Proteus lab Manual.pdf

Circuit Simulation using Proteus

Dr. V. P. Shetkari Shikshan Mandal’s

Padmabhooshan Vasantraodada Patil Institute of Technology, Budhgaon-416304

Department of Electronics Engineering

Circuit Simulation

LAB MANUAL Prepared by

Prof. K. P. Paradeshi Associate Professor, Electronics Engineering, [email protected] Department of Electronics Engg., P.V.P.I.T., Budhgaon





Circuit Simulation Experiment No.

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Name of Experiment: Roll Number

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Date Performed

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Signature (Batch In-charge)

Department of Electronics Engg., P.V.P.I.T., Budhgaon


Title: Design of Half Wave Rectifier using Proteus.

Aim: Design & stimulate the half wave rectifier.

Objectives: •

To understand principal of working of Half Wave Rectifier circuit.

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To understand the circuit arrangement of Half Wave Rectifier circuit.

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To understand the simulation using proteus.

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To observe & analyze the output waveforms

procedure of ―Half Wave Rectifier‖ circuit

Outcomes: •

Able to understand principal of working of Half Wave Rectifier circuit.

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Able to understand the circuit arrangement of Half Wave Rectifier circuit.

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Able to understand the simulaton procedure of Half Wave Rectifier circuit using proteus.

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Able to observe & analyse the circuit.

Programme Education Objective (PEO) Satisfied : 2,3 •

To enable student to analyse solve electronics engineering problem by applying basic principles of mathematics , sciences and engineering and also be able to use modern engineering techniques , skills and tools to fulfill social needs.

•

To enable to innovate, design and develop a variety of electronic and computer based components and system for applications including signal processing , communication , computer network and control system.



Introduction: The device which converts ac voltage into dc voltage is called rectifiers. In fact, rectifiers produce unidirectional & pulsating voltage from ac source. Rectifiers offers a low resistance to the flow of current in one direction & high resistance to flow of current in opposite direction that is rectifier converts AC into DC. This conversion is achieved by rectifier supplied with AC signal using step down transformer. The output of contains pulsating DC (ripple voltage) which can be removed by using filter circuit. we have to design the HWR using proteus. It is easy to design on the proteu Circuit Diagram:

Half wave rectifier

Circuit arrangement: The circuit arrangement of half wave as shown in fig. It uses one transformer for providing step down operation as well as to provide electrical isolation. In this type of arrangement diode D, load resistance RL are used, the output is taken across load resistance as shown in figure.

Theory: When a single rectifier unit is placed in series with the load across an ac supply, it converts alternating voltage into uni-directional pulsating voltage, using one half cycles of the applied voltage, the other half cycles being suppressed because it conducts only in one direction. Unless there is an inductance or battery in the circuit, the current will be zero, therefore, for half the time. This is called half-wave rectification. As already discussed, diode Department of Electronics Engg., P.V.P.I.T., Budhgaon


is an electronic device consisting of two elements known as cathode and anode. Since in a diode electrons can flow in one direction only i.e. from cathode to anode so the diode provides the unilateral conduction necessary for rectification. This is true for diodes of all types-vacuum, gas-filled, crystal or semiconductor, metallic (copper oxide and selenium types) diodes. Semiconductor diodes, because of their inherent advantages are usually used as a rectifying device. However, for very high voltages, vacuum diodes may be employed. The half-wave rectifier circuit using a semiconductor diode with a load resistance R L but no smoothing filter is given in figure. The diode is connected in series with the secondary of the transformer and the load resistance RL, the primary of the transformer is being connected to the ac supply mains. A transformer has been added to provide the desired voltage into the rectifier. For high voltage DC power supplies this would obviously be a step-up transformer,but with many solid-state equipment application voltages of 4.5,6,9,22.5,40volts are used. Thus necessitating a step-down transformer. Hence,in HWR the step-down transformer is used.

Working of a Half wave rectifier: The ac voltage across the secondary winding changes polarities after every half cycle. During the positive half-cycles of the input ac voltage i.e. when upper end of the secondary winding is positive w.r.t. its lower end, the diode is forward biased and therefore conducts current. When forward voltages crosses barrier potential (0.7v for silicon diode) the diode works as closed switch,the large current flows through RL.This current produces dc voltage across the load RL. If the forward resistance of the diode is assumed to be zero (in practice, however, a small resistance exists) the input voltage during the positive half-cycles is directly applied to the load resistance RL, making its upper end positive w.r.t. its lower end. The waveforms of the output current and output voltage are of the same shape as that of the input ac voltage. During the negative half cycles of the input ac voltage i.e. when the lower end of the secondary winding is positive w.r.t. its upper end, the diode is reverse biased and so does not conduct that is diode D acts as open switch.Thus during the negative half cycles of the input ac voltage the current through and voltage across the load remains zero if the reverse current, being very small in magnitude, is neglected. Thus for the negative half cycles no power is delivered to the load. Thus the output voltage developed across load resistance R L (VL) is a series of positive half cycles of alternating voltage, with intervening very small constant negative voltage levels, It is obvious from the figure that the output is not a steady dc, but only a pulsating dc wave. Since only half-cycles of the input wave are used, it is called a half-wave rectifier. The output voltage is unidirectional,pulsating & intermittent.



DC OUTPUT VOLTAGE:Vdc = Vm/3.14

(with no load)

Vm= maximum voltage DC OUTPUT CURRENT:Idc = Im/3.14

Im=maximum current

Significance of half wave rectifier using proteus: The most important characteristics which are required to be specified for a power supply are given below : 1. The required output dc voltage. 2. The average and peak currents in the diode. 3. The peak inverse voltage (PIV) of each diode. 4. The regulation. 5. The ripple factor

Advantages: •

Simple circuit and low cost.

Disadvantages: •

The output current in the load contains, in addition to dc component, ac components of basic frequency equal to that of the input voltage frequency. Ripple factor is high and an elaborate filtering is, therefore, required to give steady dc output.

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The power output and, therefore, rectification efficiency is quite low. This is due to the fact that power is delivered only half the time.

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Transformer utilization factor is low.

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DC saturation of transformer core resulting in magnetizing current and hysteresis losses and generation of harmonics.

The type of supply available from a half-wave rectifier is not satisfactory for general power supply. This type of supply can be satisfactory for some particular purposes such as battery charging.



Procedure using Proteus: 1. From main page of Proteus, click on ‗P‘ to pick device from library. 2. In pick device, insert the component which has to be selected & click on ‗OK‘. 3. Place the all components on the Proteus screen. 4. Right click and select edit properties to change the values. 5. Add voltage probe at input and output. 6. Select the ‗graph mode‘& choose analouge Analysis graph window. 7. Right click and select ‗add traces‘ (input & output trace). 8. Right click & activate simulation graph. 9. Observe and analyse input & output waveforms.

Circuit simulation



Conclusion: By using Proteus we can understand working of half wave rectifier. During the negative half cycles of the input ac voltage, diode D acts as open switch. Thus for the negative half cycles no power is delivered to the load. The input voltage during the positive half-cycles is directly applied to the load resistance R L and waveforms appears across diode D.







: 02


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Title: To study bridge full wave rectifier. Aim: To design and simulate bridge full wave rectifier circuit using proteus and observe the waveforms. Objectives: •

To understand principal of working of bridge full wave rectifier.

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To understand the circuit arrangement of bridge full wave rectifier.

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To understand the simulation procedure of bridge full wave rectifier circuit using proteus.

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To observe & analyze the output waveforms

Outcomes: •

Able to understand principal of working of bridge FWR circuit.

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Able to understand the circuit arrangement of bridge FWR circuit.

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Able to understand the simulation procedure of bridge FWR circuit using Proteus.

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Able to observe & analyze the circuit.

Programme Education Objective (PEO) Satisfied: 2,3 •


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To enable to innovate, design and develope a variety of electronic and computer based components and system for applications including signal processing , communication , computer network and control system.



Introduction:The Full Wave Bridge Rectifier Another type of circuit that produces the same output waveform as the full wave rectifier circuit above is that of the Full Wave Bridge Rectifier. This type of single phase rectifier uses four individual rectifying diodes connected in a closed loop "bridge" configuration to produce the desired output. The main advantage of this bridge circuit is that it does not require a special centre tapped transformer, thereby reducing its size and cost. The single secondary winding is connected to one side of the diode bridge network and the load to the other side as shown below. Working of full wave bridge rectifier: The circuit arrangement made for bridge FWR is as shown in fig. It uses one transformer for providing step down operation as well as to provide electrical isolation. In this type of arrangement, four diodes (D1, D2,D3,D4) are connected in such a way , to form a bridge. Hence it is called as ―bridge FWR‖. Here, two diodes are conducting at a time and remaining two diodes are non-conducting. The output is taken across load resistance as shown in fig. During positive half cycle of AC input, the diode D1 and D3 becomes forward bias, at the same time, diodes D2 and D4 becomes reverse bias. Thus, current flows through the path : diode D1, load resistance RL , diode D3, to the second end of transformer‘s secondary winding. In this situation, current through RL flows from top to bottom producing positive output voltage across it. During negative half cycle of AC input, diode D2 and D4 becomes forward biased, at the same time, diodes D1 and D3 becomes reverse bias. Thus, current flows through the path : diode D2, load resistance RL , diode D4, to the second end of transformer‘s secondary winding. In this situation, current flows in the same direction producing positive output voltage across load resistance as shown in waveforms.



