INSTITUTE OF AERONAUTICAL ENGINERING DUNDIGAL, HYDERABAD

Transcription

1 INSTITUTE OF AERONAUTICAL ENGINERING DUNDIGAL, HYDERABAD Digital Signal Processing Lab Work Book Name: Reg.No: Branch: Class: Section:

2 IARE-ECE Department CERTIFICATE This is to certify that it is a bonafide record of practical work done by Mr./Ms., Reg. No. in the Digital Signal Processing Laboratory in semester of year during LAB Incharge 2

3 ELECTRONIC CIRCUIT ANALYSIS LAB Part: Minimum eight experiment to be conducted Design and Simulation in Simulation Laboratory using Multisim OR Pspice OR Equivalent Simulation Software. (Any six):. Common Emitter amplifier 2. Common Source amplifier 3. Two Stage RC Coupled Amplifier 4. Current shunt and voltage series Feedback Amplifier 5. Cascode Amplifier 6. Wien Bridge Oscillator using Transistors 7. RC Phase Shift Oscillator using Transistors 8. Class A Power Amplifier (Transformer less) 9. Class B Complementary Symmetry Amplifier 0. Common base (BJT) / Common gate (JFET) Amplifier. II) Testing in the Hardware Laboratory ( Experiments: 2):. Class A Power Amplifier (with transformer load) 2. Class C Power Amplifier 3. Single Tuned Voltage Amplifier 4. Hartley and Colpitts oscillator 5. Darlington Pair 6. MOS amplifier 3

4 PULSE CIRCUITS LAB Part: 2 Minimum eight experiment to be conducted: : Linear wave shaping i) RC low pass circuit for different time constants ii) RC high pass circuit for different time constants 2. Non- linear wave shaping a) Transfer characteristics and response of Clippers i) Positive and Negative Clippers ii) Clipping at two independent levels b) The steady state output waveform of clampers for a square wave input i) Positive and Negative Clampers ii) Clamping at reference voltage 3. Comparison Operation of Comparators 4. Switching characteristics of a transistor 5. Design a Bistable Multivibrator and draw its waveforms 6. Design an Astable Multivibrator and draw its waveforms 7. Design a Monostable Multivibrator and draw its waveforms 8. Response of Schmitt Trigger circuit for loop gain less than and greater than one 9. UJT relaxation oscillator 0. The output voltage waveform of Bootstrap sweep circuit 4

5 Equipments required for Laboratories: For software simulation of Electronic circuits Computer Systems with latest specifications Connected in Lan (Optional) Operating system (Windows XP) Simulations software (Multisim/TINAPRO) Package For Hardware simulations of Electronic Circuts & Pulse circuits RPSs 0 30 v CROs 0 20 M Hz Functions Generators 0 M hz Multimeters Components Win XP/LINUX etc 5

6 INDEX ELECTRONIC CIRCUIT LAB SIMULATION LAB (Any six) HARDWARE LAB (Any two) S.No. Name of the Experiment Page No: Class A power amplifier 9 2 Class C Power Amplifier 4 3 Single Tuned Voltage Amplifier 9 4 Hartley and Colpitts oscillator 26 5 Darlington Pair 35 6 MOS amplifier 40 SIMULATION LAB (Any six) S.No. Name of the Experiment Page No: Common Emitter amplifier 45 2 Common Source amplifier 50 3 Two Stage RC Coupled Amplifier 55 4 Current shunt and voltage series Feedback Amplifier 6 5 Cascode Amplifier 66 6 Wien bridge Oscillator 7 7 RC Phase Shift Oscillator using Transistors 75 8 Class A Power Amplifier (Transformer less) 79 9 CLASS B Complementary Symmetry Amplifier 84 0 Common base (BJT) / Common gate(jfet) Amplifier 6

7 PULSE CIRCUITS(Any eight) S.No. Name of the Experiment Page No: Linear wave shaping a. RC low pass circuit for different time constants b. RC high pass circuit for different time constants 2 Non- linear wave shaping Transfer characteristics and response of Clippers i) Positive and Negative Clippers ii) Clipping at two independent levels The steady state output waveform of clampers for a square wave input i) Positive and Negative Clampers ii) Clamping at reference voltage 3 Comparison Operation of Comparators 4 Switching characteristics of a transistor 3 5 Design a Bistable Multivibrator and draw its waveforms 7 6 Design an Astable Multivibrator and draw its waveforms 2 7 Design a Monostable Multivibrator and draw its waveforms 26 8 Response of Schmitt Trigger circuit for loop gain less than and greater than one 32 9 UJT relaxation oscillator 38 0 The output voltage waveform of Bootstrap sweep circuit 43 7

8 8 Electronic Circuit Analysis & Pulse and Digital circuit Lab

9 EXPERIMENT NO- CLASS A POWER AMPLIFIER (Transformer coupled) AIM:. To study and plot the frequency response of a Class A Power Amplifier. 2. To calculate efficiency of Class A Power Amplifier. COMPONENTS & EQUIPMENT REQUIRED: S.No Apparatus Range/ Rating Quantity (in No.s). Trainer Board containing a) DC Supply voltage. b) NPN Transistor. c) Resistors. 2 V BC Ω 00KΩ 470Ω d) Capacitor. 22 F. e) Inductor. 50mH 0-00mA 2. D.C milli ammeter Cathode Ray Oscilloscope. (0-20)MHz 3. Function Generator. 0. Hz-0 4. MHz BNC Connector 2 9

10 5. 6. Connecting Wires 5A 5 THEORY: Power amplifiers are mainly used to deliver more power to the load. To deliver more power it requires large input signals, so generally power amplifiers are preceded by a series of voltage amplifiers. In class-a power amplifiers, Q-point is located in the middle of DC-load line. So output current flows for complete cycle of input signal. Under zero signal condition, maximum power dissipation occurs across the transistor. As the input signal amplitude increases power dissipation reduces. The maximum theoretical efficiency is 50%. CIRCUIT DIAGRAM: 0

11 EXPECTED GRAPH: Bandwidth=fH fl TABULAR FORM: Vin = 50 mv S.No Frequency Vo Gain A = Gain(dB) Av = (in Hz) (in volts) Vo/ Vi 20 log(vo/ Vi ) K 6 2K 7 4K 8 8K

12 9 0K 0 20K 40K 2 80K 3 00K 4 200K CALCULATIONS: When signal is removed, Vi=0 Input Power: Pin=Vcc x Ic = P Vo / 8R 2 Zero signal current, Ic = Output Power: out L OutputPower Inputpower Efficiency: η = x00 = PROCEDURE:. Connect the circuit as shown in figure. 2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without distortion. 3. By keeping input signal voltage, say at 50 mv, vary the input signal frequency from 0- MHz as shown in tabular column and note the corresponding output voltage. 4. Measure and note down the zero signal dc current by disconnecting the function generator from the circuit. 5. Calculate the efficiency according to the expressions given. 6. Plot the graph between the o/p gain and frequency and calculate the bandwidth. 2

13 PRECAUTIONS:. No loose contacts at the junctions. 2. Check the connections before giving the power supply 3. Observations should be taken carefully. RESULT:. Frequency Response of CLASS-A Power amplifier is plotted. 2. Efficiency of CLASS A Power amplifier is found to be 3. Bandwidth fh fl = VIVA QUESTIONS:. Differentiate between voltage amplifier and power amplifier 2. Why power amplifiers are considered as large signal amplifier? 3. When does maximum power dissipation happen in this circuit?. 4. What is the maximum theoretical efficiency? 5. Sketch wave form of output current with respective input signal. 6. What are the different types of class-a power amplifiers available? 7. What is the theoretical efficiency of the transformer coupled class-a power amplifier? 8. What is difference in AC, DC load line?. 9. How do you locate the Q-point? 0. What are the applications of class-a power amplifier?. What is the expression for the input and output power in class A power amplifier? 3

14 EXPERIMENT NO- 2 CLASS C POWER AMPLIFIER AIM: To study class C power amplifier and to determine its efficiency. Electronic Circuit Analysis & Pulse and Digital circuit Lab APPARATUS:. Physitech class C power amplifier trainer. 2. Signal generator 3. Milli ammeter(0-50ma). 4. BNC probes and connecting wires. THEORY : Class-C amplifiers conduct less than 50% of the input signal and the distortion at the output is high, but high efficiencies (up to 90%) are possible. The usual application for class-c amplifiers is in RF transmitters operating at a single fixed carrier frequency, where the distortion is controlled by a tuned load on the amplifier. The input signal is used to switch the active device causing pulses of current to flow through a tuned circuit forming part of the load. The class-c amplifier has two modes of operation: tuned and unturned. The diagram shows a waveform from a simple class-c circuit without the tuned load. This is called untuned operation, and the analysis of the waveforms shows the massive distortion that appears in the signal. When the proper load (e.g., an inductive-capacitive filter plus a load resistor) is used, two things happen. The first is that the output's bias level is clamped with the average output voltage equal to the supply voltage. This is why tuned operation is sometimes called a clamper. This allows the waveform to be restored to its proper shape despite the amplifier having only a one-polarity supply. This is directly related to the second phenomenon: the waveform on the center frequency becomes less distorted. The residual distortion is dependent upon the bandwidth of the tuned load, with the center frequency seeing very little distortion, but greater attenuation the farther from the tuned frequency that the signal gets. The tuned circuit resonates at one frequency, the fixed carrier frequency, and so the unwanted frequencies are suppressed, and the wanted full signal (sine wave) is extracted by the tuned load. The 4

15 signal bandwidth of the amplifier is limited by the Q-factor of the tuned circuit but this is not a serious limitation. Any residual harmonics can be removed using a further filter. The active element conducts only while the drain voltage is passing through its minimum. By this means, power dissipation in the active device is minimized, and efficiency increased. Ideally, the active element would pass only an instantaneous current pulse while the voltage across it is zero: it then dissipates no power and 00% efficiency is achieved. However practical devices have a limit to the peak current they can pass, and the pulse must therefore be widened, to around 20 degrees, to obtain a reasonable amount of power, and the efficiency is then 60-70% CIRCUIT DIAGRAM: 5

16 EXPECTED GRAPH: Bandwidth=fH fl TABULAR FORM: Vin = 50 mv S.No Frequency Vo Gain A = Gain(dB) Av = (in Hz) (in volts) Vo/ Vi 20 log(vo/ Vi ) K 6 2K 7 4K 8 8K 9 0K 0 20K 6

17 40K 2 80K 3 00K 4 200K CALCULATIONS: When signal is removed, Vi=0 Input Power: Pin=Vcc x Ic = P Vo / 8R 2 Zero signal current, Ic = Output Power: out L Efficiency: OutputPower η = x00 Inputpower Pout = x00 P in PROCEDURE:. Connect the circuit as shown in figure 2. Connect the input signal(say 5 to 8v) from signal generator 3. Connect the mili ammeter to the ic terminals. 4. By keeping input voltage constant, vary the frequency in regular steps. 5. Note down the corresponding output voltage from CRO for each frequency. 6. Plot the graph between gain (db) and frequency. 7. Calculate bandwidth from the graph. 8. Calculate the resonant frequency using 2 L T C 9. Calculate the efficiency according to the expressions given. 0. Plot the graph between the o/p gain and frequency and calculate the bandwidth. 7

18 RESULT:. Frequency Response of CLASS-C Power amplifier is plotted. 2. Efficiency of CLASS C Power amplifier is found to be 3. Bandwidth fh fl = VIVA QUESTIONS:. What is the maximum theoretical efficiency of class-c PA? 2. Sketch wave form of output current with respective input signal. 3. How do you locate the Q-point in class-c PA? 4. What are the applications of class-c power amplifier? 5. What is the expression for the input and output power in class AC power amplifier? 8

