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DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 1. COMMON EMITTER TRANSISTOR CHARACTERISTICS Aim: 1. To plot the input and output static characteristics. 2. To calculate the input dynamic resistance from the input characteristics and output dynamic resistance and current gain from the output characteristics of the given transistor. Apparatus Required: Name of the S.No Specifications Equipment/Component Quantity 1 Transistor (BC 107) Icmax=100mA PD=300mw Vceo=45V Vbeo=50V 2 Resistors-39KΩ,1KΩ Power rating=0.5w Carbon type 1 3 Regulated Power Supply 0-30V,1A 1 4 Volt meters 0-1V, 0-10V 1 5 Ammeters 0-300µA, 0-10mA 1 1 Theory: In common emitter configuration the emitter is common to both input and output. For normal operation the Base-Emitter junction is forward biased and basecollector junction is reveres biased .The input characteristics are plotted between IB and VBE keeping the voltage VCE constant. This characteristic is very similar to that of a forward biased diode. The input dynamic resistance is calculated using the formula ri = ∆ VBE / ∆ IB at constant VCE The output characteristics are plotted between IC and VCE keeping IB constant. These curves are almost horizontal. The output dynamic resistance is given by, ro = ∆VCE / ∆ IC at constant IB At a given operating point, we define DC and AC current gains (beta) as follows DC current gain βdc = IC / IB ELECTRONIC DEVICES & CIRCUITS LAB at constant VCE 1 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING AC current gain βac = ∆ IC/ ∆ IB at constant VCE. Circuit diagram: Fig A: Transistor Common Emitter Configuration Procedure: a) Input Characteristics: 1. Connect the circuit as shown in fig A. 2. Keep the voltage VCE as constant at 2V by varying VCC. 3. Vary the input voltage, VBB in steps of 1V up to 10V 4. Measure the voltage, VBE from voltmeter and current, IB through the ammeter for different values of input voltages 5. Repeat the step 3 and 4 for VCE values of 5V and 10V 6. Draw input static characteristics for tabulated values 7. At suitable operating point, calculate input dynamic resistance. b) Output Characteristics: 1. Fix input base current, IB at constant value say at 10µA. 2. Vary the output voltage, VCC in steps of 1V from 0V up to10V. 3. Measure the voltage, VCE from voltmeter and current IC through the ammeter for different values. 4. Repeat above steps 2and 3 for various values of IB=20µA and 30µA. 5. Draw output static characteristics for tabulated values ELECTRONIC DEVICES & CIRCUITS LAB 2 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Tabular forms: a) Input Characteristics: Applied S.No Voltage VBB(V) VCE = 2V VCE = 5V VCE = 10V VBE(V) IB(µA) VBE(V) IB(µA) VBE(V) IB(µA) 1 0 0 0 0 0 0 0 2 0.2 0.258 0 0.279 0 0.231 0 3 0.4 0.461 35 0.460 45 0.474 40 4 0.6 0.562 60 0.620 60 0.592 60 5 0.8 0.609 90 0.629 90 0.620 90 6 1.0 0.625 110 0.670 110 0.662 110 7 2.0 0.648 140 0.679 140 0.682 140 8 3.0 0.654 160 0.681 160 0.692 160 9 4.0 0.669 190 0.684 185 0.724 190 10 5.0 0.690 210 0.689 210 0.726 218 b) Output Characteristics: S. No Applied voltage Vcc (V) IB = 10µA VCE(V) IC(mA) IB = 20µA VCE (V) IC(mA) IB = 30µA VCE (V) IC(mA) 1 0 0 0 0 0 0 0 2 0.2 0.02 0 0.02 0 0.02 0 3 0.4 0.06 0 0.05 0 0.04 0 4 0.6 0.08 1.0 0.08 2.2 0.05 2.6 5 0.7 0.1 3.2 0.09 4.5 0.06 4.6 6 0.8 0.12 5.0 0.1 6.2 0.07 6.5 7 1.0 0.21 6.2 0.15 7.0 0.12 7.5 8 2.0 0.31 6.5 0.18 7.5 0.17 7.8 9 3.0 0.51 6.7 0.29 7.7 0.28 9.9 10 4.0 0.68 6.8 0.34 8.5 0.33 10.0 11 5.0 0.88 6.9 0.49 8.9 0.39 10.5 ELECTRONIC DEVICES & CIRCUITS LAB 3 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Model graphs: Fig B: Input Characteristics Fig C: Output Characteristics Calculations: a) Input Characteristics: Input Resistance, ri = ∆ VBE / ∆ IB at VCE constant = (0.654-0.647) / (90-30) X 10-6 = 116Ω. b) Output Characteristics: Output dynamic resistance, ro = ∆VCE / ∆ IC at IB constant = (0.9-0.15) / (9.25-7.2) X10-3 = 365.85Ω. Current gain, β = ∆ IC / ∆ IB at VCE constant = (8.8-6.8)10-3/10X10-6 = 200 Precautions: 1. Connections must be done very carefully. 2. Readings should be noted without parallax error. 3. The applied voltage, current should not exceed the maximum rating of the given transistor. Result: Input and output characteristics are observed for the given transistor in common emitter configuration. The input resistance, output resistance and the current gain are calculated. ELECTRONIC DEVICES & CIRCUITS LAB 4 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Inference: It is observed from the input characteristics that as VCE increases, the curves are shifted towards right side. This is due to the Early effect. Questions & Answers: 1. List various operating regions of Transistor A. Active region, cut-off region, and saturation region. 