Theory:.

The Diode Bridge Rectifier The four diodes labelled D1 to D4 are arranged in "series pairs" with only two diodes conducting current during each half cycle. During the positive half cycle of the supply, diodes D1 and D2 conduct in series while diodes D3 and D4 are reverse biased and the current flows through the load as shown below.

The Positive Half-cycle

During the negative half cycle of the supply, diodes D3 and D4 conduct in series, but diodes D1 and D2 switch "OFF" as they are now reverse biased. The current flowing through the load is the same direction as before.



The Negative Half-cycle

As the current flowing through the load is unidirectional, so the voltage developed across the load is also unidirectional the same as for the previous two diode fullwave rectifier, therefore the average DC voltage across the load is 0.637Vmax. However in reality, during each half cycle the current flows through two diodes instead of just one so the amplitude of the output voltage is two voltage drops ( 2 x 0.7 = 1.4V ) less than the input VMAX amplitude. The ripple frequency is now twice the supply frequency (e.g. 100Hz for a 50Hz supply). Although we can use four individual power diodes to make a full wave bridge rectifier, pre-made bridge rectifier components are available "off-the-shelf" in a range of different voltage and current sizes that can be soldered directly into a PCB circuit board or be connected by spade connectors.The image to the right shows a typical single phase bridge rectifier with one corner cut off. This cut-off corner indicates that the terminal nearest to the corner is the positive or +ve output terminal or lead with the opposite (diagonal) lead being the negative or -ve output lead. The other two connecting leads are for the input alternating voltage from a transformer secondary winding.

The Smoothing Capacitor We saw in the previous section that the single phase half-wave rectifier produces an output wave every half cycle and that it was not practical to use this type of circuit to produce a steady DC supply. The full-wave bridge rectifier however, gives us a greater mean DC value (0.637 Vmax) with less superimposed ripple while the output waveform is twice that of the frequency of the input supply frequency.



Full-wave Rectifier with Smoothing Capacitor

The smoothing capacitor converts the full-wave rippled output of the rectifier into a smooth DC output voltage. Generally for DC power supply circuits the smoothing capacitor is an Aluminum Electrolytic type that has a capacitance value of 100uF or more with repeated DC voltage pulses from the rectifier charging up the capacitor to peak voltage. However, there are two important parameters to consider when choosing a suitable smoothing capacitor and these are its Working Voltage, which must be higher than the noload output value of the rectifier and its Capacitance Value, which determines the amount of ripple that will appear superimposed on top of the DC voltage.

Too low a capacitance value and the capacitor has little effect on the output waveform. But if the smoothing capacitor is sufficiently large enough (parallel capacitors can be used) and the load current is not too large, the output voltage will be almost as smooth as pure DC. As a general rule of thumb, we are looking to have a ripple voltage of less than 100mV peak to peak.

The maximum ripple voltage present for a Full Wave Rectifier circuit is not only determined by the value of the smoothing capacitor but by the frequency and load current, and is calculated as: Department of Electronics Engg., P.V.P.I.T., Budhgaon


Bridge Rectifier Ripple Voltage

Where: I is the DC load current in amps, ƒ is the frequency of the ripple or twice the input frequency in Hertz, and C is the capacitance in Farads.

The main advantages of a full-wave bridge rectifier is that it has a smaller AC ripple value for a given load and a smaller reservoir or smoothing capacitor than an equivalent halfwave rectifier. Therefore, the fundamental frequency of the ripple voltage is twice that of the AC supply frequency (100Hz) where for the half-wave rectifier it is exactly equal to the supply frequency (50Hz).

The amount of ripple voltage that is superimposed on top of the DC supply voltage by the diodes can be virtually eliminated by adding a much improved π-filter (pifilter) to the output terminals of the bridge rectifier. This type of low-pass filter consists of two smoothing capacitors, usually of the same value and a choke or inductance across them to introduce a high impedance path to the alternating ripple component.

Another more practical and cheaper alternative is to use an off the shelf 3-terminal voltage regulator IC, such as a LM78xx (where "xx" stands for the output voltage rating) for a positive output voltage or its inverse equivalent the LM79xx for a negative output voltage which can reduce the ripple by more than 70dB (Datasheet) while delivering a constant output current of over 1 amp.



FWR Circuit Using Proteus Components required: Name

Description

Number of components required

RES

Resistor(3.3K)

1

DIODE

Diode 1N4007

4

TRAN

Transformer 280-280)

1

Vac

Ac voltage source

1

GND

Ground

3

Procedure using proteus: 1. First of all, pick a bridge of diodes 1N4007 from the components list, and place it properly on the screen 2. Pick the transformer having rating of 12-0-12 and place it as per the circuit diagram. 3. Pick a resistance of 3.3kilo ohm, and place it properly. 4. Connect the AC Source of 230V(freq.50Hz) volts source . 5. With the help of wires, connect all these components in a proper way, as shown in fig. 6. Connect voltage probes to the input & output positions. 7. open

graph option and activate analogue analysis option to observe the output

waveforms. 8. By right clicking on the window, add the traces of each component. And then simulate the graph. After that, you will get desired output waveforms of the bridge FWR. Advantages: 1.Smaller size of transformer is required due to higher value of transformer utilization factor. 2.The need of centre tap transformer is eliminated. 3.PIV rating of diode is less as compared to centre tap FWR. 4.There is no problem of transformer core saturation. Department of Electronics Engg., P.V.P.I.T., Budhgaon


Disadvantages: 1. It requires two additional diodes as compared to centre tap FWR. CIRCUIT SIMULATION

RESULT



Conclusion: By studying bridge FWR, we conclude that bridge FWR is most efficient electronic circuit which converts AC signal into DC signal. As the current flowing through the load is unidirectional, so the voltage developed across the load is also unidirectional the same as for the previous two diode full-wave rectifier, therefore the average DC voltage across the load is 0.637Vmax. However in reality, during each half cycle the current flows through two diodes instead of just one so the amplitude of the output voltage is two voltage drops ( 2 x 0.7 = 1.4V ) less than the input VMAX amplitude. The ripple frequency is now twice the supply frequency (e.g. 100Hz for a 50Hz supply).







: 03


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Title : Low Pass Filter & High Pass Filter . Aim : To Study & simulate frequency response of low pass & high pass filter.

Objectives:     

To understand principal of working of low pass & high pass filter. To understand the circuit arrangement of low pass & high pass filter. To understand the procedure of low pass & high pass filter using capacitor& register circuit using proteus. To observe the frequency response waveforms and analyze the circuit.

Outcomes:    

Students will be able to study low pass & high pass filter using capacitor& register. Students will be able to understand the circuit arrangement of pass & high pass filter Students will be able to understand the procedure of pass & high pass filter using capacitor & register circuit using proteus. Students will be able to observe the simulation and analyse the working of circuit.

Programme Education Objective (PEO) Satisfied : 2,3 



To enable student to analyse solve electronics engineering problem by applying basic principles of mathematics , sciences and engineering and also be able to use modern engineering techniques , skills and tools to fulfill social needs To enable to innovate , design and develop a variety of electronic and computer based components and system for applications including signal processing , communication , computer network and control system.



Introduction Low Pass Filter : A low-pass filter is an electronic filter that passes low-frequency signals and attenuates (reduces the amplitude of) signals with frequencies higher than the cutoff frequency. The actual amount of attenuation for each frequency varies from filter to filter. It is sometimes called a high-cut filter, or treble cut filter when used in audio applications. A low-pass filter is the opposite of a high-pass filter. A band-pass filter is a combination of a low-pass and a high-pass. Low-pass filters exist in many different forms, including electronic circuits (such as a hiss filter used in audio), anti-aliasing filters for conditioning signals prior to analog-todigital conversion, digital filters for smoothing sets of data, acoustic barriers, blurring of images, and so on. The moving average operation used in fields such as finance is a particular kind of low-pass filter, and can be analyzed with the same signal processing techniques as are used for other low-pass filters. Low-pass filters provide a smoother form of a signal, removing the short-term fluctuations, and leaving the longer-term trend. An optical filter could correctly be called low-pass, but conventionally is described as "longpass" (low frequency is long wavelength), to avoid confusion. Electronic low-pass filters Passive electronic realization

Passive, first order low-pass RC filter One simple electrical circuit that will serve as a low-pass filter consists of a resistor in series with a load, and a capacitor in parallel with the load. The capacitor exhibits reactance, and blocks low-frequency signals, causing them to go through the load instead. At higher frequencies the reactance drops, and the capacitor effectively functions as a short circuit. The combination of resistance and capacitance gives the time constant of the filter (represented by the Greek letter tau). The break frequency, also called the turnover frequency or cutoff frequency (in hertz), is determined by the time constant:



or equivalently (in radians per second):

One way to understand this circuit is to focus on the time the capacitor takes to charge. It takes time to charge or discharge the capacitor through that resistor: At low frequencies, there is plenty of time for the capacitor to charge up to practically the same voltage as the input voltage. At high frequencies, the capacitor only has time to charge up a small amount before the input switches direction. The output goes up and down only a small fraction of the amount the input goes up and down. At double the frequency, there's only time for it to charge up half the amount. Another way to understand this circuit is with the idea of reactance at a particular frequency: Since DC cannot flow through the capacitor, DC input must "flow out" the path marked (analogous to removing the capacitor). Since AC flows very well through the capacitor — almost as well as it flows through solid wire — AC input "flows out" through the capacitor, effectively short circuiting to ground (analogous to replacing the capacitor with just a wire). The capacitor is not an "on/off" object (like the block or pass fluidic explanation above). The capacitor will variably act between these two extremes. It is the Bode plot and frequency response that show this variability. Active electronic realization

An active low-pass filter Another type of electrical circuit is an active low-pass filter. In the operational amplifier circuit shown in the figure, the cutoff frequency (in hertz) is defined as: Department of Electronics Engg., P.V.P.I.T., Budhgaon


or equivalently (in radians per second):

The gain in the passband is −R2/R1, and the stopband drops off at −6 dB per octave (that is −20 dB per decade) as it is a first-order filter. Discrete-time realization Many digital filters are designed to give low-pass characteristics. Both infinite impulse response and finite impulse response low pass filters as well as filters using fourier transforms are widely used. Simple infinite impulse response filter The effect of an infinite impulse response low-pass filter can be simulated on a computer by analyzing an RC filter's behavior in the time domain, and then discretizing the model.

