Linear Circuits Analysis - MIT OpenCourseWare

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Linear Circuits Analysis. Superposition, Thevenin /Norton Equivalent circuitsSo far we have explored time-independent (resistive) elements that are also linear.A time-independent elements is one for which we can plot an i/v curve. The current isonly a function of the voltage, it does not depend on the rate of change of the voltage.We will see latter that capacitors and inductors are not time-independent elements. Timeindependent elements are often called resistive elements.Note that we often have a time dependent signal applied to time independent elements.This is fine, we only need to analyze the circuit characteristics at each instance in time.We will explore this further in a few classes from now.LinearityA function f is linear if for any two inputs x1 and x2f (x1 x 2 ) f (x1 ) f (x 2 )Resistive circuits are linear. That is if we take the set {xi} as the inputs to a circuit andf({xi}) as the response of the circuit, then the above linear relationship holds. Theresponse may be for example the voltage at any node of the circuit or the current throughany element.Let’s explore the following example.iVs1RVs2KVL for this circuit givesVs1 Vs 2 iR 0Ori Vs1 Vs 2R6.071/22.071 Spring 2006. Chaniotakis and Cory(1.1)(1.2)1

And as we see the response of the circuit depends linearly on the voltages Vs1 and Vs 2 .A useful way of viewing linearity is to consider suppressing sources. A voltage source issuppressed by setting the voltage to zero: that is by short circuiting the voltage source.Consider again the simple circuit above. We could view it as the linear superposition oftwo circuits, each of which has only one voltage source.i1i2Vs1RRVs2The total current is the sum of the currents in each circuit.i i1 i 2Vs1 Vs2 (1.3)RRVs1 Vs 2 RWhich is the same result obtained by the application of KVL around of the originalcircuit. If the circuit we are interested in is linear, then we can use superposition to simplify theanalysis. For a linear circuit with multiple sources, suppress all but one source andanalyze the circuit. Repeat for all sources and add the results to find the total responsefor the full circuit.6.071/22.071 Spring 2006. Chaniotakis and Cory2

Independent sources may be suppressed as follows:Voltage sources: Vsv Vs-suppress shortv 0-Current sources:i IsIsi 0suppressopen6.071/22.071 Spring 2006. Chaniotakis and Cory3

An example:Consider the following example of a linear circuit with two sources. Let’s analyze thecircuit using superposition.R1R2i1i2 IsVs-First let’s suppress the current source and analyze the circuit with the voltage sourceacting alone.R1R2i1vi2v Vs-So, based on just the voltage source the currents through the resistors are:i1v 0Vsi 2v R2(1.4)(1.5)Next we calculate the contribution of the current source acting aloneR1i1i v1 -R2i2iIsNotice that R2 is shorted out (there is no voltage across R2), and therefore there is nocurrent through it. The current through R1 is Is, and so the voltage drop across R1 is,6.071/22.071 Spring 2006. Chaniotakis and Cory4

v1 IsR1(1.6)And soi1 IsVsi2 R2How much current is going through the voltage source Vs?(1.7)(1.8)Another example:For the following circuit let’s calculate the node voltage v.R1vR2VsIsNodal analysis gives:Vs vv Is 0R1R2(1.9)orv R2R1R 2Vs IsR1 R 2R1 R 2(1.10)We notice that the answer given by Eq. (1.10) is the sum of two terms: one due to thevoltage and the other due to the current.Now we will solve the same problem using superpositionThe voltage v will have a contribution v1 from the voltage source Vs and a contributionv2 from the current source Is.6.071/22.071 Spring 2006. Chaniotakis and Cory5

R1VsR1v1v2R2R2Isv1 VsR2R1 R 2(1.11)v 2 IsR1R 2R1 R 2(1.12)AndAdding voltages v1 and v2 we obtain the result given by Eq. (1.10).More on the i-v characteristics of circuits.As discussed during the last lecture, the i-v characteristic curve is a very good way torepresent a given circuit.A circuit may contain a large number of elements and in many cases knowing the i-vcharacteristics of the circuit is sufficient in order to understand its behavior and be able tointerconnect it with other circuits.The following figure illustrates the general concept where a circuit is represented by thebox as indicated. Our communication with the circuit is via the port A-B. This is a singleport network regardless of its internal complexity.i AR4VnInR3 vBIf we apply a voltage v across the terminals A-B as indicated we can in turn measure theresulting current i . If we do this for a number of different voltages and then plot them onthe i-v space we obtain the i-v characteristic curve of the circuit.For a general linear network the i-v characteristic curve is a linear functioni mv b6.071/22.071 Spring 2006. Chaniotakis and Cory(1.13)6

