 51 51 E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 Ex.29 Transistor with = 75 is connected to common-base configuration. What will be the maximu collector current for an emitter current of 5 mA ? Sol. = 75, I = 5 mA e 1 75 = 1 75 – 75 = or 76 = 75 or 75 = 76 Ic I = Ie = 75 × 5 = 4.93 mA Ie c 76 E 25
JEE-Physics Ex.30 The base current is 100 A and collector current is 3 mA. (a) Calculate the values of , I and e (b) A change of 20 A in the base current produces a change of 0.5 mA in the collector current. Calculate . a.c. Sol. I = 100 A = 0.100 mA, I = 3 mA b c (a) Ic = 3 = 30 Ib 0.100 1 1 30 30 = 0.97 and I = Ic 3 31 = 3.1 mA 30 31 e 30 (b) Ib = 20 A = 0.02 mA , Ic = 0.5 mA ac Ic 0.5 25 I b 0.02 Ex.31 In npn transistor circuit, the collector current is 10 mA. If 95% of the electrons emitted reach the collector, what is the base current ? S o l . I = 95% I = 0.95 I c ce 100 100 10mA = 10.53 mA ( I = 10 mA) I = Ic c 95 95 e Now I = I + I I = I – I = 10.53 – 10 = 0.53 mA ecb bec Ex.32 In an NPN transistor 1010 electrons enter the emitter in 10–6 s and 2% electrons recombine with holes in base, then current gain and are : Ne 1010 1.6 1019 = 1.6 mA S o l . Emitter current I = = et 10 6 Base current I = 2 1.6 = 0.032 mA b 100 but I = I + I I = I – I = 1.6 – 0.032 = 1.568 mA e c b ceb Ic 1.568 0.98 Ie 1.6 and Ic 1.568 49 E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 Ib 0.032 FEEDBACK Feedback are two types : Positive feedback When input and output are in the same phase then positive feedback is there. It is used in oscillators. Negative feedback If input and output are out of phase and some part of that is feedback to input is known as negative feedback. It is used to get constant gain amplifier. 26 E
JEE-Physics TR ANSISTOR AS AN OSCILLATOR Oscillator is device which delivers a.c. output wave form of desired frequency from d.c. power even without input singal excitation. C T1 mutual inductance 1 (coupling through The electric oscillations are produced by L– C circuit n-p-n 2 magnetic field) (i.e. tank circuit containing inductor and capacitor). These 3 T'2 output oscillations are damped one i.e. their amplitude decrease with the passage of time due to the small resistance of the T2 inductor. In other words, the energy of the L – C oscillations L 4 decreases. If this loss of energy is compensated from outside, E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 then undamped oscillations (of constant amplitude) can be S1 (switch) obtained. This can be done by using feed back arrangement and a transistor in the circuit. L – C circuit producing L – C oscillations consists of an inductor of inductance L and capacitor of variable capacitance C inductor of inductance L' is connected in the collector-emitter circuit through a battery and a tapping key (K). Inductors L and L' are inductively coupled (Figure) Working When key K is closed, collector current begins to flow through the coil L. As this current grows, magnetic flux linked with coil L increase (i.e. changes). Since coil L is inductively coupled with L, so magnetic flux linked with coil L' also changes. Due to change in magnetic flux, induced e.m.f. is set up across the coil L'. The direction of induced e.m.f. is such that the emitter-base junction is forward biased. As a result of this biasing, emitter current I increases which in turn increases the collector current I [ I = I + I ]. e c e b c With the increase in collector current, magnetic flux linked with coil L also increases. This increases the e.m.f. induced in the coil L'. The increased induced e.m.f. increases the forward bias of emitter-base junction. Hence emitter current is further increased which in turn increases the collector current. The process of increasing the collector current continues till the magnetic flux linked with coil L' becomess maximum (i.e. constant). At this stage, the induced e.m.f. in coil L' becomes zero. The upper plate of the capacitor C gets positively charged during this process. When induced e.m.f. becomes zero, the capacitor C starts discharging through the inductor L. The emitter current starts decreasing resulting in the collector current. With decreasing collector current which flows through L', e.m.f. is again induced in the coil L' but in the opposite direction. It opposes the emitter current and hence collector current ultimately decreases to zero. The change in magnetic flux linked with coil L' stops and hence induced e.m.f. in the coil L becomes zero. At this stage, the capacitor gets discharged through coil L but now in the opposite direction. Now the emitter current and hence collector current increase but now in the opposite direction . This process repeats and the collector current oscillates between maximum and minimum values. f 1 27 2 LC E
