ECT 202 Analog Circuits Module 1 PDF

ECT 202 Analog Circuits MODULE 1 Joseph George K N JG 1 Syllabus Module 1: Wave shaping circuits: First order RC differentiating and integrating circuits, First order RC low pass and high pass filters. Diode Clipping circuits - Posit...

ECT 202 Analog Circuits MODULE 1 Joseph George K N JG 1 Syllabus Module 1: Wave shaping circuits: First order RC differentiating and integrating circuits, First order RC low pass and high pass filters. Diode Clipping circuits - Positive, negative and biased clipper. Diode Clamping circuits - Positive, negative and biased clamper. Transistor biasing: Need, operating point, concept of DC load line, fixed bias, self bias, voltage divider bias, bias stabilization. Module 2: BJT Amplifiers: RC coupled amplifier (CE configuration) – need of various components and design, Concept of AC load lines, voltage gain and frequency response. Small signal analysis of CE configuration using small signal hybrid-pi model for mid frequency and low frequency. (gain, input and output impedance). High frequency equivalent circuits of BJT, Miller effect, Analysis of high frequency response of CE amplifier. JG 2  Module 3: MOSFET amplifiers: MOSFET circuits at DC, MOSFET as an amplifier, Biasing of discrete MOSFET amplifier, small signal equivalent circuit. Small signal voltage and current gain, input and output impedance of CS configuration. CS stage with current source load, CS stage with diode-connected load. Multistage amplifiers - effect of cascading on gain and bandwidth. Cascode amplifier. Module 4 : Feedback amplifiers: Effect of positive and negative feedback on gain, frequency response and distortion. The four basic feedback topologies, Analysis of discrete BJT circuits in voltage-series and voltage-shunt feedback topologies - voltage gain, input and output impedance. Oscillators: Classification, criterion for oscillation, Wien bridge oscillator, Hartley and Crystal oscillator. (working principle and design equations of the circuits; analysis of Wien bridge oscillator only required). Module 5: Power amplifiers: Classification, Transformer coupled class A power amplifier, push pull class B and class AB power amplifiers, complementary-symmetry class B and Class AB power amplifiers, efficiency and distortion (no analysis required) Regulated power supplies: Shunt voltage regulator, series voltage regulator, Short circuit protection and fold back protection, Output current boosting. JG 3 JG 4  Module 1: Objectives To differentiate between analog circuits and digital circuits. To understand the charging/discharging of capacitor, voltage and current transients, Time constant To study pulse response of RC circuits- Wave shaping circuits -Differentiator, Integrator, LPF, HPF To become familiar with the analysis of diode clipping, clamping circuits - to predict the output response of a clipper and clamper circuits. To analyze and design different BJT biasing circuits. To perform load-line analysis of the common BJT configurations. To study the effect of stability factors of a BJT configuration JG 5 Wave Shaping Circuits JG 7 The different signal waveforms are to be properly shaped before applying to a signal processing circuits. Convert one waveform to another o Generate sharp narrow pulses / ramp waveform from sine / rectangular waveform  Differentiating / Integrating circuits  RC Filter  Clipping Circuits  Clamping circuits JG 8 The time-stationary (or constant-value) elements the resistor, inductor, and capacitor are called passive elements, since none of them can continuously supply energy to a circuit. The voltage v and current i, relationships are: JG 9 Capacitor A circuit component designed to store electrical charge. If a capacitor is connected to a DC source, it will charge to the voltage of the source. If then the source is disconnected, the capacitor will remain charged (assuming no leakage), until it is connected to some circuit through which it can discharge. Capacitance value ranges from picofarads (pF) to microfarads (μF). where, C is the capacitance, Q the charge stored and The larger the value of C, the more V is the voltage charge that the capacitor can hold for a given voltage. The energy stored in a capacitor, J Capacitors are widely used in electronic circuits for signal conditioning and timing. JG 10 Capacitor charging A capacitor will charge when it is connected to a dc voltage source. Before the switch is closed, the capacitor is uncharged - plate A and plate B have equal numbers of free electrons. When the switch is closed, the source moves electrons away from plate A through the circuit to plate B. Plate A becomes positive with respect to plate B. This charging process continues, the voltage across the plates builds up rapidly until it is equal to the source voltage, but opposite in polarity. When the capacitor is fully charged, there is no current JG 11 Capacitor discharging When the charged capacitor is disconnected from the source, it remains charged