Here are some examples of i-v characteristicsii v-RvIn general the i-v characteristic does not pass through the origin. This is shown by thenext circuit for which the current i and the voltage v are related byiR Vs v 0ori (1.14)v VsR(1.15)iiVsR v-Vsv-Vs/RSimilarly, when a current source is connected in parallel with a resistor the i-vrelationship isvi Is Ropen circuitivoltage(1.16)iIsR v-6.071/22.071 Spring 2006. Chaniotakis and Cory-IsRIsvshort circuitcurrent7

Thevenin Equivalent Circuits.For linear systems the i-v curve is a straight line. In order to define it we need to identifyonly two pints on it. Any two points would do, but perhaps the simplest are where theline crosses the i and v axes.These two points may be obtained by performing two simple measurements (or make twosimple calculations). With these two measurements we are able to replace the complexnetwork by a simple equivalent circuit.This circuit is known as the Thevenin Equivalent Circuit.Since we are dealing with linear circuits, application of the principle of superpositionresults in the following expression for the current i and voltage v relation.i m0 v m jV j b j I jj(1.17)jWhere V j and I j are voltage and current sources in the circuit under investigation andthe coefficients m j and b j are functions of other circuit parameters such as resistances.And so for a general network we can writei mv b(1.18)m m0(1.19)WhereAndb m jV j b j I jj(1.20)jThevenin’s Theorem is stated as follows:A linear one port network can be replaced by an equivalent circuit consisting of a voltagesource VTh in series with a resistor Rth. The voltage VTh is equal to the open circuitvoltage across the terminals of the port and the resistance RTh is equal to the open circuitvoltage VTh divided by the short circuit current IscThe procedure to calculate the Thevenin Equivalent Circuit is as follows:1. Calculate the equivalent resistance of the circuit (RTh) by setting all voltage andcurrent sources to zero2. Calculate the open circuit voltage Voc also called the Thevenin voltage VTh6.071/22.071 Spring 2006. Chaniotakis and Cory8

The equivalent circuit is nowR4VnInRThi A v-R3iA v-VocBBEquivalent circuitOriginal circuitIf we short terminals A-B, the short circuit current Isc isIsc VThRTh(1.21)Example:Find vo using Thevenin’s theorem2k Ω6k Ω12 V6k Ω vo1k Ω-The 1kΩ resistor is the load. Remove it and compute the open circuit voltage Voc orVTh.2k Ω6k Ω12 V6k Ω Voc-Voc is 6V. Do you see why?Now let’s find the Thevenin equivalent resistance RTh.6.071/22.071 Spring 2006. Chaniotakis and Cory9

2k Ω6k ΩRTh6k ΩRTh 6k Ω // 6k Ω 2k Ω 5k ΩAnd the Thevenin circuit is5k Ω5k Ω1k Ω6V 1k Ω6Vvo-And vo 1 Volt.Another example:Determine the Thevenin equivalent circuit seen by the resistor RL.R2R1 VsRLR3R4Resistor RL is the load resistor and the balance of the system is interface with it.Therefore in order to characterize the network we must look the network characteristicsin the absence of RL.6.071/22.071 Spring 2006. Chaniotakis and Cory10

R1 VsAR2BR3R4First lets calculate the equivalent resistance RTh. To do this we short the voltage sourceresulting in the circuit.R1AR2 BR1ABR3R3R2R4R4The resistance seen by looking into port A-B is the parallel combination ofR13 R1R3R1 R3(1.22)R2R4R2 R4(1.23)In series with the parallel combinationR 24 RTh R13 R 24(1.24)The open circuit voltage across terminals A-B is equal to6.071/22.071 Spring 2006. Chaniotakis and Cory11

R2R1 VsAvA-R3BvBR4VTh vA vBR4 R3 Vs R1 R3 R 2 R 4 (1.25)And we have obtained the equivalent circuit with the Thevenin resistance given by Eq.(1.24) and the Thevenin voltage given by Eq. (1.25).6.071/22.071 Spring 2006. Chaniotakis and Cory12

The Wheatstone Bridge Circuit as a measuring instrument.Measuring small changes in large quantities – is one of the most common challenges inmeasurement. If the quantity you are measuring has a maximum value, Vmax, and themeasurement device is set to have a dynamic range that covers 0 - Vmax, then the errorswill be a fraction of Vmax. However, many measurable quantities only vary slightly, andso it would be advantageous to make a difference measurement over the limited range ,Vmax- Vmin. The Wheatstone bridge circuit accomplishes this.R1 VsR2AB vu --R3RuThe Wheatstone bridge is composed of three known resistors and one unknown, Ru, bymeasuring either the voltage or the current across the center of the bridge the unknownresistor can be determined. We will focus on the measurement of the voltage vu asindicated in the above circuit.The analysis can proceed by considering the two voltage dividers formed by resistor pairsR1, R3 and R2, R4.R2R1 VsR3AB vA vB-Ru-The voltage vu is given byvu vA vB(1.26)Where,6.071/22.071 Spring 2006. Chaniotakis and Cory13