JEE-Physics ADVANTAGES OF SEMICONDUCTOR DEVICES OVER VACUUM TUBES Advantages Semiconductor devices are very small in size as compared to the vacuum tubes. Hence the circuits using semiconductor devices are more compact. In vacuum tubes, current flows when the filament is heated and starts emitting electrons. So, we have to wait for some time for the operation of the circuit. On the other hand, in semiconductor devices no heating is required and the circuit begins to operate as soon as it switched on. Semiconductor devices require low voltage for their operation as compared to the vacuum tube. So a lot of electrical power is saved. Semiconductor devices do not produce any humming noise which is large in case of vacuum tube. Semiconductor devices have longer life than the vacuum tube. Vacuum tube gets damaged when its filament is burnt. Semiconductor devices are shock proof. The cost of production of semiconductor-devices is very small as compared to the vacuum tubes. Semiconductor devices can be easily transported as compared to vacuum tube. Disadvantages Semiconductor devices are heat sensitive. They get damaged due to overheating and high voltages. So they have to be housed in a controlled temperature room. The noise level in semiconductor devices is very high. Semiconductor devices have poor response in high frequency range. Q . 3 3 Why is a transistor so called ? A n s . The word Transistor can be treated as short form of two words 'transfer resistor'. In a transistor, a signal is introduced in the low resistance circuit and output is taken across the high resistance circuit. Thus, a transistor helps to transfer the current from low resistance part to the high resistance part. Q . 3 4 The base region of a transistor is lightly doped. Explain why ? E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Electronics\\Eng\\1. Electronics-Semi Conductor Theory.p65 A n s . In a transistor, the majority carriers (holes or electrons) from emtter region move towards the collector region through base. If base is made thick and highly doped, then majority of carriers from emitter will combine with the carriers in the base and only small number of carriers will reach the collector. Thus the output or collector current will be considerably small. To get large output or collector current, base is made thin and lightly doped so that only few electron-hole combination may take place in the base region. Q . 3 5 Explain why the emitter is forward biased and the collector is reverse biased in a transistor ? A n s . In a transistor, the charge carriers move from emitter to collector. The emitter sends the charge carriers and collector collects them. This can happen only if emitter is forward biased and the collector is reverse biased so that it may attract the carriers. 28 E
JEE-Physics METHOD OF IMPULSE AND MOMENTUM Until now, we have studied kinematics of particles, Newton’s laws of motion and methods of work and energy. Newton’s laws of motion describe relation between forces acting on a body at an instant and acceleration of the body at that instant. Therefore, it only helps us do analyze what is happening at an instant. The work kinetic energy theorem is obtained by integrating equation of motion (F ) over a path. Therefore, methods of work and energy ma help us to in exploring change in speed over a position interval. Now, we direct our attention on another principle – principle of impulse and momentum. It is obtained when equation of motion (F ) is integrated with respect to ma time. Therefore, this principle facilitates us with method to explore what is happening over a time interval. Impulse of a Force Net force applied on a rigid body changes momentum i.e. amount of motion of that body. A net force for a longer duration cause more change in momentum than the same force acting for shorter duration. Therefore duration in which a force acts on a body together with magnitude and direction of the force decide effect of the force on the change in momentum of the body. Linear impulse or simply impulse of a force is defined as integral of the force with respect to time. a its t to t is given following equation. If a force F acts on body, impulse in a time interval from by the i f tf Fdt Imp ti If the force is constant, its impulse equals to product of the force vector F and time interval t. I mp F t F For one-dimensional force, impulse equals to area between force-time graph and the time axis. In the given figure is shown how a force F along x-axis varies with time t. Impulse of this force in time interval ti to ti tf t tf equals to area of the shaded portion. If several forces F1 , F2 , F3 ..... Fn act on a body in a time interval, the total impulse I mp of all these forces equals to impulse of the net force. tf tf tf tf dt ti F1dt ti F2dt ti Fndt ti Imp .............. F1 F2 ............. Fn NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 Impulse is measured in newton-second. Dimensions of impulse are MLT1 Example F Calculate impulse of force 3t 2ˆi 2t 1 ˆj 2kˆ N over the time interval from t = 1 s to t = 3 s. Solution. tf Fdt Imp Imp 3 3t 2ˆi 2t 1 ˆj 2kˆ dt t 3ˆi t 2 t ˆj 2tkˆ13 ˆi t 3 3 ˆj t 2 t 3 kˆ2t 13 1 1 ti 1 26ˆi 6 ˆj 4kˆ N-s E 1