for long periods of time, depending on its leakage resistance. When the switch is closed, the excess electrons on plate B move through the circuit to plate A. The energy stored by the capacitor is dissipated in the resistance of the conductor. The charge is neutralized when the numbers of free electrons on both plates are again equal. At this time, the voltage across the capacitor is zero, the current is zero and the capacitor is completely discharged. JG 12 Capacitor current and voltage during charging A capacitor in series with a resistor can be connected to a dc voltage source. First, assume the capacitor is uncharged and that the switch is open Now move the switch to the charge position at t = 0. Time t is measured from the instant of switching  t = 0 is defined as the instant the switch is moved to the charge position Note: It is also defined as the instant the switch is moved to discharge. JG 13 The current abruptly jumps from 0 to E/R amps, then decays to zero, i.e., the current is discontinuous, while the voltage, which is zero at the instant the switch is closed, gradually climbs to E volts, i.e., the voltage is continuous JG 14  Capacitor voltage capacitor voltage cannot change instantaneously, that is, it cannot jump abruptly from one value to another. Instead, it climbs gradually and smoothly as illustrated. JG 15  Since capacitor voltage cannot change instantaneously, its value just after the switch is closed will be the same as it was just before the switch is closed, namely 0 V. Since the voltage across the capacitor just after the switch is closed is zero (even though there is current through it), the capacitor looks momentarily like a short circuit. Capacitor Current That is, an uncharged capacitor looks like a short circuit at the instant of switching. Applying Ohm’s law yields iC = E/R amps. JG 16 Steady State Conditions JG 17 Capacitor discharging The capacitor is charged to E volts prior to switching to position 2 Since the capacitor voltage cannot change instantaneously, it will still have E volts across it just after switching JG 18  The capacitor therefore looks momentarily like a voltage source The current thus jumps immediately to –E/R amps. (a) The capacitor momentarily looks like a voltage source. Note: The current is negative since it is opposite in direction to the reference arrow.) JG 19 Capacitor Voltage and current waveforms during discharging The voltage and current decay to zero from the instant t = 0 Time t = 0 s is defined as the instant the switch is moved to the discharge position. JG 20  Capacitor Charging Equations Kirchhoff’s voltage law (KVL) yields Equation can be solved for vC as, Resistor voltage Also, JG 21 Charging Waveforms The voltage and current waveforms are exponential. JG 22 Capacitor Discharging Equations Kirchhoff’s voltage law (KVL) yields Equation can be solved for vC Resistor voltage and current JG 23 Discharging Waveforms The voltage and current waveforms are exponential. JG 24  Time Constant The rate at which a capacitor charges depends on the product of R and C. This product is known as the time constant of the circuit and is given the symbol τ Has units of seconds.  Resistance in Ohms  capacitance in Farads JG 25  Duration of a Transient The length of time that a transient lasts (i.e., the transient duration) depends on the exponential function, i.e., Thus, for all practical purposes, transients can be considered to last for only five time constants JG 26  Universal Time constant Curves Fig: Universal voltage and current curves for RC circuits RC = 1 to 5 JG 27 Fig: The voltage and current waveforms in an RC circuit for different values of time constant. The larger the time constant, the longer the duration of the transient, the longer the capacitor charging time. JG 28 Pulse Response to RC circuits JG 29 Rise and fall Times Fig: Practical pulse waveforms JG 30 RC Differentiating and Integrating Circuits JG 31  The effect of Pulse width The width of a pulse relative to a circuit’s time constant determines how it is affected by an RC circuit. Consider the circuits. Circuit 1: the output voltage across C Circuit 2: the output voltage across R. JG 32 Consider the circuit with output across C. (1) When the pulse width is very long compared with the circuit time constant, the capacitor gets enough time to charge and discharge fully. Note that charging and discharging occur at the transitions of the pulse. JG 33  (2) Since the pulse width is 5τ, the capacitor fully charges and discharges during each pulse. JG 34  (3) The capacitor does not get enough time to charge and discharge fully between pulses.  