19 AIM: EXPERIMENT NO- 3 SINGLE TUNED VOLTAGE AMPLIFIER. To study & plot the frequency response of a Single Tuned voltage amplifier. 2. To find the resonant frequency. 3. To calculate gain and bandwidth. COMPONENTS & EQUIPMENT REQUIRED: S.No Apparatus. Trainer Board containing a) DC Supply voltage. b) NPN Transistor. c) Resistors. Range/ Rating 2 V BC KΩ 50Ω KΩ 0 KΩ Quantity (in No.s) 2 d) Capacitor. 0 F 22 F F F. 2 e) Inductor. mh 2. Cathode Ray Oscilloscope. (0-20)MHz 3. Function Generator. 0. Hz-0MHz 4. BNC Connector 2 9

20 5. Connecting Wires 5A 5 Electronic Circuit Analysis & Pulse and Digital circuit Lab CIRCUIT DIAGRAM: EXPECTED WAVEFORM: 20

21 TABULAR COLUMN : C=0.022μF Vin = 50 mv C== 0.033μF Vin = 50 mv S.No Frequency Vo Gain Gain(dB) Frequency Vo Gain Gain(dB) (in Hz) (V) A = 20 (in Hz) (V) A = 20 log(vo/ Vo/ Vi log(vo/ Vi Vo/ Vi Vi ) ) K 6 2K 7 4K 8 8K 9 0K 0 20K 40K 2 80K 2

22 3 00K 4 200K THEORY: Tuned amplifiers are amplifiers involving a resonant circuit, and are intended for selective amplification within a narrow band of frequencies. Radio and TV amplifiers employ tuned amplifiers to select one broadcast channel from among the many concurrently induced in an antenna or transmitted through a cable. Selected aspects of tuned amplifiers are reviewed in this note. Parallel Resonant Circuit An idealized parallel resonant circuit, i.e. one described by idealized circuit elements, is drawn below. input impedance of this configuration, shown below the circuit diagram, is readily obtained. A modest algebraic restatement in convenient form also is shown. The significance of the definitions of the 'quality factor' Q and the resonant frequency ωo will become clear from the discussion. The influence of the Q parameter on the tuned-circuit impedance for several values of Q is plotted below for a normalized response. 22

23 PROCEDURE:. Connect the circuit as shown in figure. 2. Connect the 0.022μF capacitor 3. Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without distortion. 4. By keeping input signal voltage, say at 50 mv, vary the input signal frequency from 0-00KHz as shown in tabular column and note the corresponding output voltage. 5. Repeat the same procedure for 0.033μF capacitor. 6. Plot the graph: gain (Vs) frequency. 7. Calculate the f and f2 and bandwidth. 8. Compare the resonant frequency with theoretical value in both the cases. PRECAUTIONS: -. No loose contacts at the junctions. 2. Check the connections before giving the power supply 3. Observations should be taken carefully. RESULT:. Frequency response of RF Tuned voltage amplifier is plotted. 2. For 0.022μF, gain = Db Bandwidth= 3. For 0.033μF, gain = db 23

24 Bandwidth= VIVA QUESTIONS:. What is the purpose of tuned amplifier? 2. What is Quality factor? 3. Why should we prefer parallel resonant circuit in tuned amplifier. 4. What type of tuning we need to increase gain and bandwidth.? 5. What are the limitations of single tuned amplifier? 6. What is meant by Stagger tuning? 7. What is the conduction angle of an tuned amplifier if it is operated in class B mode? 8. What are the applications of tuned amplifier 9. What are the different types of tuned circuits? 0. State relation between resonant frequency and bandwidth of a Tuned amplifier.. Differentiate between Narrow band and Wideband tuned amplifiers? 2. Calculate bandwidth of a Tuned amplifier whose resonant frequency is 5KHz and Q-factor is Specify the applications of Tuned amplifiers. 24

25 GRAPH: 25

26 EXPERIMENT NO-4 (A) HARTLEY OSCILLATOR Electronic Circuit Analysis & Pulse and Digital circuit Lab AIM: To find practical frequency of a Hartley oscillator and to compare it with theoretical frequency for L = 0mH and C = 0.0 F, F and F. COMPONENTS AND EQUIPMENTS REQUIRED: S.No Device Range/ Rating Hartley Oscillator trainer board containing a) DC supply voltage 2V b) Inductors 5mH c) Capacitor 0.22 F 0.0 F F F d) Resistor K 0K 47K Quantity (in No.s) 2 2 e) NPN Transistor BC 07 2 Cathode Ray Oscilloscope (0-20) MHz 3. BNC Connector 4 Connecting wires 5A 4 26

27 CIRCUIT DIAGRAM: HARTLEY OSCILLATOR EXPECTED WAVEFORM: TABULATIONS: S.No LT(mH) C ( F) Theoretical Practical Practical Vo (V) frequency (KHz) waveform time frequency (KHz) (ptp) period (Sec)

28 THEORY: The Hartley oscillator is an electronic oscillator circuit in which the oscillation frequency is determined by a tuned circuit consisting of capacitors and inductors, that is, an LC oscillator. The circuit was invented in 95 by American engineer Ralph Hartley. The distinguishing feature of the Hartley oscillator is that the tuned circuit consists of a single capacitor in parallel with two inductors in series (or a single tapped inductor), and the feedback signal needed for oscillation is taken from the center connection of the two inductors. The frequency of oscillation is approximately the resonant frequency of the tank circuit. If the capacitance of the tank capacitor is C and the total inductance of the tapped coil is L then If two uncoupled coils of inductance L and L2 are used then However if the two coils are magnetically coupled the total inductance will be greater because of mutual inductance k. PROCEDURE:. Connect the circuit as shown in figure. 2. Connect 0.0 F capacitor in the circuit and observe the waveform. 3. Note the time period of the waveform and calculate the frequency: f = /T. 28

29 4. Now connect the capacitance to F and F and calculate the frequency and tabulate the readings as shown. 5. Find the theoretical frequency from the formula f = 2 L T C Where LT = L + L2 = 5 mh + 5mH = 0 mh and compare theoretical and practical values. PRECAUTIONS:. No loose contacts at the junctions. 2. Check the connections before giving the power supply 3. Observations should be taken carefully. RESULT:. For C = 0.0 F, & LT = 0 mh; Theoretical frequency = Practical frequency = 2.For C = F, & LT = 0 mh; Theoretical frequency = Practical frequency = 3. For C = F, & LTs = 0 mh; Theoretical frequency = Practical frequency = 29

30 EXPERIMENT NO-4 (B) COLPITTS OSCILLATOR Electronic Circuit Analysis & Pulse and Digital circuit Lab AIM: To find practical frequency of Colpitt s oscillator and to compare it with theoretical Frequency for L= 5mH and C= 0.00 F, F, F respectively. COMPONENTS & EQIUPMENT REQUIRED: - S.No Device Range/ Rating Colpitts Oscillator trainer board containing a) DC supply voltage 2V b) Inductors 5mH Quantity (in No.s) c) Capacitor 0.0 F 0. F 00 F d) Resistor e) NPN Transistor K.5K 0K 47K BC 07 2 Cathode Ray Oscilloscope (0-20) MHz 3. BNC Connector 4 Connecting wires 5A 4 30

31 CIRCUIT DIAGRAM: COLPITTS OSCILLATOR EXPECTED WAVEFORM: 3

32 TABULAR COLUMN: S.NO L(mH) C ( F) C2 ( F) CT ( F) Theoretical Frequency (KHz) Practical Frequency (KHz) Vo(V) Peak to peak THEORY: A Colpitts oscillator, invented in 98 by American engineer Edwin H. Colpitts, [] is one of a number of designs for LC oscillators, electronic oscillators that use a combination of inductors (L) and capacitors (C) to produce an oscillation at a certain frequency. The distinguishing feature of the Colpitts oscillator is that the feedback for the active device is taken from a voltage divider made of two capacitors in series across the inductor.the frequency of oscillation is approximately the resonant frequency of the LC circuit, which is the series combination of the two capacitors in parallel with the inductor The actual frequency of oscillation will be slightly lower due to junction capacitances and resistive loading of the transistor.as with any oscillator, the amplification of the active component should be marginally larger than the attenuation of the capacitive voltage divider, to obtain stable operation. Thus, a Colpitts oscillator used as a variable frequency oscillator (VFO) performs best when a variable inductance is used for tuning, as opposed to tuning one of the two capacitors. If tuning by variable capacitor is needed, it should be done via a third capacitor connected in parallel to the inductor (or in series as in the Clapp oscillator). 32

33 PROCEDURE:. Connect the circuit as shown in the figure 2. Connect C2= 0.00 F in the circuit and observe the waveform. 3. Calculate the time period and frequency of the waveform (f=/t) 4. Now, fix the capacitance to F and then to F and calculate the frequency and tabulate the reading as shown. 5. Find theoretical frequency from the formula f = 2 LC T Where C T CC2 C C and compare theoretical and practical values. 6. Plot the graph o/p voltage vs time period and practical frequency 2 PRECAUTIONS:. No loose contacts at the junctions. 2. Check the connections before giving the power supply 3. Observations should be taken carefully. RESULT: Hence, the frequency of oscillations of Colpitts oscillator is measured practically and compared with theoretical values.. For C= F & L= 5mH Theoretical frequency = Practical frequency = 2. For C= F & L= 5mH Theoretical frequency = Practical frequency = 33

34 3. For C=0.00 F & L= 5mH Theoretical frequency = Practical frequency = VIVA QUESTIONS:. What are the applications of LC oscillations? 2. What type of feedback is used in oscillators? 3. What is the expression for the frequency of oscillations of Colpitt s and Hartley oscillator? 4. Whether an oscillator is dc to ac converter. Explain? 5. What is the loop gain of an oscillator? 6. What is the difference between amplifier and oscillator? 7. What is the condition for sustained oscillations? 8. How many inductors and capacitors are used in Hartley Oscillator? 9. How the oscillations are produced in Hartley oscillator? 0. What is the difference between damped oscillations undamped oscillations?. How does Colpitt s differ from Hartley? 34

35 EXPERIMENT NO 5 DARLINGTON EMITTER FOLLOWER Electronic Circuit Analysis & Pulse and Digital circuit Lab AIM: To study & plot the frequency response of a Darlington emitter follower circuit. COMPONENTS & EQUIPMENT REQUIRED: S.No Device Range/ Rating Quantity (in No.s). Trainer Board containing a) DC Supply voltage. b) NPN Transistor. c) Resistors. 2 V CL KΩ 0 KΩ 2 2. Cathode Ray Oscilloscope. (0-20)MHz 3. Function Generator. 0. Hz-0 MHz 4 BNC Connector 2 5 Connecting Wires 5A 5 35

36 CIRCUIT DIAGRAM: DARLINGTON EMITTER FOLLOWER EXPECTED GRAPH: 36

37 TABULAR COLUMN : Vin = 50 mv Frequency (in Hz) 20 Output Voltage (Vo) Gain (in db) = 20log0(Vo/Vi) K 50K 00K 200K 400K 600K 37