JEE-Physics Example A one-dimensional force F varies with time according to the given F (N) graph. Calculate impulse of the force in following time intervals. 10 (a) From t = 0 s to t = 10 s. 5 10 15 t (s) (b) From t = 10 s to t = 15 s. (c) From t = 0 s to t = 15 s. Solution. F (N) AB For one-dimensional force, impulse equals to area between force- time graph and the time axis. 10 (a) I = Area of trapazium OABC = 75 N-s CE 010 O 5 10 15 t (s) (b) I = – Area of triangle CDE = – 25 N-s D 1015 (c) I = Area of trapazium OABC – Area of triangle CDE = 50 N-s 015 Impulse Momentum Principle Consider body of mass m in translational motion. When it is moving with velocity , net external force acting v on it is F . Equation of motion as suggested by Newton’s second law can be written in the form d (mv) Fdt If the force acts during time interval from t to t and velocity of the body changes from vi to v f , integrating the if above equation with time over the interval from ti to t we have f, tfti Fdt mvf mvi Here left hand side of the above equation is impulse I mp of the net force F in time interval from ti to tf, and quantities mv i and mvf on the right hand side are linear momenta of the particle at instants t and tf. If we i denote them by symbols pi and pf , the above equation can be written as I mp pf pi The idea expressed by the above equation is known as impulse momentum principle. It states that change in the momentum of a body in a time interval equals to the impulse of the net force acting on the body during the concerned time interval. For the ease of application to physical situations the above equation is rearranged as pi Imp pf This equation states that impulse of a force during a time interval when added to momentum of a body at the NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 beginning of an interval of time we get momentum of the body at the end of the interval concerned. Since impulse and momentum both are vector quantities, the impulse momentum theorem can be expressed by there scalar equations making use of Cartesian components. p1x I mp,x p2 x p1y I mp,y p2 y p1z Imp,z p2z The impulse momentum principle is deduced here for a single body moving relative to an inertial frame, therefore impulses of only physical forces are considered. If we are using a non-inertial reference frame, impulse of corresponding pseudo force must also be considered in addition to impulse of the physical forces. 2E
JEE-Physics How to apply Impulse Momentum Principle The impulse momentum principle is deduced here for a single body, therefore it is recommended at present to use it for a single body. To use this principle the following steps should be followed. ( i ) Identify the initial and final positions as position 1 and 2 and show momenta p1 and p2 of the body at these instants. ( i i ) Show impulse of each force acting on the body at an instant between positions 1 and 2. (i i i ) Use the impulse obtained in step (ii) and momenta obtained in step (i) into equation pi I mp pf . Consider a particle moving with momentum p1 in beginning. It is acted upon by two forces, whose impulses in a time interval are Imp1 and Imp2 . As a result, at the end of the time interval, momentum of the particle becomes p2 . This physical situation is shown in the following diagram. Such a diagram is known as impulse momentum diagram. I mp1 p2 p1 p1 I mp2 I mp1 I mp2 Example A particle of mass 2 kg is moving with velocity vo 2iˆ 3 ˆj m/s in free space. Find its velocity 3 s after a constant force 3iˆ 4ˆj N starts acting on it. F Solution. mv f mv o F t pf pi Imp Substituting given values, we have 2 2iˆ 3ˆj 3iˆ 4ˆj 3 13iˆ 6 ˆj 2v f 6.5iˆ 3ˆj m/s vf NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 Example vo A particle of mass 2 kg is moving in free space with velocity 2ˆi 3ˆj kˆ m/s is acted upon by force F 2ˆi ˆj 2kˆ N. Find velocity vector of the particle 3 s after the force starts acting. Solution. mv f mv o F t pf pi Imp Substituting given values, we have E 2v f 2 2ˆi 3ˆj kˆ 2ˆi ˆj 2kˆ 3 10ˆi 3ˆj 4kˆ vf 5ˆi 1.5 ˆj 2kˆ m/s 3
JEE-Physics Example F 20 N A box of mass m = 2 kg resting on a frictionless horizontal ground is acted upon by a horizontal force F, which varies as shown. Find speed of the particle 10 N when the force ceases to act. Solution. 2s 4s t pf pi Imp mvi Fdt tf mvf ti 2v 20 1 20 4 2 v = 20 m/s Example Two boxes A and B of masses m and M interconnected by an ideal rope and ideal pulleys are held at rest as shown. When it is released, box B accelerates downwards. Find velocities of box A and B as function of time t after system has been released. AB Solution. We first explore relation between accelerations aA and aB of the boxes A and B, which can be written either by using constrained relation or method of virtual work or by inspection. T T T vA = 2vB ...(i) v A B v A B Applying impulse momentum principle to box A y Tt zero x mv A p2 y p1y Imp,y Mv A 0 Tt mgt ...(ii) A mgt Applying impulse momentum principle to box B y NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 zero 2Tt x p2 y p1y Imp,y mv B 0 Mgt 2Tt ...(iii) B Ma B Mgt From equations (i), (ii) and (iii), we have vA 2 M 2m gt and vB M 2m gt M 4m M 4m 4 E