The switching occurs on the early (nearly straight line) part of the charging and discharging curves and thus, VC is roughly triangular in shape. It has an average value of V/2. JG 35 RC Integrator When T is small, frequency is high, The reactance of C (Xc = 1/2 πf C) is less. ie., the drop across the capacitor is very small Therefore, JG 36 as, (Xc = 1/2 πf C) is very small JG 37  Design: When T is small, frequency is high, The reactance of C (Xc = 1/2 πf C) is less. The drop across the capacitor is very small Let, T = 1ms C = 2.2µF Then, R = 6.8KΩ JG 38 Case 1 Good Integrator Case 2: increase T or decrease f. Not a good Integrator Case 3: Increase T further OR decrease f further Poor Integrator JG 39  RC Differentiator Output across R. (1) The capacitor gets time to charge and discharge fully between pulses. JG 40  (2) Since the pulse width is 5t, the capacitor gets enough time to charge and discharge during each pulse and therefore VR decreases to 0V JG 41  (3) Since the pulse width is less, The capacitor does not have time to charge and discharge significantly between pulses. JG 42 RC Differentiator When T is large, frequency is small, The reactance of C (Xc = 1/2πf C) is large. i.e., the drop across the capacitor is very large. Therefore, JG 43 But, as, Substituting, JG 44  Design: Drop across the capacitor is very large. Let, T = 1ms C = 220pF Then, R = 6.8KΩ JG 45  Case 1: The capacitor gets time to charge and discharge fully between pulses. A good differentiator JG 46 Case 2 : , Reduce T, OR increase f Since the pulse width is 5t, the capacitor fully charges and discharges during each pulse and therefore VR decreases to 0V Not a good differentiator JG 47  Case 3: , Reduce T further, OR increase f Consider the circuit with output across R. Since the pulse width is less, The capacitor does not have time to charge and discharge significantly between pulses. Poor differentiator JG 48 Case 1: A good differentiator Case 2 : Reduce T, OR increase f Not a good differentiator Case 3: Reduce T further, OR increase f Poor differentiator JG 49 RC Filter JG 50 RC Circuit as a Filter JG 51  RC Circuit as Low Pass Filter (LPF) The low-pass filter is realized by taking the output across the capacitor, just as in an integrator circuit. (b) Frequency response of LPF (a) LPF circuit JG 53 Transfer function of LPF JG 54  The magnitude of the transfer function is where is the cut off frequency of the LPF JG 55  Frequency response of Low-pass filter For f > fc, Vo /Vi decreases at the rate of -20dB/ decade. JG 56  Fig: Frequency response of Low-pass filter This graph, called a response curve, shows that the output voltage is equal to input voltage at the lower frequencies and decreases as the frequency increases. The frequency scale is logarithmic. JG 57  RC Circuit as High Pass Filter (HPF) The high-pass Filter is realized by taking the output across the resistor, as in a differentiator circuit. (a) HPF circuit (b) Frequency response of HPF JG 58  Transfer function of HPF This is the Transfer Function of first order High pass RC filter. JG 59  Frequency response of High Pass Filter For f < fc, Vo /Vi increases at the rate of 20dB/ decade. JG 60  Fig: Frequency response of First order High-pass filter This graph, called a response curve, shows that the output voltage is less at the lower frequencies and increases as the frequency increases. The frequency scale is logarithmic. JG 61 Cut off frequency and Bandwidth of Filters JG 62 Diode Clipping Circuits JG 63  Semiconductor Diode Formed by joining an p-type and a n-type material together.  joining of one material with a majority carrier of electrons to one with a majority carrier of holes. Fig: PN junction diode Acceptor atoms - Diffused impurities with three valence electrons- boron, gallium, and indium Donor atoms -Diffused impurities with five valence electrons- antimony, arsenic, and phosphorus In p-type material hole is the majority carrier and electron is the minority carrier. In n-type material electron is the majority carrier and hole is the minority carrier. JG 64  Diode with no Applied Bias At the instant the two materials are joined, the electrons and the holes in the region of the junction will combine, resulting in a lack of free carriers in the region near the junction Depletion region - the region of uncovered positive and negative ions. So called due to the depletion of free carriers in the region. Fig: Applied bias = 0V JG 65  Diode with Forward Bias Established by applying the positive potential to the p-type material and the negative potential to the n-type material Electrons in the n-type material and holes in the p-type material recombine with the ions near the boundary and  the width of the depletion region will be decreased  a large majority carrier current Imajority and a small reverse saturation current Is will flow. Fig: Forward JG bias V = VD V 66  Diode with Reverse Bias Established by applying the negative potential to the p-type material and the positive potential to the n-type material The number of uncovered ions in the depletion region will increase due to the large number of free electrons and holes drawn to the applied voltage.  