A simple low-pass RC filter Finite impulse response Finite impulse response filters can be built that approximate to the ideal sinc time domain response. In practice the time domain response must be time truncated and is often of a simplified shape; in the simplest case, a running average can be used giving a square time response.[3] High Pass Filter A high-pass filter (HPF) is an electronic filter that passes high-frequency signals but attenuates (reduces the amplitude of) signals with frequencies lower than the cutoff frequency. The actual amount of attenuation for each frequency varies from filter to filter. A high-pass filter is usually modeled as a linear timeinvariant system. It is sometimes called a low-cut filter or bass-cut filter.[1] HighDepartment of Electronics Engg., P.V.P.I.T., Budhgaon

Circuit Simulation using Proteus pass filters have many uses, such as blocking DC from circuitry sensitive to nonzero average voltages or RF devices. They can also be used in conjunction with a low-pass filter to make a bandpass filter.

First-order continuous-time implementation

Figure 1: A passive, analog, first-order high-pass filter, realized by an RC circuit The

simple

first-order

electronic

high-pass

filter

shown

in

Figure 1

is

implemented by placing an input voltage across the series combination of a capacitor and a resistor and using the voltage across the resistor as an output. The product of the resistance and capacitance (R×C) is the time constant (τ); it is inversely proportional to the cutoff frequency fc, that is,

where fc is in hertz, τ is in seconds, R is in ohms, and C is in farads.

Figure 2: An active high-pass filter Figure 2 shows an active electronic implementation of a first-order high-pass filter using an operational amplifier. In this case, the filter has a passband gain of -R2/R1 and has a corner frequency of



Because this filter is active, it may have non-unity passband gain. That is, highfrequency signals are inverted and amplified by R2/R1.

Advantages: Low pass filter 1. The gain of integrater decreases with frequency whereas gain of differentiator increases linearly with frequency,so it is easier to stabilize the output of integrators. 2. Due to limited B.W,integrator are less sensitive to noise voltage than differentiators. 3.It is more convenient to introduce the initial conditions in an integrator circuit. Disadvantages: For lower frequency reactance offered by capacitor C is very high & hence these frequencies are not passed at output. Low pass filter circuit using proteus:

Procedure using Proteus: 10. From main page of Proteus, click on ‗P‘ to pick device from library. 11. In pick device, insert the component which has to be selected & click on ‗OK‘. 12. Place the all components on the Proteus screen. 13. Right click and select edit properties to change the values. 14. Add voltage probe at input and output. 15. Select the ‗graph mode‘ & choose analouge Analysis graph window. 16. Right click and select ‗add traces‘(input & output trace). 17. Right click & activate simulation graph. 18. Observe and analyse input & output waveforms.



High pass filter using proteus

Result



Low pass filter using proteus:



Result



Conclusion: We can understand working of low pass & high pass filter. By using proteus we can observe simulation of circuit & analyses the waveform taken across capacitor C1. In Low pass filter the frequency is passes below cut off region And in high pass filter the frequency passes above the cut off region.







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Title: Design ± 5V and ±12V regulated power supply using Proteus.

Aim: Design and simulate the ± 5V and ±12V regulated power Supply.

Objectives:    

To understand principal of working of regulated power supply. To understand the circuit arrangement regulated power supply. To understand the procedure of regulated power supply circuit using proteus. To observe the simulation of circuit.


Able to understand principal of working of regulated power supply circuit. Able to understand the circuit arrangement of regulated power supply circuit. Able to understand the procedure of regulated power supply.‖ circuit using proteus. Able to observe the simulation of circuit.



•

To enable to innovate, design and develop a variety of electronic and computer based components and system for applications including signal processing , communication , computer network and control system.



Principle: 7805 is a 5V fixed three terminal positive voltage regulator IC. The IC has features such as safe operating area protection, thermal shut down, internal current limiting which makes the IC very rugged. Output currents up to 1A can be drawn from the IC provided that there is a proper heat sink. A 9V transformer steps down the main voltage, 1A bridge rectifies it and capacitor C1 filters it and 7805 regulates it to produce a steady 5voltDC. Introduction: The stepdown transformer, down-converts the high voltage AC input (230V,50Hz) to a 9V,2A; because the transformer we used here having a specification of 9V/A. The alternating voltage from secondary terminal of the transformer is given to a bridge rectifier. The bridge rectifier converts alternating voltage to unidirectional voltage with the switching action of diodes. This voltage is finally fed to a 5V regulator IC through a 470uF,50v electrolytic capacitor, which eliminates the ripples and make the output stable. After regulation we get a 5V DC voltage at the output of 7805 IC.

Circuit Diagram:



Circuit arrangement: The circuit arrangement of IC 7805 is as shown in fig. It uses one IC 7805 to generate +5V output . In this type of arrangement two resistors and two capacitors are used, the output is taken across pin no 3 of IC 555 as shown in figure. Theory: Theory of IC 78XX Transformer:A bridge rectifier coupled with a step down transformer is used for our design. The voltage rating of transformer used is 0-12V and the current rating is 500mA. When AC voltage of 230V is applied across the primary winding an output AC voltage of 12V is obtained. One alteration of input causes the top of transformer to be positive and the bottom negative. The next alteration will temporarily cause the reverse. Rectifier:In the power supply unit, rectification is normally achieved using a solid state diode. Diode has the property that will let the electron flow easily at one direction at proper biasing condition. Bridge rectifiers of 4 diodes are used to achieve full wave rectification. Two diodes will conduct during the negative cycle and the other two will conduct during the positive half cycle. Filtering unit:Filter circuit which is usually a capacitor acts as a surge arrester always follows the rectifier unit. This capacitor is also called as a decoupling capacitor or a bypass capacitor, is used not only to short the ripple with frequency to ground but also leave the frequency of the DC to appear at the output. Regulators:The voltage regulators play an important role in any power supply unit. The primary purpose of a regulator is to aid the rectifier and filter circuit in providing a constant DC voltage to the device. Power supplies without regulators have an inherent problem of changing DC voltage values due to variations in the load or due to fluctuations in the AC line voltage. With a regulator connected to DC output, the voltage can be maintained within a close tolerant region of the desired output. IC 7805 and 7812 regulators are used in this project for providing a DC voltage of +5V and +12V respectively. Technical Details:Transformer: 230/12 volts step down transformer, 1 ampere Diodes: IN 4007 7805: The 7805 supplies 5 volts at 1 amp maximum with an input of 7-25 volts Electrolytic Capacitors: 100pF, 330pF and 100μF, power rating of 25V. Features:· Gives a well regulated +12V and +5V output voltages · Built in overheating protection shuts down output when regulator IC gets too hot. · Very stable output voltages, reliable operation Department of Electronics Engg., P.V.P.I.T., Budhgaon


· The circuit has overload and thermal protection. ADVANTAGES: 



78xx series ICs do not require additional components to provide a constant, regulated source of power, making them easy to use, as well as economical and efficient uses of space. Other voltage regulators may require additional components to set the output voltage level, or to assist in the regulation process. Some other designs (such as a switched-mode power supply) may need substantial engineering expertise to implement. 78xx series ICs have built-in protection against a circuit drawing too much power. They have protection against overheating and short-circuits, making them quite robust in most applications. In some cases, the current-limiting features of the 78xx devices can provide protection not only for the 78xx itself, but also for other parts of the circuit.

DISADVANTAGES: 







The input voltage must always be higher than the output voltage by some minimum amount (typically 2 volts). This can make these devices unsuitable for powering some devices from certain types of power sources (for example, powering a circuit that requires 5 volts using 6-volt batteries will not work using a 7805). As they are based on a linear regulator design, the input current required is always the same as the output current. As the input voltage must always be higher than the output voltage, this means that the total power (voltage multiplied by current) going into the 78xx will be more than the output power provided. The extra input power is dissipated as heat. This means both that for some applications an adequate heatsink must be provided, and also that a (often substantial) portion of the input power is wasted during the process, rendering them less efficient than some other types of power supplies. When the input voltage is significantly higher than the regulated output voltage (for example, powering a 7805 using a 24 volt power source), this inefficiency can be a significant issue. Even in larger packages, 78xx integrated circuits cannot supply as much power as many designs which use discrete components, and are generally inappropriate for applications requiring more than a few amperes of current. Each specific model of 78xx is designed to produce only one fixed voltage output, so they may not be suitable for applications requiring a configurable or varying output (For such applications, the LM317 series of ICs are available, which are similar to 78xx ICs but can produce a configurable voltage). ± 5V and ±12V regulated power supply using Proteus.

