EEELab Manual 375

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Basic aws of Electrical Engineering TheoryIntroduction to Network TheorySinusoidal Steady State ResponseUsing Laplace Transform for ime Domain AnalysisIntroduction Operational Amplifiers

Rutgers The State University of ew JerseySchool of ngineeringepartment of Electrical and omputer Engineering

Procedure Manual forElements of Electrical Engineering Laboratory

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B i 6RUTGERS UNIVERSITY

The State University of New JerseyCollege of EngineeringDepartment of Electrical and Computer Engineering

330:375 Elements of Electrical Engineering LaboratoryFxperiment Basic Laws of Electrical Engineering Theory

The purpose of this first lab is to familiarize yourself with some of theequipment you will be using throughout the remainder of the semester, and tointroduce you to the basic laws which govern electrical engineering theory1 1 Voltage and Current Relations in Unear and Nonlinear Passive Elements

Passive elements resistors, inductors, and capacitors are those whichcannot generate electrical energy. In a passive linear element, such as aresistor, voltage and current are directly and linearly related. The relationshipbetween voltage and current in a resistor may be expressed by Ohms Lawwhere:

V=IR.Only one reading of voltage and current need be taken to graph the V/Icharacteristics of a passive linear element. With a nonlinear passive element anumber of voltage and current readings must be made in order to graph thenonlinear V/I characteristics.

The nonlinear passive element that Will be studied in this lab is a diodeA diode acts like a voltage controlld switch When the positive voltage dropacross the diode, known as the forward voltage, is greater than somethreshold voltage, the diode has zero resistance and will conduct current Atvoltages less than this threshold voltage the diode acts as an open circuit withinfinite resistance, and no current will flow


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1.2 oltage Divider RuleThe voltage dMder rul is based upon the fundamental laws developed

by Gustav Klrchholf in 1848 and by George Simon Ohm in the early 1800s.From Kirchhoffs laws we know that resistors o r more generally, elements insenes carry the same current and that voltage drop is t same across parallelelements. According to Kirchholfs laws the current In figure must flowthrough both series rslstnces Ri and R2.

Figure ording to Ohms law Vj

We are able to find an expression forVoiR2.Using Ohms law we are also able to express in terms of the input sourcevoltage, V:

Ri.R2Putting these two equations together, weobtain the voltage divider rule namelythat: V R2 Ri R2 R2 Ri R2 V.This states that the voltage appeanng across one series resistor R2 is equal to afraction of the input source voltage, where that fraction is equal to the ratio of R2to the sum of Ri and R2.

VRI

R2

0


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Figure 2For a morecompIated network, -such as the one in figure 2, we still may

apply the volt ge dMder rule to find V In this casewe must first find theequivalent resistance Req between and B across which ppersRepV_RReqV

R R2and R1+R2The voltage divider princ iple is a simple way to find the voltage

appea ring across some resistor within a circuit.1.3 Current Divider Rule

When cu rrent reaches a node at two parallel resistances the cu rren t willsplit according to the ratio of the two resistance values so that the voltage dropsacross the two parallel elements are equal, referring to figure 3:VI V2 V

122IReq. R1 R2

ii

RVI RIV2Figure


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To write an equation for and another for i2 we can see from the last equationthat: R2

R1+R2 R2Ri

Ri R2These equations represent the current divide r rule for parallel resistances., wecan apply th is concept to more difficult circu its such as in figure 2. We must findan equivalent resistance between the terminals A and B then we n apply thecurrent divider rule to find: Rl Ri R2

Ri12iR1+R2

The current divider rule provides a means with which we are able todetermine the individual currents in a circuit.1.4 Storage Devices

Capacitors are based on electric field phenomena which is developedwhen a voltage drop or net charge is separated by the distance between thetwo plates filled with some dielectric material. When a time varying voltagesuch as an AC signal is applied across the tw o plates of the capacitor a tim evariant electric field develops across the plates. This time variant efleld causesa displacement current to flow between the capacitor plates. Displacementcurrent is not a current flow made up of electrons like that which passes th rougha resistor. Displacement current is more closely related to the displacement ofcharge rather than the movement of electrons. Capacitance current is afunction of the rate of change of the voltage across the plates, ordViwhere current found by this equation is measured in amps, C in farads, V involts and t in seconds. From this equation, it should be evident that when aconstant voltage such as a DC voltage is applied across the capacitor, nocurrent will flow. n other words the capacitor will act as an open circuit to aconstant DC voltage. Another important deta il derived from this equation is thatthe voltage may not change instantaneously across a capacitor. This would


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produce an infinite corresponding to n in finite current through thecapacitor which is not physically possible.

From the differential eqution for the cpcttlve current an equationmay be derived to describe the voltage cross capacitor namely that:

vt dtThe unit of cpcitnce is the fr F. Since farad capacitor would

be about the size of Hill Center most practical applications use capacitors in thepicofarad 10 -12 to ILicrofarad 10 6 range.

2. Prelab CalculationsComplete these calculations before you come to perform the first lab.

The results of these calculations will be collected before the lab begins so besure to enter your results on the dt sheets provided.

a Using the voltage divider rule, determine the voltage cross the resistor the 330 resistor and the output voltage appearing cross terminals A-B from figure 5.

b Using the current divider rule , determine the currents i nd i2 from

figure 5.Enter your reSults from part a and b onto the dt sheets provided.

Show all the calculations on seprte ppr which is to be turned in at thest t of class.


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3. ExpenmentThe purpose of this experiment is to demonstrate a few of the

fundamental laws of electrical engineering, while being introduced to somestandard pieces of laboratory equipment.Equipment:

Variable AC/DC Power SupplyOscilloscopeDigital MultimeterResistor BlocksDecade R-C SubstituterDiode1/4 Watt, 5 Resistors

3.1 Using the DC Power SupplyThe DC power supply delivers variable negative voltage levels lrom -20

V to V and variable positive voltages from 0V to +20 V. The negative supplyis delivered between the green terminal and the black terminal markedcommon and the positive supply is from the black common terminal to the redterminal. You may vary these output voltages with the knobs above theterminals marked by the V or the +V. The dials voltage readings on theseknobs are not very accurate . There is a third separate supply below the 20 Vterminals with the white and red terminals. This supplies only a single constantpositive 5 V output

To get an output from the supply, both the black power switch markedPWR on the left side of the supply and the square white button markedOUTPUT on the DC supply must be on. Again, dont trust the voltage settingon the variable voltage knobs you should always check your actual outputvoltage with a digital voltmeter to insure that you use the correct power duringthe experiment.

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32 VI Characteristics of Linear Element Construct the circuit from figure a using an unknown resistance R value

Measure 5 poInts on the V curve by varying the input voltage magnitudeand recording the corresponding

voltage n current measurementsDetermine the value of the unknown resistance R from the slope of the Vplot.