2. List various biasing circuits A. Fixed bias, collector to base bias, and self bias. 3. Give Transistor current equation in CE configuration A. IC =β IB + (1+β) ICEO. ELECTRONIC DEVICES & CIRCUITS LAB 5 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 2. FULL WAVE RECTIFIER Aim: To observe the working of full wave rectifier with and without filter & calculate its ripple factor Apparatus Required: S. No Name of the Equipment/ Specifications Quantity Component 1. Diode(1N4001) VR (max)=1000V 1 IR(max)=50mA 2. Resistor(1KΩ) Power rating=0.5W Carbon type 6-0-6V,500mA 1 3. Transformer 4. Capacitor(1000µF/25V) 1 Cathode Ray Oscilloscope Electrolytic type, Voltage rating= 1.6v 20MHz 5. 6. Digital Multi meter 4 ½ digit 1 1 1 Theory : In the full wave rectifier circuit the transformer has a center-tap in its secondary winding. It provides out of phase voltages to the two diodes. During the positive half cycle the input, the diode D2 is reverse biased it does not conduct. But diode D1 is in forward bias and it conducts. The current flowing through D1 is also passes through the load resistor, and a voltage is developed across it. During negative half cycle diode D2 is forward biased and diode D1 is reverse biased. Now the current flows through diode D2 and load resistor. The current flowing thought the load resistor RL passes in the both half cycles. The DC voltage obtained at the output is given by Vdc = 2Vm / π. Where Vm is peak AC voltage between center-tap point and one of the diodes. It can be proved that the ripple factor of a full- wave rectifier is 0.482.The output of the full-wave rectifier contains an appreciable amount of AC voltage in addition to DC voltage. But, the required output is pure DC with out any AC voltage in it. The AC variation can be filtered by a shunt capacitor filter connected in shunt with the load. The capacitor offers low impedance path to the AC components of current. Most of the AC current passes through the shunt capacitor. All the DC current passes ELECTRONIC DEVICES & CIRCUITS LAB 6 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING through the load resistor. The capacitor tries to maintain the output voltage constant at Vm. Circuit Diagrams: Fig A: Full wave Rectifier without Filter Fig B: Full wave Rectifier with Filter Procedure : 1. Connect the circuit as shown in Fig A. 2. Apply the supply voltage 230V, 50Hz at the primary winding of the transformer. 3. Connect the CRO at the secondary winding of the transformer and measure the maximum voltage (Vm) and time period (T) at the input. Calculate the RMS input voltage using Vrms=Vm/√2. 4. Now connect the multimeter at the secondary and measure the rms voltage of the input signal. The rms voltage measured by both CRO and multimeter must be same. ELECTRONIC DEVICES & CIRCUITS LAB 7 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 5. Now connect the CRO across the load resistor and measure the maximum voltage, Vm and time period, T of the output voltage. Calculate the rms and average (dc) values of the output signal using V rms = Vm / √2 and Vdc=Vavg = 2Vm / π. 6. Measure the AC and DC voltages across the load resistor using multimeter and calculate the ripple factor as r = Vac / V dc 7. While finding ripple factor using CRO, use r =[ [(Vrms / Vdc )2 – 1]]1/2 8. Compare the measured ripple factor value with theoretical value. 9. Now close the switch ‘s’ to connect the capacitor filter across the load resistor, RL then connect the CRO at output terminals and measure the both ripple AC voltage and DC voltages. Calculate the ripple factor. Also measure the time period T of ripple AC voltage. 10. Tabulate the values with filter and without filter. Observations : Fig C : Input Waveform Fig D: Output Wave Form Without Filter ELECTRONIC DEVICES & CIRCUITS LAB 8 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Fig E: Output Wave form with filter Tabular form: Full-Wave With Out Filter With Filter Rectifier CRO Multimeter CRO Multimeter Vrms (V) 6.36 6.38 0.34 0.32 Vdc (V) 5.72 13.29 10 9.1 Ripple Factor, r 0.4808 0.4808 0.034 0.035 Precautions : 1. Connections must be given very carefully. 2. Readings should be taken with out any parallax error. 3. The applied voltage and current should not exceed the maximum ratings of the diode. Result: Input and output waveform with and without filter of a full wave rectifiers are observed. The ripple factor with and with out filter are calculated. Inference: The ripple factor of Full wave rectifier with filter is less compared to that with out filter. Questions & Answers: 1.What are the limitations of half wave rectifier A. Poor efficiency, less ripple factor 2. Give theoretical values for ripple factor and efficiency of center tapped full wave rectifier. A. r=0.48,η=81.2% ELECTRONIC DEVICES & CIRCUITS LAB 9 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 3. SCR CHARACTERISTICS Aim: a) To obtain the forward characteristics of SCR. b) To identify the break over voltage at different gate voltages. Apparatus Required: Name of the S. No Equipment/Component Specifications Quantity IH (max.)=4A 1 SCR(TYN 604) P (max.)=10W 1 VH (max.)