vA VsR3R1 R3(1.27)vB VsRuR 2 Ru(1.28)AndAnd vu becomes:Ru R3vu Vs R1 R3 R 2 Ru (1.29)A typical use of the Wheatstone bridge is to have R1 R2 and R3 Ru. So let’s takeRu R3 ε(1.30)Ru R3vu Vs R1 R3 R 2 Ru R3 ε R3 Vs R1 R3 R1 R3 ε (1.31)Under these simplifications,As discussed above we are interested in the case where the variation in Ru is small, that isin the case where ε R1 R3 . Then the above expression may be approximated as,vu VsεR1 R36.071/22.071 Spring 2006. Chaniotakis and Cory(1.32)14

The Norton equivalent circuitA linear one port network can be replaced by an equivalent circuit consisting of a currentsource In in parallel with a resistor Rn. The current In is equal to the short circuit currentthrough the terminals of the port and the resistance Rn is equal to the open circuit voltageVoc divided by the short circuit current In.The Norton equivalent circuit model is shown below:i InRnv-By using KCL we derive the i-v relationship for this circuit.i In v 0Rn(1.33)ori v InRn(1.34)For i 0 (open circuit) the open circuit voltage isVoc InRn(1.35)Isc In(1.36)And the short circuit current isIf we choose Rn RTh and In Vocthe Thevenin and Norton circuits are equivalentRTh6.071/22.071 Spring 2006. Chaniotakis and Cory15

iiRThA v-VocInRThv-B Thevenin CircuitNorton CircuitWe may use this equivalence to analyze circuits by performing the so called sourcetransformations (voltage to current or current to voltage).For example let’s consider the following circuit for which we would like to calculate thecurrent i as indicated by using the source transformation method.3V3Ωi6Ω6Ω3Ω2ABy performing the source transformations we will be able to obtain the solution bysimplifying the circuit.First, let’s perform the transformation of the part of the circuit contained within thedotted rectangle indicated below:3V3Ωi6Ω6Ω3Ω2AThe transformation from the Thevenin circuit indicated above to its Norton equivalentgives0.5 A6Ω3Ωi6Ω6.071/22.071 Spring 2006. Chaniotakis and Cory3Ω2A16

Next let’s consider the Norton equivalent on the right side as indicated below:3Ωi0.5 A6Ω6Ω3Ω2AThe transformation from the Norton circuit indicated above to a Thevenin equivalentgives3Ωi0.5 A6Ω3Ω6Ω6VWhich is the same as6Ωi0.5 A6Ω6Ω6VBy transforming the Thevenin circuit on the right with its Norton equivalent we havei0.5 A6Ω6Ω6Ω1AAnd so from current division we obtain1 3 1i A3 2 26.071/22.071 Spring 2006. Chaniotakis and Cory(1.37)17

Another example: Find the Norton equivalent circuit at terminals X-Y.XR1R3R2IsVsR4YFirst we calculate the equivalent resistance across terminals X-Y by setting all sources tozero. The corresponding circuit isXR1R3RnR2R4YAnd Rn isRn R 2( R1 R3 R 4)R1 R 2 R3 R 4(1.38)Next we calculate the short circuit currentXR1R3R2IsIscVsR4Y6.071/22.071 Spring 2006. Chaniotakis and Cory18

Resistor R2 does not affect the calculation and so the corresponding circuit isXR1R3IscIsVsR4YBy applying the mesh method we haveIsc Vs IsR3 InR1 R3 R 4(1.39)With the values for Rn and Isc given by Equations (1.38) and (1.39) the Nortonequivalent circuit is definedXInRnY6.071/22.071 Spring 2006. Chaniotakis and Cory19

Power Transfer.In many cases an electronic system is designed to provide power to a load. The generalproblem is depicted on Figure 1 where the load is represented by resistor RL.linearelectronicsystemRLFigure 1.By considering the Thevenin equivalent circuit of the system seen by the load resistor wecan represent the problem by the circuit shown on Figure 2.RThi vLVThRL-Figure 2The power delivered to the load resistor RL isP i 2 RL(1.40)The current i is given byi VThRTh RL(1.41)And the power becomes2 VTh P RL RTh RL (1.42)For our electronic system, the voltage VTh and resistance RTh are known. Therefore if wevary RL and plot the power delivered to it as a function of RL we obtain the generalbehavior shown on the plot of Figure 3.6.071/22.071 Spring 2006. Chaniotakis and Cory20