Measure the value of R with the multimeter Make sure the power deliveredby the source is off while measuring resistance

V

V

Ammeter

IN Characteristics of Nonlinear ElementConstruct the circuit shown in

figure b where the band on the diode is thenegative - terminal shown in the figure. Measure 10 points on the INcurve for both positive and negative voltages Make sure to record

number of voltage and current readings near the transition region where

the diode just starts to conduct current Be careful not to exceed 15mA

through the diode

Figure a

Figure 4b


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3.4 Volta;. DMder/ Current Divider Circuit the output of the DC power supply to 10 V Confirm the magnitudeof the output voltag i with the digital voltmeter then shut the power supply off

Construct the circuit shown in figure 5 usIng the resistance values given. Afteryouve checked your circu it configuration turn the power back on

FIgure 5 Measure the voltage Vo appearing across terminals AS with the voltmeter. Toread a voltage across some element with the meter you must have theleads of the voltmeter in para llel with that element. Record the voltageappearing across the 220 resistor and across the 330 fl resistor Measure the currents i ii and using the digita l ammeter. To read currentthrough some element with the multimeter the meter must be in series withthat element. How closely do your experimental results agree with thosefound theoretically?

3.5 Capacitance CircuitThe ac supply is the supply to the far right of the power supply. t is able

to deliver up to 13 V at a frequency as high as MHz with output waveshapesof a sine square or triangular wave. You must use a special connection boxthat enables you to see various waveforms on the oscilloscope This box alsoallows you to use standard wire leads to construct a circuit even though theoutput connection from the ac supply is a BNC type connector. This box alsoallows you to connect an element in your circuit to the oscilloscope in order toview its output voltage waveform. Your instructor will describe how to use theN onn tion box correctly


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Construct the circuit from figure Using the ac supply at about Hz

as the input signal verify using the oscilloscope that the output waveform

across the capacitor is the integral of the input signal for the following:

Square wave simulates a constant voltage DC at a low signal frequency

Sine wave riangular wave simulates a linear signal at a low sig

nal frequency

Draw the input and output waveforms on the graph provided on thedata sheets

Ii

Figure 6

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4. Data Sheets

iner Element

V voltsmililamp

Nonlinear ElementV voltsmiulamp

V voltsmilliamp

Voltage Divider Volts Volts o VoltsTheoretical

ExperimentalPercent Error:

urrent Divider milliarnps i milliamps milliampsTheoreticalExperimentalPercent Error:


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apacitanceCircuit

Constaflt DCi input square wave

input

output

jpe wavR input

input

output

jpar inout triangular wave

input

ouput

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5. ReportEach oerson is to submit the follow ing :5.1 A one page summary of the entire lab which reviews what was learned inthis experiment.As a grouo submit the fo llow ing :5.2 Data sheets from the experiment, which include theoretical andexperimental results.5.3 Graph of the data from 32 plot the voltage as a function of currentDetermine the value of the unknown resistance from the slope of the V/I curveCalculate the percent error of your experimental results, and enter these resultson the data sheets.

Graph of the data from 3.3 plot the urr nt as a function of voltage for thenonlinear device.5.4 Prove the theoretical capacitor voltage waveshape (magnitude A,frequency rom the r u t of figure 6 for the three input signals (constantsinusoid, and linear inputs using:

vt dtwhere the u n tion of is identical in o rm tothat of the input voltage signal fromthe supply5.5 Submit completed prelab calculations for lab showing all of your work.Enter your theoretical results into the lab 1 data sheets.NOTE:The final r port must be completed. assembled. and stapled together beforecoming to p r orm the nxt lab. The completed report should include:

An individual summary for each group member. One opy of the prefab calculations for the entire groupOne copy of any graphs, proofs etc. for the entire group

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B-127RUTGERS UNIVERSITY

The State University of New JerseyCollege of Engineenng

Department of Electrical and Computer Engineering33:375 Elem ents of Electrical Engineer ing Laboratory

Experiment 2 Introduction to Network Theory

There exist a number of techniques which enable us to analyze anycomplicated network of resistances voltage supplies and current sources andto reduce the network into a more simple yet accurate representation. n thisexperimentwe will study four such techniques of network reduction

11 Theory of SuperpositionWhen there is more than one independent source of energy such as

voltage or current sources in a network one often uses the technique ofsuperposition to reduce the circuit Superposition theory states that a totalresponse may be determined by finding the response to each source separatelyand then summing the individual responses. This is accomplished by removingall but one of the sources in a network and finding the currents produced by thatone source alone within the network. This is continued until the currentcontributions produced have been found in response to each individual sourcealone. These currents are then summed together to find the total current in theoriginal network. Although the use of superposition is limited to resistivenetwork application n this experiment the theory of superposition may be usedto analyze any linear system including networks which contain capacitanceand inductance.

n the example given below in figure there are two independent energysources one voltage supply and one current source. First we remove all thesources except one. To remove a current source from the network t must beopen circuited. To remove a voltage source t must be short circuited Letsfirst remove the current source as in figure 2 and find the currents in the networkproduced by the voltage supply alone.


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Figure 2There are a num r of different methods to find the currents ii, I2 13

and 14 In this method, we first find the equivalen t resistance Req of the entirene twork to find ii, where the symbol will be used to indicate a parallelcombination between elements:

Req 4 2 3 n +62Req.8n12 V .15AFrom figure 2, one can see that all th current passing through the

resistance must continue into the 4 resistor current th rough seriesresistances are equal , or:i3=i4Therefore we can write an equation for 2 and using th cu rren t divider rule

2n+4n ii2n4=bOAand

12A

Figure

12 V

I 20 V 3 open circuit


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3ii3 4 Q 4 5ANow we must remove the voltage source and analyze the network with

the current source alone as seen in figure 3.

VReq 12 A 2 VReq 24 V

Since Req is in paral lel to theiden tical where:

V42 VReq -24 V-24V =-6A

2Q

2A

From figure 3 it should be ev ident that the following expression is true: j 12 A

i3 6 ASince the 6 and the 3 resistors are in parallel they are at the samepoten tial V but at opposite polarities see Figure 3.

il=i-V V6 +6AV--12V

Therefore replacing the value of 12 V into the equations for individual currents- -12V =2A

Figure 3Req 6 fl 2 ll4Req 2

Req 2A

Figure 4

4 resistor, the voltage across them are

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vAs an example we will

section on superposition.

12V2 -4A.To complete the superposition technique one now must sum the individualcon tributions of currents from each energy source considered separately.

ii il+il.15+2.17Ai2i2+i2.1O-4.6A13.13+i3.5+6.liA4i4+i4.5+6 -1ASuperposition is a useful technique for network analysis. It enables us to

not only determine the current in each resistor in the network but also thevoltage across each element and the power dissipated within each resistancewith only a bit more math.1.2. Theory of Mesh Current Analysis

Yet another technique of finding the currents within a network is by meshcurrent analysis. A mesh current is defined as the current which flows onlywithin the perimeter of a particular window or mesh of the network. Thesewindows or meshes are identified by the dotted lines in figure 5. This particularexample has 3 meshes. The mesh currents are identified as Ii 12 and 13

Mesh current analysis is based upon Kirchhoffs voltage law, whichstates that the algebraic sum of all the voltages around any closed path in acircuit equals zero, or that

analyze the same ne twork as was done in the

A

6a

120 V

Figure 5


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Lets sum the voltage drops around loop 1:-120 V li 6 I1-12 3 0 Mesh Equation

The direction of current flow In the mesh determines the polarity of the voltagedrop across an element. or a positive voltage crop current must flow from ahigh potential + to a lower potential - . In the voltage supply in figure 5,current traveling clockwise f lows from a low - to a high potential + , thereforethe voltage difference across the supply is a negative voltage drop. Torepresent this negative voltage drop the potential difference is written theqution as a -120 V.Summing the voltage drops around loop 2:

l2-l1 3 l2 2 I2-l3 41 0 Mesh Equation 2The current 3 in mesh 3 Is equal to the current supplied by the source, or:

3 12 A. Mesh Equation 3By combining .like terms in mesh equations and 2, and replacing the solutionfor 3 it can be found that the mesh currents are equal to :

Ii 17A12=11A 3 12 A

Using the notation from figure to find the individual currents it is found that=11 =17A

i2=I1 -12=6Ai3 12=11 Ai4=l2 -l3 -1A

These exactly match those currents found by the method of superposition.