=5V 2 Variable resistor 0-10KΩ 1 3 Resistor - 1KΩ 1 4 Regulated Power Supply Power rating=0.5w Carbon type 0-30V,1A 5 Ammeters 0-50mA 2 6 Digital multimeter 4 ½ digit 1 1 Theory: SCR acts as a switch when it is forward bias. When the gate is kept open IG = 0 and the operation of SCR is similar to PNPN diode. When IG < 0 the break over voltage required to allow the current through SCR is large. When IG > 0 less amount of break over voltage is sufficient. With very large positive gate currents break over may occur at a very low voltage such that the characteristic of SCR is similar to ordinary PN diode. As the voltage at which SCR is switched ON can be controlled by varying gate current. Once the SCR is turned ON, the gate losses control and cannot be used to switch the device OFF. One way to turn the device OFF is by lowering the anode current below the holding current by reducing the supply voltage below the holding voltage, keeping the gate open. At this point even if the gate signal is removed the device keeps ON conducting, till the current level is maintained to a minimum level of holding current. ELECTRONIC DEVICES & CIRCUITS LAB 10 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Circuit Diagram: Fig A: SCR Characteristics Procedure: 1. Connect the circuit as shown in Fig A. 2. Initially some gate current is applied by varying the V2. 3. Voltage V1 is slowly varied and different reading of ammeter (IA) and voltmeter (VAK) are taken. 4. The voltage at which the SCR is triggered and heavy current flows is noted as VBO, forward breakdown voltage. 5. Now apply the gate current more than IG. 6. Steps 3 & 4 are repeated and note down corresponding currents and voltages. 7. Draw the graph between VAK and IA at different gate currents. ELECTRONIC DEVICES & CIRCUITS LAB 11 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Tabular form: Applied S. No IG1 = 3mA IG > IG1 Voltage,V1 (V) VAK (V) IA (mA) VAK (V) IA (mA) 1 0 0 0 0 0 2 1 1 1 1 1 3 2 2 1 2 1 4 3 3 1 3 1 5 4 4 1 4 1 6 5 5 1 1 4 7 6 6 1 1 6 8 7 1 6 1 8 9 8 1 12 1 10 10 9 1 14 1 14 11 10 1 16 1 16 12 11 1 20 1 18 13 12 1 22 1 20 14 13 1 24 1 22 Model Graph: ELECTRONIC DEVICES & CIRCUITS LAB 12 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Result: The VI-characteristics of SCR are observed and Break over Voltage at different gate currents is noted. Inference: It is observed that as applied gate current increases, the forward break over voltage reduces & the device conducts early. Questions: 1. What are the advantages of Thyristor Family? A. Low power dissipation 2. Define the following terms a) Holding current b) Forward break over voltage A. The minimum current at which SCR turns from OFF state to ON state is called holding current. The maximum forward voltage at which the current through SCR increases and voltage across SCR drops is called forward break over voltage. 3. What are the different operating regions of SCR? A. Forward breakdown region, Reverse breakdown region, and Forward conduction region. 4. List the applications of SCR? A. High power applications, switching applications, and controlled device. ELECTRONIC DEVICES & CIRCUITS LAB 13 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 4. RC COUPLED AMPLIFIER Aim: To plot the frequency response characteristics of two stages RC coupled amplifier. Apparatus Required: S. No Name Of The Specifications Quantity. ____ 1 Component/ Equipment 1 Two stage RC coupled amplifier circuit board 2 Cathode Ray Oscilloscope 20 MHz 1 3 Signal Generator 0 -1MHZ 1 4 Regulated Power Supply 0-30V,1A 1 Theory: To improve gain characteristics of an amplifier, two stages of CE amplifier can be cascaded. While cascading, the output of one stage is connected to the input of another stage. If R and C elements are used for coupling, that circuit is named as RC coupled amplifier. Each stage of the cascade amplifier should be biased at its designed level. It is possible to design a multistage cascade in which each stage is separately biased and coupled to the adjacent stage using blocking or coupling capacitors. In this circuit each of the two capacitors c1 & c2 isolate the separate bias network by acting as open circuits to dc and allow only signals of sufficient high frequency to pass through cascade. ELECTRONIC DEVICES & CIRCUITS LAB 14 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Circuit Diagram: Fig A: Two stage RC Coupled Amplifier Procedure: 1. Connect the circuit as per the circuit diagram. 2. Apply supply voltage, Vcc= 12V. 3. Now feed an ac signal of 20mV peak-peak at the input of the amplifier with different frequencies ranging from 20Hz to 1MHz and measure the amplifier output voltage, Vo. 4. Now calculate the gain in dB for various input signal frequencies using AV = 20 log10 (V0/VS). 