G.H.RAISONI COLLEGE OF ENGINEERING, NAGPUR Department:-Electronics & Communication Engineering

Branch:-7thSemester[Electronics] Subject:- Electronic System Design

List of Experiment

CYCLE 1

1 A) To design a Transistor Shunt Voltage Regulator. B) Simulate the Transistor Shunt Voltage Regulator and observe the waveform

using MICROCAP. 2 A) To design a Emitter Follower type of Voltage Regulator. B) Simulate the Emitter Follower type of Voltage Regulator and observe the

waveform using MICROCAP. 3 A) To design a RC phase shift oscillator. B) Generate the oscillations in microcap using RC Phase Shift Circuit. 4 A) To design a Wein Bridge oscillator.

B) Generate the oscillations in microcap using Wein Bridge oscillator. 5 A) To design a Common Emitter Amplifier. B) To obtain the Amplifier output on MICROCAP using Transistor.

CYCLE 2

6 A) Study of Low Voltage Regulator using IC 723. B) Simulate and observe the regulated waveform on MICROCAP. 7 A) To Design a Series Voltage Regulator using Pre-Regulator circuit. B) Simulate and observe the regulated waveform from SVR on microcap. 8 A) To design 2nd order Low Pass Filter.

B) To study the frequency Vs gain characteristic of Low Pass Filter. 9 A) To design 2nd order High Pass Filter. B) To study the frequency Vs gain characteristic of High Pass Filter. 10A) To design a Diode Function Generator. B) To obtain the waveform for Diode Function Generator on MICROCAP.

Experiment No.-1 Aim :- Design Transistor Shunt Regulator. Problem :- Design transistor shunt regulator for Vi = 15 + 10%V , IZmin = 10mA, ILmax = 200mA, Vo = 7.5V , hfe = 40. Tool :- Microcap software. Circuit Diagram:-

Analysis :- Vo = Vz + VBE

IDmin = ILmax + IZmin + ICmax RD = (Vimin – V0) / IDmin

IBmin = ICmin / hfe

RL = Vo / IL I1 = IZmin - IBmin R1 = VBE / I1 Power Dissipation across zener = Vz x IZmax

Power Dissipation across transistor = (Vimax – Vo) ILmax

Simulation Results:-

Result :- Various parameters have been calculated theoretically and have been

found in accordance with those obtained through microcap software. Also graph against Vi and Vo has been obtained.

Viva Questions :-

1) Classify linear voltage regulators. 2) What is a shunt regulator?

Experiment No.-2

Aim :- Design and simulate Emitter Follower type of Voltage Regulator. Problem :- Design transistor shunt regulator for Vi = 15 + 10%V ,

IZmin = 10mA, ILmax = 500mA, Vo = 8.2V , hfe = 40. Tool :- Microcap software. Circuit Diagram :-

Theory :- A series voltage regulator is shown in the figure. It consists of a transistor as a pass element and Zener diode as a reference source. The Zener diode used as reference element in a series voltage has breakdown voltage always less than the required output voltage. In this circuit a fraction of the output voltage V0 is compared with reference voltage Vz. The difference is amplified by the transistor. If the input voltage Vi increases by ∆Vi, then V0 increases only slightly and yet transistor may cause a large current change in R2. Thus it is possible for almost all of ∆Vi to appear across R3 and for output voltage to remain constant. Thus for all variation in input voltage , the output voltage remains constant. Analysis :-

Vo = Vz - VBE

IBmax = ILmax / hfe

IDmin = IBmax + IZmin

RD = (Vimin – Vz) / IDmin IBmin = ICmin / hfe

RL Or R1 = Vo / IL Power Dissipation across zener = Vz x IZ Power Dissipation across transistor = (Vimax – Vo) ILmax

Simulation Results :-

Result :-

Various parameters have been calculated theoretically and have been found in accordance with those obtained through microcap software. Also graph against Vi and Vo has been obtained.