Component: 230V ac transformer, diodes, capacitor and IC 7805.



Procedure using Proteus: 1. From main page of Proteus, click on ‗P‘ to click device from library. 2. In pick device, insert the component which has to be selected & click on ‗OK‘. 3. Then place the all components on the Proteus screen. 4. Right click the component properties to change the required value 5. Add voltage probe at input and output of diagram. 6. Select the graph mode & choose AC sweep analysis option 7. Right click in the graph window and select add traces(input & output trace). 8. Right click & activate simulate graph 9. Observe input & output waveforms and analysis the results.

CIRCUIT SIMULATION

Result



Conclusion: By using ± 5V and ±12V regulated power supply using Proteus; we understand that the voltage regulators play an important role in any power supply unit. The primary purpose of a regulator is to aid the rectifier and filter circuit in providing a constant DC voltage. To avoid variations in the load or due to fluctuations in the AC line voltage, IC 7805 and 7812 regulators are used.







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Title: Study and Analysis of Non inverting amplifier.

Aim: Design, simulate and analyze non inverting amplifier using IC 741 using following tools 1) Analogue analysis 2) Frequency analysis 3) Noise analysis 4) Distortion analysis Objectives: •

To understand principal of working of Non inverting amplifier

•

To understand the circuit arrangement of Non inverting amplifier

•

To understand the simulation using proteus.

•

To observe & analyze the output waveforms using different analysis tools.

procedure of Non inverting amplifier circuit

Outcomes: •

Able to understand principal of working of Non inverting amplifier circuit.

•

Able to understand the circuit arrangement of Non inverting amplifier circuit.

•

Able to understand the simulation procedure Non inverting amplifier circuit using Proteus.

•

Able to observe & analyse the circuit.


To enable student to analyse solve electronics engineering problem by applying basic principles of mathematics , sciences and engineering and also be able to use modern engineering techniques , skills and tools to fulfill social needs

•

To enable to innovate , design and develop a variety of electronic and computer based components and system for applications including signal processing , communication , computer network and control system.



Introduction: Non-inverting amplifier is one of the most popular op amp circuits similar to Op amp inverting amplifier circuit. It provides a gain to the input signal without any change in the polarity. If a sine wave is fed to the input of this op amp non inverting amplifier, the output will be an amplified sine wave with zero phase shift. Here the input is applied to the non inverting terminal of the op amp. The non inverting amplifier gain is given by the expression A=1+R3/R2 where R2 is the feedback resistance and R3 is the input resistance. The input impedance of non-inverting amplifier is extremely large, typically 100MΩ.

Working of Non- inverting Amplifier:





The second basic configuration of an operational amplifier circuit is that of a Noninverting Amplifier. In this configuration, the input voltage signal, (Vin ) is applied directly to the non-inverting ( + ) input terminal which means that the output gain of the amplifier becomes "Positive" in value. The result of this is that the output signal is "in-phase" with the input signal. Feedback control of the non-inverting amplifier is achieved by applying a small part of the output voltage signal back to the inverting ( - ) input terminal via a R3- R2 voltage divider network, again producing negative feedback. This closed-loop configuration produces a non-inverting amplifier circuit with very good stability, a very high input impedance, R1 approaching infinity, as no current flows into the positive input terminal, (ideal conditions) and a low output impedance, Rout as shown below. The working of non inverting amplifier is similar to that of inverting amplifier except that the output has no phase shift here.



The resistors R2 and R3 form a voltage divider network.



A negative feedback is provided by applying a little of output voltage to the inverting input terminal through the potential divider network R2 and R3.



The voltage gain of the amplifier is determined by the ratios of R3 and R2 since Gain, A=1+R3/R2

So the amplitude of the output voltage signal can be varied by varying either of the resistors R2 or R3



Theory:•

Closed loop configuration 1) Non-inverting amplifier 2) Inverting amplifier 3) Differential amplifier

1. Non-inverting amplifier Non-inverting amplifier is voltage series feedback amplifier. The input signal is given to the non-inverting terminal of op-amp & output signal is in same phase with input. So it is called as non inverting amplifier.

•

Closed loop voltage gain

The closed loop voltage gain is given by, Af=Vo/Vin . . . . . . . . . . . .(1) The output voltage Vo is given by, Vo=A*(Vid) =A*(V1-V2) . . . . . . . . . (2) where V1=Vin V2=Vf Therefore feed back voltage Vf is given by, Vf=(R2/R2+Rf)*Vo . . . . . . . . . . (3) Department of Electronics Engg., P.V.P.I.T., Budhgaon


Therefore output voltage Vo is given by, Vo=A*(Vin-Vf) Vo=A*(Vin-R2*Vo/R2+Rf) Vo(1+AR2/R2+Rf)=AVin Vo/Vin=A/(R2+Rf+AR2/R2+Rf) Af=A(R2+Rf)/(R2+Rf+AR2) . . . . .(4) Generally A is very large. Therfore AR2>>R2+Rf & R2+Rf+AR2=AR2 Therefore Af=(R2+Rf)/R2 Af=1+Rf/R2 . . . . . . . . . . .(5) The gain of voltage seriers feedback amplifier is determined by the two resistances R2& Rf. •

Closed loop voltage gain (Af) in terms of open loop gain(A)

The feedback voltage is given by, Vf=(R2/R2+Rf)*Vo Vf/Vo=R2/R2+Rf Vf/Vo=B=R2/R2+Rf . . . . . . . . . . . .(6) The closed loop gain (Af) is expressed as follows from Eq (4) Af=A(R2+Rf)/(R2+Rf+AR2) . . . . . . (7) Rearranging Eq (7),we get, Af=A/(1+AB) where B=R2/R2+Rf •

Input resistance with feedback

Input resistance of op-amp with feedback is the equivalent resistance which observed from the non-inverting input. Let Rf is the feedback resistance & Ri is the input resistance of op-amp.So the input resistance of op-amp with feedback is given by,



Rif=Vin/Iin . . . . . . . . . . . . . .(8)

where Iin=Vid/Ri Rif=Vin/(Vid/Ri) . . . . . . . . .(9) However, Vid=Vo/A & Vo=(A/1+AB)*Vin . . . . . .(10)

substitute the Eq (10) in Eq (9), we get, Rif=Vin/(Vo/ARi) =Ri*Vin/(Vo/A) =Ri*Vin/((A/1+AB)*(Vin/A)) =Ri*(1+AB) . . . . . . . . .. . . (11)

So from Eq (11), we conclude that the input resistance of non-inverting amplifier with feedback is (1+AB) times Ri. •

The output resistance (Ro) with feedback

The output resistance (Ro) wiyh feedback is the resistance that is observed backward from the output terminal. The analysis is done by using thevenin's theorem of circuit. In this the independant source is reduced to zero and apply external voltage Vo and current Iois calculated. Rof=Vo/Io . . .. . . . . . . .(12) By applying kirchoff's current law, Io=Ia+Ib . . . . . . . . .. . (13) since (Rf+R2)||Ri>>Ro & Ia>>Ib Therefore by applying kirchoff's voltage law for output loop, vo-RoIo-AVid=0 Io=Vo-AVid/Ro . . . . . . . .(14) However, Vid=V1-V2 Department of Electronics Engg., P.V.P.I.T., Budhgaon


where V1=0 V2=Vf therefore Vid=-Vf =-(R2/R2+Rf)*Vo =-BVo . . . . . . . . . . . . . (15)

substituting the value of Vid in Eq (14), we get, Io=Vo-A(-BVo)/Ro Io=Vo+ABVo/Ro . . . . . . . . (16)

substitute the value of Io from Eq (16) in Eq (12), Rof=Vo/(Vo+ABVo/Ro) =Ro/1+AB . . . . . . . . . . (17)

From Eq (17), we can conclude that the output resistance of voltage series feedback amplifier is 1/1+AB times Ro.

Non-inverting amplifier circuit using Proteus Components Required : Name

IC Multi battery

Description

Number required

IC 741

1

cell Multi battery(12V)

cell 2

RES

Resistor(10K)

3

PULSE

As Input

1

GND

Ground

1


of

components


Procedure using Proteus: Step 1:-Selection of components 1) Selection of OP-AMP:    

Activate component mode Go to pick components (click on ‗P‘) Type 741 in search section Select IC 741 Place the component in circuit

2) Selection of Resistors:    

Again go to pick component section And select three generic resistors Place these 3 resistors in circuit at required place Then go to edit properties by right clicking on that component Set each resistor to 10 kilo ohm

3) Selection of batteries:    

Again go to pick component section Then go to Miscellaneous And select multi cell batteries Place those batteries in circuit Choosing edit properties option set batteries to 12V

4) Select sine generator from generator section 5) Do the connections (wiring)as per circuit diagram 6) Add voltage probe at the 6 th and pin of operational amplifier (741) Step 2:-Simulation Graph Analysis:    

Go to graphs option and select analogue Drag that graph in circuit section Insert voltage probe in graph section Go to ‗add trace‘ and add the traces of input and output Right click and activate simulate graph option

Repeat the same steps for noise , frequency , distortion analysis.