5. Draw a graph with frequencies on X- axis and gain in dB on Y- axis. From graph calculate bandwidth. ELECTRONIC DEVICES & CIRCUITS LAB 15 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Tabular Form: Input voltage, VS = 20mV peak-peak Output Input Frequency Voltage (Hz) peak-peak S. No Vo (mV) Gain, Av = 20log(Vo/Vs) (dB) 3 50 100 200 22 25.5 58 0.82 02.11 09.24 4 5 6 7 8 9 10 11 12 13 14 15 500 1K 5K 10K 20K 50K 100K 200K 300K 500K 700K 1M 152 330 8500 8500 8500 8500 8500 8500 8500 8500 4000 900 17.61 24.34 52.56 52.56 52.56 52.56 52.56 52.56 52.56 52.56 46.02 33.06 1 2 Model Graph: ELECTRONIC DEVICES & CIRCUITS LAB 16 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Observations: Maximum gain (Av) = 52.56dB Lower cutoff frequency (Fl) = 4.5 KHz Upper cutoff frequency (FH) =580 KHz Band width (B.W) = (FH – FL) = 575.5 KHz Gain bandwidth product = Av (B.W) = 30.24M Hz Precautions: 1. Connections must be done very carefully. 2. Readings should be noted without any parallax error. 3. The applied voltage and current should not exceed the maximum ratings of the given transistor. Result: Frequency response of RC Coupled Amplifier Characteristics of was observed. Inference: The bandwidth of RC coupled amplifier is large compared to CE amplifier. Questions & Answers: 1. List different coupling methods A. a) Direct coupled b) Transformer coupled c) RC coupled 2. Define lower cut-off and upper cut-off frequencies A. Lower cutoff frequency is defined as the frequency at which magnitude of voltage gain in low frequency range falls to 1/√2 or 0.707 of magnitude of gain in mid frequency range. Upper cutoff frequency is defined as the frequency at which magnitude of voltage gain in high frequency range falls to 1/√2 or 0.707 of magnitude of gain in mid frequency range. 3. Define Band width A. Bandwidth is defined as difference between upper & lower cutoff frequencies. ELECTRONIC DEVICES & CIRCUITS LAB 17 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 5. VOLTAGE SERIES FEED BACK AMPLIFIER Aim: To find the gain of the Voltage Series feed back amplifier with & without feedback. Apparatus Required: S. No Name of the Specifications Quantity Component/Equipment 1 Transistor( BC 107) 2 Resistors(100KΩ,47KΩ, 68KΩ,10KΩ,4.7KΩ) Icmax=100mA PD=300mw Vceo=45V Vbeo=50V Power rating=0.5w Carbon type 2 1 2 5 3 Capacitors(0.1µF,100µF) Electrolytic type, Voltage rating= 1.6v 4 1 4 Function generator 0 -1MHZ 1 5 Cathode Ray Oscilloscope 20 MHz 1 Theory: The other name of voltage series feedback amplifier is shunt derived series fed feedback amplifier. The fraction of output voltage is applied in series with input voltage through feedback circuit. Feedback circuit shunt the output but in series with input. So the output impedance is decreased while input impedance is increased. The input & output impedance of an ideal voltage series feedback amplifier is infinite & zero respectively. The resistor RE & capacitor CE are used to provide necessary biasing for the amplifier with voltage series feed back gain of the amplifier decreases. ELECTRONIC DEVICES & CIRCUITS LAB 18 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Circuit Diagram: Fig A: Voltage Series Feed Back Amplifier Procedure: 1. Make sure that the switch S is opened. 2. Apply a sine wave of 40mv peak to peak amplitude at 1 kHz from signal generator to the input of amplifier circuit. 3. Measure the output amplitude VO (p-p) and Calculate the gain of amplifier without feedback by using A =VO/VS. 4. Provide the voltage series feed back by closing the switch S and repeat steps 2 and 3 to find the gain with feed back AF = VO / VS. 5. Calculate the feedback factor β using AF = A / 1+Aβ. 6. Calculate theoretically β value from β = RE / (RE +R). ELECTRONIC DEVICES & CIRCUITS LAB 19 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Frequency Response: Input ac voltage VS = 40 mV peak-peak Frequency (Hz) 1K Output Voltage V0 (V) Gain =20 log10(VO / VS) (dB) With out With With out With Feedback Feedback Feedback Feedback 0.48 0.32 12 8 Model Graph: ELECTRONIC DEVICES & CIRCUITS LAB 20 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Precautions: 1. Connections must be given very carefully. 2. Readings should be noted without parallax error. 3. The applied voltage, current should not exceed the maximum rating of the given transistor. Result: The theoretical and practical values of Gain with and without feed back and the Factor (β) of voltage series feedback amplifier are determined. Inference: It is observed that the gain of the amplifier reduces with feed back and the band width increases by the same amount. Questions & Answers: 1. Why Feed back used in amplifiers? A. To improve the amplifier characteristics with required manner. 