1.3. Nodal nalysis TheoryA third method commonly used to analyze a network is referred to as

nodal analysis. This technique is based on another of Kirchhoffs basic laws Kirchhoffs current law, which states that the algebraic sum of all the currents atany node a circuit equals zero. Using the circuit shown in figure 1, we will

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find the voltage drops across the four resistances using nodal analysis. A

12V 12A

Figure 6n figure 6 the two nodes are labeled as node A and node B. As drawn on thefigure there are three currents that enter node A and also three currents thatenter into node B. From KirchhoflS law the sum of all the currents into a nodemust equal zero so for the two nodes:

ia ib ic=0 NodeAid if =0 Node B.We now must wrt equations for these currents in terms of the. voltage at

the node where Va is the voltage difference from node A to ground and Vb thevoltage difference from node B to ground120Vaa.

Vaib ybVa

2 Q VaVbid Vb

4if=12A

Replacing these currents into the two nodal equations we find thatVa=18VVb=4V

Replacing the known voltages into the node current equations it is found that:ia 17 Alb 6 AIc d 11 Aie=1Aif=12A

Therefore the branch currents as drawn in figure would equal


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iiia17A ib 6 A3idil Ai4-ie.-1 AThese currents are identical to those currents found with the other three

methods of network nlysis1.4. Thevenin quivalents

If we are interested in finding the current delivered to one particularelement in a network c alled the load resistance RL sometimes it is best torepresent the circuit equivalently as a single voltage source in series with tworesistances depicted in figure 7.

RIRL

Voc

The thevenin voltge source Voc is equal to the volt ge which appearsacross the term inls of the open circuited load resistance RL. The resistance is the equivalent resistance of the entire network that appears across theterminals of the open circuited load resistance when the energy sources areremoved. As an example we will again use the network from figure 1.Choosing the lod resistance to be the 2 Q resistor we open circuit the loadand find Voc appearing across the terminals marked A nd in figure 8.

A

120V 4fl 12A Figure 8

Figure 7


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Using Kirchhoffs volt ge law the sum of the voltages in the center loop equalszero,

V 3c2 Voc V4c.oTo find an expression for we will sum the voltages around the l ft

loop. Since there is an open circuit where current cn not flow between pointsA and B all the current Ii flowing th rough the 6 fl resistor will continue downinto the 3 resistor to ground.

120 V+li 6c1+l 3.0Ii. AV3fll1 3.40V

The current flowing in the right loop as drawn by the dotted arrow on figure 8 isequal to -12 A. he voltage across the 4 fl resistor is then:

V4.-12A 4a-48VTherefore the open circuit volt ge is equal to:

VoC=V3-V4188V.To find the equivalent resistance of the network appearing across

terminals A B we remove the sources and keep the load resistance opencircuited

C

Figure 9From figure 9 is easy to calcu late the equivlent thevenin resistance Rj asRI=6II31+4=6ep lacing the clcultedvalues for and Rj into figure 7 is seen that thecurrent passing into the 2 load resistance, is equal to0 881= R j RL =6 2 =1

This may also be drawn as:


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Again we find that the current through the 2 Q resistance is 11 A which exactlymatches the current found with the other techniques descnbed earlier

2 Prelab CalculationsInclude these calculations along with your group report for experiment 2These calculations ut be completed before you begin this lab .

2 1 SuperposItionUsing the technique of superposition calculate the voltages across each

resistor and currents through each resistor for the network given in figure 10

2 2 esh CurrentUsing the technique of mesh current calculate the three mesh currentsfor the network in figure 10

2 3 Node Voltagesing nodal analysis find the voltage from node A to grbund and the

voltage from node B to ground for the circuit in figure 10

24 TheveninUsing standard thevenin equivalents find and RI if the 510 resistor

is considered to be the load resistance


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c. Connect both Ei and E2 and measure the voltage appearing acrosseach resistance due to the combined effects of the voltage sources Howclosely do these voltages compare to those found by superposition?32 Mesh Currents

Measunng current is more difficult than measuring voltages. To measurecurrent flowing into a resistor you must place the digital ammeter in series withthe resistor. N ote: When finding voltage you place the voltmeter in pa ral lelwith the element you are studying . Therefore to find the current flowing in the220 resistor you must reconfigure the circuit to place the ammeter in s6riesbetween the voltage supply and the resistor. The left-most mesh current will bethe current in the 220 fl resistor. The center mesh current will be the currentthrough the 510 resistor The right-most mesh current will be the currentwhich passes through the 820 resistor. Using the circuit in figure 10 measurethese three mesh currents using the ammeter How closely do they match thetheoretical mesh currents?

Using the circu it in figure 10 measure the w node voltages as referredto ground. The voltage Va will appear from node A to ground and Vb willappear from node B to ground. How closely do these match your predictednode voltages?

3 4 Thevenina Agin using the circuit in figu re 10 find the open circuit voltage acrossthe load resistance of 510 by measuring the voltage which appears acrossthe open circu it terminals where the load resistance used to be. Then removethe sources short El and E2 and measure with the digital ohmmeter theequivalent thevenin resistance that appears across the open circuited terminalsof RL. How well does this compare with what you calculated?

b. Now replace the load resistance of 510 with the decade resistorbox. At the lower resistance values take measurements qu ick ly to avoidcausing any damage to the resistance box due to prolonged exposure toexcessive amounts current. Vary the resistance from 10 to 10 k2 whilemeasuring the voltage across RL. Simultaneously calculate the powerdelivered to the load at each resistance value

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4 ata Sheets

Superposition V Oc V33 fl V p V75 fl V8 flEl aloneTheoreticalExperimentalPercent Error:E aloneTheoreticalExperimentalPercent Error:Ei and ETheoreticalExperimen talPercent Error:

Mesh urrents Ii 2 3Thea retical:ExperimentalPercent Error:

Node Voltages Va VbTheoreticalExperimentalPercent Error:

Thevenin Equivalents RTheoreticalExperimentalPercent Error:


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Thevenin Equivalents and the Maximum Power delivered to a Load Resistance

5. Reportach person is to submit the following:5.1 A one page summary of the entire lab which reviews what was learned inthis experimentAs a group subm it the following:5.2 Data sheets from the experiment, which include theoretical andexperimental results5.3 Graph of the data from 3 4 b on one graph plot PL, IL and VL as afunction of RL. Determine the value of RL where PL is maximum . Note: It maybe best to use three differen t yaxes scalings for the plot of t IL and VL.