Conclusion: we have studied operation of non-inverting amplifier. We have observed input and output waveforms for the given gain parameter. Also with the help of analogue, frequency, noise and distortion options and different graphs we have analysed the noninverting amplifier.







: 06


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Date Checked

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Title: Transient response analysis using RLC components.

Aim: Design a simple circuit using RLC Component to observe transient response using Proteus.

Objectives: •

To understand principal of working of Transient Response of RLC Circuit

•

To understand the circuit arrangement of Transient Response of RLC Circuit

•

To understand the procedure of Transient Response of RLC Circuit circuit using proteus.

•

To observe the simulation of circuit

Outcomes: •

Able to study Transient Response of RLC Circuit.

•

Able to understand the circuit arrangement of Transient Response of RLC Circuit.

•

Able to understand the procedure of Transient Response of RLC Circuit in proteus.

•

Able to observe the simulation of circuit.



•

To enable to innovate ,design and develop a variety of electronic and computer based components and system for applications including signal processing , communication , computer network and control system.



Introduction-

An RLC circuit (or LCR circuit or CRL circuit or RCL circuit) is an electrical circuit consisting of a resistor, an inductor, and a capacitor, connected in series or in parallel. The RLC part of the name is due to those letters being the usual electrical symbols for resistance, inductance and capacitance respectively. The circuit forms a harmonic oscillator for current and will resonate in a similar way as an LC circuit will. The main difference that the presence of the resistor makes is that any oscillation induced in the circuit will die away over time if it is not kept going by a source. This effect of the resistor is called damping. The presence of the resistance also reduces the peak resonant frequency somewhat. Some resistance is unavoidable in real circuits, even if a resistor is not specifically included as a component. A pure LC circuit is an ideal which really only exists in theory. There are many applications for this circuit. They are used in many different types of oscillator circuits. Another important application is for tuning, such as in radio receivers or television sets, where they are used to select a narrow range of frequencies from the ambient radio waves. In this role the circuit is often referred to as a tuned circuit. An RLC circuit can be used as a band-pass filter, band-stop filter, low-pass filter or high-pass filter. The tuning application, for instance, is an example of band-pass filtering. The RLC filter is described as a second-order circuit, meaning that any voltage or current in the circuit can be described by a second-order differential equation in circuit analysis Circuit diagram –



Circuit arrangement: The circuit arrangement of An RLC circuit as shown in fig. It uses one Resistor, Inductor and Capacitor connected in series or in parallel. Output is taken across the capacitor of 10uF.

Theory: Transient responsea transient response or natural response is the response of a system to a change from equilibrium. The transient response is not necessarily tied to "on/off" events but to any event that affects the equilibrium of the system. The impulse response and step response are transient responses to a specific input (an impulse and a step, respectively

Plot showing underdamped and overdamped responses of a series RLC circuit. The critical damping plot is the bold red curve. The plots are normalised for L=1, C=1 and The differential equation for the circuit solves in three different ways depending on the value of . These are underdamped ( ), overdamped ( ) and critically damped ( ). The differential equation has the characteristic equation. The roots of the equation in s are,[6]

The general solution of the differential equation is an exponential in either root or a linear superposition of both,



The coefficients A1 and A2 are determined by the boundary conditions of the specific problem being analysed. That is, they are set by the values of the currents and voltages in the circuit at the onset of the transient and the presumed value they will settle to after infinite time.[7] Properties-

Typical second order transient system properties Rise time Rise time refers to the time required for a signal to change from a specified low value to a specified high value. Typically, these values are 10% and 90% of the step height. Overshoot Overshoot is when a signal or function exceeds its target. It is often associated with ringing. Settling time Settling time is the time elapsed from the application of an ideal instantaneous step input to the time at which the output has entered and remained within a specified error band. Delay time The delay time is the time required for the response to reach half the final value the very first time. Peak time The peak time is the time required for the response to reach the first peak of the overshoot. RLC circuit using Proteus Components required: Name

Description

Number required

RES

Resistor

1

CAP

Capacitor

1

INDUC

Inductor

1


of

components


PULSE

As Input

1

GND

Ground

1

Procedure using Proteus: 1. From main page of Proteus, click on ‗P‘ to pick device from library. 2. In pick device, insert the component which has to be selected & click on ‗OK‘. 3. Place the all components on the Proteus screen. 4. Right click and select edit properties to change the values. 5. Add voltage probe at input and output. 6. Select the ‗graph mode‘ & choose analouge Analysis graph window. 7. Right click and select ‗add traces‘ (input & output trace). 8. Right click & activate simulation graph. 9. Observe and analysis input & output waveforms.

Circuit Stimulation Using Proteus

RESULT Department of Electronics Engg., P.V.P.I.T., Budhgaon


Conclusion: We understand the transient response using RLC circuit in proteus software. We observe the simulation of circuit & study the transient response.







: 07


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Date Performed

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Title: Simulation of Astable Multivibrator using transistor using Proteus.

Aim: Design & simulate the Astable Multivibrator using transistor.

Objectives: •

To understand principal of working of Astable Multivibrator.

•

To understand the circuit arrangement of Astable Multivibrator

•

To understand the procedure of Astable Multivibrator using transistor circuit using proteus.

•

To observe the waveforms and analyze the circuit

Outcomes: •

Able to study Astable Multivibrator using transistor.

•

Able to understand the circuit arrangement of Astable Multivibrator

•

Able to understand the procedure of Astable Multivibrator using transistor circuit using proteus.

•

Able to observe the simulation and analysis of circuit.



•


Introduction: An astable multivibrator is a regenerative circuit consisting of two amplifying stages connected in a positive feedback loop by two capacitive-resistive coupling networks. The amplifying elements may be junction or field-effect transistors, vacuum tubes,operational amplifier or other types of amplifier. The example diagram shows bipolar junction transistors.



The circuit is usually drawn in a symmetric form as a cross-coupled pair. Two output terminals can be defined at the active devices, which will have complementary states; one will have high voltage while the other has low voltage, (except during the brief transitions from one state to the other)

Circuit Diagram:

Theory: Operation: The circuit has two stable states that change alternatively with maximum transition rate because of the "accelerating" positive feedback. It is implemented by the coupling capacitors that instantly transfer voltage changes because the voltage across a capacitor cannot suddenly change. In each state, one transistor is switched on and the other is switched off. Accordingly, one fully charged capacitor discharges (reverse charges) slowly thus converting the time into an exponentially changing voltage. At the same time, the other empty capacitor quickly charges thus restoring its charge (the first capacitor acts as a timesetting capacitor and the second prepares to play this role in the next state). The circuit operation is based on the fact that the forward-biased base-emitter junction of the switchedon bipolar transistor can provide a path for the capacitor restoration State 1 (Q1 is switched on, Q2 is switched off): In the beginning, the capacitor C1 is fully charged (in the previous State 2) to the power supply voltage V with the polarity shown in Figure 1. Q1 is on and connects the lefthand positive plate of C1 to ground. As its right-hand negative plate is connected to Q2 base, a maximum negative voltage (-V) is applied to Q2 base that keeps Q2 firmly off. C1 begins discharging (reverse charging) via the high-value base resistor R2, so that the voltage of its Department of Electronics Engg., P.V.P.I.T., Budhgaon


right-hand plate (and at the base of Q2) is rising from below ground (-V) toward +V. As Q2 base-emitter junction is backward-biased, it does not conduct, so all the current from R2 goes into C1. Simultaneously, C2 that is fully discharged and even slightly charged to 0.6 V (in the previous State 2) quickly charges via the low-value collector resistor R4 and Q1 forwardbiased base-emitter junction (because R4 is less than R2, C2 charges faster than C1). Thus C2 restores its charge and prepares for the next State 2 when it will act as a time-setting capacitor. Q1 is firmly saturated in the beginning by the "forcing" C2 charging current added to R3 current; in the end, only R3 provides the needed input base current. The resistance R3 is chosen small enough to keep Q1 (not deeply) saturated after C2 is fully charged. When the voltage of C1 right-hand plate (Q2 base voltage) becomes positive and reaches 0.6 V, Q2 base-emitter junction begins diverting a part of R2 charging current. Q2 begins conducting and this starts the avalanche-like positive feedback process as follows. Q2 collector voltage begins falling; this change transfers through the fully charged C2 to Q1 base and Q1 begins cutting off. Its collector voltage begins rising; this change transfers back through the almost empty C1 to Q2 base and makes Q2 conduct more thus sustaining the initial input impact on Q2 base. Thus the initial input change circulates along the feedback loop and grows in an avalanche-like manner until finally Q1 switches off and Q2 switches on. The forward-biased Q2 base-emitter junction fixes the voltage of C1 right-hand plate at 0.6 V and does not allow it to continue rising toward +V. State 2 (Q1 is switched off, Q2 is switched on): Now, the capacitor C2 is fully charged (in the previous State 1) to the power supply voltage V with the polarity shown in Figure 1. Q2 is on and connects the right-hand positive plate of C2 to ground. As its left-hand negative plate is connected to Q1 base, a maximum negative voltage (-V) is applied to Q1 base that keeps Q1 firmly off. C2 begins discharging (reverse charging) via the high-value base resistor R3, so that the voltage of its left-hand plate (and at the base of Q1) is rising from below ground (-V) toward +V. Simultaneously, C1 that is fully discharged and even slightly charged to 0.6 V (in the previous State 1) quickly charges via the low-value collector resistor R1 and Q2 forward-biased base-emitter junction (because R1 is less than R3, C1 charges faster than C2). Thus C1 restores its charge and prepares for the next State 1 when it will act again as a time-setting capacitor...and so on... (the next explanations are a mirror copy of the second part of Step 1). Output pulse shape The output voltage has a shape that approximates a square waveform. It is considered below for the transistor Q1. During State 1, Q2 base-emitter junction is backward-biased and the capacitor C1 is "unhooked" from ground. The output voltage of the switched-on transistor Q1 changes rapidly from high to low since this low-resistive output is loaded by a high impedance load (the series connected capacitor C1 and the high-resistive base resistor R2). During State 2, Q2 base-emitter junction is forward-biased and the capacitor C1 is "hooked" to ground. The output voltage of the switched-off transistor Q1 changes exponentially from



low to high since this relatively high resistive output is loaded by a low impedance load (the capacitance C1). This is the output voltage of R1C1 integrating circuit. To approach the needed square waveform, the collector resistors have to be low resistance. The base resistors have to be low enough to make the transistors saturate in the end of the restoration (RB < β.RC).