2. List various advantages of negative feedback A. a) stabilizes the gain. b) Increases bandwidth, input impedance. c) Reduces output impedance, noise & distortions. ELECTRONIC DEVICES & CIRCUITS LAB 21 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 6. CURRENT SERIES FEED BACK AMPLIFIER Aim: To find the gain of the Current Series feed back amplifier with & without feedback. Apparatus Required: S. No Name of the Specifications Quantity Icmax=100mA PD=300mw Vceo=45V Vbeo=50V Power rating=0.5W Carbon type 1 Component/ Equipment 1 Transistor( BC 107) 2 Resistor(470Ω,4.7kΩ,10KΩ) 1 2 3 Capacitors(0.1µF, 1µF) Electrolytic type, Voltage rating=0.6V 1 4 Function generator 0 -1MHZ 1 5 Cathode Ray Oscilloscope 20 MHz 1 Theory: In Current series feedback amplifier, a feedback voltage is developed which is proportional to the output current. This is called current feedback even though it is a voltage that subtracts from the input voltage. One of the most common methods of applying the current series feedback is to place RE between the emitter lead of a common emitter amplifier and ground. When RE is properly bypassed with a large capacitor CE , the output voltage is V0 and the voltage gain without feedback is A. Resistor RE provides d.c bias stabilization, but no AC feedback. When the capacitor CC is removed, an a.c voltage will be developed across RE due to the emitter current flowing through it and this current is approximately equal to output collector current. This voltage drop across RE will serve to decrease the input voltage between base and emitter, so that the output voltage will decrease to V0I. The gain of amplifier with negative feedback is now AF. With current series feed back both input and output resistances increases. ELECTRONIC DEVICES & CIRCUITS LAB 22 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Circuit Diagram: Fig A: Current Series Feed Back Amplifier Procedure: 1. Make sure that the switch S is closed. 2. Apply a sine wave of 40mv peak to peak amplitude at 1 kHz from signal generator to the input of amplifier circuit. 3. Measure the output amplitude VO (p-p) and Calculate the gain of amplifier without feedback by using A =VO/VS. 4. Provide the current series feed back by open the switch S and repeat steps 2 and 3 to find the gain with feed back AF = VO / VS. 5. Calculate the feedback factor β using AF = A / 1+Aβ. 6. Calculate theoretically β value from β = RE / (RE +R). ELECTRONIC DEVICES & CIRCUITS LAB 23 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Observations: Input voltage VI = 40 mV. Output Voltage ,Vo ( V) Frequency (Hz) Gain =20 Log(VO / VI) (dB) With out With With out With Feedback Feedback Feedback Feedback 0.6 0.36 15 9 1K Model Graph: Precautions: 1. Connections must be given very carefully. 2. Readings should be noted without parallax error. 3. The applied voltage, current should not exceed the maximum rating of the given transistor. ELECTRONIC DEVICES & CIRCUITS LAB 24 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Result: The theoretical and practical values of Gain, feed back Factor of current series feedback amplifier was determined. Inference: It is observed that the gain of the amplifier reduces with feed back and the band width increases by the same amount. Questions & Answers: 1. What is the difference between voltage series & current series feedback amplifiers? A. The output impedance of current series feedback amplifier is high when compared with voltage series feedback amplifiers. 2. What is the other name of current series feedback amplifier? A. The other name of current series feedback amplifier is Series derived series fed feedback amplifier. ELECTRONIC DEVICES & CIRCUITS LAB 25 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 7. RC PHASE SHIFT OSCILLATOR Aim: To determine the frequency of oscillations of an RC Phase shift oscillator. Apparatus Required: S. No Name of the Specifications Quantity Component/Equipment 1 Transistor( BC107) Icmax=100mA PD=300mw Vceo=45V Vbeo=50V 1 2 Resistors - Power rating=0.5w Carbon type 1 56KΩ,2.2KΩ,100KΩ,10KΩ 3 Capacitors(10µF/25V),0.01µF Electrolytic type 3 2 Voltage rating=1.6v 3 4 Potentiometer 0-10KΩ 1 5 Regulated Power Supply 0-30V,1A 1 6 Cathode Ray Oscilloscope 20 MHz 1 Theory: In the RC phase shift oscillator, the combination RC provides self-bias for the amplifier. The phase of the signal at the input gets reverse biased when it is amplified by the amplifier. The output of amplifier goes to a feedback network consists of three identical RC sections. Each RC section provides a phase shift of 600. Thus a total of 1800 phase shift is provided by the feedback network. The output of this circuit is in the same phase as the input to the amplifier. The frequency of oscillations is given by F=1/2π RC (6+4K)1/2 Where, R1=R2=R3=R, C1=C2=C3=C and K=RC/R. ELECTRONIC DEVICES & CIRCUITS LAB 26 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Circuit Diagram: Fig A. RC Phase shift Oscillator Procedure: 1. Connect the circuit as shown in Fig A. 2. Switch on the power supply. 3. Connect the CRO at the output of the circuit. 4. Adjust the RE to get undistorted waveform. 5. Measure the Amplitude and Frequency. 6. Compare the theoretical and practical values. 