SuperpositionExample: Analyzing a Circuit with SuperpositionSuperposition Example (2)Superposition Example (3)Thevenin Equivalent CircuitThevenin DerivationThevenin Source ResistanceThevenin Equivalent ExampleThevenin Example (2)Non-Linear Components and SourcesLinear RelationsCalculating $R_{th}$ DirectlyExample: By Direct MethodThevenin Example (2)Norton Equivalent CircuitSource TransformationsNorton Equivalent ExampleNorton Example (2)Norton Example (3)Voltage Amplifier ExampleTransconductance Amplifier ExampleMaximum Power TransferMaximum Power Transfer (cont)Example of Maximum Power TransferThe Wheatstone BridgeWheatstone (2)Wheatstone (3)Wheatstone ExampleEE 42/100Lecture 6: Network TheoremsELECTRONICSRev C 3/1/2012 (8:30PM)Prof. Ali M. NiknejadUniversity of California, BerkeleyCopyrightc 2012 by Ali M. NiknejadA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 1/?? – p. 1/Superposition•If a circuit is linear, then by the principal of superposition, we can analyze thecircuit one source at a time. The total response is the sum of the outputs due tothe individual sources.•This is clear if we re-write the matrix equation as followsAx = b = b1+ b2+ · · ·•Note that we have partitioned the source terms so that each bkonly contains asingle source. Clearly, the solution is given byx = A−1b1+ A−1b2+ · · · = x1+ x2+ · · ·•where xkis the solution with source k turned on and all other sources set to zero.That means that other voltage sources are short circuited (zero voltage) and othercurrent sources are open circuited (zero current).A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 2/?? – p. 2/Example: Analyzing a Circuit with SuperpositionR1R2R3vs1is1is2v1•In this example there are three independent sources. When we analyze the circuitsource by source, the circuit is often simple enough that we can solve theequations directly by inspection.•First turn of is1and is2. Zero current means that we replace these sources withopen circuits. The node voltage v1is therefore by inspectionvvs11=R2R1+ R2vs1A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 3/?? – p. 3/Superposition Example (2)R1R2R3vs1is1is2v1•Next turn off vs1(short circuit) and is2(open circuit). The current is1will thereforedivide between r2and r1and establish a voltage at node v1(equivalently, it see’sa parallel combination of R1and R2vis11=R2R1R1+ R2is1•Finally, we turn off all sources except is2. Now only R1and R2remain (R3isdangling)vis21=R2R1R1+ R2is1A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 4/?? – p. 4/Superposition Example (3)•By superposition, the node voltage v1is the sum of the three node voltages due toeach sourcev1= vis11+ vis21+ vvs11=R2R1+ R2(vs1+ R1(is1+ is2))•We can verify the solution by performing KCL directly at node 1(v1− vs1)G1+ v1G2− is1− is2= 0v1(G1+ G2) = vs1G1+ is1+ is2•The answer here is just as fast but we don’t have any intuition about the operationof the circuit. We’re perhaps more likely to make an algebraic error if it’s all mathwithout any thinking.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 5/?? – p. 5/Thevenin Equivalent Circuit+++vthRth“Black Box”•A powerful theorem in circuit analysis is the Thevenin equivalent theorem, whichlet’s us replace a very complex circuit with a simple equivalent circuit model.•In the black box there can be countless resistors, voltage sources (independentand dependent), current sources (independent and dependent), and yet theterminal behavior of the circuit is captured by two elements.•How can this be? Well, there is a big assumption in that all the resistors are linear(follow Ohm’s Law) and all dependent sources are also linear.•The equivalent circuit representation is often called a “black box", since the detailsof the circuitry are hidden.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 6/?? – p. 6/Thevenin Derivation+++vthRth“Black Box”+voc−+vth−•Since a circuit is linear, then no matter how complicated it is, it’s response to astimulus at some terminal pair must be linear. It can therefore be represented by alinear equivalent resistor and a a fixed constant source voltage due to thepresence of independent sources in the circuit.•To find the equivalent source value, called the Thevenin voltage source vth, simplyobserve that the open-circuit voltage of both the “black box" and the original circuitmust equal, which meansvth= voc•In other words, open-circuit the original circuit, find its equivalent output voltage atthe terminals of interest, and that’s vthA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 7/?? – p. 7/Thevenin Source Resistance+++vthRth“Black Box”iscvthRth•To find equivalent Thevenin source resistance Rth, notice that in order for theterminal behavior of the two circuits to match, the current flow into a load resistorhas to be the same for any load value. In particular, take the load as a short circuit.•The output current of the Thevenin equivalent under a short circuit is given byvthRt•Equating this to the short-circuit current of the original circuitry, we haveisc=vthRtor equivalentlyRth=vthisc=vociscA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 8/?? – p. 8/Thevenin Equivalent ExampleR1R2R3R4vsv2v3•In the above circuit, we will calculate the Thevenin equivalent circuit.•We begin by finding the open-circuit voltage. In this case, it’s a simple applicationof the voltage divider.v3= v2R4R3+ R4v2=R2||(R3+ R4)R1+ R2||(R3+ R4)vsvoc= v3=R4R3+ R4R2||(R3+ R4)R1+ R2||(R3+ R4)vs=R4R2(R3+ R4)(R3+ R4)(R1(R2+ R3+ R4) + R2(R3+ R4))A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 9/?? – p. 9/Thevenin Example (2)R1R2R3R4vsv2isc•Next we find the short-circuit current in the original circuit. The resistance loadingthe source under this condition is given byisc= isR2R2+ R3is=vsR1+ (R2||R3)=vs(R2+ R3)(R2+ R3)R1+ R2R3isc=R2vsR1R2+ R1R3+ R2R3Rth=vociscA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 6 p. 10/?? – pNon-Linear Components and Sources•A non-linear resistor has a non-linear I-V relation. For exampleV = r1I + r2I2+ r3I3orV = cos(I · Rx)•A non-linear dependent source is a non-linear function of one or more independentcurrents/voltages in the circuit. Some examples of
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