5.4 Draw the thevenin equivalent circuit for the network studied in theexper im ent using experimen tal values obtained for .Ioc and Rj.

5.5 Prove that power delivered to the load resistance from the theveninequivalent circuit is a maximum when RL R. Start with the basic definitionthat:

IL) RL

6 Submit completed prelab calculations for lab 2 showing all of your work.Enter your theo retical results into the lab 2 data sheets.


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B- 128 RUTGERS UNIVERSYThe State University of New JerseyCollege of Eng ineeringDepartment of Electrical and Computer Engineer ing33 :375 Elements of Electrical Eng ineering LaboratoryExpe riment 3 Sinusoidal Steady State Response

This lab demonstrates techniques of circuit analysis using phasorrepresentation. The phenomenon of resonance in a series RLC circuit willalso be investigated

1. Phasor Representation of a Sinusoidal SignsPhasor representation enables us to analyze a circuit in which theinput signal is sinusoidal, written as:

vrVmSifl wt+eor sin wt 0where the Vm or lm is the amplitude of the sinusoidal Wave, w is theangular frequency of the sinuso id in radians/sec, and a is the phase shift ofthe sinusoid in degrees. Figure provides an example of how to developsuch an equat ion for any sinuso idal signal . This particular general form ofthe equation given above is referred to as the instantaneous voltage orcurren t equa tion.

wt radians

VtVmiVm2Vrri

vim2 sin wt

Figure

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t is often important to compare the ability of a time vary ing voltage orcurrent to deliver energy to a resistor with the ability of a constant or dcvoltage or current to deliver energy to a resistor To make th is comparisonwe must use the effective or mis value of the sinuso id The effective valueof voltage Is derived from the definition

VeffFor a sinusoidal input voltage v the effective value reduces toVeff .fl 707 Vmwhich is equal to the amplitude of the sinusoidai wave divided by Y

By conven tion phasor notation uses the effective value to represent asinusoidal wave which is then expressed as a capital letter. The phasorrepresentation of the sinusoidal voltage signal would then be:veThe sinusoidal current would be equal to:

Phasor representation gives a simple way to analyze a circuits sinusoidalsteady state response without complicated mathematics.1.2 Phasor Represent?tion of Circuit ElementsThe impedance of a capacitor may be represented as:

zc=j=To represent th is impedance as a phasor it is important to first understandthe role of the imaginary term The imaginary term is equal to f andit gives us information about the phase shift produced by an element Apositive corresponds to a phase shift of positive 900 and a negative corresponds to a phase shift of negative 900 Therefore the impedance of acapacitor may be represented in phasor no tation as:

ZC=td 900wCThe impedance of an inductor is written as:

jwL


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nphasor notation, t s is equal to:

W Since resistance is purely positive and real because there is no term thephase shift associated with a resistor is 0 This is represented in phasornotation as:

ZRRA phasor diagram of these three elements is presented in figure 2.

The y axis is the imaginary axis, and the x x s is the real axis.Imaginary Axis

wL

R r-eai xts1/ wC

Figure A phasor diagram like the example in figure 2, shows the impedance ofeach element as a vector and these phasors follow the same rules asvectors in the way that they add and subtract. Phasor diagrams are alsoused to show phase and mgnitude relations between the elements in acircuit. Although the example used here was the phasor impedancediagram you may also draw phasor diagrams for voltages or currents for theindividual elements.


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1.3 Polar to Rectangular and BackTo change from phasor representation in polar form to that of a

rectangular form Is performed as:Y.x.o

Y cose singYou will no tice that in the rectangular form the expression is of two separateparts a real term and an Imaginary term n the rectangular form of theexpression, you may add and subtract a number of terms together by addingreal terms together to find the to tal real term and then adding all theimaginary terms together to obtain one complete imaginary term

To change from rectangular to polar phasor notation you do thefollowing:

YAjBMagnitude ofYYATo perform any multiplication or division of complex terms, each term mustbe in polar notation When perform ing multiplication the magnitudes ofeach term get multiplied togetherand the angles of each term sum together.

Acz Bp ABa+f3To perform division divide the magnitudes of the two terms, and subtract theangle of the denominator from the angle of the numerator:

B4 B

1.4 Series RL and RC NetworksOnce a capacitance or inductance is in its complex impedance form it

will follow the same rules as does a resistor in terms of series and parallelcircuit reduc tion namely elements in series ll add together, and etementsin parallel will sum as: owever keep in mind that addition/subtraction must be performed on

variables in rectangular form and that real terms and imaginary terms mustremain separate.

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15Series RLC ResonanceSince capacitors and inductors have impedances that are dependent

on angular frequency you might also guess that the voltages across theseelements are also frequency dependent. Therefore in a RLC circuit therewill be a frequency at which the output voltage will be in phase w

ith the inputvoltage At this frequency known as the resonance frequency there will beno phase shift between the input and output voltages. At resonance themagnitude of the voltage across the capacitor exactly matches themagnitude of the voltage across the inductor and since these elements are1800 out of phase on the imaginary axis these two voltages w cancel oneanother. The only voltage drop in the series RLC circuit at resonance woccur across the resistor and therefore the output will be at its greatestmagnitude at resonance. Since a resistance will not produce any phasechange at the output the output voltage will be in phase with the inputsignal. At any other frequency there is a voltage drop not only across theresistor but also across the inductor and capacitor as well and the outputvoltage magnitude is less than at resonance and there will be somewhat ofa phase shift between input and ou tput.

For a series RLC circuit the resonance frequency ceo can bedetermined by the equation:

A typical graph of the output voltage as a function of angularfrequency is given in figures 3a and 3b. To describe this curve it isimportant to locate the resonance frequency to find out informationregard ing power delivery and also to determine the sharpness of the curvesurrounding resonance compare figures 3a and 3b . For a series RLCcircuit frequency range of useful output signal amplitude called thebandwidth is defined as the range of frequency for which the powerdelivered to R is greater than or equal to times the power delivered to R atresonance P. By virtue of the shape of the frequency response curvethere are two frequencies at which the power delivered to R is equal to

These two points are known as the half power points. These pointsmay also be found using output voltage relations where the halt power


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points are located by the two points where the output voltage is equ l to:VV out

where V is equal to the maximum voltage at resonance.he b ndwidth is the angular frequency range between the upperand lower half power points which can also be found by :

As you can see, the bandwidth of a series RLC circuit depends only on theratio of R to L The resonance frequency of a series RLC circuit dependsonly on the values selected for and C

The relative sharpness of a frequency response is determined by avalue known as the quality f ctor 0 , equal to:

Wbw

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I;

I andwidth

l

cal 00 w2Figure

Vm

.7O7Vm

V Volts

Vm

.7O7Vm

V Voltsandwidth

ci i O02 i

Figure


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2. Prelab CalculationsComplete these calculations and submit your results along with your

lab write up for experiment 3. These must be completed before beginningthis lab.2.1 RC Network

Refer to figure 4 below. Compute values for R and C such that thetotal series impedance 2 5 36.1 a when 10 kHz. Record the valueson the data sheet provided for lab 3