Astable Multivibrator using transistor circuit using proteus.

Component:

Name

Description


RES

Resistor

4

CAP

Capacitor

2

VDC

Dc source

TRANS.

Transistor

2

GND

Ground

1


voltage 1


Procedure using Proteus: 1. From main page of Proteus, click on ‗P‘ to click device from library. 2. In pick device, insert the component which has to be selected & click on ‗OK‘. 3. Then place the all components on the Proteus screen. 4. Right clik the component properties to change the required value 5. Add voltage probe at input and output of diagram. 6. Select the graph mode & choose AC sweep analysis option 7. Right click in the graph window and elect add traces(input & output trace). 8. Right click & activate simulate graph 9. Observe input & output waveforms and analysis the results.

CIRCUIT SIMULATION USING PROTES

RESULT



Conclusion: We can understand the working of Astable Multivibrator . By using proteus we can observe the simulation of circuit & analyse the waveforms taken across capacitor C2. Astable multivibrator has no stable state. It generates square wave of predetermined frequency. Hence it is a square wave generator.







: 08


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Date Performed

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Date Checked

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Title: Simulation of Astable Multivibrator using 555 using Proteus.

Aim: Design & stimulate the Astable Multivibrator using 555.

Objectives: 

To understand principal of working of Astable Multivibrator.



To understand the circuit arrangement of Astable Multivibrator To study Astable Multivibrator using IC 555.

 

To understand the procedure of ―Astable Multivibrator using IC 555‖ circuit using proteus.



To observe the simulation of circuit.

Outcomes: 

Able to study Astable Multivibrator using IC 555.



Able to understand the circuit arrangement of Astable Multivibrator



Able to understand the procedure of ―Astable Multivibrator using IC 555‖ circuit using proteus.




Programme Education Objective (PEO) Satisfied : 2,3 







Introduction:

Astable Multivibrator can be designed by using 555 timer IC, Op Amps and also using transistors. The 555 IC provide accurate time delay from mille seconds to hours. The frequency of oscillation can be controlled manually by simple modification. It suitable for circuit designers with a relatively stable, cheap, and user-friendly integrated circuit for both monostableandAstableapplications. The 555 timer IC was first introduced around 1971 by the Signetics Corporation as the SE555/NE555. This is a simple 555 timer circuit project. Astable Multivibrator is simply an oscillator circuit that produces continuous pulses. The frequency can be controlled by changing the values of R1, R2 and C1. You can construct Astable multivibrator using transistors also, but the 555 circuit is comparatively simple. we have to design the Astable Multivibrator using proteus. It is easy to design on the proteus. The Astable multivibrator generates a square wave, the period of which is determined by the circuit external to IC 555. The Astable multivibrator does not require any external trigger to change the state of the output. Hence the name free running oscillator. The time during which the output is either high or low is determined by the two resistors and a capacitor which are externally connected to the 555 timer.



Circuit Diagram:

INTERNAL DIAGRAM OF IC 555



Circuit arrangement: The circuit arrangement of Astable Multivibrator as shown in fig. It uses one IC 555 as an Astable multivibrator. In this type of arrangement two resistors and two capacitors are used, the output is taken across pin no 3 of IC 555 as shown in figure.

Theory: The circuit diagram for the Astable multivibrator using IC 555 is shown here. The astable multivibrator generates a square wave, the period of which is determined by the circuit external to IC 555. The Astable multivibrator does not require any external trigger to change the state of the output. Hence the name free running oscillator. The time during which the output is either high or low is determined by the two resistors and a capacitor which are externally connected to the 555 timer. The above figure shows the 555 timer connected as an Astable multivibrator. Initially when the output is high capacitor C starts charging towards Vcc through RA and RB. However as soon as the voltage across the capacitor equals 2/3 Vcc , comparator1 triggers the flip-flop and the output switches to low state. Now capacitor C discharges through RB and the transistor Q1. When voltage across C equals 1/3 Vcc, comparator 2‘s output triggers the flip- flop and the output goes high. Then the cycle repeats. The capacitor is periodically charged and discharged between 2/3 Vcc and 1/3 Vcc respectively. The time during which the capacitor charges from 1/3 Vcc to 2/3 Vcc is equal to the time the output remains high and is given by Tc =0.693(RA+RB)C Where, RA and RB are in ohms and C is in Farads. Similarly the time during which the capacitor discharges from 2/3 Vcc to 1/3 Vcc is equal to the time the output is low and is given by td =0.693(RB) C

Thus the total time period of the output waveform is T=td+td=0.693(RA+2RB) C

Therefore the frequency of oscillation f=1/T=1.45/ (RA+2RB) C

The output frequency, f is independent of the supply voltage Vcc



Working of Astable Multivibrator: Principle of working: The Astable multivibrator generates a square wave, the period of which is determined by the circuit external to IC 555. The Astable multivibrator does not require any external trigger to change the state of the output. The time during which the output is either high or low is determined by the two resistors and a capacitor which are externally connected to the 555 timer. 

Consider the flip flop is initially cleared, when the power is switched on, then the output of inverter will be HIGH.



Now the capacitor C1 starts charging through R1 and R2. (Discharge transistor Q1 is OFF)



When the capacitor voltage exceeds 2/3 Vcc, the upper comparator output will be High, it Reset the control flip flop.



So the Q output of control flip flop will be LOW and Q‘ will be High. So the final output from Inverter is LOW



At the same time, the discharge transistor Q1 turns ON and the capacitor starts discharge through R2



When the capacitor voltage less than 1/3 Vcc, the lower comparator output will be high, then the control flip flop get set to High. (Q=1, Q‘=0, Final output=1)



Now the discharge transistor Q1 if OFF and then capacitor starts charging. This process continues.



The LED connected at the output will glows according to the output status.



4th pin is Reset pin, a Low voltage at this pin resets the IC. The Low signal is applied to the base terminal of reset transistor Q2. Then it turns ON followed by Discharge capacitor Q1 and capacitor discharges.



Charging

Discharging

Resetting

Advantages and Disadvantages of Astable Multivibrator:

(i)Advantages: 1. It is easy and quick to design an Astable using a 555 (also easy to troubleshoot). 2. The 555 timer can source or sink large amounts of current (some handle200mA). 3. It may require fewer devices or connections (1 IC, a cap and two resistors). 4. The 555 may have less frequency drift with temperature changes. 5. Some 555 timers have a wide input voltage range (3.5 to 18V). 555s have a large frequency range.

(ii)Disadvantages: 1.A disadvantage of the 555 is a poor rise and fall time which may cause problems for some edge-triggered devices. Department of Electronics Engg., P.V.P.I.T., Budhgaon


Astable Multivibrator using IC 555 circuit using proteus. Component:

Name

Description


RES

Resistor

2

CAP

Capacitor

2

555

Timer

1

VDC

Dc source

voltage 1

GND

Ground

3

Specifications: The most important characteristics which are required to be specified for a circuit are given below :

Supply voltage (VCC)

4.5 to 15 V

Supply current (VCC = +5 V) 3 to 6 mA Supply current (VCC = +15 V) 10 to 15 mA Output current (maximum)

200 mA

Maximum Power dissipation 600 mW Operating temperature

0 to 70 °C




CIRCUIT SIMULATION



Result

Conclusion: We can understand the working of Astable Multivibrator as well as the detailed circuit of IC 555.By using proteus we can observe the simulation of circuit & study the waveforms taken across pin no.3 of IC 555.







: 09


:

Date Performed

:

Date Checked

:




Title: Simulate simple RC circuit using AC sweep analysis. Aim: To simulate RC circuit (low pass filter) using Ac sweep analysis. Objectives: •

To understand principal of working of RC circuit.

•

To understand the circuit arrangement of RC circuit.

•

To study simple RC circuit using Ac sweep analysis.

•

To understand the procedure of simple RC circuit using Ac sweep circuit using proteus.

•

To observe the simulation of circuit.

Outcomes: •

Able to understand principal of working of simple RC circuit.

•

Able to understand the circuit arrangement of simple RC circuit.

•

Able to study simple RC circuit using Ac sweep analysis.

•

Able to understand the procedure of simple Rc circuit using Ac sweep analysis circuit using proteus.