Viva Questions :-

1) What is a series voltage regulator? 2) What is the other name of a simple series voltage regulator?

Experiment No.-3 Aim :- Design and simulation of RC Phase Shift Circuit Problem :- Design RC Phase Shift Oscillator for output frequency of 2 KHz. Tool :- Microcap software. Circuit Diagram :-

Theory :- An oscillator circuit thrives on feedback circuit and has no input source. The three conditions for sustained oscillations are :-

1) It should consist of positive feedback network. 2) A total phase shift of 0 or 3600 is required. 3) BarkHausen’s criteria must be fulfilled |A.�| = 1

R-C Phase Shift Oscillator Op-amp is used here in inverting configuration as amplifier , with feedback loop consisting of a R-C phase shift network. The phase shift network consists of three R-C voltage dividers. The capacitor reactance and phase shift introduced by each divider varies with freq. between 0 to 90 degrees . At the described operating freq., each R-C network introduces a phase shift of 600, giving a total of 1800 phase shift. Another 1800 phase shift is introduced at the input of inverting amplifier making total phase shift around the loop 3600. From calculation it is found that f0 = 1/ 2��6 RC Rf / R1 = 29 This circuit will produce the sinusoidal waveform of frequency f0, if the gain is 29 & total phase shift around the loop is 3600 .In most of the cases amplifier gain is kept more than 29. So that Gain � does not become less than unity and the oscillations does not die out.

Analysis :-

f0 = 1/ 2��6 RC Rf / R1 � 29

Assume the value of C. Simulation Results:-

Result :-

The above circuit is implemented using microcap software and its results are found in agreement with that of theoretical

results. Viva Questions :-

1) What is Barkhausen’s criteria? 2) How oscillations are generated in RC Phase Shift Oscillator? Explain.

Experiment No.-4 Aim :- To Design Wein Bridge Oscillator. Problem :- Design a Wein Bridge Oscillator for output frequency of 1 KHz. Tool :- Microcap software. Circuit Diagram :-

Theory :- An oscillator circuit thrives on feedback circuit and has no input source. The three conditions for sustained oscillations are :-

1) It should consist of positive feedback network. 2) A total phase shift of 0 or 3600 is required. 3) Barkhausen’s criteria must be fulfilled |A.�| = 1

Wein Bridge Oscillator An oscillator circuit in which a balanced bridge is used as the feedback network is the Wein Bridge Oscillator. The active element is an operational amplifier, which has a very large positive voltage gain, negligible output resistance and very high input resistance. We assume further that Av is constant over the range of frequencies of operation of this circuit. The frequency is computed as

f0 = 1/ 2�RC In this case the loop gain must equal unity and must have a zero phase shift but the magnitude must not be zero. It can be done if Vi / Vo = 1/3 and A = 3.

Analysis :- 1 + Rf / R1 =3 f0 = 1/ 2�RC Assume the value of C. Simulation Results:-

Result :-

The above circuit is implemented using microcap software and its results are found in agreement with that of theoretical results.

Viva Questions :-

1) Classify different types of Oscillators. 2) What is the gain of a Wein Bridge Oscillator?

Experiment No.- 5 Aim :- Design and simulation Transistor Amplifier in CE mode. Problem:- Design a transistor amplifier stage to satisfy the following requirements: RL = 1K,hfe = 100,Vcc = 9V and frequency response = 50 to 50000 Hz. Apparatus :-Power supply, Bread board, Resistor, CRO, Connecting wires, CRO

probes etc.