R

C

VRFigure 4

22 RL NetworkRefer to figure 5 below. Compute the value for R when L 5 mH suchthat the magnitude of the series impedance Z at 10 kHz. Record

the value of R on the data sheets provided for lab 3

VL

Figu re 5 0

Vs

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3. xperimen tThe purpose of this experiment is to study the phase relationships of

elements in a passive network containing resistors and energy storagedevices

Equipment:Variable ACDC Power Supp lyDigital ultimeterResistance apacitance SubstituterInductance SubstituterOscilloscope

3.1 ConventionsThe oscilloscope shows the actual or instantaneous voltage as afunction of time however the digital voltmeter records RMS or effective

voltages equal to the instantaneous voltage divided by Within these labinstructions a subscript of RMS will be included i that measurement isintended as an effective or RMS value. there is not a subscript on thevalue it is assumed to be given as an instantaneous value. You will berequired to record both instantaneous and effective voltages on the datasheets

3.2 R-C NetworkSet up the circuit from figure 4 using the component values determined

in your prelab calculations Adjust the magnitude of the supply voltage sothat you deliver mA of current through the series resistance R. This isaccomplished changing the input signal voltage until VRmax = R X mA onthe oscilloscope. Then adjust input signal frequency to be 10 kHz.

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Measure the magnitudes of VR, VS. and VC using the oscilloscope. Animportant thing to realize is that in order to measure the voltage across anelement with the oscilloscopethat element you wish to record from must beconnected to the - terminal of the supply voltage Therefore if the circuit isin the configuration n figure 7, only VS and VR may be recorded. Tomeasure Vc one must switch the order of the series components so that theresistor is connected to the + term inal of the supply and the capacitor connected to the - terminal nother subtle point is that you must onlyconnect the positive terminal of the BNC connection box to the element youwish to measure voltage across seefigure. 7 . Connecting both the positiveand negative terminals of the BNC box will cause that element to be shortcircuited and thus removed from the network.

Remember that these magnitudes read off the oscilloscope areinstantaneous values.

Figure 7

oscilloscope

supply

to oscilloscope


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Measure the phase angle between yR and VS where we know thatthe phase angle of VR 0 and is therefore the reference to obtain VS.Record this angle on the data sheets The phase angle is measured usingthe oscilloscope in the dual trace mode, where you may view bo th imagessimultaneously. Adjust the volts/division until both images just il l the screenvertically. Then adjust the time/division until you see on ly one half of thesinusoidal wave, corresponding to 1800. You then n count the number ofdivisions that make up this 180 in order to calculate the number of degreesper division. Since there are 5 smaller hash marks for each division, forhigher accuracy yoi.t may also want to calculate the number of degrees perhash mark.

t is important to record the sign of the angle either or as well as

the magnitude of the phase shift. On the oscilloscope tracing, a shift to theright indicates a negative angle a lag, whereas a positive angle a leadwill appear as a left shifted trace This follows from the fac t that the x axis ofthe oscilloscope tracing is a standard graph of time, therefore time increasesto the right. A trace to the right isat a later tim e than a trace to the left. Thismeans that the trace to the right is lagging the left tracing corresponding to anegative phase angle.

3.3 RL NetworkSet the inductance substituter to the inductance value determined in

your prelab calculations Using the digita l multimeter as a DC ohmmetermeasure the inherent DC series resistance of the inductor. Record thisvalue on the data sheets provided.

Set up the circuit tram figure 5 using the component valuesdetermined in your prelab calculations. Readjust the resistance of the R Cbox to compensate for the series DC resistance of t

he inductor so that theto tal series resistance is still equal to R. Adjust the magnitude of the supplyvoltage so that you deliver mAmax of current through the newcompensated resistance R.

Measure the magnitudes of VR VS and VL using the oscilloscope.Record these values on the data sheets

Measure only the phase angle inc lude the sign between VR andVS Record th is value on the data sheets


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3.4 Series R-L-C NetworkArrange a series circuit with R 100 compensate this value with

the D resist ne of the inductance , using the values for C and L usedabove, and f 10 kHz. Adjust the series current to be equal to 10 mAmax.

Measure the four voltages and enter on the data sheets. Measure only the phse shift include sign between VS and VR.

3.5 Resonance Using the sme series R-L-C circuit s in 3.4, measure VR, VC and

VL s functions of frequency from 2 kHz to 20 kHz. The source voltage willtend to hnge s the frequency is changed, therefore to sve tediouscalculations later, vary the source voltage to keep VSmax 1.0 V This isdone using the digital voltmeter connected in parallel to VS remember thatthe multimeter reads in RMStherefore to maintain an instantaneous 1.0 Vthe multimeter must read .707 VRMS.This section proeeds more quickly if you set up one circuit tomesure VR, then vary the frequency from 2 kHz to 20 kHz recording onlyVR. Then switch the circuit to measure, say VC, and vary the frequency sdescribed while recording only VC, and so on for VL

Determine the frequency of the one half power points by finding thetwo points above and below resonance at which VR .707 VRmax. Recordthese values on the dt sheets and then calculate bandwidth from the twopower points, and also determine the quality factor from your data.4. DataSeries R C Network

From Prelab Calculations: R

C Network Magnitude Volts VRIVsIIVcI

Phase_Angle__VSMeasuredCalculated Calculated Calculated

These calculated values are determined later in the tab report write up.

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Series R L NetworkFrom Prelab Calculation: DC Resistance of the Inductor: AL

Magnitude VoltsRL NetworkVRVsVL

Phase_ ngle__VSMeasured

Calculated_1Calculated_1Calculated_1

Series -L- C Network

R L C Network Magnitude VoltsVRVsVcVL

Phase_ ngle_ VS esured

alculated Calculated_1Calculated_1

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R-L-C Series w VR Volts VC Volts VL Volts

kHz krad/s

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eries RLCVR h Bandwidth Q

easured alculatedhese values are theoretical calculated from the values of R arid C

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5. ReportEach person is to submit the following:5.1 A one page summary of the entire lab which reviews what was learnedin this experiment.As a group submit the following:5.2 Data sheets from the experiment which include any theoreticai andexperimental results.

5.3 Draw an accurate phasor diagram of the three voltages measured inthe series R C circuit. Remember that VR has a 0 phase angle and VC isat 90 from VR. Using this diagram and your knowledge of trigonometryderive 3 expressions for the phase angle between VS and VR such as:

e=sin1JEnter the results of these three calculated phase angles into your datatables.

Draw an accurate phasor diagram of the three voltages measured inthe series R circuit. Using this diagram and your knowledge oftrigonometry derive 3 expressions for the phase angle between VS and VR.Enter the results of these three calculated phase angles into your datatables.

5.4 Draw an accurate phasor diagram of the four voltages from the series R C circuit. Derive three expressions for the phase angle between VSand VR from the phasor diagram. Enter these results onto the data sheets.VR5.5 From the section of the lab on resonance plot on separate graphsVL VC W

and as a function of where0= LC VR

Locate and label on the graph the resonance frequency one half powerpoints and bandwidth.5.5 Submit completed prelab calculations for lab 3 showing all of yourwork. Enter your theoretical results into the lab 3 data sheets.