JEE-Physics Impulsive Motion Sometimes a very large force acts for a very short time interval on a particle and produces finite change in momentum. Such a force is known as impulsive force and the resulting motion as impulsive motion. When a batsman hits a ball by bat, the contact between the ball and the bat lasts for a very small duration t, but the average value of the force F exerted by the bat on the ball is very large, and the resulting impulse Ft is large enough to change momentum of the ball. During an impulsive motion, some other forces of magnitudes very small in comparison to that of an impulsive force may also act. Due to negligible time interval of the impulsive motion, impulse of these forces becomes negligible. These forces are known as non-impulsive forces. Effect of non-impulsive forces during an impulsive motion is so small that they are neglected in analyzing impulsive motion of infinitely small duration. Non-impulsive forces are of finite magnitude and include weight of a body, spring force or any other force of finite magnitude. When duration of the impulsive motion is specified, care has to be taken in neglecting any of the non-impulsive force. In analyzing motion of the ball for very small contact duration (usually in mili-seconds), impulse of the weight of the ball has to be neglected. Unknown reaction forces may be impulsive or non-impulsive; their impulse must therefore be included. Example A 100 gm ball moving horizontally with 20 m/s is struck by a bat, as a result 35 m/s 20 m/s it starts moving with a speed of 35 m/s at an angle of 37° above the horizontal in the same vertical plane as shown in the figure. (a) Find the average force exerted by the bat if duration of impact is 0.30 s. (b) Find the average force exerted by the bat if duration of impact is 0.03 s. (c) Find the average force exerted by the bat if duration of impact is 0.003 s. (d) What do you conclude for impulse of weight of the ball as duration of contact decreases? Solution. The impulse momentum diagram of the ball is shown in the figure below. Here F, mg, and t represent the average value of the force exerted by the bat, weight of the ball and the time interval. mgt = 1.0t y P = 3.5 f p = 2 p = 2.1 ix x 37° fy Ft p = 2.8 F t fx x F t y NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 Applying principle of impulse and momentum in x- direction, we have p fx p ix I mp,x 2 Fx t 2.8 Fx 4.8 N ...(i) t Applying principle of impulse and momentum in y- direction, we have p fy piy I mp,y 0.0 Fy t 1.0t 2.1 Fy 2.1 1.0 N ...(ii) t E5
JEE-Physics (a) Substituting t = 0.30 s, in equations (i) and (ii), we find (b) Substituting t = 0.03 s, in equations (i) and (ii), we find F 16iˆ 8ˆj N (c) Substituting t = 0.003 s, in equations 1 and 2, we find F 160ˆi 71ˆj N F 1600ˆi 701ˆj N (d) It is clear from the above results that as the duration of contact between the ball and the bat decreases, effect of the weight of the ball also decreases as compared with that of the force of the bat and for sufficiently short time interval, it can be neglected. Momentum and Kinetic Energy A moving particle possesses momentum as well as kinetic energy. If a particle of mass m is moving with velocity v, magnitude of its momentum p and its kinetic energy K bear the following relation. p2 1 K pv 2m 2 Example An object is moving so that its kinetic energy is 150 J and the magnitude of its momentum is 30.0 kg-m/s. With what velocity is it traveling? Solution. K p 2 1 pv v 2 150 10.0 m/s 2m 2 30.0 Internal and external Forces and System of interacting Particles Bodies applying forces on each other are known as interacting bodies. If we consider them as a system, the forces, which they apply on each other, are known as internal forces and all other forces applied on them by bodies not included in the system are known as external forces. Consider two blocks A and B placed on a frictionless horizontal floor. Their weights W and W are counterbalanced by normal A B 12 W W reactions N and N on each of them from the floor. Push F 1 2 12 FA N NB by the hand is applied on A. The forces of normal reaction on-A on-B Non-A and Non-B constitute Newton’s third law action-reaction N N NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 pair, therefore are equal in magnitude and opposite in 1 direction. Among these forces weights W and W applied by 2 12 the earth, normal reactions N and N applied by the ground 12 and the push F applied by the hand are external forces and normal reactions Non-A and Non-B are internal forces. If the blocks are connected by a spring and the block A is either pushed or pulled, the forces W , W , N and 1 21 N still remain external forces for the two block system and the forces, which the spring applies on each other 2 are the internal forces. Here force of gravitational interaction between them being negligible has been neglected. We can conceive a general model of two interacting particles. In the m 1 F12 F21 m figure is shown a system of two particles of masses m and m . Particle 2 12 m attracts m with a force F12 and m attracts (or pulls) m with a 12 21 Particles attracting each other force F21 . These forces F12 and F21 are the internal forces of this two- F12 m 1 m particle system and are equal in magnitude and opposite in directions. 2 F21 Instead of attraction may repeal each other. Such a system of two particles repealing each other is also shown. Particles repealing each other 6E