Circuit Diagram:-

Theory :- The Transistor Amplifier in Common Emitter configuration is used to amplify both current as well as voltage level of the signal and hence it is most preferred configuration. The circuit diagram of common emitter configuration is as shown in figure. C2 and C3 are coupling capacitor while C1 is emitter bypass capacitor. V1 is the sine source of 1MHz, which is applied at the base emitter junction and output is obtained at the collector of transistor. Formula :- Select Vce = Vcc/2 and Ic = 1mA. Assume suitable R5. Vcc = Ic*Rc + Ie*R1 + Vce and find R1. Select suitable R3 VB = Ie*R1 +VBE and I3 = VB/R3 Assume suitable hfe and calculate base current IB I2 = I3 + IB and VR2 = Vcc – VB R2 = VR2/I2 RLeff = Rc*R3/(Rc + RL) Ro = Rs*R3/(Rs + R3) XCe = 10% of R1 at 50Hz. Procedure:-

1) Open Microcap software. 2) Open and save new file . 3) Using tools, Draw the circuit Shown in fig . 4) Obtain o/p by using the circuit.

5) Construct the circuit on breadboard and observe waveforms on CRO.

6) Draw i/p and o/p waveforms. Simulation Results :-

Result :- CE amplifier has been designed and the input has been amplified.

Viva Questions :- 1) Why do we need amplifiers? 2) Classify the different types of amplifiers

Experiment No.- 6

Aim :- Study of low Voltage Regulator using IC 723. Problem :- Design a LVR to given Vo=5V at 2A maximum load current. Connect

a foldback protection circuit. Components:- Resistor , Capacitor, IC 723 Apparatus :- Breadboard , Power Supply component Circuit Diagram:-

Theory :- The basic operation of any regulator remain same but they differ in their construction .They also differ in their applications . The most important function of any regulator is to regulate or control the input voltage from aberrations and also feed regulated or controlled voltage to bad circuit . Regulators generally are classified as SVR and VSR ie series VR and shunt VR In series voltage regulator input current is equal to out put current. There fore it is widely used in high voltage and low current applications . In shunt voltage regulator input voltage is equal to the out put voltage there fore it is widely use in high current low voltage applications. Series voltage regulator are widely used because of their high % efficiency , improved stability over shunt voltage regulators . Series voltage regulator can also be implemented using IC723 . Consequently we can use IC723 as low voltages SVR as well as high voltage SVR . In low voltage regulator out put voltage is less than input voltage and vice versa is the case with high voltage regulators. We design a low voltage regulator to obtain output voltage between 2 to 7 volts and HVR for 7 to 32 volts output.

Formula :- For LVR:- R1 = VR –Vo ; Rv = 20%(R1+R2) , assume IA R2 = Vo / IA ; R1’= R1 -Rv /2 ; R2’ = R2 -Rv/2 Hfe = Ic/IB = Icmax / Icout ; Vin = VBE + Vsense +Vo

Vinmin= Vout +3 ; Vinominal = Vinmin + 10% of Vinmin Icmax = Vsense (RA+RB) / RSC-RB + RA / RSC-RB

S/C condn , IL= ILSC Vsense (Ra +RB )/ RSC-RB RA+RB = VA/ IA P. Dissipation = VCE *IC=(Vin –Vo) IL =(Vinmax-Vo) Icmax

Procedure :-1) Open the MICROCAP create & save a new file 2) Draw the circuit diagram using tool & components 3) Observe the o/p by executing it 4) Draw the same circuit. on breadboard & observe the o/p on CRO. 5) Plot the corresponding o/p for the same . Simulation Results:-

Result :- The LVR using IC 723 has been designed and simulated successfully. Viva Questions :-

1) What are the different pins of IC723? 2) What is the output voltage range for high and low voltage

regulators using IC723?

Experiment No.-7 Aim:- Design & simulation of Series Voltage Regulator with Pre-regulator. Problem :- Design a SVR using Pre-Regulator to give V0 = 15V at 500mA. The input is 27 ± 3V with source resistance of 3�. The transistors available are with hfe = 40 and 100. Pre regulator has a P-N-P transistor with hfe = 100 and 1/hoe = 10k.