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B-129 RUTGERS UNIVERSITYThe State University of New JerseyCollege of EngineeringDepartm ent of Electrical and Computer Engineering

330:375 Elements of Electrical Engineering LaboratoryExperiment 4 Using Laplace Transform for Time Domain Analysis

This lab demonstrates the use of Laplace transform analysis to determ inethe transient time response of a network.1. Introduction to the Laplace Transform

The Laplace transform is a mathematical transformation that simplifiescomplicated functions, such as differentiation and integration, Into simpleralgebraic functions such as multiplication and division.

The Laplace transform of a function is given by the expressionL{ft = 1t t0

where the symbol L f t means the Lap lace transform of the function ft. TheLaplace transform is often denoted by the notation of Fs , or,L{ft =Fs.This notation helps to remind us that once the Laplace transform integral hasbeen evaluated, the resulting expression is a function of the complex variable swhich has the dimension of frequency and is no longer a function of time. TheLaplace transform thus transforms a function from the time domain 1 t into thefrequency domain Fs.

Although you may prove for yourself the Laplace transform for any givenft , most of the more common transforms are more easily determined byreferring to a Laplace transform pair table such as the one given in your text.This table consists of a number of functional transforms such as sin wI, I, eat,etc. that have been already transformed into the Laplace domain. Operationaltransforms, those transforms concerned with mathematical operations on f tsuch as the transform of the derivative of 1t , have also been proven for you inyourtext.

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2. Using the aplace TransformAs an example of using the Laplace transform to determine the response

of a network to an input driving function, we will analyze the series R-L-C circuitgiven in figure 1. CII

itv

RFigure

If we wish to describe the current i t which flows through this network, wewrite an integ rodifferen tia l equation which describes the voltage across eachelement in terms of 1t after the switch is closed at 1=0,

V i tR fi t dt.We wish to find a single equation that describes 1t from this complicatedmathematical expression. To solve for 1 t , we must use the technique ofLaplace transformation in order to reduce the complexity of the mathematicsinvolved To transform this complicated expression into the Laplace domainwe take the Laplace trans form of each side of the equation to find an expressionfor Is .

L v i tR L fi tdt.This is evaluated by finding the Lap lace transform of each individual term

v =L it R +LL dIilj l fitd t .he driving function of DC voltage is applied to the circuit only after the switchhas been thrown at 1=0. This driving function is a step function of magnitude V S

S

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V

L V _ .that begins at time equal to zero. According to the Laplace transform table, thetransform of a step input is equal to

he next term Is the current 1t times a constant R, which transforms to I t R R i s.The th ird term contains an operational transform of the derivative of the currentmultiplied by a constant L This term has a Laplace transform equal to

LL dLil L si s 0where 10 Is the initial current through the circuit when t 0. The last termcontains the integrl of the current, which transforms into

L fi tdt_ I 1 i10he term 10 is the initial value of the integ ral term just before applying thesource function.

The complete expression in the Laplace domain is then=RIs+Lsl s iO +.1.+ 110

Solving th is expression for is, we obtain0 V i1QsC r +si0- LC = R

R+sL+ s2+s +This expression may be simplified if we say that the initial values 10 and i1 0are equal to zero. The Laplace domain current I s is thenV V

Is s-sl s-s2

When the coefficient of the s term is equal to in the denominator of theLaplace domain equation, the denominator is referred to as theharacteristicequation. The characteristic equation must be factored into s-s s-s2 in orderto determine the time solution of it from solution of is in the Laplace domain.By partial fractions, I s may be written as

i s = s-si s-s2where Ki and 1


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it . K1 eSrom this result, one n see that the transient tim e sou ton of 1 t depends on

the roots i and S2 of the characteristic equation. The roots Si and S2 may bedetermined from the quadratic formula,

-b 4acS1,S2 2aThe so u t on the term gb2 4ac determines whether the roots s and S2 willbe real or complex. Zeta is known as the damping coefficient of thequadratic term. The critical value of is one. If is less than one isunderdamped , the solution of gb2 4ac is negative and the the roots of thequadratic factor are complex. If is equal to one is cntically damped, theso ut on of 4b2 4ac is equal to zero, and the roots of the quadratic arerepeated real roots si s2 If Is greater than one is overdamped thesolution of -,db2 4ac is positive , and the quadratic may be factored with its tworeal roots, s+sl s+s2.

For less than one, we can rewrite the characteristic equation in terms ofand i fl the uridamped natural radian frequency, where

s2+s +j2wns+ wn2=s+on2+wdWhere COd is the damped frequency of oscillation and is equivalent to

COd wnJi2By comparison of like terms on either side of this equation, it can be seen that

and2Wn=.

The expression for Is may be rewritten in terms of and COd asVL V

s+ O 3d OdL s+ O WdUsing the Laplace transform tables, the corresponding time domain so u t on isequal to

i t sincOd t .coL

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2. Prelab CalculationsComplete these calculations and submit your results along with your labwrite up for experiment 4. These must be completed before beginning this lab.

2.1 Step Response of an Underdamped RLC NetworkFor the network n figure 2, calculate the Laplace domain so lution for thestransfer function, vs Using a vlue of 40 mH and C .2 l.F , calculate

the undamped natural radian frequency on and the damping coeffic ient inte rmsof R. From this equation determine the value of R for 07 0.1, and001 Using these values of R and assuming a step input of 10 V solve for thetime after the step when the output will reach the first maximum and firstminimum. Using the complete so lution of VoUtt), determine the voltage at thesemaxima and minima when the input step is 10 volts nter your theoreticalresults onto the data sheets of lab 5. R

VinIOV Vout

Laplace transformation has enabled us to find a complete time domainso lution for it ) from a set of complicated integ rodifferential equations of i t).

Figure 2

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22 Step Response of a Critically Damped RLC NetworkFor the network in figure 3 calculate the Laplace domain solution for thestransfer function vj s Again using L 4 mH and C 2 iiF calculate n

and in terms of R From these equations calculate the value of R which willresult in a critica lly damped response Enter this value on the data sheets forlab 5

Vin1OV A C Vout

Figure

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3 ExperimentThe purpose of th is experiment is to learn of the uses of the Laplace

transform to predict the response of a circuit to a step Input

Equipment:Variable AC/DC Power SupplyDigital MultimeterResistance apacitance SubstituterInductance SubstituterOscilloscope

31 Step Response of an L C NetworkConstruct the circuit shown in figure 2 and connect the functiongenerator to Vr and the scope to both and Vout While a function generator

cannot produce a true step functlona square wave with a sufficiently longperiod is generally adequate The square wave that is produced by the functionVgenerator typically oscillates between n order to produce the desired stepinput you must i rst adjust the DC offset of the AC signal generator such that theinput square wave mimis a step input that starts at zero and steps up to VVolts on the oscilloscope

Starting with an input square wave of about 200 Hz monitor theresponse of the circuit on the scope and adjust the input frequency so that mostof the transient response is over before the next square wave begins

For each value of R calculated in part 21 carefully measure the time andvoltage values f the irst maximum and minimum nter the data on the sheetsprovided

32 Critically Damped Networkonstruct the network shown in figure 3 and adjust to the value

calculated in part 22 Using the same step input Vin as was used in part 31vary R to obtain a critically damped response where the output signal followsthe step as close ly as possible Measure this new Ntrimmedw value of R with thedigital ohmmeter and enter this value on the data sheets Using the ohmmetermeasure the DC resistance of the inductor and enter RL on your data sheets

