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Linear Circuits Analysis. Superposition, Thevenin /Norton Equivalent circuits So 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 is only 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. Time-independent 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. Linearity A function f is linear if for any two inputs x1 and x2 fx1+x2()=fx1()+fx2() Resistive circuits are linear. That is if we take the set {xi} as the inputs to a circuit and f({xi}) as the response of the circuit, then the above linear relationship holds. The response may be for example the voltage at any node of the circuit or the current through any element. Let’s explore the following example. RVs1Vs2i KVL for this circuit gives 12Vs Vs iR 0+−= (1.1) Or 1Vs Vsi2R+= (1.2) 6.071/22.071 Spring 2006. Chaniotakis and Cory 1And as we see the response of the circuit depends linearly on the voltages and . A useful way of viewing linearity is to consider suppressing sources. A voltage source is suppressed by setting the voltage to zero: that is by short circuiting the voltage source. 1Vs 2Vs Consider again the simple circuit above. We could view it as the linear superposition of two circuits, each of which has only one voltage source. RVs1i1 Ri2Vs2 The total current is the sum of the currents in each circuit. 12112ii iVs Vs2RRVs VsR=+=++= (1.3) Which is the same result obtained by the application of KVL around of the original circuit. 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 2Independent sources may be suppressed as follows: Voltage sources: Vs+-v=Vssuppress+-v=0short Current sources: i=IssuppressIsi=0open 6.071/22.071 Spring 2006. Chaniotakis and Cory 3An example: Consider the following example of a linear circuit with two sources. Let’s analyze the circuit using superposition. R1VsR2Isi1 i2+- First let’s suppress the current source and analyze the circuit with the voltage source acting alone. R1VsR2i1v i2v+- So, based on just the voltage source the currents through the resistors are: 1iv 0= (1.4) 22VsivR= (1.5) Next we calculate the contribution of the current source acting alone R1 R2i1i i2i+-Isv1 Notice that R2 is shorted out (there is no voltage across R2), and therefore there is no current through it. The current through R1 is Is, and so the voltage drop across R1 is, 6.071/22.071 Spring 2006. Chaniotakis and Cory 41vIsR1= (1.6) And so 1iIs= (1.7) 22VsiR= (1.8) How much current is going through the voltage source Vs? Another example: For the following circuit let’s calculate the node voltage v. R1R2VsIsv Nodal analysis gives: 012Vs v vIsRR−+−= (1.9) or 21212 12RRRvVsIsRRRR=+++ (1.10) We notice that the answer given by Eq. (1.10) is the sum of two terms: one due to the voltage and the other due to the current. Now we will solve the same problem using superposition The voltage v will have a contribution v1 from the voltage source Vs and a contribution v2 from the current source Is. 6.071/22.071 Spring 2006. Chaniotakis and Cory 5R1R2Vsv1 R1R2Isv2 2112RvVsRR=+ (1.11) And 12212RRvIsRR=+ (1.12) Adding 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 to represent a given circuit. A circuit may contain a large number of elements and in many cases knowing the i-v characteristics of the circuit is sufficient in order to understand its behavior and be able to interconnect it with other circuits. The following figure illustrates the general concept where a circuit is represented by the box as indicated. Our communication with the circuit is via the port A-B. This is a single port network regardless of its internal complexity. iVnInR3R4 ABv+- If we apply a voltage v across the terminals A-B as indicated we can in turn measure the resulting current i . If we do this for a number of different voltages and then plot them on the 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 function imvb=+ (1.13) 6.071/22.071 Spring 2006. Chaniotakis and Cory 6Here are some examples of i-v characteristics Rv+-i iv In general the i-v characteristic does not pass through the origin. This is shown by the next circuit for which the current i and the voltage v are related by 0iR Vs v+−= (1.14) or vVsiR−= (1.15) v+-iVsR ivVs-Vs/R Similarly, when a current source is connected in parallel with a resistor the i-v relationship is viIsR=−+ (1.16) Rv+-iIs ivRIs-Isopen circuitvoltageshort circuitcurrent 6.071/22.071 Spring 2006. Chaniotakis and Cory 7Thevenin Equivalent Circuits. For linear systems the i-v curve is a straight line. In order to define it we need to identify only two pints on it. Any two points would do, but perhaps the simplest are where the line crosses the i and v axes. These two points may be obtained by performing two simple measurements (or make two simple calculations). With these two measurements we are able to replace the complex network 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 superposition results in the following expression for the current i and voltage v relation. 0jjjjimv mV bI=+ +jj∑∑ (1.17) Where jV and jI are voltage and current sources in the circuit under investigation and the coefficients jm and jb are functions of other circuit parameters such as resistances. And so for a general network we can write imvb=+ (1.18) Where 0mm= (1.19) And jjjjjbmV b=+jI∑∑ (1.20) Thevenin’s Theorem is


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MIT 6 071J - Study Notes

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