JEE-Physics In similar way we may conceive a model of a system of n interacting particles having masses m m ,... mi. .. .mj.. ... and mn respectively. The m m 1, 2 i Fij n forces of interaction Fij and Fji between mi and mj are shown in the m m 1 3 figure. Similar to these other particles may also interact with each m2 F ji other. These forces of mutual interaction between the particles are m j internal forces of the system. Any of the two interacting particles always System of n interacting particles. apply equal and opposite forces on each other. Here fore simplicity only the forces Fij and Fji are shown. Principle of Conservation of linear momentum The principle of conservation of linear momentum or simply conservation of momentum for two or more interacting bodies is one of guiding principles of the classical as well as the modern physics. To understand this principle, we first discuss a system of two interacting bodies, and then extend the ideas developed to a system consisting of many interacting bodies. Consider a system of two particles of mass m and m . Particle m 1 2 1 F12 F21 m m attracts m with a force F12 and m attracts m with a force F21 . 2 12 21 These forces have equal magnitudes and opposite directions as shown Particles attracting each other in the figure. If the bodies are let free i.e. without any external force acting on any of them, each of them move and gain momentum equal F12 m 1 m to the impulse of the force of interaction. Since equal and opposite 2 F21 interaction forces act on both of them for the same time interval, the momenta gained by them are equal in magnitude and opposite in Particles repealing each other direction resulting no change in total momentum of the system. However, if an external force acts on any one of them or different forces with a nonzero resultant act on both of them, the total momentum of the bodies will certainly change. If the system undergoes an impulsive motion, total momentum will change only under the action of external impulsive force or forces. Internal impulsive forces also exist in pairs of equal and opposite forces and cannot change the total momentum of the system. Non-impulsive forces if act cannot change momentum of the system by appreciable amount. For example, gravity is a non-impulsive force, therefore in the process of collision between two bodies near the earth the total momentum remains conserved. NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 The total momentum of a system of two interacting bodies remains unchanged under the action of the forces of interaction between them. It can change only if a net impulse of external force is applied. In similar way we may conceive a model of a system of n interacting particles having masses m m, ....mi....mj.....and m respectively. The m m 1, 2 n i n forces of interaction Fij and Fji between mi and mj are shown in the m Fij m 1 3 F ji figure. Since internal forces exist in pairs of equal and opposite forces, m in any time interval of concern each of them have a finite impulse but j m2 their total impulse is zero. Thus if the system is let free, in any time System of n interacting particles. interval momentum of every individual particle changes but the total momentum of the system remains constant. It can change only if external forces are applied to some or all the particles. Under the action of external forces, the change in total momentum of the system will be equal to the net impulse of all the external forces. E7
JEE-Physics Thus, total momentum of a system of particles cannot change under the action of internal forces and if net impulse of the external forces in a time interval is zero, the total momentum of the system in that time interval will remain conserved. p final p initial The above statement is known as the principle of conservation of momentum. It is applicable only when the net impulse of all the external forces acting on a system of particles becomes zero in a finite time interval. It happens in the following conditions. • When no external force acts on any of the particles or bodies. • When resultant of all the external forces acting on all the particles or bodies is zero. • In impulsive motion, where time interval is negligibly small, the direction in which no impulsive forces act, total component of momentum in that direction remains conserved. Since force, impulse and momentum are vectors, component of momentum of a system in a particular direction is conserved, if net impulse of all external forces in that direction vanishes. Example Two blocks of masses m and M are held against a compressed spring on a frictionless horizontal floor with the help of a light thread. When the thread is cut, the smaller block leaves the spring with a velocity u relative to the larger block. Find the recoil velocity of the larger block. Mm Solution. NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 When the thread is cut, the spring pushes both the block, and impart them momentum. The forces applied by the spring on both the block are internal forces of the two-block system. External forces acting on the system are weights and normal reactions on the blocks from the floor. These external forces have zero net resultant of the system. In addition to this fact no external force acts on the system in horizontal direction, therefore, horizontal component of the total momentum of the system remains conserved. Velocities of both the objects relative to the ground (inertial frame) are shown in the adjoining figure. v uv Mm Since before the thread is cut system was at rest, its total momentum was zero. Principle of conservation of momentum for the horizontal direction yields n Mv m(u v ) 0 p horizontal 0 i 1 v mu M m 8E