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4. t SheetsStec Resoonse of an Underdamoed R C Network

R First Maximum First Minimum jisec Volts t psec Volts

CaIc u latd 7

TheoreticalCalculated

0.1TheoreticalCalculated

0.01Theoretical

Critically Damped NetworkTheoretical Resistance for Critical Damping

R=Actual Resistance Required for Critical Damping

R=Actual DC Resistance of the Induc torRL=

t


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5. epor tEach person is to submit the following:5.1 A one page summary of the entire lab which reviews what was learned inthis experiments a group submit the following:52 Data sheets from the experiment which include any theoretical andexperimental results53 In the calculations of part 22 the inductor was assumed to be idealalthough it has a significant non zero series DC resistance RL. lter thenetwork of figure 3 to include this component and derive the characteristicequation for this altered network sing the value of inductive resistancemeasured in part 3 2 and the resistance actual value required for criticldamping determine whether this departure from idel significntly alters thedamping factor or the undamped natural radian frequency On


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8-130 RUTGERS UNIVERSITYThe State University of New JerseyCollege of Engineering

Department of Electrical and Computer Engineering

330:375 Elements of Electrical Engineering LaboratoryExperiment 5 Introduction to Operational Amplifiers

The operational amplifier isa fundamental electronic circuit element iscapable of perform ing mathematical operations such as addition subtractionmultiplication, division, integration and differentiation on input signals. Theoperational amplifier is also commonly used to isolate different stages in anelectronic circuit.1.1 Theory

The best approach to understanding the ideal operational amplif ier opamp is to forget about the thousands of internal components such astransistors, diodes, resistors and capacitors, and consider the device simply interms of its inputs and output We therefore will stress its behavior rather thanits internal operation

Currently op amps are made by a number of companies around theworld, and are available in several different packages containing from one tofour op amps. The op amp that will be used this experiment, the i.A741,contains only one amplifier, and is fabricated into an eight pin package asshown in figure 1. The terminals of primary interest are

1. the inverting input,2. the rioninverting input,3. the outpu t,4. the positive power supply V+, and5. the negative power supply Vj.

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2 7

The other terminals are of little or no concern to us . The two balance inputs areprimarily used to compensate for pro lm s caused by aging of the amplifier andtemperature induced errors which are in most cases negligible. The otherterminal is of no interest because t is an unused terminal no t connected to theinternal amplifier circuit: NC stands for no connection.

The circuit symbol for the idealized operational amplifier containing thefive terminals of primary interest is shown in figure 2. Because t is inconvenientto use word labels for the terminals terminal designation is simplified with theuse of symbols The noninverting input terminal Is labeled with a plus sign +and the Inverting terminal is labeled with a minus sign - . The positive andnegative power supplies to the amplifier are marked V+ and V respectivelyOn most op amp circuit representations the power supply labels are no tincluded in the symbol but are understood to be included in the design Thesymbol for the op amp ut ut is drawn at the apex of the triangle.

VP

n

V

Vo

Balance

Inverting Input

Noninverting Input 3

NC

6 OutputBalanceg

Figure

3

4

Figure 2

5q.


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A positivegoing signal t the inverting - input produces a negativegoing signal at the output whereas the same signal at the noninverting + inputproduces a positivegoing output signa l. he amplifier responds only to thedifference of the input volt ge to the noninverting terminal Vp and the inputvoltage to the inverting terminal v,., . This voltage difference between the twoinput terminals is known as the differential voltage where:

vdvpvnWith a differential-input voltage v the output voltage V will be Avd, where isthe open loop gain of the amplifier, or that:V A dAlthough the value of the open loop gain A is quite large on the order ofthe magnitude of the output is limited by the range of the supply voltagestypically 15 volts.To understand the oper tionl amplifier it is importan t to lern its idealproperties seen in figure 3, which are:

1. he open loop voltage gain is infinite 2. he input resistance between the input termina ls is infinite r. 3. he output resistance is zero r 0.4. There is zero input offset voltage V =0 if 0.

Vd

Figure 3From these ideal characteristics two very important additional properties result,namely that:1. With the feedback loop closed between the output terminal nd the

invrtin input terminal, will be driven to the potential of Vp p2. There is no current flow into either input terminal, 0.

V

vn Vo


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1.2 Basic Op-Amp ConfigurationsAs mentioned before, the open loop g in A of the op amp is extremely

large, typically 100,000. he output signal v is therefore 100,000 times largerthan the differential Input voltage Vd. There are times that such large gain of thedifferential input signal is not required. We can control the gain of the op ampby the use of external elements such as a feedback resistor that connects fromthe output termin l to the inverting - terminal of the op amp. Operationalamplifiers are connected in two basic modes: 1 the inverting and 2 thenaninverting configurations.1.2.1 The Inverting Amplifier

he most basic circuit for the inverting amplifier is shown in figure 4where the + terminal is grounded and the input signal is applied to the -terminal through a resistor with a feedback resistor R connecting the outputto the - terminal Using the properties stated earlier we are able to determinethe gain of the inverting amplifier.

yin

RI

RI n

vo

Figure 4Since the + terminal is at zero potential it is grounded the - terminal

must lso be t zero potential, or t a virtual ground. If V is zero the inputvolt ge signal Vj must appear across Rin, making the input current in equal


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yin

ince no current can flow into the terminal of the op amp in must also flowthrough Aj, or that

f inThe output voltage appears across Rf and is negative due to the sign

inversion of the input signal in the amplifier. The current through Rf isV

Since in and are equal we can state thatyin vRin

Therefore the gain or the outputlinput ratio for the inverting amplifier must equal

he gain can be varied by adjusting either or1.2.2 The Noninverting Amplifier

In this configuration shown in figure 5, the input voltage signal is appliedto the + terminal of the op amp. The external feedback components thatdetermine the gain are always connected to the inverting - terminal. The gainfor the noninverting amplifier is equal to

inThe gain may be varied by changing the values for Rf and Rin. lithe feedbackresistor is shortcircuited as in figure 6, or that 0, the gain is equal to 1, andthe output signal is an exact replica of the input voltage. This mode where thegain is equal 1, or unity gain is called the voltage follower configuration andis often used to isolate between different sections of a circuit.

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Rf

S

Rin

vo

Figure

Ri n

vo

Figure


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2. Preliminary Calculations

Complete all theoretical results and submit your prelab calculations withthe report from the first lab. Be sure to enter yo

ur results on the data sheetsprovided

2.1 Determine the theoretical output voltage for the voltage follower circuit fromfigure 6 with Rj 47 k when

a VDCb VI 5 AC peak to peak at kHz.

22 Calculate the theoretical output voltage and current for a noninvertingamplifier shown in figure 5 when the following is true:a Vjn =5 DC Ri 47 kQ and

I R 100 calculate and Iii R 2.2 calculate and fiii Rf 47 calculate and lliii Rf 75 k calculate and l

b Let Rj 47 and Rf 75 kfl. Calculate the theoretical outputvoltage when the input voltage is an AC signal 5 volts peak to peak at kHz.