38 800K 000K THEORY: The Darlington transistor (often called a Darlington pair) is a compound structure consisting of two bipolar transistors (either integrated or separated devices) connected in such a way that the current amplified by the first transistor is amplified further by the second one. This configuration gives a much higher common/emitter current gain than each transistor taken separately and, in the case of integrated devices, can take less space than two individual transistors because they can use a shared collector. Integrated Darlington pairs come packaged singly in transistor-like packages or as an array of devices (usually eight) in an integrated circuit. A Darlington pair behaves like a single transistor with a high current gain (approximately the product of the gains of the two transistors). In fact, integrated devices have three leads (B, C and E), broadly equivalent to those of a standard transistor. A general relation between the compound current gain and the individual gains is given by: If β and β2 are high enough (hundreds), this relation can be approximated with: Darlington pairs are available as integrated packages or can be made from two discrete transistors; Q (the left-hand transistor in the diagram) can be a low power type, but normally Q2 (on the right) will need to be high power. The maximum collector current IC(max) of the pair is that of Q2. A typical integrated power device is the 2N6282, which includes a switch-off resistor and has a current gain of 2400 at IC=0A. A Darlington pair can be sensitive enough to respond to the current passed by skin contact even at safe voltages. Thus it can form the input stage of a touch-sensitive switch. 38

39 PROCEDURE: a) Connect the circuit diagram as shown fig. b) Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without distortion. c) By keeping input signal voltage, say at 50 mv, vary the I/P signal frequency from 50Hz to MHz in step as shown in tabular column and note the corresponding O/P voltage. d) Plot the graph between gain in (db) vs frequency. e) Calculate the maximum gain and bandwidth. BW=f2-f PRECAUTIONS:. No loose contacts at the junctions. 2. Check the connections before giving the power supply 3. Observations should be taken carefully. RESULT: Hence the frequency response Darlington emitter follower circuit has been studied and plotted Gain: Bandwidth: fh fl = VIVA QUESTIONS:. What is the difference between emitter follower and Darlington pair ckt? 2. What is the value of i/p impedance of a typical Darlington pair ckt? 3. What is the value of o/p impedance of a typical Darlington pair ckt? 4. What is the value of current and voltage gain of a typical Darlington pair ckt? 5. Mention the applications of a Darlington pair ckt 39

40 EXPERIMENT NO: 6 MOS AMPLIFIER AIM: a) To Plot the frequency response of a common source amplifier. b) Calculate gain. c) Calculate bandwidth. COMPONENTS & EQUIPMENTS REQUIRED: S.No Device Range/Rating Quantity (in No.s). FET amplifier Trainer Board with (a) DC supply voltage (b) FET (c) Capacitors (d) Resistors 2V BFW 0. F 47 F.5K 4.7 K 2 M 2. Signal generator 0.Hz-MHz 40

41 CIRCUIT DIAGRAM: THEORY: A common-source amplifier is one of three basic single-stage field-effect transistor (FET) amplifier topologies, typically used as a voltage or transconductance amplifier. The easiest way to tell if a FET is common source, common drain, or common gate is to examine where the signal enters and leaves. The remaining terminal is what is known as "common". In this example, the signal enters the gate, and exits the drain. The only terminal remaining is the source. This is a common-source FET circuit. The analogous bipolar junction transistor circuit is the common-emitter amplifier. The common-source (CS) amplifier may be viewed as a transconductance amplifier or as a voltage amplifier. As a transconductance amplifier, the input voltage is seen as modulating the current going to the load. As a voltage amplifier, input voltage modulates the amount of current flowing through the FET, changing the voltage across the output resistance according to Ohm's law. However, the FET device's output resistance typically is not high enough for a reasonable transconductance amplifier (ideally infinite), nor low enough for a decent voltage amplifier (ideally zero). Another major drawback is the amplifier's limited high-frequency response. Therefore, in practice the output often is routed through either a voltage follower (common-drain or CD stage), or a current follower (common-gate or CG stage), to 4

42 obtain more favorable output and frequency characteristics. The bandwidth of the common-source amplifier tends to be low, due to high capacitance resulting from the Miller effect. The gate-drain capacitance is effectively multiplied by the factor. TABULAR COLUMN : Vin = 50 mv Frequency (in Hz) 20 Output Voltage (Vo) Gain (in db) = 20log0(Vo/Vi) K 50K 00K 200K 400K 42

43 600K 800K 000K PROCEDURE:. Connect the circuit diagram as shown in figure. 2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without 3.distortion. 3. By keeping input signal voltage, say at 50mV, vary the input signal frequency from 0 to MHz in steps as shown in tabular column and note the corresponding output voltages. 4. Save the circuit and simulate. 5. Calculate the maximum gain and bandwidth using bode plotter. Compare the values with the practical circuit values PRECAUTIONS: Check whether the connections are made properly or not. RESULT: Hence, the frequency response of FET (CS) amplifier is plotted VIVA:. Draw the character sites of mosfet. 2. Define varies region of mosfet character sties. 3. Right the current equation for mosfet for varies region. 4. Define second order effect of a mosfet. 43

44 SIMULATION LAB 44

45 EXPERIMENT NO: CE AMPLIFIER AIM: To plot the frequency response of CE amplifier and calculate gain bandwidth. SOFTWARE REQUIRED: MultiSim Analog Devices Edition 0.0 COMPONENTS & EQUIPMENTS REQUIRED: - S.No Apparatus Range/ Rating Quantity (in No.s). CE Amplifier trainer Board with DC power supply DC power supply NPN transistor Carbon film resistor (e)carbon film resistor (f) Capacitor. Cathode Ray Oscilloscope. 2. 2V 5V BC 07 00K, /2W 2.2K, /2W 0.µF 2 (0-20)MHz Function Generator. BNC Connector 0. Hz-0 MHz 2 45

46 5. Connecting Wires 5A 5 CIRCUIT DIAGRAM: CE AMPLIFIER 46

47 EXPECTED GRAPH: Bandwidth = fh-fl TABULAR COLUMN: Input voltage: Vi = 50mV Frequency (in Hz) Gain (in db) = 20 log 0 VO/ Vi K 2K 4K 8K 0K 47

48 20K 30K 40K 50K 60K 80K 00K 250K 500K 750K 000K THEORY: The CE amplifier provides high gain & wide frequency response. The emitter lead is common to both input and output circuits and is grounded. The emitter base is forward biased. The collector current is controlled by the base current rather than emitter current. The input signal is applied to base terminal of the transistor and amplifier output is taken across collector terminal. A very small change in base current produces a much larger change in collector current. Frequency response of an amplifier is defined as the variation of gain with respective frequency. The gain of the amplifier increases as the frequency increases from zero till it becomes maximum at lower cut-off frequency and remains constant till higher cut-off frequency and then it falls again as the frequency increases. At low frequencies the reactance of coupling capacitor CC is quite high and hence very small part of signal will pass through from one stage to the next stage. 48

49 At high frequencies the reactance of inter electrode capacitance is very small and behaves as a short circuit. This increases the loading effect on next stage and service to reduce the voltage gain due to these reasons the voltage gain drops at high frequencies. At mid frequencies the effect of coupling capacitors is negligible and acts like short circuit, where as inter electrode capacitors acts like open circuit. So, the circuit becomes resistive at mid frequencies and the voltage gain remains constant during this range. PROCEDURE:. Connect the circuit diagram as shown in figure. 2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without distortion. 3. By keeping input signal voltages at 50mV, vary the input signal frequency from 0 to MHz in steps as shown in tabular column and note the corresponding output voltages. 4. Save the circuit and simulate. 5. Calculate the maximum gain and bandwidth using bode plotter. Compare the values with the practical circuit values. PRECAUTIONS: Check whether the connections are made properly or not. RESULT: Frequency response of CE amplifier is plotted. Gain, AV = db. Bandwidth= fh--fl = Hz. VIVA QUESTIONS:. What are the advantages and disadvantages of single-stage amplifiers? 2. Why gain falls at HF and LF? 3. Why the gain remains constant at MF? 4. Explain the function of emitter bypass capacitor, CE? 5. How the band width will effect as more number of stages are cascaded? 6. Define frequency response? 7. What is the phase difference between input and output waveforms of a CE amplifier? 49

50 EXPERIMENT NO: 2 COMMON SOURCE AMPLIFIER AIM: a) To Plot the frequency response of a common source amplifier. b) Calculate gain. c) Calculate bandwidth. SOFTWARE REQUIRED: MultiSim Analog Devices Edition 0.0 COMPONENTS & EQUIPMENTS REQUIRED: S.No Device Range/Rating Quantity. FET amplifier Trainer (in No.s) Board with (a) DC supply voltage (b) FET (c) Capacitors (d) Resistors 2V BFW 0. F 47 F.5K 4.7 K M 2 2. Signal generator 0.Hz-MHz THEORY: The FET is a type of transistor commonly used for weak signal amplification. The device can amplify analog or digital signals. It can also switch DC or function as an oscillator. In the FET current flows along a semiconductor path called the channel. At one end of the channel, there is an electrode called source. At the other end of the channel there is an electrode called the drain. 50

51 Frequency response of an amplifier is defined as the variation of gain with respective frequency. The gain of the amplifier increases as the frequency increases from zero till it becomes maximum at lower cut-off frequency and remains constant till higher cut-off frequency and then it falls again as the frequency increases. At low frequencies the reactance of coupling capacitor CC is quite high and hence very small part of signal will pass through from one stage to the next stage. At high frequencies the reactance of inter electrode capacitance is very small and behaves as a short circuit. This increases the loading effect on next stage and service to reduce the voltage gain due to these reasons the voltage gain drops at high frequencies. At mid frequencies the effect of coupling capacitors is negligible and acts like short circuit, where as inter electrode capacitors acts like open circuit. So, the circuit becomes resistive at mid frequencies and the voltage gain remains constant during this range CIRCUIT DIAGRAM: COMMON SOURCE AMPLIFIER 5

52 EXPECTED GRAPH: TABULAR COLUMN: Input = 50mV Frequency (in Hz) 20 Gain (in db) = 20log0(Vo/Vi)

53 5000 0K 50K 00K 200K 400K 600K 800K PROCEDURE:. Connect the circuit diagram as shown in figure. 2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without 3.distortion. 3. By keeping input signal voltage, say at 50mV, vary the input signal frequency from 0 to MHz in steps as shown in tabular column and note the corresponding output voltages. 4. Save the circuit and simulate. 5. Calculate the maximum gain and bandwidth using bode plotter. Compare the values with the practical circuit values PRECAUTIONS: Check whether the connections are made properly or not. 53

54 RESULT: Hence, the frequency response of FET (CS) amplifier is plotted. Gain = db (maximum). 3. Bandwidth= fh--fl = Hz. VIVA QUESTIONS:. What is the difference between FET and BJT? 2. FET is unipolar or bipolar? 3. Draw the symbol of FET? 4. What are the applications of FET? 5. FET is voltage controlled or current controlled? 6. How does FET acts as an amplifier? 7. What are the advantages of FET over BJT? 8. What is the region of FET so that it acts as an amplifier? 9. What are the differences between JFET and MOSFET? 54

55 EXPERIMENT NO- 3 TWO STAGE RC COUPLED AMPLIFIER Electronic Circuit Analysis & Pulse and Digital circuit Lab AIM:. To plot the frequency response of a RC coupled amplifier with a pair of shunted emitter capacitors of 0 μf and 00μF. 2. To calculate gain. 3. To calculate bandwidth. SOFTWARE REQUIRED: MultiSim Analog Devices Edition 0.0 COMPONENTS & EQUIPMENT REQUIRED: S.No Device Range/ Rating Quantity (in No.s). Trainer Board containing a) DC Supply voltage. b) NPN Transistor. c) Resistors. d) Capacitors. 2 V BC KΩ 2.2 KΩ KΩ 0 KΩ 00 F 0 F Bode Plotter 3. Function Generator. 0. Hz-0 MHz 55