Figure 3.The curve has a maximum which occurs at RL RTh.In order to show that the maximum occurs at RL RTh we differentiate Eq. (1.42) withrespect to RL and then set the result equal to zero.2dP2 ( RTh RL ) 2 RL ( RTh RL ) VTh dRL( RTh RL) 4 (1.43)dP 0 RL RTh 0dRL(1.44)andand so the maximum power occurs when the load resistance RL is equal to the Theveninequivalent resistance RTh.1Condition for maximum power transfer:RL RTh(1.45)The maximum power transferred from the source to the load isP max 1VTh 24 RTh(1.46)d 2Pd 2PBy taking the second derivativeand setting RL RTh we can easily show that 0 , therebydRL2dRL2the point RL RTh corresponds to a maximum.6.071/22.071 Spring 2006. Chaniotakis and Cory21

Example:For the Wheatstone bridge circuit below, calculate the maximum power delivered toresistor RL.R1 VsR2RLR3R4Previously we calculated the Thevenin equivalent circuit seen by resistor RL. TheThevenin resistance is given by Equation (1.24) and the Thevenin voltage is given byEquation (1.25). Therefore the system reduces to the following equivalent circuitconnected to resistor RL.iRTh vLVThRL-For convenience we repeat here the values for RTh and VTh.R4 R3 VTh Vs R1 R3 R 2 R 4 RTh (1.47)R1R3R2R4 R1 R3 R 2 R 4(1.48)The maximum power delivered to RL isR4 R3Vs 2VThR1 R3 R 2 R 4 P max R2R4 4 RTh R1R3 4 R1 R3 R 2 R 4 226.071/22.071 Spring 2006. Chaniotakis and Cory(1.49)22

In various applications we are interested in decreasing the voltage across a load resistorby without changing the output resistance of the circuit seen by the load. In such asituation the power delivered to the load continues to have a maximum at the sameresistance. This circuit is called an attenuator and we will investigate a simple example toillustrate the principle.Consider the circuit shown of the following Figure.attenuatorRThRsa VThRpRLvobThe network contained in the dotted rectangle is the attenuator circuit.The constraints are as follows:1. The equivalent resistance seen trough the port a-b is RTh2. The voltage vo k VThDetermine the requirements on resistors Rs and Rp.First let’s calculate the expression of the equivalent resistance seen across terminals a-b.By shorting the voltage source the circuit for the calculation of the equivalent resistanceisattenuatorRThRsaRpReffbThe effective resistance is the parallel combination of Rp with Rs RTh.Reff ( RTh Rs ) // Rp( RTh Rs ) Rp RTh Rs Rp6.071/22.071 Spring 2006. Chaniotakis and Cory(1.50)23

Which is constrained to be equal to RTh.RTh ( RTh Rs ) RpRTh Rs Rp(1.51)The second constraint giveskVTh VThRpRp RTh Rs(1.52)And so the constant k becomes:k RpRp RTh Rs(1.53)By combining Equations (1.51) and (1.53) we obtainRs 1 kRThk(1.54)Rp 1RTh1 k(1.55)AndThe maximum power delivered to the load occurs at RTh and is equal tok 2VTh 2Pmax 4 RTh6.071/22.071 Spring 2006. Chaniotakis and Cory(1.56)24

Representative Problems:P1.Find the voltage vo using superposition.(Ans. 4.44 Volts)vo2Ω3Ω1Ω2V4Ω6VP2.Calculate io and vo for the circuit below using superposition(Ans. io 1.6 A, vo 3.3 V)2A4Ω3Ωio12 VP3.2Ω1Ω3Ω4Ω1Ausing superposition calculate vo and io as indicated in the circuit below(Ans. io 1.35 A, vo 10 V) vo 4Ω24 Vio3Ω1Ω3Ω2A6.071/22.071 Spring 2006. Chaniotakis and Cory3Ω12 V25

P4.Find the Norton and the Thevenin equivalent circuit across terminals A-B of thecircuit. (Ans. In 1.25 A , Rn 1.7 Ω , VTh 2.12V )A4Ω3Ω1Ω5AP5.3ΩBCalculate the value of the resistor R so that the maximum power is transferred tothe 5Ω resistor. (Ans. 10Ω)R5Ω24 V10 Ω12 VP6.Determine the value of resistor R so that maximum power is delivered to it fromthe circuit connected to it.R1R2Vs R-R3P7R4The box in the following circuit represents a general electronic element.Determine the relationship between the voltage across the element to the currentflowing through it as indicated.iR1R2 VsvR3-6.071/22.071 Spring 2006. Chaniotakis and Cory26

If the circuit we are interested in is linear, then we can use superposition to simplify the analysis. For a linear circuit with multiple sources, suppress all but one source and analyze the circuit. Repeat for all sources and add the results to find the total response for the full circuit. 6.071/22.071 Spring 2006. Chaniotakis and Cory 2

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