3 Calculate the theoretical ut ut voltage and current for an invertingamplifier shown in figure 4 when the following is true:

a Vin5VDC Rjn47kafldI R =100 calculate and lii R 22 calculate and Iiii R 7 kfl calculate and fliii R 75 kfl calculate and l

b Let Rj 47 k and Rf 75 k Calculate the theoretical outputvoltage when the input voltage is an AC signal 5 volts peak to peak at kHz.

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3. ExperimentEquipment:

Variable AC/DC Power upplyDigital Multimeter2Channel Oscilloscope1/4 Watt, 5 ResistorsProtoboard

3.1 Connecting the oscilloscopeUse the convention that channel I Is the Input signal and channel 2

Is the output signal and that the trace for channel should be above the tracefor channel 2.32 Connecting the protoboard

lease refer to figure 7. Vertical lines are connected internally in groupsof live holes which are not connected internlly to the neighboring set of 5vertical holes There is no connection between horizontal lines. The op amp isplugged into the protoboard by straddling the wide band that separates thegroups of 5 holes This provides a convenient means to plug the op ampdirectly into the protoboard and do the necessary wiring.

The DC power supplies V and V- are connected to the red and blackinput plugs and these then are wired to the power supply buses This giveseasy access to the power supplies to later connect to the op amp

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On the power supp ly buses holes areconnected horizontally but not vertically

i

G s

5 vertical holes areconnected internally yet

The op amp plugsneighbonng sets of 5 directly into thever tical holes are no t protoboardinternally connected

Figure 3 3 Voltage Follower

DO NOT TURN THE POWER SUPPLY ON UNTIL YOUR INSTRUCTORCHECKS YOUR WIRING. Construct the voltage follower of figure 6 onto theprotoboard with Rj 47 kfl After the circuit has been checked and beforepowering up the op amp verify that the DC supply voltage is 15 Volts and thatthe current limit on your voltage supply is turned down to a minimum to protectthe op amp

a Record the output voltage seen on the oscilloscope when the inputvoltage is 5 volts DC.

b Record the output voltage seen on the oscilloscope when the inpu tvoltage is an AC signal 5 volts peak to peak at kHz.

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34 Noninverting AmplifierConstruct the noninverting amplifier of figure 5 on the protoboard with the

feedback resistor Rf 100 D and input resistance 47 kfl After your circuit

has been checkeda Record the output voltage seen on the oscilloscope with the inputvoltage j equal to 5 V DC and compare this measured output voltage withthat expected with the gain of the noninverting amplifier 1 . Place thedigital ammeter in series with 1 Record the vlue of current f

ii Replace by k2 and measure and liii Replace Rf by 47 kQ and measure and lliii Replace by 75 and measure and I

b With Rf equal to 75 k record the output voltage seen on theoscilloscope when the input voltage is an AC signal 5 volts peak to peak at kHz.35 Inverting Amplifier

Construct the inverting mplifier of figure 4 on the protoboard with thefeedback resistor Rf 100 and input resistance j 47 kG. After your circuithas been checked

Record the output voltage seen on the oscilloscope with the inputvoltage j equal to 5 V DC and compare this

measured output voltage withthat expected with the gain of the inverting amplifier - . Place the digitalammeter in series withR Record the value of current If

ii Replace Rf by 22 kG and measure and fiii Replace by 47 kG and measure and Iiiii Replace Rf by 75 kG and measure and I

b With equal to 75 kG record the output voltage seen on theoscilloscope when the input voltage is an AC signal 5 volts peak to peak at kHz.


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3.6 Op Amp circuit for addition and subtractionRf

Figure 8We wish to build an op amp circuit to pertorm the mathematic summation:

V x2 y zWe need to onstru t the summing amplifier of figure 8 on the protoboard toimplement a portion of the desired equation. The inverting summing amplifiershown in figure 8 has an output which follows the equation:--x.yWith careful selection of the resistance values, we n get the output:1 =-.1x+2y-.1x-2yWhich is similar to the first part of our mathematic equation. Using a secondsumming amplifier in the same configuration as in figure with inputs equal to1 and z, the output will then be:V zwhich wilt produce the result:

V -.1 x -2y zWhich results in the desired output namely:.lx2y z .

volR R

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a ecord the output voltage when the hipta voltages are toCalVoltsDCyeSVolteDCzSVoIaDCb ecord th output voltage when the put voltages r equal toXa VOItSOCy VoltsDCz VoltsDC


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4. Data Sheets all theoretical results must be completed before beginning thisexperiment.

Voltage Follower VVin=5VOltDC

Experimental:_Theoretical:

V = 5 Volts peak to peakExperimental:Theoretical: peak to peak

Noninverting Amplifier Vo fVj=5VDC Rf=1OOExperimental:Theoretical:

Vjn5VDC Rf kExperimental:Theoretical:

Vj =5 DC, R = 4.7 k1Experimental:Theoretical:

Vi=5VDC R =7 5 kExperimental:Theoretical:

Vj =5 V pp AC, kHzRf = 7Experimental: Not measuredTheoretical: Not Calculated

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Inverting Amplifier VV=5VDC R= OO2

Experimental:Theoretical:

Vj 5VDC Rf=2 2kL2Experimental:Theoretical:

Vjn5VDC Rf4.7k2Experimental:Theoretical:

VI=5VDC Rf=7.5kExperimental:Theoretical:

Vj =5 V pp AC, kHzRf=7.5kExperimental: Not measuredTheoretical: Not Calculated

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Summing Amplifier Design:First StageiR

Second StageRR

ummin Amplifier Vx Volts DC Experimental:y =5 Volts DCz =5 Volts DC Theoretical:x 5 Volts DC Experimental:y Volts DCz Volts DC Theoretical:

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5 eport

Each person is to submit the following:

5 A one page summary of the entire lab which reviews what ws learned inthis lab.As a group, submit the following:

5.2 Derive the gain equation shown below for the noninverting amplifierused in this experiment:VGain tr 1 for a noninverting amplifiery in

ii Derive an expression which describes the current l through for thenoninverting amplifier used in this experiment. How does the current Imeasured experimentally in section 3.5 compare to that found theoretically withthe expression derived?

5.3 Derive the output voltage equation for an inverting summing amplifierwhere the output is negative of the scaled sum of and V2 input voltages:

Vo=-j- V1+jV2)

5.4 Complete the data tables of experimental and theoretical values

Submit completed prelab calculations for Lab 5 Enter your results ontothe data sheets for lab 5.

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5 ReportEach person is to submit the following:51 A one page summary of the entire lab which reviews what was learned nthis lab As a group submit the following:5 2 er ive the gain equation shown below for the noninvertLng am plifierused in this experiment:VGain for a non inverting amplifier Rin

Derive an expression which describes the current I th rough Rf for thenoninverting am plifier used in this experiment How does the current ljmeasured experimentally in section 35 compare to that found theoretically withthe expression derived? 53 Derive the output voltage equation for an inverting summing amplifierwhere the output is negative of the scaled sum of and V2 input voltages :

Vo j V54 Complete the data tables of expe rimental and theo retical values

55 Submit completed prelab calculations for Lab 5 Enter your results ontothe data sheets for lab 5

EEELab Manual 375

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Transcript of EEELab Manual 375