JEE-Physics Example A shell fired vertically up, when reaches its highest point, explodes North into three fragments A, B and C of masses mA = 4 kg, mB = 2 kg 3 m/s and mC = 3kg. Immediately after the explosion, A is observed A 4.5 m/s moving with velocity vA = 3 m/s towards north and B with a velocity B East C v = 4.5 m/s towards east as shown in the figure. Find the velocity B v of the piece C. C Solution. Explosion takes negligible duration; therefore, impulse of gravity, which is a finite external force, can be neglected. The pieces fly off acquiring above-mentioned velocities due to internal forces developed due to expanding gases produced during the explosion. The forces applied by the expanding gases are internal forces; hence, momentum of the system of the three pieces remains conserved during the explosion and total momentum before and after the explosion are equal. Assuming the east as positive x-direction and the north as positive y-direction, the momentum vectors p A and p B of pieces A and B become mAv A ˆj 12 kg-m/s and mBv B iˆ 9 kg-m/s pA pB Before the explosion, momentum of the shell was zero, therefore from the principle of conservation of momentum, the total momentum of the fragments also remains zero. (9iˆ 12ˆj ) pA pB pC 0 pC From the above equation, velocity of the piece C is (3iˆ 4 ˆj ) = 5 m/s, 53° south of west. vC pC mC Example In free space, three identical particles moving with velocities v oˆi , 3v o ˆj and 5v okˆ collide successively with each other to form a single particle. Find velocity vector of the particle formed. Solution. Let m be the mass of a single particle before any of the collisions. The mass of particle formed after collisions must be 3m. In free space, no external forces act on any of the particles, their total momentum remains conserved. Applying principle of conservation of momentum, we have m v o iˆ 3 m v o ˆj 5 m v o kˆ 3 m pfinal v p initial NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 v 1 v o ˆi 3ˆj 5kˆ m/s 3 Example A bullet of mass 50 g moving with velocity 600 m/s hits a block of 600 m/s mass 1.0 kg placed on a rough horizontal ground and comes out of Block the block with a velocity of 400 m/s. The coefficient of friction between the block and the ground is 0.25. Neglect loss of mass of the block as the bullet pierces through it. (a) In spite of the fact that friction acts as an external force, can you apply principle of conservation of momentum during interaction of the bullet with the block? (b) Find velocity of the block immediately after the bullet pierces through it. (c) Find the distance the block will travel before it stops. E9
JEE-Physics Solution. (a) There is no net external force in the vertical direction and in the horizontal direction, only external force friction is non-impulsive, therefore momentum of the bullet-block system during their interaction remains conserved. (b) Let us denote velocities of the bullet before it hits the block and immediately after it pierces through the block by vbo and vb, velocity of the block immediately after the bullet pierces through it is vB and masses of the bullet and the block by m and M respectively. These are shown in the adjacent figure. v v bo B v mM b Mm Immediately before the bullet hits the block Immediately after the bullet pierces the block Applying principle of conservation of momentum for horizontal component, we have mvbo mv b Mv B vB m vbo vb M Substituting the given values, we have vB = 10 m/s (c) To calculate distance traveled by the block before it stops, work kinetic energy theorem has to be applied. K1 1 M v 2 Mg K =0 2 B N = Mg 2 M v F = Mg B k M x During sliding of the box on the ground only the force of kinetic friction does work. W12 K 2 K 1 Mgx 0 1 M v 2 2 B x v 2 B 2g Substituting given values, we have x = 20 m Example Ballistic Pendulum : A ballistic pendulum is used to measure speed of bullets. It consists of a wooden block NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 suspended from fixed support. A wooden block of mass M is suspended with the help of two threads to prevent rotation while swinging. A bullet of mass m moving v horizontally with velocity vo hits the block and becomes embedded in o the block. Receiving momentum from the bullet, the bullet-block system swings to a height h. Find expression for speed of the bullet in terms of given quantities. Solution. When the bullet hits the block, in a negligible time interval, it becomes embedded in the block and the bullet- block system starts moving with horizontally. During this process, net force acting on the bullet-block system in vertical direction is zero and no force acts in the horizontal direction. Therefore, momentum of the bullet-block system remains conserved. 10 E