56 CIRCUIT DIAGRAM: TWO STAGE RC COUPLED AMPLIFIER EXPECTED GRAPH: 56

57 TABULAR FORM: Vin = 50 mv S.No Frequency (in Hz) 00 C=0μF Gain(dB) 20 log(vo/ Vi ) C=00μF Frequency Gain(dB) (in Hz) 20 log(vo/ Vi ) K 6 2K 7 4K 8 8K 9 0K 0 20K 40K 2 80K 3 00K 4 200K 57

58 5 300K 6 500K 7 700K 8 900K 9 M THEORY: As the gain provided by a single stage amplifier is usually not sufficient to drive the load, so to achieve extra gain multi-stage amplifier are used. In multi-stage amplifiers output of one-stage is coupled to the input of the next stage. The coupling of one stage to another is done with the help of some coupling devices. If it is coupled by RC then the amplifier is called RC-coupled amplifier. Frequency response of an amplifier is defined as the variation of gain with respective frequency. The gain of the amplifier increases as the frequency increases from zero till it becomes maximum at lower cut-off frequency and remains constant till higher cut-off frequency and then it falls again as the frequency increases. At low frequencies the reactance of coupling capacitor CC is quite high and hence very small part of signal will pass through from one stage to the next stage. At high frequencies the reactance of inter electrode capacitance is very small and behaves as a short circuit. This increases the loading effect on next stage and service to reduce the voltage gain due to these reasons the voltage gain drops at high frequencies. At mid frequencies the effect of coupling capacitors is negligible and acts like short circuit, where as inter electrode capacitors acts like open circuit. So, the circuit becomes resistive at mid frequencies and the voltage gain remains constant during this range. 58

59 PROCEDURE:. Connect the circuit as shown in figure for 0 μf. 2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without distortion. 3. By keeping input signal voltage, say at 50 mv, vary the input signal frequency from 0- MHz as shown in tabular column and note the corresponding output voltage. 4. Save the circuit and simulate. 5. Calculate the maximum gain and bandwidth using Bode plotter. Compare the values with the practical circuit values 6. Repeat the same procedure for C=00μF. PRECAUTIONS: Check whether the connections are made properly or not. RESULT: Hence, the frequency Response of RC coupled (2 stage) amplifier for 0μF and 00 μf is plotted.. For C=0 μf, Gain= Bandwidth =fh fl = 2. For C=00μF Gain= Bandwidth =fh fl = 59

60 VIVA:. What is the need for Cascading? 2. What are the types of Coupling Schemes for Cascading? 3. What are the advantages of RC coupling 4. What is the effect of bypass Capacitor on frequency response 5. What is the effect of Coupling Capacitors 60

61 EXPERIMENT NO-4 CURRENT SHUNT AND VOLTAGE SERIES FEEDBACK AMPLIFIER AIM: To study and plot the frequency response of a current shunt and voltage series feedback amplifier. SOFTWARE REQUIRED: MultiSim Analog Devices Edition 3.0 COMPONENTS & EQUIPMENT REQUIRED: S.No Apparatus Range/ Rating Quantity (in No.s). a) DC Supply voltage. b) NPN Transistor. c) Resistors. d) Capacitor. 2 V BC 07 47kΩ 2.2KΩ 0kΩ k 0. F. 22 F Bode plotter 4. Function Generator. 0. Hz-0 MHz 6

62 CIRCUIT DIAGRAM: Current shunt (with out capacitor) 62

63 Current shunt (with capacitor) EXPECTED GRAPH: THEORY: Feedback plays a very important role in electronic circuits and the basic parameters, such as input impedance, output impedance, current and voltage gain and bandwidth, may be altered considerably 63

64 by the use of feedback for a given amplifier. A portion of the output signal is taken from the output of the amplifier and is combined with the normal input signal and thereby the feedback is accomplished. There are two types of feedback. They are i) Positive feedback and ii) Negative feedback. Negative feedback helps to increase the bandwidth, decrease gain, distortion, and noise, modify input and output resistances as desired. A current shunt feedback amplifier circuit is illustrated in the figure. It is called a series-derived, shunt-fed feedback. The shunt connection at the input reduces the input resistance and the series connection at the output increases the output resistance. This is a true current amplifier. TABULAR FORM: Input voltage = 50mv Voltage series feedback Frequency Out Hz put k 2k 5k 8k 0k 20k 40k 60k 00k Current shunt (without capacitor) Current shunt(with capacitor) gain output gain Output Gain 64

65 400k 600k 800k M PROCEDURE:. Connect the circuit as shown in figure 2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without distortion. 3. By keeping input signal voltage, say at 50 mv, vary the input signal frequency from 0- MHz as shown in tabular column and note the corresponding output voltage. 4. Save the circuit and simulate. 5. For current shunt feedback amplifier with shunt capacitor (with and without capacitor) voltage series feedback amplifier (with and without feedback resistance). Repeat the above procedure. 6. Calculate the maximum gain and bandwidth using Bode plotter. Compare the values with the practical circuit values PRECAUTIONS:. No loose contacts at the junctions. 2. Check the connections before giving the power supply 3. Observations should be taken carefully. RESULT: Frequency responses for voltage series (with and without feedback amplifier),current shunt (with and without capacitor are plotted) 65

66 EXPERIMENT NO-5 CASCODE AMPLIFIER AIM:.To plot the frequency response of Cascode amplifier. 3. To calculate bandwidth. SOFTWARE REQUIRED: MultiSim Analog Devices Edition 3.0 COMPONENTS & EQUIPMENT REQUIRED: S.No Device Range/ Rating Quantity (in No.s). Trainer Board containing a) DC Supply voltage. b) NPN Transistor. c) Resistors. d) Capacitors. 2 V KΩ 4.7 KΩ 0nF Bode Plotter 3. Function Generator. 0. Hz-0 MHz 66

67 CIRCUIT DIAGRAM: EXPECTED WAVEFORM: 67

68 THEORY: Cascode amplifier is a cascade connection of a common emitter and common base amplifiers. It is used for amplifying the input signals. The common application of cascade amplifier is for impedance matching. The low impedance of CE age is matched with the medium of the CB sage. TABULAR COLUMN: Input = 50mV Frequency (in Hz) 20 Gain (in db) = 20log0(Vo/Vi) K 68

69 50K 00K 200K 400K 600K 800K PROCEDURE:. Connect the circuit as shown in figure 2. Adjust input signal amplitude in the function generator and observe an amplified voltage at the output without distortion. 3. By keeping input signal voltage, say at 50 mv, vary the input signal frequency from 0- MHz as shown in tabular column and note the corresponding output voltage. 4. Save the circuit and simulate. 5. Calculate the maximum gain and bandwidth using Bode plotter. Compare the values with the practical circuit values PRECAUTIONS: Check whether the connections are made properly or not. RESULT: Hence, the frequency Response of cascode amplifier is plotted. 69

70 Gain= Bandwidth =fh fl = VIVA QUESTIONS:. What is effect of coupling capacitor. 2. Draw the h parameter equivalent circuit for cascode amplifier 3. What short circuit current gain for cascode amplifier. 4. What are the characters ties of cascode 5. What are the application of cascade amplifier 70

71 EXPERIMENT NO-6 WEIN BRIDGE OSCILLATOR Electronic Circuit Analysis & Pulse and Digital circuit Lab AIM: To find practical frequency of a wein bridge oscillator and to compare it with theoretical frequency R=0k R2= 8.2k and C,C2 for 0.0uf,0.022uf & 0.033uf. SOFTWARE REQUIRED: MultiSim Analog Devices Edition 0.0 COMPONENTS AND EQUIPMENTS REQUIRED: S.No Device Range/ Rating wein bridge oscillator trainer board containing a) DC supply voltage 2V b) Capacitor 0.0 F F F F---- K c) Resistor 0K K K d) NPN Transistor BC e) Zener diode 5.v Quantity (in No.s) CRO 7

72 CIRCUIT DIAGRAM: EXPECTED WAVEFORM THEORY: The Wien bridge oscillator employs a balanced wien bridge as the feedback network. Two stage CE amplifier provides 360 o phase shift to the signal. So the wien bridge need not introduce any phase shift to satisfy Barkausen criterion.the attenuation of the bridges calculated to be /3 at resonant frequency. So the amplifier stage should provide a gain of exactly 3 to make loop gain unity. Since the gain of two stage amplifier is the product of individual stages, overall gain becomes very high. But the gain will be trimmed down to 3 by negative feedback network. The emitter resistors of both stages are kept unbypassed. This provides a current series feedback which ensures the stability of operating point and reduction of gain. Frequency of oscillation is given by f = /2πRC 72

73 TABULATIONS: S.No R R2 C C2 Theoretical Practical Vo (V) kω kω µf µf frequency (KHz) frequency (KHz) (ptp) PROCEDURE:. Connect the circuit as shown in figure. 2. Connect C& C2 to 0.0 F capacitor in the circuit and observe the waveform. 3. Note the time period of the waveform and calculate the frequency: f = /T. 4. Now connect the capacitance to F and F and calculate the frequency and tabulate the readings as shown. 5. Find the theoretical frequency from the formula 6. f = R=0k R2= 8.2k and C,C2 for 0.0uf,0.022uf & 0.033uf. Compare theoretical and practical values. PRECAUTIONS:. No loose contacts at the junctions. 2. Check the connections before giving the power supply 3. Observations should be taken carefully. 73

74 RESULT:. For C = 0.0 F, & C2=0.0 Theoretical frequency = Practical frequency = 2. For C = F, &; C2=0.0 Theoretical frequency = Practical frequency = 3. For C = F, & 0.033µF Theoretical frequency = Practical frequency = VIVA QUESTIONS:. Give the formula for frequency of oscillations. 2. What is the total phase shift provided by the oscillator 3. What is the condition for wien bridge oscillator to generate oscillations 4. What is function of lead-lag network in wein bridge oscillator 5. Which type of feedback is used in wein bridge oscillator. 6. What gain of wein bridge oscillator. 7. What are the application of wein bridge oscillator. 8. What is the condition for oscillation. 9. What is the difference between damped oscillations undamped oscillations. wein bridge oscillator is either LC or RC oscillator 74

75 EXPERIMENT NO-7 RC PHASE SHIFT OSCILLATOR Electronic Circuit Analysis & Pulse and Digital circuit Lab AIM: To find practical frequency RC phase shift oscillator and to compare it with theoretical frequency for R=0K and C = 0.0 F, F & F respectively SOFTWARE REQUIRED: MultiSim Analog Devices Edition 0.0 COMPONENTS AND EQUIPMENTS REQUIRED: S.No Device Range/ Rating RC phase shift oscillator trainer board containing a) DC supply voltage 2V b) Capacitor 000 F F F F F---- c) Resistor K K K K d) NPN Transistor BC Quantity (in No.s) CRO 75

76 CIRCUIT DIAGRAM: RC PHASE SHIFT OSCILLATOR EXPECTED WAVEFORM: THEORY: RC phase shift oscillator has a CE amplifier followed by three sections of RC phase shift feedback networks. The output of the last stage is return to the input of the amplifier.the values of R and C are chosen such that the phase shift of each RC section is 600.thus,the RC ladder network produces a total phase shift of 800 between its input and output voltage for the given frequencies since CE amplifier produces 800 phase shift the total phase shift from the base of the transistor around the circuit and back to the transistor will be exactly 3600 or 00.The frequency of oscillation is given by 76

