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<p>DEPT. OF ELECTRONICS AND COMMUNICATION ENGINEERINGSUB: EC 2151 ELECTRIC CIRCUITS AND ELECTRON DEVICES (Common to ECE, CSE and IT Branches)</p> <p>SEM / YEAR: II / IUNIT I CIRCUIT ANALYSIS TECHNIQUES Kirchoffs current and voltage laws series and parallel connection of independent sources R, L and C Network Theorems Thevenin, Superposition, Norton, Maximum power transfer and duality Stardelta conversion. UNIT II TRANSIENT RESONANCE IN RLC CIRCUITS Basic RL, RC and RLC circuits and their responses to pulse and sinusoidal inputs Frequency response Parallel and series resonances Q factor single tuned and Double tuned circuits. UNIT III SEMICONDUCTOR DIODES Review of intrinsic &amp; extrinsic semiconductors Theory of PN junction diode Energy band structure current equation space charge and diffusion capacitances Effect of temperature and breakdown mechanism Zener diode and its Characteristics. UNIT IV TRANSISTORS Principle of operation of PNP and NPN transistors study of CE, CB and CC Configurations and comparison of their characteristics Breakdown in transistors Operation and comparison of N-Channel and P-Channel JFET drain current Equation MOSFET Enhancement and depletion types structure and operation Comparison of BJT with MOSFET thermal effect on MOSFET. UNIT V SPECIAL SEMICONDUCTOR DEVICES Tunnel diodes PIN diode, varactor diode SCR characteristics and two transistor equivalent model UJT Diac and Triac Laser, CCD, Photodiode, Phototransistor, Photoconductive and Photovoltaic cells LED, LCD. TEXT BOOKS: 1. Joseph A. Edminister, Mahmood, Nahri, Electric circuits Shaum series, Tata Mc Graw Hill, (2001) 2. S.Salivahanan, N. Suresh Kumar and A. Vallavanraj, Electronic devices &amp; Circuits Tata Mc Graw Hill, 2nd edition (2008) 3. David A. Bell, "Electronic Devices and circuits", Oxford University Press, 5th Edition, (2008) REFERENCES: 1. William H.Hayt, J.V.Jack, E.Kemmebly &amp; Steven M. Durbin, Engineering circuit analysis Tata Mc Graw Hill, 6nd edition (2002) 2. J. Millman and Halkins, Satyebranta jit, Electronic devices &amp; Circuits Tata Mc Graw Hill, 2nd edition (2008)</p> <p>1</p> <p>UNIT-1 CIRCUIT ANALYSIS TECHNIQUES PART-A1. Define current &amp; voltage?The flow of free electrons in a metal is electric current and its unit measure is in amperes. The ability of the charged body to do work is called as electric potential and unit is volts.</p> <p>2. State ohms law.The ratio of potential difference b/w any two points of a conductor to the current flowing, provided the physical conditions do not change.</p> <p>3. State the limitations of ohms law. Doesnt apply to non metallic conductors. Doesnt apply to non linear devices.</p> <p>4. State Kirchhoffs voltage law.The algebraic sum of the potential around a closed loop is zero</p> <p>5. State two salient points for series combinations. The current is same to all the elements The total resistance is greater than the greatest resistance in the circuit</p> <p>6. State Super position theorem.The Responses in circuit in multiple source is given by the algebraic sum of the responses due to the individual sources acting alone</p> <p>7. State Thevenins theorem.Any linear network with output terminal can be replaced by a single voltage source in series with the impedance (Thevenin resistance)</p> <p>8. State Nortons theorem.Any linear network with output terminal can be replaced by a single current source in parallel with impedance</p> <p>9. When superposition theorem can be applied?When there are more than one source acting in a network. 2</p> <p>10. Explain the uses of Thevenins theorem. Applicable to all linear circuits This Theorem is useful when it is desired to know the effect of responses in the network</p> <p>PART B1. Calculate the mesh current i1 and i2 in the following circuit</p> <p>Equation for Mesh 1</p> <p> 12V + v2 + v12 + v4 = 0 12V + 2 i1 + 12 (i1 i2 ) + 4 i1 = 0 18 i1 12 i2 = 12</p> <p>Equation for Mesh 2</p> <p>v12 + v9 + 8V + v3 = 0 12 (i1 i2 ) +9 i2 +V 8</p> <p>3 i2 0 + =</p> <p> 1 2 i1 + 2 4 i2 = 8Now we have to solve the two equations: 18 12 12 i1 12 = 24 i2 8 </p> <p>i1 0.6667 A i = 0 A 2 </p> <p>2.