JEE-Physics Before the bullet Immediately after the hits the block bullet becomes embedded in the block v p o Let us denote momentum of the bullet-block system immediately after the bullet becomes embedded in the block by p and apply principle of conservation of momentum to the system for horizontal component of momentum. p = mvo Using equation K = p2 / 2m, we can find kinetic energy K of the bullet-block system immediately after the 1 bullet becomes embedded in the block. K1 mvo 2 2M m During swing, only gravity does work on the bullet-block system. Applying work-kinetic energy theorem during swing of the bullet-block system, we have Immediately after the K K =0 bullet becomes 1 2 embedded in the block p h W12 K 2 K 1 M m gh 0 mvo 2 2 M m Rearranging terms, we have vo M m 2gh m NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 Impact between two bodies Impact or collision is interaction of very small duration between two bodies in which the bodies apply relatively large forces on each other. Interaction forces during an impact are created due to either direct contact or strong repulsive force fields or some connecting links. These forces are so large as compared to other external forces acting on either of the bodies that the effects of later can be neglected. The duration of the interaction is short enough as compared to the time scale of interest as to permit us only to consider the states of motion just before and after the event and not during the impact. Duration of an impact ranges from 1023 s for impacts between elementary particles to millions of years for impacts between galaxies. The impacts we observe in our everyday life like that between two balls last from 103 s to few seconds. Central and Eccentric Impact The common normal at the point of contact between the bodies is known as line of impact. If mass centers of the both the colliding bodies are located on the line of impact, the impact is called central impact and if mass centers of both or any one of the colliding bodies are not on the line of impact, the impact is called eccentric impact. E 11
JEE-Physics B A B A Common Normal Common Common Normal Common or Tangent or Tangent Line of Impact Line of Impact Central Impact Eccentric Impact Central impact does not produce any rotation in either of the bodies whereas eccentric impact causes the body whose mass center is not on the line of impact to rotate. Therefore, at present we will discuss only central impact and postpone analysis of eccentric impact to cover after studying rotation motion. Head–on (Direct) and Oblique Central Impact If velocities vectors of the colliding bodies are directed along the line of impact, the impact is called a direct or head-on impact; and if velocity vectors of both or of any one of the bodies are not along the line of impact, the impact is called an oblique impact. A B tB n u A u u u A B B n A Central Impact Oblique Impact In this chapter, we discuss only central impact, therefore the term central we usually not use and to these impacts, we call simply head-on and oblique impacts. Furthermore, use of the line of impact and the common tangent is so frequent in analysis of these impacts that we call them simply t-axis and n-axis. Head–on (Direct) Central Impact To understand what happens in a head-on impact let us consider two balls A and B of masses m and m A B moving with velocities uA and uB in the same direction as shown. Velocity uA is larger than uB so the ball A hits the ball B. During impact, both the bodies push each other and first they get deformed till the deformation reaches a maximum value and then they tries to regain their original shape due to elastic behaviors of the materials forming the balls. AB AB AB uu vv u u A B AB Deformation Restitution NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 Period Period Instant when Instant of maximum Instant when impact starts deformation impact ends The time interval when deformation takes place is called the deformation period and the time interval in which the ball try to regain their original shapes is called the restitution period. Due to push applied by the balls on each other during period of deformation speed of the ball A decreases and that of the ball B increases and at the end of the deformation period, when the deformation is maximum both the ball move with the same velocity say it is u. Thereafter, the balls will either move together with this velocity or follow the period of restitution. During the period of restitution due to push applied by the balls on each other, speed of the ball A decrease further and that of ball B increase further till they separate from each other. Let us denote velocities of the balls A and B after the impact by vA and vB respectively. 12 E