Components:- NPN and PNP Transistors, Resistors, Zener Diodes. Apparatus :- Bread board, power supply . Circuit Diagram :-

Theory :- In Series Voltage Regulators The stability of the regulator can be improved by increasing the value of R3 which can be done by using Darlington Pair. The stability can further be improved if we replace R3 by constant current source. It is often called as transistor pre-regulator. This pre-regulator provides the constant current to the collector of Error Amplifier transistor and base of series pass transistor. The zener diode in the pre-regulator circuit will maintain a constant voltage between the base of pre-regulator transistor and the collector of series pass transistor and generating base emitter voltage constant which is 0.6v. This will give a constant drop across Re of pre-regulator and so the current flowing through it is constant and that output is nothing but emitter output of pre-regulator transistor. Also Ie = Ic for pre-regulator transistor and so the transistor provides constant output Ic to the rest of the circuit. Thus a pre-regulator circuit improves the regulation and also reduces the output impedance of the regulator.

Formula:- For Pre-regulator Vzo= Vi’min – Vo – 1V

Vi’min = Vimin – Rs.ILmax

Assume Izmin and rz R2 = (Vi’min - Vzo)/ Izmin Re = (Vzo - VBE)/ I3min Roc = 1/hoe + hfe.Ro/hoe(Re + hie’) For Regulator Vz = 50% of Vb1 Rd = (Vo –Vz)/ Izmin R1= (Vo –Vb2)/I1 R2 = Vb2/I1 I1= 2% to 4% of ILmax and >=20 times IB2max. Procedure :- 1) Open the MICROCAP, Draw the circuit and save a new file.

2) Simulate and run the circuit diagram on MICROCAP. 3) Observe the waveforms obtained and draw the conclusion. 4) Draw the same ckt . on breadboard & observe the o/p on CRO .

5) Plot the corresponding o/p for the same . Simulation Results :-

Result :- SVR using Pre-Regulator has been simulated and designed successfully and satisfactorily.

Viva Questions :-

1) Why do we need Pre-regulators? 2) What is the other name of a pre-regulator circuit?

Experiment No.-8 Aim :- To Design 2nd Order Low Pass Filter & study the Frequency Vs Gain characteristics Problem :- Design a low pass filter of 2nd order having cut off frequency of 1KHz. Apparatus :- Op- Amp 741, Resistances, capacitor, Function generator. Circuit Diagram :-

Theory :- Filters are the circuits which are specifically designed for frequency rejection over a certain range and allow frequency over a specific range. Filters may be of three types:-

1) Low pass filter 2) High pass filter 3) Narrow and Wide band pass filter.

Active filters are designed without the use of inductors which are costly and bulky. The designing and independently as there is no coupling between input and output under active filters , filters with every high quality factor can be designed. Important parameters for filter:-

1) cut off frequency. 2) Gain variation in passband 3) Attenuation in stopband. 4) Rate of roll off beyond cut off frequency.

Low Pass Filters: Low pass filters are those filters that allow low frequency signals to pass i.e from d.c. to fc and block the higher frequency signals i.e. the signals above the cut off level fc. The basic difference between a 1st order filter and the 2nd order filter is that in a 1st order filter the roll off/on rate is 20db/decade while in 2nd order it will be twice the first order i.e. 40 db/decade. Also in 2nd order filters we require a pair of RC circuit more than that required in 1st order filters.

Analysis :- Given fc = 1KHz, Assume the value of C. Fc = 1/ 2�RC 1 + Rf / R1 =3 , Assume the value of R1.

Observation:- Sr. No. Frequency (Hz) V0 V0 / Vin Gain 1. 100 2 300 3 500 4 700 5 900 6 1.1 7 1.3 8 1.5 9 1.7 10 1.9 Procedure :-

1) Make the connections according to the circuit diagram before turning on the supply.

2) Choose appropriate standard values of resistances and capacitances. 3) For different frequencies and note the corresponding output voltage.