</p> <p>Explain about Krichoff's current and voltage laws with suitable examples.KIRCHOFFS VOLTAGE LAW (KVL)</p> <p>3</p> <p>The sum of the voltage drops around any closed loop is zero.M JO IM L A IO A R P IC T NK L tells us tha an set of elem V t y ents w hich are con ected at n both en s d carry the s e vo am ltag e . W say these e ents are e lem in p llel ara K L clo V ckw ise, s tart at to p: V V =0 b a</p> <p>.</p> <p>V =V a b</p> <p>K LE A P E V X MLa</p> <p>+ va 3</p> <p>1</p> <p>+ vb -</p> <p>2</p> <p>P th 1 a : P th2 a : P th3 a :</p> <p>v a +v 2 +v b =0v b v 3 +v c =0v a +v 2 v 3 +v c =0</p> <p>KIRCHOFFS CURRENT LAW (Sum of currents entering node) (Sum of currents leaving node) = 0M JO IM L A IO A R P IC T NK Ltells us th a of th elem ts C at ll e en in a sin le b n g ra ch carryth e sam curren e t . W say th elem ts are e ese en inser ies .</p> <p>C rren en u t terin n =C g ode urren leavin n e t g od</p> <p>i1 = i</p> <p>2</p> <p>KIRCHHOFFS CURRENT LAW EXAMPLE24 A 10 A -4 A i</p> <p>Currents entering the node: 24 A Currents leaving the node: 4 A + 10 A + i Three formulations of KCL:</p> <p>1: 2: 3:</p> <p>24 = 4 + 10 + i 24 ( 4) 10 i = 0 24 4 + 10 + i = 0</p> <p>i = 18 A i = 18 A i = 18 A</p> <p>Page no. 64-67, R.S. Sedha, A text book of Applied Electronics S.CHAND pub 2005. 4</p> <p>+</p> <p>+</p> <p>b</p> <p>E a p so x m le f th e c se re lo d p th a s:</p> <p>v2</p> <p>v3</p> <p>c</p> <p>+ vc </p> <p>3. Use nodal analysis to find v1 in this circuit.</p> <p>v1 12 v1 0 v1 10 0 + + =0 4 6 2</p> <p>3(v1 12) + 2(v1 0) + 6(v1 1 0 0) = 0</p> <p>3v1 36 + 2v1 + 6v1 60 = 0</p> <p>v1 = 8.727V4. Using superposition theorem find the voltage drop across each resistor</p> <p>Since we have two sources of power in this circuit, we will have to calculate two sets of values for voltage drops and/or currents, one for the circuit with only the 28 volt battery in effect.</p> <p>and one for the circuit with only the 7 volt battery in effect:</p> <p>When re-drawing the circuit for series/parallel analysis with one source, all other voltage sources are replaced by wires (shorts), and all current sources with open circuits (breaks). Since we only have voltage sources (batteries) in our example circuit, we will replace every inactive source during analysis with a wire. Analyzing the circuit with only the 28 volt battery, we obtain the following values for voltage and current:</p> <p>5</p> <p>Analyzing the circuit with only the 7 volt battery, we obtain another set of values for voltage and current:</p> <p>5. Determine the voltage drop across each one (applying ohms law in its proper context)</p> <p>6</p> <p>6. Find the voltage across AB using thevenin's theorem and hence draw its equivalentcircuit</p> <p>UNIT-2 TRANSIENT RESONANCE IN RLC CIRCUITS PART-A1. Define waveformInstantaneous values of voltage or current plotted against time.</p> <p>2. Define time periodTime taken to complete one full cycle. T=2/</p> <p>3. Define peak factorRatio of peak value of the wave to the rms value of the wave Peak factor = Vp/Vrms 7</p> <p>4. Define form factorRatio of rms value to the average value of the wave. Form factor = Vrms /Vavg</p> <p>5. Calculate the period for the given frequency 100 KHz?T=1/f = 1/100x103 = 10s</p> <p>6. A sinusoidal voltage is applied to a capacitor. The frequency of the sine wave is 2KHz. Determine the capacitive reactance.Xc = 1/ = 1/2fc = 7.96K</p> <p>= 1/2xx2x103x0.01x10-6</p> <p>7. For the circuit having inductance of 50mH and a voltage source of Vrms 10V and frequency 10KHz connected in series determine the rms current in the circuit?XL = 2fL = 3.141K Irms = Vrms / XL = 10/3.141x103 = 3.18mA</p> <p>8. Define quality factorQuality factor for the circuit is defined as</p> <p>9. Define resonant circuit.The circuit that treats a narrow range of frequencies very differently than all other frequencies is called a resonant circuit.