77 F = /2ΠRC 6 PROCEDURE:. Connect the circuit as shown in figure. 2. Connect the F capacitors in the circuit and observe the waveform. 3. Save the circuit and simulate. 5. Calculate the time period and frequency of the resultant wave form. Compare the values with the practical circuit values 6. Repeat the same procedure for C=0.033 F and 0.0 F and calculate the frequency and tabulate as shown. 5. Find theoretical frequency from the formula f = /2 RC 6 and compare theoretical and practical frequencies. PRECAUTIONS:. No loose contacts at the junctions. 2. Check the connections before giving the power supply 3. Observations should be taken carefully. RESULT: For C = F & R=0K Theoretical frequency= Practical frequency= For C = F & R=0K Theoretical frequency= Practical frequency= For C = 0.0 F & R=0K Theoretical frequency= Practical frequency= 77

78 TABULAR COLUMN: S.No C R Theoretical Practical Vo (p-p) ( F) ( ) Frequency Frequency (Volts) (KHz) (KHz) K K K 78

79 EXPERIMENT NO- 8 CLASS-A POWER AMPLIFIER (series fed) Electronic Circuit Analysis & Pulse and Digital circuit Lab AIM: To study the frequency response of a series fed class-a power amplifier and calculate efficiency of the amplifier circuit. SOFTWARE REQUIRED: MultiSim Analog Devices Edition 0.0 COMPONENTS & EQUIPMENT REQUIRED: S.No Apparatus Range/ Rating Quantity (in No.s). a) DC Supply voltage. b) NPN Transistor. c) Resistors. 2 V BC Ω 00KΩ 470Ω d) Capacitor. e) Inductor. 22 F. 50mH 2. D.C Milliammeter 0-00mA 3. Bode plotter 4. Function Generator. 0. Hz-0 MHz 79

80 CIRCUIT DIAGRAM: EXPECTED GRAPH: THEORY: Power amplifiers are mainly used to deliver more power to the load. To deliver more power it requires large input signals, so generally power amplifiers are preceded by a series of voltage amplifiers. In class-a power amplifiers, Q-point is located in the middle of DC-load line. So output current flows for complete cycle of input signal. Under zero signal condition, maximum power dissipation occurs across the transistor. As the input signal amplitude increases power dissipation reduces. The maximum theoretical efficiency is 25%. 80

81 PROCEDURE:.Make the connections as per the circuit diagram. 2. Measure base, emitter and collector D.C voltages of both stages and compare against estimated values. 3. Apply the input at input terminals of the circuit from the function generator. 4. Keep the input signal at constant frequency under mid frequency region and adjust the amplitude such that output voltage undistorted. 5. Calculate the power efficiency and compare it with theoretical efficiency. RESULT: The maximum input signal amplitude which produces undistorted output signal is The practical efficiency of the circuit is VIVA QUESTIONS:. Differentiate between voltage amplifier and power amplifier 2. Why power amplifiers are considered as large signal amplifier? 3. When does maximum power dissipation happen in this circuit?. 4. What is the maximum theoretical efficiency? 5. Sketch wave form of output current with respective input signal. 6. What are the different types of class-a power amplifiers available? 7. What is the theoretical efficiency of the transformer coupled class-a power amplifier? 8

82 8. What is difference in AC, DC load line?. 9. How do you locate the Q-point? 0. What are the applications of class-a power amplifier? TABULAR COLUMN: S.No Frequency (in Hz) Gain(dB) Av = 20 log(vo/ Vi ) K 6 2K 7 4K 8 8K 9 0K 0 20K 40K 2 80K 3 00K 4 200K 82

83 CALCULATIONS: Efficiency is defined as the ratio of AC output power to DC input power DC input power = Vcc x ICQ AC output power = VP-P 2 / 8RL Under zero signal condition: Vcc = IBRB + VBE IBQ =( Vcc - VBE ) / RB ICQ = β x IBQ VCE = Vcc - ICRC 83

84 EXPERIMENT - 9 Electronic Circuit Analysis & Pulse and Digital circuit Lab CLASS B POWER AMPLIFIER (COMPLEMENTARY SYMMETRY AIM: To study the CLASS B Complementary Symmetry amplifier and to calculate its efficiency. SOFTWARE REQUIRED: MultiSim Analog Devices Edition 0.0 COMPONENTS & EQUIPMENT REQUIRED: S.No Apparatus Range/ Rating Quantity (in No.s). a) DC Supply voltage. b) NPN Transistor. c) Resistors. 2 V BC KΩ KΩ Ω 8KΩ 2 2 d) Capacitor. 0. F D.C Milliammeter 0-00mA 3. Bode plotter 4. Function Generator. 0. Hz-0 MHz 84

85 CIRCUIT DIAGRAM: EXPECTED GRAPH: THEORY: Power amplifiers are designed using different circuit configuration with the sole purpose of delivering maximum undistorted output power to load. Push-pull amplifiers operating either in class-b are class- AB are used in high power audio system with high efficiency. In complementary-symmetry class-b power amplifier two types of transistors, NPN and PNP are used. These transistors acts as emitter follower with both emitters connected together. In class-b power amplifier Q-point is located either in cut-off region or in saturation region. So, that only 80o of the input signal is flowing in the output. In complementary-symmetry power amplifier, during the positive half cycle of input signal NPN transistor conducts and during the negative half cycle 85

86 PNP transistor conducts. Since, the two transistors are complement of each other and they are connected symmetrically so, the name complementary symmetry has come Theoretically efficiency of complementary symmetry power amplifier is 78.5%. PROCEDURE:. Switch ON the CLASS B amplifier trainer. 2. Connect Milliammeter to (A) terminals and DRB to the RL terminals and fix RL=50Ω. 3. Apply the input voltage from the signal generator to the Vs terminals. 4. Connect channel of CRO to the Vs terminals and channel 2 across the load. 5. By varying the input voltage, observe the maximum distortion less output waveform. And note down the voltage reading. 6. Calculate the efficiency. OBSERVATIONS: Vs=2v FREQUENCY Vo Idc (ma) Efficiency (volts) 0 KHz CALCULATIONS: Pin=Vcc x Idc Idc= Vo = RL Pout = 2 o V 8R L Efficiency= Po/Pi x00 86

87 RESULT: Thus efficiency of CLASS B (Complementary symmetry) amplifier calculated. VIVA.Classfide large signal amplifier based of operating point. 2.state the advantages of push pull class b power amplifier over class b power amplifier. 3. what is harmonic distortion how even harmonic is eliminated using push pull 4. list advantages of complementary symmetry configuration over push pull amplifier. 5. What is covertion efficiency of class B power amplifier. 87

88 88 Electronic Circuit Analysis & Pulse and Digital circuit Lab

89 PULSE CIRCUITS 89

90 PULSE CIRCUITS LAB INTRODUCTION: Practical knowledge in electronic circuits and pulse circuits plays vital role in designing and developing circuits for various applications. The PC lab was designed to give practical overview on different amplifier circuits and pulse circuits. This Lab provides students with the opportunity to gain an experience in connecting the circuits and studying the responses of these circuits. A software tool is also provided for simulating the circuit. This lab deals with the methods for the generation of pulse waveforms making use of semiconductor devices like Diodes, Transistors and UJT s. Pulse waveforms play a significant role in household appliances like television radio and digital clocks. Their industrial applications include digital instrumentation, Control Systems, digital Computers, data processing systems and CRO s. The term Pulse waveforms is popularly employed in electronics to refer to non-sinusoidal waveforms. These waveforms play a vital role in electronic communication, and they are of immense help in pulse Communication, television Engineering, radar, and telemetry. Pulse waveforms change between the LOW and HIGH levels. A positive going pulse is one that goes from a normally LOW logic level to a HIGH level and then back again. Digital waveforms are made up of a series of pulses. Clock Signal Waveform Active LOW - if the state change occurs from a HIGH to a LOW at the clock s pulses falling edge. Duty Cycle - this is the ratio of the clock width to the clock period. Clock Width - this is the time during which the value of the clock signal is equal to a logic, or HIGH. Clock Period - this is the time between successive transitions in the same direction, ie, between two rising or two falling edges. Clock Frequency - the clock frequency is the reciprocal of the clock period, frequency = /clock period 90

91 Most natural quantities that we see are analog and vary continuously. Analog systems can generally handle higher power than digital systems Timing circuits networks composed of resistors, capacitors and inductors are called linear network and they do not change the waveform of a sine wave when it is transmitted through them. On the other hand when non-sinusoidal waveforms, (e.g. step, ramp, exponential) are applied to the input of such networks the output signal may have very little resembles with the input waveform. The action of a linear network in producing a waveform at its output different from its input is called linear wave shaping. The wave shaping is used to perform any one of the following functions.. To hold the waveform to a particular d.c. level. 2. To generate one wave form the other 3. To limit the voltage level of the waveform of some presenting value and suppressing all other voltage levels in excess of the present level. 4. To cut-off the positive and negative portions of the input waveform. Shaping circuits may be either series RC or series RL circuits. The series RC and RL circuits electrically perform the mathematical operation of integration and differentiation. Therefore, the circuits used to perform these operations are called integrators and differentiator. The differentiator circuits are used to generate sharp narrow pulses either from distorted pulse waveform or from rectangular wave forms. The integrator circuits are required to generate a voltage, which are required to generate a voltage, which increases or decreases linearly with time. The semiconductor Signal Diode is a small non-linear semiconductor devices generally used in electronic circuits, where small currents or high frequencies are involved such as in radio, television and digital logic circuits. The signal diode which is also sometimes known by its older name of the Point Contact Diode or the Glass Passivated Diode, are physically very small in size compared to their larger Power Diode cousins. 9

92 EXPERIMENT NO:- LINEAR WAVE SHAPING Electronic Circuit Analysis & Pulse and Digital circuit Lab i) RC low pass circuit AIM:. To design low pass RC circuits for different time constants and verify their responses for a square wave input of given frequency. 2. To study the operation of low pass circuit as an integrator. APPARTUS REQUIRED:. Resistor (00kΩ) - No 2. Capacitors (0.uF, 0.0uF & 0.00uF) - No 3. Dual trace CRO - No 4. Bread Board - No 5. Signal Generator - No 6. CRO Probes - 2 No 7. Connecting wires CIRCUIT DIAGRAM: THEORY: Low Pass RCcircuit :The reactance of the capacitor depends upon the frequency of operation. At very high frequencies, the reactance of the capacitor is zero. Hence the capacitor in fig..2 acts as short circuit. As a result, the output will fall to zero. At low frequencies, the reactance of the capacitor is infinite. So the capacitor acts as open circuit. As a result the entire input appears at the output. Since the circuit allows only low frequencies, therefore it is called as low pass RC circuit. Low pass RC circuit as an integrator: In low pass circuit, if the time constant is very large in comparison with the time required for the input signal to make an appreciable change, the circuit is called an integrator. Under these circumstances the voltage drop across C will be very small in comparison to the drop across R and we may consider that the total input Vi appears across R. i = Vi/R 92

93 DESIGN: Choose T = msec,for RC T ; the Low Pass Circuit as an Integrator RC low pass circuit: (Design procedure for RC low pass circuit) i) Long time constant: RC > > T ; Where RC is time constant and T is time period of input signal. Let RC = 0 T, Choose R = 00kΩ, f = khz. C = 0 / 0 3 Χ 00Χ0 3 = 0.µf ii) Medium time constant: RC = T C = T/R = / 0 3 Χ00Χ0 3 = 0.0µf iii) Short time constant: RC < < T RC = T/0 C = T/0R = / 0Χ0 3 Χ00Χ0 3 = 0.00 µf. a) RC=T b) RC >>T c) RC<< T 93