•


Program Education Objective(s) (PEO) Satisfied: 2, 3 •


•

To enable to innovate, design and develop a variety of electronic and computer based components and system for applications including signal processing, communication, and computer network and control system.

Principle: An simple RC circuit in which output taken across capacitor ‘c’. At low frequencies capacitor will occur maximum reactance hence acts as an open circuit. Resulting in transferring low frequency signal to the output at high frequencies capacitive reactance is negligible hence it act as an short transfer to the output.



Circuit Diagram:

Circuit arrangement: The circuit arrangement of simple RC circuit as shown in fig. it consists of resister and capacitor. Output taken across capacitor as shown in fig. in these circuit Vr is voltage across resister and Vc is voltage across capacitor. Theory: A capacitor is a device for storing charge. The ability of a capacitor to hold a charge is measured by its capacitance C. For a capacitor, Q = CV, where Q is the charge on one of the capacitor plates, C is the capacitance of the capacitor, and V is the potential difference maintained across the capacitor plates. The unit of capacitance is the Farad (F), where one Farad equals one Coulomb per volt (1F = 1C/V). If a capacitor is connected to a battery, it will cause a charge +Q to develop on one plate and a charge -Q to develop on the other. If the battery is removed from the circuit the capacitor is connected to a resistor, then the capacitor will discharge through the resistor. The voltage across the resistor is given by V = IR, where I is the current through the resistor at a given time, and R is the resistance of the resistor. Since V = Q/C, = we can also write I = V/R = Q/RC As the capacitor discharges, Q becomes smaller, and I also become smaller. The current at any time t is given by: I = I0e-t/RC = I0e-t/τ Where I0 = initial value of the current, ti = time elapsed in seconds since the discharging began τ = RC = capacitive time constant for the RC circuit, and e = 2.71828... .

AC Sweep: In simple RC circuit we can apply AC sweep. For Ac sweep analysis is nothing but it is going to calculate to frequency response of the circuit. over range of frequencies the frequency response is nothing but the magnitude Vs frequency and phase Vs frequency. If Department of Electronics Engg., P.V.P.I.T., Budhgaon


the frequency is double it is called an octave. If the frequency is increased by the factor of 10 it is called as decay increase. The decade increase in frequency the magnitude changes by -20 and the slope of magnitude block is -20db/dc there are different type of sweep analysis. We have to consider following certain parameter in the frequency sweep. i.e. starting frequency and ending frequency. We have to specify the scale both linear, octave and decade. •

Linear sweep : The frequency is sweep linearly from the starting frequency to the ending frequency with a step (number of step) the next frequency generated by a constant to a present value.

•

Octave sweep : The frequency is sweep logarithmically by octave the next frequency is generated by octave the next frequency is generated by multiplying a present value by a constant the larger than the unity octal sweep is use when frequency range is wide.

•

Decade sweep : The frequency is sweep logarithmically by decade. Decade is used if frequency range is widest.

How to do the AC Sweep Analysis using Proteus Simulation Tool? Introduction: •

AC sweep analysis is nothing but a frequency response analysis.

•

It is a linear analysis (small-signal analysis).

•

It means it only considers the gain and phase response of the circuit; it does not limit voltages or currents.

•

The non linear devices (like transistor amplifier) must be liberalized to run the analysis. To do this conversion, we have to

a.

Compute the DC bias point for the circuit

b. Compute the complex impedance and/or transconductance values for each value for each device at this bias point. c.

Perform the linear circuit analysis at the frequencies of interest by using simplifying approximations.

•

The Proteus tool will take care of all these activities.ie,

•

The Proteus tool calculates the small-signal response of the circuit to a combination of inputs by transforming it around the bias point and treating it as a linear circuit.



Working of a Simple RC Circuit:

When the toggle switch is in the open position shown in the diagram, the capacitor is not connected the Electromotive Force, emf, and, unless the capacitor was previously charged, there will be no charges stored in the capacitor (i.e., q = 0) and the potential difference between the plates of the capacitor will correspondingly be zero as well. If the switch is toggled so that it connects the capacitor to the Electromotive Force, charges will tend to accumulate on the plates of the capacitor, + on one plate, _ on the other. This will continue until the accumulated charge creates a potential difference (Vc) between the two plates that is numerically equal to the electromotive force. That is, when: Vc = , Current flow through the connecting wires will cease (i.e., I = 0). When the switch is toggled to its alternate position, (i) the emf is bypassed, (ii) the two plates of the capacitor are connected, and (iii) the charges stored on the capacitor will tend to pass through the connecting wire to the opposite plate. In other words, the charged capacitor will discharge. There are 2 important things to remember about the electrical properties of R-C circuits: 1. When the capacitor (C) is fully charged, the following relationship holds: q = C Equation 1a Where q is the total amount of charge stored by the capacitor, C is the capacitance of the capacitor, and  is the electromotive force that is actually charging the capacitor. Note that q = 0 when the capacitor is discharged. Also note that the equation may be rearranged thus: q/C =  = Vc

Equation 1b

Meaning that placing excess + on one plate of the capacitor and excess – on the other plate will generate a potential difference between the plates. 2. Because of the presence of a resistance (R) in the circuit, current flow through the circuit is slowed. As a result, changing the amount of charge stored on the plates of the capacitor requires time. For example, if you were to start with a completely Department of Electronics Engg., P.V.P.I.T., Budhgaon


discharged capacitor (i.e., q = 0 and Vc = 0) and connect it to a battery, the charging of the capacitor would be described by an exponential equation, qt = C(1 e-t/RC)

Equation 2a

And if C = 1,  = 10 and R = 1, the graph of qt vs. time would look like this:

Simple RC Circuit using AC sweep circuit using proteus.

Component: Name

Components Used

Description


RES

RC05

Resistor

2

CAP

CAP

Capacitor

2

GND

SMB_SPL0

Ground

3



Procedure using Proteus: •

From main page of Proteus, click on ‗P‘ to click device from library.

•

In pick device, insert the component which has to be selected & click on ‗OK‘.

•

Place the all components on the Proteus screen.

•

Right click and select edit properties & select the proper value.

•

Add voltage probe at input and output of circuit.

•

Select the graph mode then add trace in graph window.

•

Right click & select add traces (input & output trace).

•

Right click & activate simulation graph

•

Observe the input & output waveforms.

CIRCUIT SIMULATION USING PROTEUS



Conclusion: We can conclude that ac sweep analysis is a linear analysis; it only considers the gain and phase response of the circuit. It gives linear frequency response. The frequency is swept linearly from the starting frequency to the ending frequency with a step the next frequency generated by a constant to a present value.







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Title: Fourier and Audio analysis using Proteus Aim: To study Fourier & Audio analysis using Proteus. Objectives:    

To understand principal of working of Fourier Transform To understand the circuit arrangement of Fourier Transform To understand the concept of Fourier Transform using proteus. To observe the simulation of circuit and analyze using Fourier Transform.

Outcomes:   

Students will be able to study Fourier Transform. Students will be able to understand the concept of Fourier Transform Students will be able to observe the simulation of circuit using Fourier Transform and its application.

Programme Educational Objective (PEO) Satisfied : 2,3 

To enable student to analyse, solve electronics engineering problem by applying basic principles of mathematics , sciences and engineering and also be able to use modern engineering techniques , skills and tools to fulfill social needs




Theory:Development of Fourier Transform From Fourier Series:The exponential form of Fourier series representation of a periodic signal is given by, x(t)=

+∞ 𝑗𝑛𝛺𝑡 −∞ 𝑐 n𝑒

....1

Where, 1

x(t)= 𝑇

𝑇

2 𝑗𝑛𝛺𝑡 −𝑇 𝑥(𝑡) 𝑒 2

d(t)


….2


In the Fourier representation using equation (1), the Cn for various values of n are the spectral components of the signal x(t), located at intervals of fundamental frequency 𝛺. Therefore the frequency spectrum is discrete in nature. The Fourier representation of a signal using equation (1) is applicable for periodic signals. For Fourier representation of non periodic signals, let us consider that the fundamental period tends to infinity, the fundamental frequency Ω tends to zero or becomes very small. Since fundamental frequency Ω is very small, the spectral components will lie very close to each other and so the frequency spectrum becomes continuous. In order obtain the Fourier representation of non -periodic signal let us consider that the fundamental frequency Ω0 is very small. Let, Ω0→ ∆ Ω0 . On replacing Ω0 by ∆ Ω0 in equation (1) we get, +∞ 𝑗𝑛 ∆𝛺𝑡 −∞ 𝑐 n𝑒

x(t)=

On substituting for Cn in the above equation (2). (By taking Ƭ as dummy variable for integration ) . we get, x(t)=

1

+∞ 𝑛=−∞.

[𝑇

𝑇 2 𝑗𝑛𝛺𝑡 −𝑇 𝑥(𝑡) 𝑒 2

d(t)] 𝑒 𝑗𝑛 ∆𝛺𝑡 ……(3).

We know that, 2𝜋

Ω0 =2𝜋f0=

𝑇

∴

;

1 𝑇

=

Ω

…(4)

2𝜋

Since, Ω0→ ∆ Ω0 .

1 𝑇

Ω

= 2𝜋

On substituting for +∞ 𝑛=−∞.

x(t)= 1

x(t)=2𝜋

1

[𝑇

+∞ 𝑛=−∞.