7. Plot the graph amplitude versus frequency Theoretical Values: f = 1 / 2 π RC √6+4K =1 / 2 π (10K) (0.01µF) √6+4(0.01) = 647.59Hz ELECTRONIC DEVICES & CIRCUITS LAB 27 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Tabular Form: S.NO 1 Theoretical Practical Frequency(Hz) Frequency(Hz) 647.59 639.23 % Error 1.2 Model Graph: Result: The frequency of RC Phase Shift Oscillator is determined. Inference: It is observed that the RC phase shift oscillator produces low frequency oscillations at audio frequencies. Questions & Answers: 1. Define oscillator A. The electronic circuit which produces the out put with out applying in put Ac 2. What is BARKHAUSEN CRITERION? A. IABI=1 and Phase shift=0 or 360 degrees. ELECTRONIC DEVICES & CIRCUITS LAB 28 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING OVERVIEW OF 8085 MICROPROCESSOR KIT 1.1 General Description: Vmc-85/9 is single board microprocessor training /development kit configured around the most widely used microprocessor of today’s world. Based on the 8085 microprocessor It can be used to train engineers to control any industrial process and to develop software for 8080and8085 based system. The kit communicates with the outside world through a key board having 28 keys and seven segment hexadecimal displays. The kit also has the capability of interacting with teletypewriter, CRT terminal and an audio cassette recorder through the interfaces provided on the board. Other devices like a serial printer or floppy drives etc. can be connected to the kit.Vmc-85/9 provides 2k byte of ram and 4k bytes of EPROM. The total on board memory can be very easily expanded to 64k bytes in an appropriate combination of RAM and ROM. The monitor is incorporated from 0000-0FFF and the necessary. 2k bytes of RAM has an address of 2000-27FF.The input/output structure of VMC-85/9 provides 24 programmable I/O lines expandable to 48 I/O lines. It has got 16 bit programmable Timer/Counter for generating any type of counting etc. The on board 8259 provides 8 level of interrupts .the onboard battery back up for RAM retains the memory content in the case of power failure. The on board resident system monitor software is very powerful and provides various software utilities. The kit provides various powerful software commands like SEND, RECEIVE, INSERT, DELETE, BLOCK, MOVE, RELOCATE, STRING, FILL& MEMORY COMPARE etc. which are very helpful in debugging/developing the software. VMC-85/9 is configured around the internationally adopted STD bus, which is the most popular Bus for process control and real time applications. All the address, Data and Control lines are available at the edge connector through the buffers. The kit is fully expandable for any kind of application. 1.2 System Specification: C.P.U - 8bit Microprocessor, the 8085-A MEMORY - Total on board capacity-64kbytes RAM - 2k bytes (6116), space for further expansion ELECTRONIC DEVICES & CIRCUITS LAB 29 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING ROM - 4k bytes of EPROM loaded with powerful monitor program (2732), space for further expansion using 2716/2732/2764/27128. TIMER - 16 bit programmable timer/counter using 8253 I/O - 24 I/O lines expandable to 48 I/O using 8255 PPI INTERRUPTS- 8 different level interrupts through 8259 KEYBOARD - 10 keys for command. 16 keys for hexadecimal data entry. 1 key for Vector interrupt and 1 key for reset LED DISPLAY- 6 seven segment displays, 4 for address field. BUS All data, address and control signals (TTL compatible) - Available at edge connector. INTERFACE - 1) Audio cassette recorder. 2)20 mA current loop through SID/SOD lines. 3) RS-232-c through SID/SOD lines with auto baud rate. 4) One RS-232-c through 8251 with a programmable baud rate. 5) EPROM programmer POWER SUPPLY - +5V, 1.5 A for the kit REQUIREMENT - +12 V ±5%, 250mA for CRT and TTY. - +24 V ±5%, 100 mA for EPROM programmer interface OPERATING TEMP - 0 to 50º C. 1.3 System Capabilities: 1) Examine the contents of any memory location. 2) Examine/modify the contents of any of the up internal register. 3) Modify the contents of any of the RAM location. 4) Move a block of data from one location to another location. 5) Insert one or more instructions in the user program. 6) Delete one or more instructions from the user program. 7) Relocate a program written for some memory area to some other memory area. 8) Find out a string of data lying at a particular address. 9) Fill a particular memory area with a constant. 10) Compare two blocks of memory. 11) Insert one or more data bytes in the user’s program/data entry. 12) Delete one or more data bytes from the user’s program/data area. 13) Transmit a program from memory to audio cassette recorder. 14) Receive a program into memory from Audio cassette Recorder ELECTRONIC DEVICES & CIRCUITS LAB 30 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING 15) Check the contents of an EPROM for blank. 16) List the contents of an EPROM into RAM area. 17) Verify the contents of an EPROM with any memory area. 18) Program an EPROM. 19) *List a program or a block of data on the SIOD or prepare a paper tape. 20) Enter a program or a block of data from the SIOD using the paper tape. 21) Execute a program at full clock speed. 22) Execute a program in single step i.e , instruction by instruction. 