94 PROCEDURE:. Connect the circuit, as shown in figure. 2. Apply the Square wave input to the circuit (Vi = 0 VP-P, f = KHz) 3. Calculate the time constant of the circuit by connecting one of the Capacitor provided. 3. Observe the output wave forms for different input frequencies (RC<<T, RC=T, RC> T) as shown in the tabular column for different time constants. 4. Plot the graphs for different input and output waveforms. PRECAUTIONS:. Avoid loose and wrong connections. 2. Avoid eye contact errors while taking the observations in CRO. OBSERVATIONS: Low pass RC circuit: Time constant in S.No. m. sec RC=T O/P Voltage levels in Volts V V2 V 2 RC>>T V2 3 RC<<T V 94

95 Low pass RC circuit: Time constant in m.sec Output Signal Voltage levels(theoretical) in Volts Output Signal Voltage levels(practical) in Volts Short time constant (RC<<T) Medium time constant (RC=T) Long time constant (RC>>T) REVIEW QUESTIONS:. Name the signals which are commonly used in pulse circuits and define any two of them? 2. Define linear wave shaping? 3. Define attenuator and types of attenuator? 4. Explain the fractional tilt of a high pass RC circuit. Write the Expression? 5. State the lower 3-db frequency of high-pass circuit? 6. Distinguish between the linear and non-linear wave shaping circuits. 7. Define Percentage Tilt and Rise time? 8. Show that a high pass circuit with a small time constant acts as differentiator? 9. Define Rise time? Give the relations between rise time and bandwidth? 0. Show that a low pass circuit with a time constant acts as Integrator?. Name a wave shaping circuit which produces a Ramp wave as an output by taking a step signal as input and draw its output for a sinusoidal wave 2. Write the expressions for the output of a low pass circuit by a step and symmetrical square waves? 3. Prove that for any periodic input wave form the average level of the steady state output signal from an RC high pass circuit is always zero. 4. Explain the response of a high pass RC circuit to a step input signal? 5. Name a wave shaping circuit which produces a Ramp wave as an output by taking a step signal as input and draw its output for a sinusoidal wave? 6. Write the expressions for the output of a low pass circuit by a step and symmetrical square waves? 95

96 GRAPH: RESULT: The characteristics of RC low pass circuit for different time constants are verified. 96

97 EXPERIMENT NO:- LINEAR WAVE SHAPING ii) RC high pass circuit AIM:. To design high pass RC circuits for different time constants and verify their responses for a square wave input of given frequency. 2. To find the % tilt of high pass RC circuit for long time constant. 3. To study the operation of high pass RC circuit as a differentiator APPARTUS REQUIRED:. Resistor (00kΩ) - No 2. Capacitors (0.uF, 0.0uF & 0.00uF) - No 3. Dual trace CRO - No 4. Bread Board - No 5. Signal Generator - No 6. CRO Probes - 2 No 7. Connecting wires CIRCUIT DIAGRAM: RC Differentiator THEORY: High Pass RC circuit :The reactance of the capacitor depends upon the frequency of operation. At very high frequencies, the reactance of the capacitor is zero. Hence the capacitor in fig.. acts as short circuit. As a result the entire input appears at the output. At low frequencies, the reactance of the capacitor is infinite. So the capacitor acts as open circuit. Hence no input reaches the output. Since the circuit allows only high frequencies, therefore it is called as high pass RC circuit. 97

98 High pass RC circuit as a differentiator: In high pass RC circuit, if the time constant is very small in comparison with the time required for the input signal to make an appreciable change, the circuit is called a Differentiator. Under these circumstances the voltage drop across R will be very small in comparison with the drop across C. Hence we may consider that the total input Vi appears across C. So that the current is determined entirely by the capacitor. i = C dvi/dt. The output signal voltage across R is Vo = RC dvi/dt. i.e. The output is proportional to the differentiation of the input. Hence the high pass RC circuit acts as a differentiator for RC << T DESIGN: RC high pass circuit: i) Long time constant: RC > > T ; Where RC is time constant and T is time period of input signal. Let RC = 0 T, Choose R = 00kΩ, f = khz. C = 0 / 0 3 Χ 00Χ0 3 = 0.µf iv) Medium time constant: RC = T C = T/R = / 0 3 Χ00Χ0 3 = 0.0µf v) Short time constant: RC < < T RC = T/0 C = T/0R = / 0Χ0 3 Χ00Χ0 3 = 0.00 µf. EXPECTED WAVEFORMS: a) RC=T b) RC >>T 98

99 c) RC <<T PROCEDURE:. Connect the circuit, as shown in figure. 2. Apply the Square wave input to the circuit (Vi = 0 VP-P, f = KHz) 3. Calculate the time constant of the circuit by connecting one of the Capacitor provided. 3. Observe the output wave forms for different input frequencies (RC<<T, RC=T, RC> T) as shown in the tabular column for different time constants. 4. Plot the graphs for different input and output waveforms. OBSERVATIONS: S.No. Time constant in m.sec. RC=T 2 RC>>T 3 RC<<T O/P Voltage levels in Volts V V V2 V 2 V V V2 V 2 V V V2 V 2 99

100 High Pass RC Circuit: Time constant in m.sec Output Signal Voltage levels(theoretical) in Volts Output Signal Voltage levels(practical) in Volts Short time constant (RC<<T) Medium time constant (RC=T) Long time constant (RC>>T) Time constant in m.sec %Tilt (Theoretical) %Tilt (Practical) Short time constant (RC<<T) Medium time constant (RC=T) REVIEW QUESTIONS:. When HP-RC circuit is used as Differentiator? 2. Draw the responses of HPF to step, pulse, ramp inputs? 3. Draw the responses of LPF to step, pulse, ramp inputs? 4. Define % tilt and rise time? 5. When LP-RC circuit is used as integrator? 6. Why noise immunity is more in integrator than differentiator? 7. Why HPF blocks the DC signal? 8. What is meant by linear wave shaping? 9. Define time constant? 00

101 0. Differences between High pass and Low pass RC circuits?. What is the working principle of high pass and low pass RC circuits for non sinusoidal signal inputs. GRAPH: RESULT: The characteristics of RC High pass circuit for different time constants are verified 0

102 AIM: EXPERIMENT NO:-2(a) NON LINEAR WAVE SHAPING CLIPPERS To study the various clipper circuits and to plot the output waveforms for a sinusoidal input signal. APPARATUS:. CRO (Dual Channel) - No. 2. Signal Generator No. 3. Breadboard - No. 4. Diode (N4007) - 2 No. 5. Resistor ( KΩ) - No. 6. D.C Power Supply (dual) - 2 No. 7. Connecting wires CIRCUIT DIAGRAMS& EXPECTED WAVEFORMS: 02

103 Negative Clipper 03

104 Negative Clipping wih positive reference voltage TRANSFER CHARACTERISTIC CURVES: Positive Clipper: Positive Clipper with positive reference voltage 04

105 Positive Clipper with Negative reference voltage Negative Clipper Negative Clipper with positive reference voltage Slicer: PROCEDURE: -. Connect the circuit as shown in figure 2. Apply the input Sine wave to the circuit. (8Vp-p, 2 KHz) 3. Switch on the power supply and adjust the output of AF generator to 8V (peak to peak) 05

106 4. Observe the input and output waveforms on CRO and note down the readings. 5. Plot the graphs of input Vs output waveforms for different clipping circuits. OBSERVATIONS: Sl No. Type of Clipper Reference Voltage 0V Theoretical Clipping Voltage levels Practical Clipping Voltage levels Series Positive Clipper 2V -2V 0V 2 Series Negative Clipper 2V -2V 0V 3 Shunt Positive Clipper 2V -2V 0V 4 Shunt Negative Clipper 2V -2V 5 Two level clipper PRECAUTIONS: -. Avoid loose and wrong connections. 2. Avoid parallax errors while taking the readings using CRO. REVIEW QUESTIONS:. Define nonlinear wave shaping? 2. Define clipping circuit? 3. What is piecewise linear mode of a diode? 4. What are the different types of clippers? 5. Which kind of a clipper is called a slicer circuit? 6. What are the disadvantages of the shunt clipper? 7. Justify that a clamping circuit is a dc inserter? 8. Define Series clipper and shunt clipper? 9. Express the meaning of transmission region and attenuation region of a Clipping Circuit? 06

107 GRAPH: RESULT: The characteristics of the different clippers using diodes are verified. 07

108 AIM: EXPERIMENT NO:-2(b) NON LINEAR WAVE SHAPING CLAMPERS Electronic Circuit Analysis & Pulse and Digital circuit Lab To study the various clamping circuits and to plot the output waveforms for a sinusoidal input of given peak amplitude. (Choose f= khz, Vp-p =8V) APPARATUS:. CRO (Dual Channel) - No. 2. Signal Generator - No. 3. Breadboard - No. 4. Diode (N4007) - No. 5. Resistor (00 KΩ) - No. 6. Capacitor (µf) - No. 7. D.C Power Supply (dual) - No. 8. Connecting wires THEORY: The process whereby the form of a sinusoidal signals are going to be altered by transmitting through a non-linear network is called non-linear wave shaping. Non-linear elements in combination with resistors and capacitors can function as clamping circuit. A Clamping circuit is one that takes an input waveform and provides an output i.e a faithful replica of its shape, but has one edge clamped to the voltage reference point. The clamping circuit introduces the d.c component at the output side, for this reason the clamping circuits are referred to as d.c restorer or d.c reinserted. Clamping circuits are classified as two types. i) Negative Clampers ii) Positive Clampers 08

109 CIRCUIT DIAGRAM: 09

110 0 Electronic Circuit Analysis & Pulse and Digital circuit Lab

111 PROCEDURE:-. Connect the circuit as shown in figures (-5) 2. Switch on the power supply and adjust the output of AF generator to 8V (peak to peak) 3. Square wave input and Observe the output waveforms on CRO and note down the Readings. 4. Plot the graphs of input Vs output waveforms for different clamping circuits. OBSERVATIONS: Sl No. Type of Clamper Reference Voltage 0V Theoretical Clamping reference Voltage level Practical Clamping reference Voltage level Positive Clamper 2V -2V 0V 2 Negative Clamper 2V -2V PRECAUTIONS: -. Avoid loose and wrong connections. 2. Avoid parallax errors while taking the readings using CRO. REVIEW QUESTIONS:. What are the applications of clamping circuits? 2. What is the synchronized clamping? 3. Why clamper is called as a dc inserter? 4. What is clamping circuit theorem. How the modified clamping circuit theorem does differs from this? 5. Differentiate ve clamping circuit from +ve clamping circuits in the above circuits? 6. Describe the charging and discharging of a capacitor in each circuit?