1 𝑇

from equation (4) in equation (3) we get ,

𝑇 2 𝑗𝑛𝛺𝑡 −𝑇 𝑥(𝑡) 𝑒 2 1

[𝑇

d(t)] 𝑒 𝑗𝑛 ∆𝛺𝑡


.

d(t)] 𝑒 𝑗𝑛 ∆𝛺𝑡 ∆ Ω.

Ω The fundamental period T tends to infinity. On letting limit T tends to infinity in the above equation we get,



x (t)= 𝑡 When T x (t)= x(t)=

𝐿𝑡

1

∞ 2𝜋

+∞ 𝑛=−∞.


1

[𝑇

∞;Σ ∫; ∆Ω

Ω.

+∞ +∞ ( −∞ 𝑥(Ƭ) 𝑒 −𝑗𝑛𝛺𝑡 −∞ +∞ 𝑥(𝑗 Ω) −∞

X(jΩ)=

+∞ 𝑥 −∞

d(t)] 𝑒 𝑗𝑛 ∆𝛺𝑡 ∆ .

dƬ ) 𝑒 𝑗𝑛𝛺𝑡 dΩ.

…..(5)

𝑒 𝑗𝑛 ∆𝛺𝑡 dΩ.

(𝑡)𝑒 −𝑗𝑛 ∞𝑡 dƬ

…(6)

The equation (6) Fourier transform of x (t) and equation (5) is inverse Fourier transform of x (t). Since the equation (6) extracts the frequency component of the signal , transformation using equation (6) is also called as analysis of the signal x (t).Since the equation (5) combines frequency components of the signal , the inverse transformation using equation (5) is also called as synthesis of the signal x (t). Definition of fourier transform :_Let, x (t) = Continuous time signal X (jΩ) = Fourier transform of x(t) The Fourier transform of continuous time signal, X(t) is defined as, X(jΩ)=

+∞ 𝑥 −∞

𝑡 𝑒 −𝑗 Ωt 𝐷𝑡

Also, x(jΩ) is denoted as F{x(t)} where ―F‖ is symbol used to denote the Fourier transform operation. 𝐹{𝑥(𝑡)} =

+∞ 𝑥(𝑡) 𝑒 −𝑗 Ωt dt −∞

Note:-Sometimes the Fourier Transform is expressed as a function of cylic frequency F, rather than radian frequency Ω .The Fourier transform as a function of cyclic frequency F, is defined as, +∞ X(jF)= −∞ 𝑥(𝑡) 𝑒 −𝑗 2FПt dt Condition for Existence of Fourier Transform :The Fourier transform of x(t) exists if it satisfies the following Dirichlet condition. 1.The x(t) be absolutely integrable . i.e.

+∞ 𝑥 −∞

𝑡 < ∞



2.The x(t) should have a finite number of Maxima and minima within any finite interval. 3. The x(t) can have a finite number of discontinuities within any interval. Definition of inverse Fourier Transform:The inverse fourier transform of X(j )is defined as, X(t)=F-1 X jΩ

1

= 2П

+∞ 𝑋 −∞

𝑗Ω 𝑒 𝑗 Ω𝑡 𝑑 Ω

The signals x(t) and X(j) are called Fourier Transform pair and can be expressed as shown below, x(t)

𝐹

𝑋(𝑗Ω)

Note:When Fourier transform is expressed as a function of cyclic frequency F,the inverse Fourier transform is defined as, +∞ X(t)=𝐹 −1 𝑋 𝑗Ω = −∞ 𝑥(𝑗𝐹) 𝑒 𝑗 2FПt dF Frequency spectrum using Fourier Transform:The X(jΩ) is a complex function of Ω .Hence it can be expressed as a sum of real part and imaginary part as given below. X(jΩ)=𝑋𝑟 𝑗Ω + 𝑗𝑋 𝑗Ω where , 𝑋𝑟 𝑗Ω = 𝑅𝑒𝑎𝑙 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑋(𝑗Ω) 𝑋𝑖 𝑗Ω = 𝐼𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑋(𝑗Ω) The magnitude of X(jΩ) is called Magnitude Spectrum. Magnitude Spectrum, 𝑋 𝑗Ω

=

𝑋𝑟 2 (𝑗Ω) + 𝑋𝑖 2 (𝑗Ω)

Or Magnitude spectrum , 𝑋(𝑗Ω) =

𝑋 𝑗Ω 𝑋 ∗ (𝑗Ω)

where , 𝑋 ∗ 𝑗Ω = 𝐶𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑜𝑓 𝑋(𝑗Ω) The phase of X(jΩ) is called Phase spectrum. Department of Electronics Engg., P.V.P.I.T., Budhgaon


𝑋 𝑗Ω

Phase spectrum=<𝑋 𝑗Ω = tan−1 𝑋 𝑖

𝑟

𝑗Ω

The magnitude spectrum will always have even symmetry and phase spectrum will have Odd symmetry .The magnitude and phase spectrum together called Frequency spectrum.

Procedure using Proteus: 1. From main page of Proteus, click on ‗P‘ to click device from library. 2. In pick device, insert the component which has to be selected & click on ‗OK‘. 3. Place the all components on the Proteus screen. 4. Right click the component properties to change the required value 5. Add voltage probe at input and output circuit points. 6. Select the graph mode & choose Fourier & Audio analysis option. 7. Right click in the graph window and select add traces (input & output trace). 8. Right click & activate simulate graphs. 9. Observe output waveforms and analysis the results.

Circuit Simulation



Fourier analsis:-

Audio analysis:-

Conclusion: We have understood the concept of Fourier Transform and audio analysis.We can observe from waveforms that Fourier Transform is useful to find out major spectral components from the given signals.







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Title: Study of Decade Counter using IC 7490 Aim: To study Decade Counter using IC-7490 Objectives:    

To understand principal of working of decade counter using IC 7490. To understand the circuit arrangement decade counter using IC 7490. To understand the procedure of ―decade counter using IC7490‖ using proteus. To observe the simulation of circuit.


Able to understand principal of working of decade counter using IC 7490 circuit. Able to understand the circuit arrangement of decade counter using IC7490 circuit. Able to understand the procedure of decade counter using IC7490 circuit using proteus. Able to observe the simulation of circuit.

Programme Educational Objective (PEO) Satisfied : 2,3 

To enable student to analyse, solve electronics engineering problem by applying basic principles of mathematics , sciences and engineering and also be able to use modern engineering techniques , skills and tools to fulfill social needs




Theory: A circuit used for counting the pulses is known as a Counter. Basically there are two types of counter: 1. Asynchronous counter (ripple counter) 2. Synchronous counter In case of asynchronous counter all the flip-flops are not clocked simultaneously, whereas ,in a Synchronous counter all the flip-flops are clocked simultaneously. A Ring counter and twisted ring counter is the examples of synchronous counter. Department of Electronics Engg., P.V.P.I.T., Budhgaon


Figure below shows the internal structure of 7490.

Fig1. Internal Structure of 7490 It consists of four flip-flop internally connected to provide a mod-2 and a mod-5 Counter. The mod-2 and mod-5counters can be used independently or in Combination. There are two reset inputs R1and R2 both of which are to be connected to logic 1level for clearing all the flip-flops. The two inputs S1 and s2 when Connected to logic 1 level, are used for setting the counter to 1001. The 7490 integrated circuit counts the number of pulses arriving at its input.The number of pulses counted (up to 9) appears in binary form on four pins of the ic.When the tenth pulse arrives at the input, the binary output is reset to zero (0000) and a single pulse appears at another output pin. So for ten pulses in there is one pulse out of this pin. The 7490 therefore divides the frequency of the input by ten.If this pulse is applied to the input of a second 7490 then this second ic will count the pulses from the first ic. It will give one pulse out after 100 pulses have been applied to the first ic.The 7490 can be connected to divide by other values. Pin Diagram

Circuit Diagram of Decade Counter



Decade Counter Truth Table Clock Count 1 2 3 4 5 6 7 8 9 10 11

Output bit Pattern Decimal Value QD QC QB QA 0 0 0 0 0 0 0 0 1 1 0 0 1 0 2 0 0 1 1 3 0 1 0 0 4 0 1 0 1 5 0 1 1 0 6 0 1 1 1 7 1 0 0 0 8 1 0 0 1 9 Counter Resets its Outputs back to Zero



Timing diagram of decade counter

Advantages of decade counter 1.

Decade counter are easier to design.

2.

With all clock inputs wired together there is no inherent propagation delay.

3.

Overall faster operation may be achieved compared to any counters

4.

Decade counter are faster and more reliable as they use the same clock signal for all flipflops.

Disadvantages of decade Counters: 1.

An extra "re-synchronizing" output flip-flop may be required.

2.

To count a truncated sequence not equal to 2n, extra feedback logic is required.

3.

Counting a large number of bits, propagation delay by successive stages may become undesirably large.

4.

This delay gives them the nickname of "Propagation Counters".

5.

Counting errors occur at high clocking frequencies. Components:- IC 7490, IC 7447,resistors.




Decade Counter using proteus

Result



Conclusion: We understand that the 7490 integrated circuit counts the number of pulses arriving at its input. The number of pulses counted (up to 9) appears in binary form on four pins of the IC When the tenth pulse arrives at the input, the binary output is reset to zero (0000) and a single pulse appears at another output pin.


Proteus lab Manual.pdf

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