23) Download HEX file to PC (These facilities are available if serial I/O device are also connected to the kit). Hardware Description: General: The system has got 8085-a as the Central processing unit. The clock frequency for the system is 3.07MHz and is generated from a crystal of 6.14MHz. 8085 has got 8 data lines and 16 address lines. The lower 8 address lines and 8 bit data lines are multiplexed. Since the lower 8 address bits appear on the bus during the first clock cycle of a machine cycle and the 8 bit data appears on the bus during the second and third clock cycle, it becomes necessary to latch the lower lower 8 address bits during the first clock cycle so that the 16 bit address remains available in subsequent cycles. This is achieved using a latch 74-LS-373. Memory: VMC-85/9 provides 2Kbytes of CMOS RAM using 6116 chip and 4Kbytes of EPROM using 2732. The total on board memory can be expanded up to 64Kbytes. The various chips which can be used are 2716, 2732, 2764, 27128, 6116 and 6264. There are 6 memory spaces provided on VMC-85/9. These six pages are divided into three blocks of two memory spaces each. Each memory space can define any address slots from 0000-FFFF depending upon the size of the memory chip. Same way any block (i.e, two memory spaces) can be defined to have any of the chips, 2716/2732/2764/ 27128/ 6116/ 6264. I/O Devices: The various I/O chips used in VMC-85/9 are 8279, 8255, 8253, 8251 and 8259. The functional role of all these chips are given below: 8279: 8279 is a general purpose programmable keyboard and display I/O interface device designed for use with the 8085 microprocessor. It provides a scanned interface to 28 ELECTRONIC DEVICES & CIRCUITS LAB 31 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING contact key matrix provided in VMC-85/9 and scanned interface for the six 7-segment displays. 8279 has got 16 X 8 display RAM which can be loaded or interrogated by the CPU. When a key is pressed, its corresponding code is entered in FIFO queue of 8279 and can now be read by the microprocessor. 8279 also refreshes the display RAM automatically. 8255: 8255 is a programmable peripheral interface (PPI) designed to use with 8085 microprocessor. This basically acts as a general purpose I/O component to interface peripheral equipments to the system bus. It is not necessary to have an external logic to interface with peripheral devices since the functional configuration of 8255 is programmed by the system software. It has got three I/O ports of 8 lines each( PortA, Port-B, and Port-C). Port C can be divided into two ports of four lines each named as Port C upper and Port C lower. Any I/O combination of Port A, Port B, Port C upper and Port C lower can be defined using the appropriate software commands. VMC-85/9 provides six I/O ports of 8 lines each using two 8255 chips. 8253: This chip is a programmable interval timer/ counters and can be used for the generation of accurate time delays under software control. Various other functions can be implemented with this chip are programmable rate generator, event counter, binary rate multiplier, real time clock etc. This chip has got three independent 16 bit counters, each having a count rate of up to 2MHz. This first timer counter (Counter 0) is being used for single step operation. The second timer counter (Counter 1) is being used for generating programmable baud rate while using 8251. For single step operation clock 0 signal of counter 0 is getting a clock frequency of 1.535MHz. 8251: This chip is a programmable communication interface and is used as a peripheral device. This device accepts data characters from the CPU in parallel format and then converts them into serial data characters for the CPU. This chip will signal the CPU whenever it can accept a new character for transmission or whenever it has received a character for the CPU. The CPU can read the complete status of it at any time. 8251 has been utilized in VMC-85/9 for CRT terminal and TTY interface. Display: VMC-85/9 provides six digits of seven segment display. Four digits are for displaying the address of any location or name of any register, where as rest of the two digits are meant for displaying the contents of memory location or of a register. All the six digits of the display are in hexadecimal notation. ELECTRONIC DEVICES & CIRCUITS LAB 32 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Buffers: Space has been provided on the board of VMC-85/9 for buffering the data, address and control lines. One just needs to put buffers chips on the space provided for these lines. All these address, data and control lines (TTL Compatible) are available to the user at the PCB edge connector in the STD bus configuration. The buffer ICS used in the VMC-85/9 are 74-LS-245 and 74-LS-240. In order to facilitate the multiprocessing operation, all address, data and necessary control lines have been made Bidirectional. Interface: VMC-85/9 provides an interface for Audio Cassette Recorder. The user can store his program into the recorder and can load back the program into the system memory as and when required. The system provides two commands namely SEND AND RECEIVE to store and to load from the Cassette tape. The subroutines required for transmission and parallel to serial and serial to parallel conversion are all incorporated into the system monitor program. VMC-85/9 