112 7. What is the function of capacitor? GRAPH: RESULT: The characteristics of the different clampers using diodes are verified 2

113 EXPERIMENT NO: 3 TRANSISTOR AS A SWITCH Electronic Circuit Analysis & Pulse and Digital circuit Lab AIM: To study and observe the switching characteristics of a transistor. APPARTUS REQUIRED:-. Resistor 2.2 KΩ, 68KΩ - No 2. Transistor - BC 07 - No 3. Dual trace CRO. - No 4. Function generator. - No 5. Probes - 2 No 6. Connecting wires. CIRCUIT DIAGRAM: 3

114 EXPECTED WAVEFORM: THEORY: The transistor Q can be used as a switch to connect and disconnect the load RL from the source VCC. When a transistor is saturated, it is like a closed switch from the collector to the emitter. When a transistor is cut-off, it is like an open switch VCE = VCC Saturation: The point at which the load line intersects the IB = 0 curve is known as cut-off. At this point, base current is zero and collector current is negligible small i.e., only leakage current ICEO exists. At cut-off, the emitter diode comes out of forward bias and normal transistor action is lost. The transistor appears like a closed switch. VCE(sat) VCC The intersection of the load line and the IB = IB(sat) is called saturation. At this point base current is IB(sat) and the collector current is maximum. At saturation, the collector diode comes out of reverse bias, and normal transistor action is again lost. PROCEDURE:-. Connect the circuit as shown in figure. 2. Switch on the power supply and observe the output of the function generator on CRO. Adjust input signal amplitude such that output signal peak-to peak value is less than the Saturation level. 3. Observe output waveforms on CRO and note down the readings. 4. Plot the graphs between input and output waveforms at a given input frequency. 4

115 Calculate the parameters: - a) Rise Time (tr) b) Fall Time (tf) c) Delay Time (td) d) Storage Time (ts) e) Turn ON Time (ton) f) Turn OFF Time (toff Note:- Rise Time: - It is the time taken to rise 0% of the Max value of the signal to 90% of the Max value of the signal. Fall Time: - It is the taken to fall 90% of the Max value of the signal to 0% of the Max value of the signal. Delay Time: - It is the time taken to rise from 0% to0% of the Max value of the signal. Storage Time: - It is the time taken to fall from 00% to 90% of the Max value of the Signal. Turn ON Time: - It is the sum of Delay time and Rise time. Turn OFF Time: - It is the sum of Storage time and fall time. PRECAUTIONS: -. Avoid loose and wrong connections. 2. Aviod parallax error while taking the readings using CRO. REVIEW QUESTIONS:. Name the devices that can be used as switches? 2. Draw the Practical and piece-wise linear diode V-I characteristics? 3. Describe the two regions of a diode? 4. Write the Expressions for the Static and dynamic resistance of a diode? 5. Define Forward recovery time and reverse recovery time? 6. Define Storage time and Transition time of a diode? 7. Explain how a Diode can be used as a switch? 8. Write short notes on Transistor switching times 5

116 GRAPH: RESULT: -The switching characteristics of a transistor are verified. 6

117 EXPERIMENT NO: 4 BISTABLE MULTIVIBRATOR AIM: To study the characteristics of bistable multivibrator using transistors. APPARATUS REQUIRED:. CRO (Dual Channel) - No. 2. Function Generator - No. 3. CDS - No. 5. Resistor ( KΩ, 39 KΩ,3.9,0) - 2 No. 6 Capacitors (00 pf) - 2 No 7. Transistor (BC 07) - 2 No. 8. Diodes 9. Regulated D.C Power Supply (dual) - No. 0. Connecting wires CIRCUIT DIAGRAM: Electronic Circuit Analysis & Pulse and Digital circuit Lab 7

118 EXPECTED WAVEFORMS: THEORY:A Bistable circuit is one which can exist indefinitely in either of two stable states and which can be induced to make an abrupt transition from one state to the other by means of external excitation. The Bistable circuit is also called as Bistable multivibrator, Eccles Jordon circuit, Trigger circuit, Scaleof-2 toggle circuit, Flip-Flop & Binary. A Bistable Multivibrator is used in a many digital operations such as counting and the storing of binary information. It is also used in the generation and processing of pulse-type waveform. They can be used to control digital circuits and as frequency dividers. There are two outputs available which are complements of one another. i.e. when one output is high the other is low and vice versa. Operation: When VCC is applied, one transistor will start conducting slightly more than that of the other, because of some differences in the characteristics of a transistor. Let Q2 be ON and Q be OFF. When Q2 is ON, The potential at the collector of Q2 decreases, which in turn will decrease the potential at the base of Q due to potential divider action of R and R2. The potential at the collector of Q increases which inturn further increases the base to emitter voltage at the base of Q2. The voltage at the collector of Q2 further decreases, which inturn further reduces the voltage at the base of Q. This action will continue till Q2 becomes fully saturated and Q becomes fully cutoff. Thus the stable state of binary is such that one device remains in cut-off and other device remains at saturation. It will be in that state until the triggering pulse is applied to it. It has two stable states. For every transition of states triggering is required. At a time only one device will be conducting. NEED OF COMMUTATING CAPACITORS (SPEED UP CAPACITORS): It is desired that the transition should take place as soon as the trigger pulse is applied but such is not the case. When transistor is in active region it stores charge in its base and when it is in the saturation region it stores 8

119 even none charge. Hence transistor cannot come out of saturation to cut-off. Until all such charges are removed. The interval during which conduction transfer one transistor to other is called as the transition. PROCEDURE:-. Connect the circuit as shown in Figure. 2. Observe the output of the square wave oscillator-using Oscilloscope. 3. Connect the output of square oscillator to the trigger input Of Bistable Circuit and observe output waveforms using Oscilloscope. 4. By varying input signal (Trigger) frequency, observe both input and corresponding output Waveforms Using Oscilloscope. 5. Plot the graph for input and output waveforms at different input (Trigger) frequencies. OBSERVATIONS: S.no Time period Voltage 2 PRECAUTIONS: -. Avoid loose and wrong connections. 2. Aviod parallax errors while taking the readings using CRO. REVIEW QUESTIONS :. What are the applications of a Bitable Multivibrator? 2. Describe the operation of commutating capacitors? 3. Why is a Binary also called a flip-flop? 4. Mention the name of different kinds of triggering used in the circuit shown? 5. What are the disadvantages of direct coupled Binary? 6. How many types of unsymmetrical triggering are there? 7. What are catching diodes? 9

120 8. Which triggering is used in binary counting circuits? GRAPH: RESULT: The characteristics of Bistable Multivibrator using Transistors are verified. 20

121 EXPERIMENT NO:-5 ASTABLE MULTIVIBRATOR AIM: To study the characteristics of Astable Multivibrator using transistors. Electronic Circuit Analysis & Pulse and Digital circuit Lab APPARATUS REQUIRED :. CRO (Dual Channel) - No. 2. Function Generator No. 3. Breadboard - No. 4. Resistor (2.2 KΩ, 47 KΩ) - 2 No. each 5. Capacitors (0.µF) - 2 No. s Transistor (BC 07) - 2 No. s 7. Regulated D.C Power Supply (dual) - No. 8. Connecting wires. CIRCUIT DIAGRAM: 2

122 EXPECTED GRAPH: THEORY: The Astable circuit has two quasi-stable states. Without external triggering signal the astable configuration will make successive transitions from one quasi-stable state to the other. The astable circuit is an oscillator. It is also called as free running multivibrator and is used to generate Square Wave. Since it does not require triggering signal, fast switching is possible. Operation: When the power is applied, due to some importance in the circuit, the transistor Q2 conducts more than Q i.e. current flowing through transistor Q2 is more than the current flowing in transistor Q. The voltage VC2 drops. This drop is coupled by the capacitor C to the base by Q there by reducing its forward base-emitter voltage and causing Q to conduct less. As the current through Q decreases, VC rises. This rise is coupled by the capacitor C2 to the base of Q2. There by increasing its base- emitter forward bias. This Q2 conducts more and more and Q conducts less and less, each action reinforcing the other. Ultimately Q2 gets saturated and becomes fully ON and Q becomes OFF. During this time C has been charging towards VCC exponentially with a time constant T = RC. The polarity of C should be such that it should supply voltage to the base of Q. When C gains sufficient voltage, it drives Q ON. Then VC decreases and makes Q2 OFF. VC2 increases and makes Q fully saturated. During this time C2 has been charging through VCC, R2, C2 and Q with a time constant T2 = R2C2. The 22

123 polarity of C2 should be such that it should supply voltage to the base of Q2. When C2 gains sufficient voltage, it drives Q2 On, and the process repeats. Design Procedure: The period T is given by T = T + T2 = 0.69 (RC + R2C2) For symmetrical circuit, with R = R2 = R & C = C2 = C T =.38 RC Let VCC = ; hfe = (for BC07), VBESat = ; VCESat = Let C = µf & T=. T =.38 RC R = KΩ Choose ICmax = 0mA, VCC VCESAT RC = = IC max TABLE: S.NO OUTPUT VOLTAGES TRANSISTOR IN ON TRANSISTOR IS OFF VC VC2 VB VB2 S.No Gate Width (Theoritical) Gate Width (Practical) 23

124 PROCEDURE:-. Connect The Circuit As Shown In Figure. 2. Observe The Output Of The Circuit Using Oscilloscope and measure the time Period Of The Signal And Compare It With Theoretical Value By Varying Dc source V (5V to 0V) in steps (take minimum two readings). 3. Plot the output waveforms on the graph paper for one set of values. 4. Repeat the steps from to 3 with timing capacitor 0.0μF. 5. Connect the circuit as shown in figure Repeat the steps from to 4. PRECAUTIONS: - Avoid loose and wrong connections. 2. Aviod parallax errors while taking the readings using CRO. REVIEW QUESTIONS:. What are the other names of Astable Multivibrator? 2. Define quasi stable state? 3. Is it possible to change time period of the waveform with out changing R & C? 4. Collector waveforms are observed with rounded edges. Explain? 5. Explain charging and discharging of capacitors in an Astable Multivibrator? 6. How can an Astable Multivibrator be used as VCO? 7. Why do you get overshoots in the Base waveforms? 8. What are the applications of Astable Multivibrator? 9. How can Astable Multivibrator be used as a voltage to frequency converter? 0. What is the formula for frequency of oscillations? 24

125 GRAPH: RESULT: -The characteristics of Astable multivibrator using transistors are verified. 25

126 EXPERIMENT NO:-6 MONOSTABLE MULTIVIBRATOR AIM : To study the characteristics of Monostable Multivibrator using transistors. Electronic Circuit Analysis & Pulse and Digital circuit Lab APPARATUS REQUIRED:. CRO (Dual Channel) - No. 2. Function Generator No. 3. CDS - No. 4. Resistor ( KΩ, 0 KΩ,00 KΩ,22 KΩ) - No. each 5. Transistor (BC 07) - 2 No. 6. Regulated D.C Power Supply (dual) - No. 7. Connecting wires CIRCUIT DIAGRAM:- 26

127 CASE : CASE 2: 27

128 THEORY:- The Monostable circuit has one permanently stable and one quasi-stable state. In the monostable configuration, a triggering signal is required to induce a transition from the stable state to the quasi-stable state. The circuit remains in its quasi-stable for a time equal to RC time constant of the circuit. It returns from the quasi-stable state to its stable state without any external triggering pulse. It is also called as one-shot a single cycle, a single step circuit or a univibrator. Operation: Assume initially transistor Q2 is in saturation as it gets base bias from VCC through R. coupling from Q2 collector to Q base ensures that Q is in cutoff. If an appropriate negative trigger pulse applied at collector of Q (VC) induces a transition in Q2, then Q2 goes to cutoff. The output at Q2 goes high. This high output when coupled to Q base, turns it ON. The Q collector voltage falls by IC RC and Q2 base voltage falls by the same amount, as voltage across a capacitor c cannot change instantaneously. The moment, a ve trigger is applied VC, Q2 goes to cutoff and Q starts conducting. There is a path for capacitor C to charge from VCC through R and the conducting transistor Q. The polarity should be such that Q2 base potential rises. The moment, it exceeds Q2 base cut-in voltage, it turns ON Q2 which due to coupling through R from collector of Q2 to base of Q, turns Q OFF. Now we are back to the original state i.e. Q2 On and Q OFF. Whenever trigger the circuit into the other state, it cannot stay there permanently and it returns back after a time period decided by R and C. Pulse width is given as T = 0.69RCsec. DESIGN:- 28