The width of the depletion region will be increased and  The majority carrier current will be reduced to zero  Only the reverse saturation current Is will flow. Fig: Reverse bias V =JG -VDV 67  Shockley’s equation JG 68 Different types of diodes JG 69 V - I Characteristics of diode (b) (a) Fig: V - I characteristics of diode (a) practical (b) ideal JG 70  Diode Clipping Circuits Clippers are networks that employ diodes to “clip” away a portion of an input signal without distorting the remaining part of the applied waveform. The half-wave rectifier is an example of the simplest form of diode clipper- depending on the orientation of the diode, the positive or negative region of the applied signal is clipped” off. Categories: series and parallel. Fig: Half Wave rectifier circuits The series configuration - the diode is in series with the load, The parallel configuration- the diode in a branch parallel to the load JG 71 The diode is in series with the load Fig: Series clipper: circuit diagram and input –output waveforms JG 72  Series Clipper with a DC supply the dc supply can  aid or work against the source voltage  be in the leg between the supply and output or in the branch parallel to the output. Q1. Find the output voltage for the network for the series clipper with power supply of +V. Fig: Series clipper – with DC supply JG 73 General procedure for analyzing networks 1. Take careful note of where the output voltage is defined. It may be directly across the resistor R, or in some cases it may be across a combination of series elements. 2. Try to develop an overall sense of the response by simply noting the “pressure” established by each supply and the effect it will have on the conventional current direction through the diode. The diode will be on for any voltage vi that is greater than V volts and off for any lesser voltage.  For the on condition, vo = vi - V  For the off condition, vo = 0 as determined by Kirchhoff’s due to the lack of current. voltage law. JG 74 3. Determine the transition voltage that will result in a change of state for the diode from the off to the on state. Suppose the diode is ideal. The voltage across the resistor is 0 V because the diode current is 0 mA. The result is, Vi - V = 0, and so is the transition voltage JG 75  Draw a line on the sinusoidal input at the supply voltage V as shown in Figure to define the regions where the diode is on and off. For the on region, the diode is replaced by a short-circuit equivalent, and the output voltage is defined by For the off region, the diode is an open circuit, ID = 0 mA, and the output voltage is JG 76 4. It is often helpful to draw the output waveform directly below the applied voltage using the same scales for the horizontal axis and the vertical axis. JG 77 Q2. Determine the output waveform for the sinusoidal input. JG 78  Solution: The positive region of vi and the dc supply are both forcing the diode to turn on. Thus, the diode is in the on state for the entire range of positive voltages for vi and a good portion of the negative region of the input signal. Once the vi goes negative, it would have to exceed the dc supply voltage of 5V before it could turn the diode off. The transition from one state to the other will occur when when the diode is on, For voltages less than -5 V, the diode is in the open-circuit state V0 = 0 V. JG 79 Q3. Find the output voltage for the network, if the applied signal is the square wave. JG 80 Solution: JG 81 Transfer Characteristics of a circuit defined as the plot of input voltage Vin along the X axis VS. output voltage Vo along the Y axis of that circuit. JG 82 Q4. Find the output voltage for the biased series clipper. Plot the transfer characteristics. JG 83 Q5. Find the output voltage for the network, for the biased series clipper shown and plot the transfer characteristics. Transfer characteristics. JG 84 Q6. Find the output voltage for the network, for the circuit and plot the transfer characteristics. Transfer characteristics. JG 85 Q7. Design a circuit to obtain the given transfer characteristics. Transfer characteristics. JG 86 Parallel Clippers The diode is in parallel with the load For vi > 0 For vi < 0 Fig: Parallel clipper – circuit and input –output waveforms JG 87 Q8. Determine vo for the network, if the diode is ideal. JG 88 Solution: The output is defined across the series combination of the 4 V supply and the diode. The diode will be in the on state for the entire negative region, in addition to a small portion of the positive region of the input signal. When the diode is in the on state, the output voltage will be directly across the 4V dc supply. To find the transition voltage: When the diode is on the When the diode is in output will be 4 V. the off state, vo = vi JG 89 Q9. Determine vo for the network, if the voltage drop across the diode is 0.7 V. JG 90  Solution: The output is defined across the series combination of the 4 V supply and the diode drop(0.7V). To find the transition voltage: When the diode is on When the diode is in the output will be 3.3V. the off state, vo = vi JG 91 Q10. Find the output voltage for the network, for the biased parallel clipper with +V and plot