provides (0-20mA) current loop and RS-232-C interface through the SID and SOD lines of 8085. The selection of (0-20mA) current loop and RS-232-C is done by a switch at the left bottom side of the kit. An EPROM programmer interface is provided on the board of kit to facilitate the programming of the 2716/2732/2732A/2764/27128-EPROMS. An additional RS232-C interface is provided through 8251 with programmable baud rate. Any serial device like printer, floppy drive or CRT terminal etc. can be connected to it. Battery back up: The VMC-85/9 provides a battery back up for the onboard RAM area. The battery back up circuitry is based around LM-393. Since each socket can be defined to have 6264 or 6116 chip also, the VCC to each memory socket is given through the Black box-2. The VCC of all the memory sockets are brought at the Black box-2 named as VCC M0 to VCC M5. Any RAM area to be backed up by battery, its corresponding VCC M point should be connected to CMOS +5V point in the Black box-2 and other points should be connected to VCC point. ELECTRONIC DEVICES & CIRCUITS LAB 33 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Simple Programs using 8085 microprocessor: Addition of two 8 – bit numbers: Address Hex code 5100 5101 5102 5103 3A 00 41 47 5104 5105 5106 5107 3A 01 41 80 5108 5109 510A 510B 32 05 41 76 Mnemonic Operand LDA 4100 MOV B, A LDA 4101 ADD B STA 4105 HLT Comments Load the contents of accumulator with the contents of memory location 4100 Move the contents of accumulator into register B Load the contents of accumulator with the contents of memory location 4101 Add the contents of accumulator with the contents of register B Store the contents of accumulator in 4105 memory location Stop the program. Masking of lower nibble: Address Hex code 5100 5101 5102 5103 5104 3A 00 41 E6 F0 5105 5106 5107 5108 32 01 41 76 Mnemonic Operand LDA 4100 ANI F0 (H) STA 4101 ELECTRONIC DEVICES & CIRCUITS LAB HLT Comments Load the contents of accumulator with the contents of memory location 4100 The contents of accumulator are logically AND ed with an immediate data F0 (H) Store the contents of accumulator in 4101 memory location Stop the program. 34 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Subtraction of two 8 – bit numbers: Address Hex code 5100 5101 5102 5103 3A 00 41 47 5104 5105 5106 5107 3A 01 41 80 5108 5109 510A 510B 32 05 41 76 Mnemonic Operand LDA 4100 MOV B, A LDA 4101 SUB B STA 4105 HLT Comments Load the contents of accumulator with the contents of memory location 4100 Move the contents of accumulator into register B Load the contents of accumulator with the contents of memory location 4101 Subtract the contents of register B from the contents of accumulator Store the contents of accumulator in 4105 memory location Stop the program. 1’s Complement of a given Byte: Address Hex code 5100 5101 5102 5103 3A 00 41 2F 5104 5105 5106 5107 32 05 41 76 Mnemonic Operand LDA 4100 CMA STA 4105 HLT Comments Load the contents of accumulator with the contents of memory location 4100 Complement the contents of accumulator Store the contents of accumulator in 4105 memory location. Stop the program. 2’s Complement of a given Byte: Address Hex code Mnemonic Operand 5100 5101 5102 5103 3A 00 41 2F LDA 4100 CMA 5104 3C INR A 5105 5106 5107 5108 32 05 41 76 STA 4105 ELECTRONIC DEVICES & CIRCUITS LAB HLT Comments Load the contents of accumulator with the contents of memory location 4100 Complement the contents of accumulator The contents of accumulator are incremented by ‘1’ Store the contents of accumulator in 4105 memory location. Stop the program. 35 DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING Addition of two 16 – bit numbers: Address Hex code 5100 5101 5102 5103 2A 00 41 EB 5104 5105 5106 5107 2A 02 41 19 5108 5109 510A 510B 22 05 41 76 Mnemonic Operand Comments LHLD 4100 Load HL pair the contents of memory location 4100 LHLD 4102 Exchange the contents of HL pair with the contents of DE pair Load HL pair the contents of memory location 4102 DAD D SHLD 4105 XCHG HLT Add the contents of HL register pair with the contents of DE register pair Store the contents of HL pair in 4105 memory location Stop the program. Subtraction of two 16 – bit numbers: Address Hex code Mnemonic Operand Comments 5100 5101 5102 5103 2A 00 41 EB LHLD 4100 Load HL pair the contents of memory location 4100 LHLD 4102 Exchange the contents of HL pair with the contents of DE pair Load HL pair the contents of memory location 4102 5104 5105 5106 5107 2A 02 41 7B MOV A, E 5108 95 SUB L 5109 510A 510B 510C 32 04 41 7A STA 4104 MOV A, D 510D 94 SUB H 510E 510F 5110 5111 32 05 41 76 STA 4105 ELECTRONIC DEVICES & CIRCUITS LAB XCHG HLT Move the contents of register E into accumulator Subtract the contents of register L from the contents of accumulator Store the contents of accumulator in 4104 memory location Move the contents of register D into accumulator Subtract the contents of register H from the contents of accumulator Store the contents of accumulator in 4105 memory location Stop the program. 36