129 OBSERVATIONS: S.NO OUTPUT VOLTAGES TRANSISTOR IN ON TRANSISTOR IS OFF VC VC2 VB VB2 S.No Gate Width (Theoretical) Gate Width (Practical) PROCEDURE:-. Connect the circuit as shown in figure. 2. Observe the output of the Square wave generator-using oscilloscope. 3. Connect the output of square oscillator to the trigger input of monostable circuit and also Connect diode to the collector of the Q.observe trigger spikes at Q collector using Oscilloscope. 4. Connect one of the timing capacitor C to the circuit (say C=0.0μF) and observe the monostable at collector of Q2 using oscilloscope. 5. Measure and note the pulse width of output signal and compare with the theoretical value (T=.RC). 6. By varying trigger input frequency, observe the corresponding output waveforms. 7. Plot the graph for input and output waveforms at different input frequencies. 8. Repeat the steps from 4 to 6 for timing capacitor C=0.μF. PRECAUTIONS: -. Avoid loose and wrong connections. 2. Avoid parallax while taking the readings using CRO. REVIEW QUESTIONS:. What are applications of Monostable Multivibrator? 2. Why a Monostable Multivibrator is called a gating circuit? 29

130 3. Explain the waveform of VB? 4. Describe the operation of the capacitor C3 in the circuit? 5. Why is the time period T also called Delay time? 6. Justify, Why Monostable Multivibrator is called one-shot circuit? 7. Why is the ve voltage given at the base of Q transistor? 8. What is the no of quasi & stable states of Monostable Multivibrator? 9. What is meant by quasi stable state? 0. What is the function of commutating capacitors? GRAPH: 30

131 RESULT: -The characteristics of Monostable multivibrator using transistors are verified 3

132 EXPERIMENT NO: 7 SCHMITT TRIGGER AIM: To observe and note down the output waveforms of Schmitt trigger using transistors.. APPARATUS REQUIRED:. CRO (Dual Channel) - No. 2. Function Generator - No. 3. CDS - No. 4. Resistor (0 KΩ,50Ω) - No each 5. Resistor (820Ω) 6. Resistor ( KΩ) No. s 3 No. s 7. Capacitors (0.022 µf,00 µf) - No. 8. Transistor (BC07) - 2 No. 9. Regulated D.C. Power Supply (dual) - No. 0. Connecting wires CIRCUIT DIAGRAM: 32

133 EXPECTED WAVEFORMS: THEORY: In digital circuits fast waveforms are required i.e, the circuit remain in the active region for a very short time (of the order of nano seconds) to eliminate the effects of noise or undesired parasitic oscillations causing malfunctions of the circuit. Also if the rise time of the input waveform is long, it requires a large coupling capacitor. Therefore circuits which can convert a slow changing waveform(long rise time) in to a fast changing waveform (small rise time) are required. The circuit which performs this operation is known as Schmitt Trigger. In a Schmitt trigger circuit the output is in one of the two levels namely low or high. When the output voltage is raising the levels of the output changes. When the output passes through a specified voltage V known as Upper trigger level, similarly when a falling output voltage passes through a voltage V2 known as lower triggering level. The level of the output changes V is always greater than V2.The differences of these two voltages is known as Hysteresis. 33

134 TABLE: S.NO OUTPUT VOLTAGES TRANSISTOR IN ON TRANSISTOR IS OFF VC VC2 S.NO LTP UTP VH PROCEDURE: Observation of UTP and LTP:. Connect the circuit as per the circuit diagram. 2. Apply the square wave input of KHz to the circuit. 3. Switch on the power supply and note down the amplitude and time period for the input square wave. 4. Observe the output waveform and note down the amplitude and time period. 5. Keep Re and Re2 in minimum condition (extremely in anticlockwise direction) 6. Initially keep DC source voltage at zero and observe the output of the Schmitt trigger (it will be in low state i.e. around 6V). 7. Vary the DC source output (i.e input voltage of the Schmitt trigger) slowly from zero. 8. Note down the input voltage value at which the output of the Schmitt trigger goes to high (UTP). Still increase (upto 0V) the input voltage and observe that the output is constant. 9. Now slowly decrease the input voltage and note down the value at which the output of the Schmitt trigger comes back to the original state (LTP). 0. Compare the values LTP and UTP with theoretical values. Schmitt Trigger as a Squaring circuit:. Connect a triangle wave signal from an external function generator to the input of the level changer. 34

135 2. Connect the output of the level changer to the input of the Schmitt trigger. 3. Connect CH input of CRO to the input signal and CH2 to the output of the schmitt trigger. 4. Adjust the amplitude of the input signal to such a level that we observe square wave at the output. 5. Note down the points of input where the output is high (UTP) and low (LTP) and note that both the levels are not one and the same. 6. Find Re value and compare it with the theoretical value. 7. Repeat the steps 3 to 6 with different types of signals (sine, ramp etc). 8. From the above observations we can notice that Schmitt trigger converts any arbitrary waveform into a square/rectangle wave. CALCULATIONS: Calculation of UTP: UTP V EN V r (Vr is cut in voltage i.e. 0.6 V) ( V ' VBE2) Re ( hfe ) V EN (hfe of 2N 2369 is 50) R R ( h ) b e fe V and Rb is the Thevenins equivalent voltage and resistance between base of Q2 and ground when Q is in cut-off. V ' R c VccR2 R R 2 = R b R2( Rc R R R R c 2 Calculation of LTP: LTP V R ( V ' V 2) e BE r ar Re 35

136 a R R R 2 (a=voltage ratio from collector of Q to base of Q2 ) 2 R R2( R R R R R c (Where R is the Thevenins Equivalent Resistance when Q2 is c 2 in cut-off) PRECAUTIONS:. Avoid loose and wrong connections. 2. Avoid parallax errors while taking the readings using CRO. REVIEW QUESTIONS:. What are the applications of Schmitt Trigger? 2. Define hysteresis action? 3. Why is Schmitt Trigger called a squaring circuit? 4. What is UTP & LTP? 6. What is the difference between a Binary and Schmitt Trigger? 36

137 GRAPH: RESULT: The output waveforms of Schmitt trigger are observed. 37

138 EXPERIMENT NO: 9 UJT RELAXATION OSCILLATOR AIM: To obtain a sawtooth waveform using UJT and test its performance as an oscillator. APPARATUS REQUIRED:. CRO (Dual Channel) - No. 2. Function Generator - No. 3. Bread Board - No. 4. Resistor (47 KΩ,00Ω) - No each 7. Capacitors (0.µF) - No. 8. Transistor (2N2646) - No. 9. Regulated D.C. Power Supply (dual) - No. 0. Connecting wires CIRCUIT DIAGRAM : R 47 KΩ E B 2 V CC =2V 2N2646 B C 0. µf R 2 00 Ω Figure:5. Emitter 2. Base 3. Base2 38

139 THEORY: A Unijunction transistor (UJT), as the very implies, has only one p-n junction, unlike a BJT which has two p-n junctions. The equivalent circuit of the UJT + is as shown in figure. I E R B2 V BB V E R B V RB is the resistance between base B and the emitter, and it is basically a variable resistance, its value being dependent upon the emitter current IE. RB2 is the resistance between base B2 and the emitter, and the value is fixed. Consider the circuit as shown in figure. Let IE = 0. Due to the applied voltage VBB a current I. Form the equivalent circuit, it is evident that the diode cannot conduct unless the emitter voltage VE = V? + V, where V? is the cutin voltage of the diode. This value of the emitter voltage which makes the diode conduct is termed as peak voltage, and it is denoted as VP. Figure: We have VE = V? + V, It is obvious that if VE < VP, the UJT is OFF, and if VE < VP, the UJT is ON. - Figure 3. shows the emitter characteristics of a UJT (plot of VE vs IE) 39

140 V E V P Peak Point Negative resistance region V V Valley Point 0 I P I V I E The main application of UJT is in switching circuits wherein rapid discharging of capacitor is very essential. Having understood the basic of UJT, we shall next study the working of UJT relaxation oscillator. Working of UJT relaxation oscillator (OR UJT sweep circuit) V or VBB Figure 3. R B2 R + B 2 E B R B C V s Figure:4 The UJT sweep circuit shown in the figure 4 consists of a UJT, a capacitor and a resistor arranged as shown. We studied that a UJT is OFF as long as VE < VP, the peak voltage. Hence initially when the UJT is OFF, the capacitor C charges through the resistance R from the supply voltage V. Let VS = capacitor voltage. It is seen that when the capacitor voltage VS rises to the value VP the UJT readily conducts. When the UJT becomes ON, the capacitor discharges and its voltage falls. When the voltage falls to the valley point VV, the UJT becomes OFF and the capacitor charges again to VP. 40

141 This cycle of charging and discharging of the capacitor C repeats, and as a result, a saw tooth wave form of voltage across C is generated. PROCEDURE:. Connect the circuit as shown in figure with designed values. 2. Note down the voltages and frequencies across C& R2. 3. The time period of the output wave form is noted and is compared with theoretical value T = R? C[ ln {(VBB VV) / (VBB VP)}] Plot the graphs of Vc and VR. REVIEW QUESTIONS:. Describe some important applications of a UJT? 2. Is the name UJT appropriate? 3. Write short notes on UJT as a relaxation oscillator? 4. Discuss the concept of ve resistance? 5. Define the intrinsic stand off ratio and explain its importance? 4

142 GRAPH: RESULT: Sawtooth waveform using UJT and test its performance as an oscillator is verified. 42

143 EXPERIMENT NO: 0 BOOTSTRAP SWEEP GENERATOR AIM: To observe and note down the output waveforms of bootstrap sweep generator. APPARATUS REQUIRED:. Resistors (0KΩ) 4 No. 2. Resistors (KΩ) - No 3. Capacitor (000µF) - No 4. Capacitor (0.0µF) - No 5. Diode (IN448) - No 6. Transistors ( 2N2222) - 2 No 7.. Dual CRO - No. 8. Connecting wires. Electronic Circuit Analysis & Pulse and Digital circuit Lab CIRCUIT DIAGRAM: 43

144 THEORY: The input to Q is the gating waveform. Before the application of the gating waveform, at t = 0, transistor Q is in saturation. The voltage across the capacitor C and at the base of Q2 is VCE(sat). To ensure Q to be in saturation for t = 0, it is necessary that its current be at least equal to ice / hfe so that Rb < hfer. With the application of the gating waveform at t = 0, Q is driven OFF. The current ic now flow into C and assuming units gain in the emitter follower V0? VCC t. When the sweep RC starts, the diode is reverse biased, as already explained above, the current through R is supplied by C. The current VCC / R through C and R now flows from base to emitter of Q2.if the output V0 reaches the voltage VCC in a time TS / Tg, then from above we have TS = RC. EXPECTED WAVEFORMS: F = 800Hz OBSERVATIONS: F = 3KHz S.No Frequency( Hz) Sweep time(ts) Restoration time(tr) F = 800Hz F = 3KHz Sweep error(es) PROCEDURE:. Connect the circuit as per the circuit diagram. 2. Apply the square wave input of KHz to the circuit. 3. Switch on the power supply and note down the amplitude and time period for the input square wave. 4. Observe the output waveform and note down the amplitude and time period. 44