</p> <p>10. Give the expression for quality factor of series RLC circuit.Q=</p> <p>1 L/C R</p> <p>11. Give the expression for quality factor of parallel RLC circuit.</p> <p>Q= R C / L PART B1. The RL circuit has an emf of 5 V, a resistance of 50 W, an inductance of 1 H, and no initial current. Find the current in the circuit at any time t. Distinguish between the transient and steady-state current.8</p> <p>Answer:After substituting: Solving:</p> <p>The formula is:</p> <p>I.F. =</p> <p>When This gives us:</p> <p>,</p> <p>so</p> <p>The transient current is:</p> <p>A</p> <p>A</p> <p>2. A series RL circuit with R = 50 W and L = 10 H has a constant voltage V = 100 V applied at t = 0 by the closing of a switch. Find (a) The equation for i (you may use the formula rather than DE), (b) The current at t = 0.5 s (c) The expressions for VR and VL (d) The time at which VR = VL Answer We solve it using the formula:</p> <p>9</p> <p>We have: (b) At (c)</p> <p>A</p> <p>(d)</p> <p>VTC for this example is: 3. A series RC circuit with R = 5 W and C = 0.02 F is connected with a battery of E = 100 V. At t = 0, the voltage across the capacitor is zero. (a) Obtain the subsequent voltage across the capacitor. (b) As t , find the charge in the capacitor. Answers We will solve this 3 ways, since it has a constant voltage source: 1 and 2: Solving the DE in q, as: </p> <p>a linear DE and variables separable</p> <p>10</p> <p>3. Using the formulae</p> <p>and</p> <p>Method 1 - Solving the DE in q From the formula: , we obtain:</p> <p>On substituting, we have:</p> <p>Solving this differential equation as a linear DE, we have: IF = So So Now, since So As Now,</p> <p>, (that is, when</p> <p>,</p> <p>) this gives:</p> <p>,</p> <p>C.</p> <p>11</p> <p>For comparison, here is the solution of the DE using variables separable:</p> <p>Since</p> <p>,</p> <p>we have</p> <p>Method 2: NowSo:</p> <p>We use the formulae</p> <p>and</p> <p>.</p> <p>Now</p> <p>From here, we use</p> <p>and obtain:</p> <p>12</p> <p>So</p> <p>, as before. Also, as</p> <p>,</p> <p>C.</p> <p>4. Find the charge and the current for t &gt; 0 in a series RC circuit where R = 10 W, C = 4 103 F and E = 85 cos 150t V. Assume that when the switch is closed at t = 0, the charge on the capacitor is -0.05 C. Answer: [We cannot use the formulae constant.] From the formula: Since , , we obtain: and , we have: and , since the voltage source is not</p> <p>Now, we can solve this differential equation in q using the linear DE process as follows: IF = ,</p> <p>Then we use the integration formula (found in a standard integral table):</p> <p>We obtain:</p> <p>So, dividing throughout by</p> <p>gives:</p> <p>13</p> <p>We now need to find K Means So this gives us:</p> <p>We are also asked to find the current. We simply differentiate the expression for q:</p> <p>5. In the RC circuit shown below, the switch is closed on position 1 at t = 0 and after 1 is moved to position 2. Find the complete current transient.</p> <p>Answer: At t = 0, the switch is at Position 1. We note that</p> <p>Using SNB to solve this differential equation, we have:</p> <p>NOTE: By differentiating, this gives us:14</p> <p>We need to find : . Now, at , the charge will be:</p> <p>At</p> <p>, switch at Position 2: again:</p> <p>Applying the formula</p> <p>NOTE: The negative voltage is because the current will flow in the opposite direction through the resistor and capacitor. Exact solution is:</p> <p>So the current transient will be:</p> <p>This expression assumes that time starts at , so we need:</p> <p>. However, we moved the switch to Position 2 at</p> <p>So the complete current transient is: For15</p> <p>For</p> <p>UNIT III SEMICONDUCTOR DIODES PART -A1. What are Semiconductors? Give examples?The materials whose electrical property lies between those of conductors and insulators are known as Semiconductors. Ex germanium, silicon.