TUTORIALS KE

 (I) Velocity of the electrons Soln Velocity of electrons V = {√(2eEa)/(m)}      = {√ (2 x 1.6 x10^-19 x 2000)/(9.1 x 10-31)}       = 26519741.77 m/s (II) Acceleration of the electron Soln   Acceleration = change in velocity/time                 = (26519741.77 - 0)/(0.26x 10-6)                  = 1.01999 x 10^14 m/s^2 (III) Loss in potential energy     Loss in potential energy P.E = eEa                = (1.6 x10^-19) x ( 2000)                 = 3.2 x 10^-16 Joules (III) Gain in Kinetic energy       K.E = 1/2(m x V^2)      = 1/2( 9.1 x 10^-31 x 26519741.77^2)       = 3.2 x 10^-16 Joules

 Avalanche breakdown is a phenomenon which occurs in lightly doped junctions which occur when a high reverse voltage is applied across the junction which tend to cause an increase in electric field across the junction. This generated electric field exerts a force on the electrons at junction and frees them from the covalent bonds. These free electrons will gain acceleration and it will start moving across the junction with high velocity. This results in collosion with neighbouring atoms. These collisions in high velocity tend to generate more free electrons. These free electrons will start drifting and electrons -hole pairs recombination occurs across the junction which results in a rapid increase in current uncontrollably. <a href="https://kol.jumia.com/api/click/link/f0647e0d-9d90-4183-b025-e2cf08db8f34/42461daf-6d6a-4cf2-b0ca-7e066511e571"><img src="https://kol.jumia.com/banners/OK6RTn7jyXBJ8ORsrvFNC6o0L8wUYB5zNJOYzNEI.jpg" alt="PHONES CATEGORY"

 (I) Hole : It is the vacant space that occur due to absence of an electron in an atom. (II) Forbidden gap : It is the gap getween the valence band and conduction band. (III) Donor atoms : They are pentavalent impurity atoms which adds electrons to form N-type semiconductor.

 (I) Filter :       - used to remove the A.C component and allow only D.C component to reach the load from the rectified output. (II) Transformer:        - It isolates the rectifier circuit from power line and thus reduces the risk of electric shock.         - It allows us to step up or step down the A.C input voltage .

  A zener diode is like an ordinary diode but it is suitably doped so as to have a sharp breakdown voltage. It is always connected in a reverse biased state and it has a sharp breakdown volatage called zener volatge Vz. A zener diode is used as a voltage regulator to provide a constant voltage from the source whose voltage may vary over sufficient range. When the circuit is properly designed, the load voltage Eo becomes equal to Vz even if the input voltage Ei and load resistance RL vary. When the input voltage increases, the ouput voltage remains constant at Vz = Eo and the excess voltage is dropped across the series resistance R. This causes an increase in the total current I and as a result the zener will conduct the increase of current in I while the load current remains constant. Hence the output voltage Eo remains constant even if the input voltage Ei changes. When the input voltage becomes constant and the load resistance RL decreases this causes an increase in load current henc

 (I) Voltage regulation Soln % voltage regulation = {(no load voltage - full load voltage)/(full load voltage)} x 100  ={ (60-56)/60} x 100 = 7.1425% (II) ripple Soln Ripple factor = V ac(rms) /V dc                         = 1.25 v/ 56V                          = 2.232%

 When a drain to source voltage Vds is applied between the drain and source terminals and the voltage on the gate is zero, the two PN junctions at the sides of the bar tend to establish a depletion layer. The holes will flow from the source to the drain through a channel between the depletion layers. The size of the depletion layer determines the width of the channel and hence the current conduction through the channel. When a reverse voltage Vgs is applied between the gate and source terminals , the width of the depletiom layer is increased this reduces the width of the conducting channel hence imcreasing the resistance of the p-type  bar. Hence reducing the current flowing from the drain to source. When the reverse bias voltage Vgs on the gate is reduced the width of the depletion also decreases and hence increasing the crossectional area of the Channel and hence flow of current by holes increases.

 (I) FETs are voltage controlled devices while bipolar junction transistors are current controlled devices (II) FETs are less affected by radiation compared to BJTs (III) The noise produced by FETs is lower compared to that produced by BJTs. (IV) FETs are easier to fabricate compared to BJTs (v) FETs has high input impedance compared to BJTs.

 (I) Base voltage Vb Soln   Base voltage Vb = Voltage drop across resistor R2  Base voltage Vb = { (R2/(R1+R2)) x Vcc}                 = { ( 4000/(4000+ 40000)) X 22 v}                  = 2 Volts (II) Emitter voltage Ve Emitter voltage Ve = Vb - Vbe                                   = 2 - 0.7                                    = 1.3 Volts (III) Emitter current Ie         Emitter current Ie = Ve/Re                                          = 1.3V/ 1500 ohms                                           = 8.667 x 10^ -4 A (III) Collector voltage Vc        Collector voltage Vc = Ic Rc                          = (8.667 x 10^ -4 x 10000)                           = 8.667 Volts (V) Collector emitter voltage Vce Collector emitter voltage Vce = Vcc -Ic(Rc+Re) = 22 - 8.667x 10^-4(10000+1500) = 22- 9.967 = 12 Volts      

 (I) Ionization potential : It is the energy needed to remove an electron from an isolated atom. (II) Covalent bonding: It is a force holding atoms of two non-metals together which share a pair of electron. (III) Doping: process of adding impurities to intrinsic semiconductor to increase conducting properties.