Equation of Impulse and Momentum during impact JEE-Physics Impulse momentum principle describes motion of ball A during deformation period. ...(i) Ddt ...(ii) m u ...(iii) AA ...(iv) mu A mAuA Ddt mAu Impulse momentum principle describes motion of ball B during deformation period. mu Ddt mu BB B mBuB Ddt mBu Impulse momentum principle describes motion of ball A during restitution period. m u Rdt m v A A A m Au Rdt m Av A Impulse momentum principle describes motion of ball B during restitution period. mu Rdt mv B B B mBu Rdt mBv B Conservation of Momentum during impact From equations, (i) and (ii) we have mAuA mBuB m A mB u ...(v) ...(vi) From equations, (iii) and (iv) we have mA mB u mAv A mBv B ...(vii) From equations, (v) and (vi) we obtain the following equation. mAv A mBv B mAuA mBuB The above equation elucidates the principle of conservation of momentum. Coefficient of Restitution Usually the force D applied by the bodies A and B on each other during period differs from the force R applied by the bodies on each other during period of restitution. Therefore, it is not necessary that magnitude of impulse Ddt of deformation equals to the magnitude of impulse Rdt restitution. The ratio of magnitudes of impulse of restitution to that of deformation is called the coefficient of restitution and is denoted by e. NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 e Rdt ...(viii) Ddt Now from equations (i), (ii), (iii) and (iv), we have e vB vA ...(ix) uA uB Coefficient of restitution depend on various factors as elastic properties of materials forming the bodies, velocities of the contact points before impact, state of rotation of the bodies and temperature of the bodies. In general, its value ranges from zero to one but in collision where kinetic energy is generated its value may exceed one. Depending on values of coefficient of restitution, two particular cases are of special interest. E 13
JEE-Physics Perfectly Plastic or Inelastic Impact For these impacts e = 0, and bodies undergoing impact stick to Perfectly Elastic Impact each other after the impact. For these impacts e = 1. Strategy to solve problems of head-on impact Write momentum conservation equation mAv A mBv B mAuA mBuB ...(A) Write rearranging terms of equation of coefficient of restitution v B v A e uA uB ...(B) Use the above equations A and B. Example A ball of mass 2 kg moving with speed 5 m/s collides directly with another of mass 3 kg moving in the same direction with speed 4 m/s. The coefficient of restitution is 2/3. Find the velocities after collision. Solution. Denoting the first ball by A and the second ball by B velocities immediately before and after the impact are shown in the figure. u= u =4 vv A B AB AB AB Immediately after Immediately before impact starts impact ends Applying principle of conservation of momentum, we have mBv B mAv A mAuA mBuB 3v B 2v A 2 5 3 4 3v B 2v A 22 ...(i) Applying equation of coefficient of restitution, we have v B v A e uA uB vB vA 2 5 4 3 3v B 3v A 2 ...(ii) From equation (i) and (ii), we have vA = 4 m/s and vB = 4.67 m/s A n s . Example A block of mass 5 kg moves from left to right with a velocity of 2 m/s and collides with another block of mass 3 kg moving along the same line in the opposite direction with velocity 4 m/s. (a) If the collision is perfectly elastic, determine velocities of both the blocks after their collision. (b) If coefficient of restitution is 0.6, determine velocities of both the blocks after their collision. NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 Solution. Denoting the first block by A and the second block by B velocities immediately before and after the impact are shown in the figure. A B v A Bv u =2 u =4 A B A B Immediately before Immediately after impact starts impact ends Applying principle of conservation of momentum, we have mBv B m Av A m AuA mBuB 3v B 5v A 5 2 3 4 3v B 5v A 2 ...(i) 14 E
JEE-Physics Applying equation of coefficient of restitution, we have v B v A e uA uB v B v A e 2 4 v B v A 6e ...(ii) (a) For perfectly elastic impact e = 1. Using this value in equation (ii), we have vB – vA = 6 ...(iia) Now from equation (i) and (iia), we obtain vA = – 2.5 m/s and vB = 3.5 m/s (b) For value e = 0.6, equation 2 is modified as vB – vA = 3.6 (iib) Now from equation (i) and (iib), we obtain v = – 1.6 m/s and v = 2.0 m/s A B Block A reverse back with speed 1.6 m/s and B also move in opposite direction to its original direction with speed 2.0 m/s. Example Two identical balls A and B moving with velocities uA and uB in the same direction collide. Coefficient of restitution is e. (a) Deduce expression for velocities of the balls after the collision. (b) If collision is perfectly elastic, what do you observe? Solution. Equation expressing momentum conservation is vA vB uA uB ...(A) Equation of coefficient of restitution is v B v A euA euB ...(B) (a) From the above two equations, velocities v and v are AB vA 1 e u 1 e u ...(i) 2 2 A B vB 1 e u 1 e u ...(ii) 2 2 A B (b) For perfectly elastic impact e = 1, velocities vA and vB are vA = uB ...(iii) NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Centre of mass & collision\\Eng\\Theory.p65 v = u ...(iv) B A Identical bodies exchange their velocities after perfectly elastic impact. Conservation of kinetic energy in perfectly elastic impact For perfectly elastic impact equation for conservation of momentum and coefficient of restitution are mAv A mBv B mAuA mBuB ...(A) vB vA uA uB ...(B) Rearranging the terms of the above equations, we have mA v A uA mB uB v B uA vA vB uB E 15
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