</p> <p>2. What are Intrinsic and Extrinsic Semiconductors? Pure form of semiconductors are said to be intrinsic semiconductor. Ex: germanium, silicon. If certain amount of impurity atom is added to intrinsic semiconductor the resulting semiconductor is Extrinsic or impure Semiconductor. 3. What is forward bias and reverse bias in a PN junction?When positive of the supply is connected to P type and negative to N type then it is forward bias. When positive of the supply is connected to N type and negative to P type then it is reverse bias.</p> <p>4. What is Reverse saturation current?The current due to the minority carriers in reverse bias is said to be reverse saturation current.</p> <p>5. Give two applications of PN junction diode. As rectifier in power supplies. As switch in logic circuits 6. Define Transition or space charge capacitance. The parallel layers of oppositely charged immobile ions on the two sides of the junction form the Transition or space charge capacitance. CT =A/W Where = permitivity of the material, A=cross sectional area of the junction and W=width of the depletion region 7. Define Diffusion capacitance. The capacitance that exists in the forward biased junction is called a diffusion or storage capacitance. It is also defined as the rate of change of injected charge with applied voltage. CD=dQ/dV</p> <p>16</p> <p>8. What is peak inverse voltage? It is the maximum reverse voltage that can be applied to the pn junction without damaging the junction 9. What is Zener breakdown? The breakdown potential can be brought to lower levels by increasing the doping levels in the ptype and n-type materials of the diode. In this case breakdown is initiated through a direct rupture of the covalent bonds due to the existence of the strong electric field at the junction. 10. Draw the V-I characteristics of a pn diode</p> <p>PART B1. Explain the theory of pn junction diode.Refer (Electronic devices and circuits by S.Salivahanan) Page no: 86</p> <p>2. Explain about Zener diode characteristics.Refer (Electronic devices and circuits by S.Salivahanan) Page no: 122</p> <p>3. Explain about the energy band structure.Refer (Electronic devices and circuits by S.Salivahanan) Page no: 93</p> <p>4. Discuss the diode current equation.Refer (Electronic devices and circuits by S.Salivahanan) Page no: 98</p> <p>5. Explain briefly about the VI Characteristics of pn junction diode.Refer (Electronic devices and circuits by S.Salivahanan) Page no: 103</p> <p>UNIT-IV TRANSISTORS Part: A1. What are PNP and NPN transistors?PNP Transistor: The figure below shows a PNP Transistor. 17</p> <p>A block representation of a layer of n-type material between two layers of p-type material is known as a pnp transistor. NPN Transistor: The figure below shows an NPN Transistor.</p> <p>A block representation of a p-type material between two layers of an n-type material is known as npn transistor.</p> <p>2. What is a CE configuration?CE Configuration-In CE Configuration, the input voltage is applied between the B and E terminals and the output is taken at the C and E terminals. 3. What is a CB Configuration? In CB Configuration- the base terminal is common to both the input (EB) voltage and output (CB) Voltage.</p> <p>4. What is a CC Configuration?In CC Configuration- the collector terminal is common to both input CB Voltage and CE Voltage.</p> <p>5. What are the characteristics of CE configuration?The various characteristics of CE Configuration are: Input Characteristics: To determine the input characteristics, Vce is held constant and IB levels are recorded for several values of VBE. Output Characteristics: IB is maintained constant a...</p>