  When the external voltage applied is zero, the potential barrier at the junction does not allow current to flow and therefore circuit current is zero as shown by point O. When the junction is forward biased, the potential barrier is reduced. At some forward voltage of i.e 0.7 v , the potential barrier is eliminated completely and the current starts flowing in the circuit and as a result the current begins to increase with corresponding increase in forward voltage as shown by region AB on the curve where the PN junction behaves as an ordinary resistor. When the junction is reverse biased, the potential barrier is increased and hence the resistance of the junction and hence practically no current flows through the circuit , a very small current flows through the circuit with reverse bias as shown by the reverse characteristic. This is called reverse saturation current and it is due to flow of minority charge carriers.

 In his study on atoms ,Rutherford proposed Rutherford atomic model( Rutherford planetary model of an atom). The following are the salient features his model: (I) Every atom consist of tiny central core called nucleus which is entirely positively charged and it is a zone in which almost whole mass of the atom is concentrated. (II) Most of an atom is empty space (III) In free space around the nucleus there are electrons which move in orbits just like the planets around the sun and the centripetal force  needed for orbital motion of electrons is provided by electrostatic forces due to positively charged nucleus. (III) An atom as a whole is electrically neutral hence the total posively charge of the nucleus is exactly equal to total negative charge of all the electrons orbiting in an atom

 (I) Current in the material Soln  t = 1 sec V = 12 v 1coulomb = 6.25 x 10^ 8 electrons ?                     10^ 8 electrons   = (1 x 10^ 8 electrons)/(6.25 x 10^8)     = 0.16 coulombs Charge Q = current x time  Current = charge (q)/time(t)                 = 0.16/1 sec                 = 0.16 amperes (II) Power dissipated Soln Power(P) = I ^ 2 x R             R  = V / I                  = 12/0.16                   = 75 ohms Power dissipated = I ^2 R                                  = 0.16^2 x 75                                   = 1.92 Watts

 Ld = length of deflecting plates, m  d = distance between deflecting plates,m m = mass of the electron ,kg L = distance between screen and the centre of the deflecting plates, m Ea = accelerating voltage to the plates, volts Ed = deflecting potential of the plates (I) Velocity of the electron Soln Velocity of electron,  Vox = √{(2eEa)/(m)} Ea = 2500 v Ed = 150v Ld = 1.5cm L = 55cm d = 6 mm m = 9.1094 x 10^-31 kg e = 1.6022 x 10^ -19 Coulombs = {√ (2 x 1.6022 x 10^-19 x 2500)/(9.1094x 10^-31)} = √ 8.794x 10^ 14 = 29655037.64 m/s (II) Amount of deflection D = {((L x Ld x Ed)/(2 x d x Ea)) x (m)}     = {((0.55 x 0.015 x 150)/(2 x 6 x 10^-3 x 2500))}     = 0.04125 m (III) Deflection sensitivity              = { ( Ld x L )/(2 x d x Ea)}           = { ( 1.5 x 10^ -2 x 0.55)/(2 x 6 x 10^ -3 x 2500)}           = 2.75 x 10^ -4 m/ volt (IV) Deflection factor            = 1/deflection sensitivity             = 1/ ( 2.75 x 10^ -4)             = 3636.3636 volt/meter

It contains 4 diodes d1 ,d2, d3, and d4 connected to form a bridge. The A.c supply to be rectified is applied to the opposite ends of the bridge through the transformer. And between the ends of the bridge the load resistance RL is connected.  During the positive half-cycle of secondary voltage, the end P of the secondary winding becomes positive and end Q negative. This makes the diodes D1 and D3 to be forward biased while diodes D2 and D4 to be reverse biased. These two diodes will be in series through the RL hence current flows from point A to B through the load RL. During the negative half-cycle of secondary voltage the end P becomes negative and end Q positive. This makes diodes D2 and D4 forward biased whereas diodes D1 and D3  are reverse biased. Therefore, only diodes D2 and D4 conducts. These two diodes will be in series through the load RL.

 (I) Space charge : It is the collection or distribution of electrons which are emitted from the metal surface on a three dimensional region. (II) Work function : It is the minimum amount of energy required to remove an electron from a solid ( metal surface).

 (I) Active region : EB junction is forward biased and CB junction is reversed biased. Transistor in this region operates as an amplifier. (II) Saturation region : Both junctions of the transistor are forward biased. In this state transistor behaves as a close switch. (III) Cut-off region : Occurs when both junctions of the transistor are reverse biased. In this state transistor behaves as an open switch.

  (a) Gate voltage, Vg Soln Gate voltage Vg = Voltage drop across resistor R2 Vg = {R2/(R1+R2) } x Vcc       = {280000/(2000000+280000)} × 16 volts       = 1.9649 volts (b) Source Voltage Vs Soln Source voltage, Vs = Id ×Rs          = (0.0025 x 1500)          = 3.75 volts (c) Gate to Source Voltage Vgs Soln Vgs = Vg - Id Rs         = 1.9649 - 3.75         = - 1.79 volts (d) Drain Voltage, Vd Soln Drain voltage Vd = Id Rd                                = (0.0025 x 2500)                                = 6.25 volts (e) Drain to source voltage ,Vds Vds = Vdd - Id(Rd + Rs)         = 16 - 0.0025( 2500 + 1500)          = 6 volts

  It has high drain resistance Moderate input impedance It is slower in operation

 It should have a low work function. A low work function helps to emit electrons from the cathode surface in a comparatively lower temperature It should have a high melting point. The temperature required to emit an electron from the cathode surface is quite high compared to the melting point of normal metals. A lower melting point causes vaporization of metals before they emit electrons. It should have high mechanical strength.