4) Calculate the Gain using the formula and plot the graph between freq. And Gain

Simulation Results :-

Result :- Thus a low pass circuit is designed with cut off frequency 1KHz and the is plotted between freq. and Gain. Viva Questions :-

1) What is the significance of filters? 2) Classify the different types of filters?

Experiment No.-9 Aim :- To Design & simulate 2nd Order High Pass Filter & obtain the Frequency Vs Gain Curve. Problem :- Design a High pass filter of 1st order having cut off frequency of 2KHz. Apparatus :- Op- Amp 741, Resistances, capacitor, Function generator. Circuit Diagram :-

Theory :-

Filters are the circuits, which are specifically designed for frequency rejection over a certain range and allow frequency over a specific range. Filters may be of three types:-

1) Low pass filter 2) High pass filter 3) Narrow and Wide band pass filter.

They may be classified as 1) Active filters 2) Passive filters.

Active filters are designed without the use of inductors, which are costly and bulky. The designing is done independently as there is no coupling between input and output under active filters and hence filters with very high quality factor can be designed. Important parameters for filter:-

1) Cut off frequency. 2) Gain variation in Passband 3) Attenuation in Stopband. 4) Rate of roll off beyond cut off frequency.

High Pass Filters: High pass filters are those filters that allow high frequency signals to pass and block the lower frequency signals i.e. the signals below the cut off level. The basic difference between a 1st order filter and the 2nd order filter is that in a 1st order filter the roll off/on rate is 20db/decade while in 2nd order it will be twice the first order i.e. 40 db/decade. Also in 2nd order filters we require a pair of RC circuit more than that required in 1st order filters.

Analysis :- Given fc = 2KHz, Assume the value of C. Fc = 1/ 2�RC 1 + Rf / R1 =3 , Assume the value of R1 Observation :- Sr. No. Frequency (Hz) V0 V0 / Vin Gain 1. 100 2 300 3 500 4 800 5 1K 6 2K 7 3K 8 5K 9 7K 10 10K … …. Procedure :- 1) Make the connections according to the circuit diagram before turning on the supply. 2) Choose appropriate standard values of resistances and capacitances. 3) For different frequencies and note the corresponding output voltage.

4) Calculate the Gain using the formula and plot the graph between freq. and Gain

Simulation Results:-

Result :-

Thus a high pass circuit is designed with cut off frequency 2KHz and graph is drawn between freq. and Gain.

Viva Questions :-

1) What is the difference between the 1st order and 2nd order filters? 2) What do you mean by Cut off frequency?

Experiment No.-10

Aim :- To design a Diode function generator. Problem :- Realize the function V0 = Vi2 / 2 , Vi = 0 to 5V. Select suitable power supply . Assume drop across diode = 0.6 V , Also Rf = 100K Apparatus:- Microcap software. Circuit Diagram :-

Theory :- The circuit of diode function generator is shown in figure. It basically consists of number of diodes which are connected in parallel where each one of them is provided with different inputs. This is done in order to get multiple slopes. From the figure , it can be seen that – Vxn = VRn x RAn / (RAn + RBn ) + Vi x RBn / (RAn + RBn ) For increasing slope:-

S0 = - Rf / R1 When diodes are OFF. S1 = - Rf / R1 || RA1 When D1 is ON. S2 = - Rf / R1 || RA1 || RA2 When D1 and D2 are ON.

S3 = - Rf / R1 || RA1 || RA2 || RA3 When D1, D2 and D3 are ON. S4 = - Rf / R1 || RA1 || RA2 || RA3 || RA4 When all diodes are ON. Observation :-

Sr.No. Vi(volts) Vo(theoretical) Vo( practical) slope 1 0 2 1 3 2 4 3 5 4 6 5

Result :- The circuit of diode function generator is designed and both the theoretical and Practical values of V0 are found. Viva Questions :-

1) What is a Diode Function Generator? 2) What are the applications of a Diode Function Generators?