the transfer characteristics. JG 92 Q11. Find the output voltage for the network, for the biased parallel clipper and plot the transfer characteristics. Transfer characteristics. JG 93 Q13. Find the output voltage for the network, for the biased parallel clipper with +V. JG 94 Q12. Find the output voltage for the network, for the biased parallel clipper with -V. JG 95 Q12. Find the output voltage for the double side clipper circuit. Also plot the transfer characteristics. JG 96  Zener Diode Clipping Circuits A silicon semiconductor diode designed to conduct in the  forward direction like an ordinary diode  reverse direction when a certain specified voltage is reached. Consists of a heavily doped p-n junction Has a well-defined reverse-breakdown voltage, at which it starts conducting current, without getting damaged. Manufactured with a wide range of voltages and can be used to give different voltage references on each half cycle. Available with breakdown voltages, VZ ranging from 2.4 to 33 volts Fig: Zener Diode characteristics JG 97 Voltage drop across the zener diode remains constant over a wide range of voltages, thus suitable for use in  clipping circuits,  voltage regulators etc. Fig: Zener Diode characteristics Fig: Zener Diode transfer characteristics JG 98 Zener Diode Clipping Circuits Used to clip the top, or bottom, or both of a waveform at a particular dc level and pass it to the output without distortion. Fig: Zener Diode clipper Acts like a biased diode clipping circuit with the bias voltage being equal to the zener breakdown voltage. Fig: Input and output waveforms JG 99 Double side clipper The output waveform will be clipped at the zener voltage plus the 0.7V (forward volt drop of the other diode). Fig: Zener Diode double side clipper Fig: Transfer characteristics Fig: Input and output waveforms JG 100 Diode Clamping Circuits JG 101  Diode Clamping Circuits Consists of a diode, a resistor, and a capacitor that shifts a waveform to a different dc level without changing the appearance of the applied signal. Additional shifts can be obtained by introducing a dc supply to the basic structure. Fig: Diode clamping circuit The resistor and capacitor must be chosen such that the time constant determined by τ = RC is sufficiently large to ensure that the voltage across the capacitor retains its voltage during the interval the diode is non conducting. For all practical purposes, it is assumed that the capacitor fully charges or discharges in five time constants JG 102 General procedure for analyzing the clamping circuits Step 1: Start the analysis by examining the response of the portion of the input signal that will forward bias the diode. Step 2: Fig: Clamping circuit During the period that the diode is in the on state, assume that the capacitor will charge up instantaneously to a voltage level determined by the surrounding network. τ = RC is very small, as R is shorted out The result is that the capacitor will quickly Fig: The equivalent circuit when charge to the peak value of V volts with the the diode is on polarity indicated. Also, Vo = 0V (since the diode is shorted). JG 103 Step 3: When the diode is in the off state, the capacitor holds on to V, as RC time constant is very large. Fig: The equivalent circuit when the diode is off Step 4: Throughout the analysis, maintain a continual awareness of the location and defined polarity for vo to ensure that the proper levels are obtained. JG 104  Applying Kirchhoff’s voltage law around the input loop results in Step 5: Check that the total swing of the output matches that of the input. Fig: Input and output waveforms JG 105 Q1. Determine v o for the network. Fig: Diode clamping circuit JG 106 Solution. For the period T1 to T2,When the diode is on and Vo = 5V Fig: Equivalent circuit when the diode is in the on state. Applying Kirchhoff’s voltage law around the input loop results in For the period T2 to T3, the diode is in the off state. results in Fig: Equivalent circuit when the diode is in the off state. JG 107 Input and output waveforms Fig: Input and output waveforms of the clamping circuit JG 108  Q2. Determine Vo for the network. Fig: Positive clamping with -VR JG 109 Solution. The diode is in the on state during the negative half cycle, only when Vi crosses -VR Applying Kirchhoff’s voltage law around the input loop results in Fig: Equivalent circuit when the diode is on The diode is in the off state for the entire positive half cycle and for the portion of negative half cycle above -V results in Fig: Equivalent circuit when the diode is off JG 110 Fig: Input and output waveforms for Positive clamping with -VR JG 111  Q3. Determine v o for the network. Fig: Positive clamping JG 112  Q4. Determine v o for the network. Fig: Positive clamping with +VR JG 113 Q5. Determine v o for the network. Fig: Negative clamping JG 114 Q6. Determine v o for the network. Fig: Negative clamping with +VR JG 115 Q7. Determine v o for the network. Fig: Negative clamping with -VR JG 116