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Transcript
EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross Linear Circuits -Special Properties • Circuits consisting only of linear elements are linear circuits. • There are simple “equivalent circuits” for “one-port” linear circuits. 1 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross TWO-TERMINAL LINEAR RESISTIVE NETWORKS (“One Port” Circuit) Interconnection of two-terminal linear resistive elements with only two “accessible” terminals + a b 2 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross I-V CHARACTERISTICS OF LINEAR TWO-TERMINAL NETWORKS i i 5K + + + 5V 5V v Associated v Unassociated Apply v, measure i, or vice versa Apply v, measure i, or vice versa i(mA) 1 + i(mA) 1 Associated (i defined in) .5 Unassociated (i defined out) If V = 2.5V .5 v(V) 5 5 -.5 -.5 -1 -1 v(V) If R = 2.5K 3 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross BASIS OF THÉVENIN THEOREM • All linear one-ports have linear I-V graph • A voltage source in series with a resistor can produce any linear I-V graph by suitably adjusting V and I THUS • We define the voltage-source/resistor combination that replicates the I-V graph of a linear circuit to be the Thévenin equivalent of the circuit. • The voltage source VT is called the Thévenin equivalent voltage • The resistance RT is called the Thévenin equivalent resistance 4 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross FINDING VT, RT BY MEASUREMENT 1) VT is the open-circuit voltage VOC (i.e., i = 0) RT + + VT VOC 2a) RT is the ratio of VOC to iSC, the short-circuit current iSC V R T OC ISC 2b) If VT = 0, you need to apply test voltage, then RT i Note direction of i ! + VTEST i VTEST 5 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross FINDING VT, RT BY ANALYSIS 1) Calculate VOC. VT = VOC 2a) If have only resistors and independent voltage/current sources, turn off independent voltage/current sources and simplify remaining resistors OR, if VOC IS NONZERO, Find ISC, then RT=VOC/ISC Finish this later when we look at dependent sources… 6 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross How to turn off voltage/current sources • Disable a voltage source: make it have zero voltage Turn voltage source into short circuit (wire) • Disable a current source: make it have zero current Turn current source into open circuit (air) 7 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross NORTON EQUIVALENT CIRCUIT Corollary to Thévenin: IN ISC (short - circuit current) RN is found the same way as for Thévenin equivalent IN RN 8 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross EXAMPLE Find the Thévenin and Norton equivalents of: A 25 K Find VAB = VOC from voltage divider. Left to right: (4 V rise across 25K + 75K) 75 K 3 V across 75K, 1 V across 25k, + at bottom. + So, VAB = 3 + 6 = 3V = VOC 3 RTH 18.8K 3 .16 10 ISC = 6V/75K+2V/25K = 0.16 mA 2V + 6V B equivalent to and equivalent to A 18.8K A 3V + Thévenin 0.16 mA B Norton 18.8 K B 9 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross Dependent Voltage and Current Sources A linear dependent source is a voltage or current source that depends linearly on some other circuit current or voltage. Example: you watch a certain voltmeter V1 and manually adjust a voltage source Vs to be 2 times this value. This constitutes a voltage-dependent voltage source. Circuit A + V1 - Vs=2V1 + - Circuit B This is just a (silly) manual example, but we can create such dependent sources electronically. We will create a new symbol for dependent sources. 10 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross Dependent Voltage and Current Sources • A linear dependent source is a voltage or current source that depends linearly on some other circuit current or voltage. • We can have voltage or current sources depending on voltages or currents elsewhere in the circuit. Here the voltage V is proportional to the voltage across the element c-d . c + Vcd + - V = A v x Vcd - d A diamond-shaped symbol is used for dependent sources, just as a reminder that it’s a dependent source. Circuit analysis is performed just as with independent sources. 11 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross EXAMPLE OF THE USE OF DEPENDENT SOURCE IN THE MODEL FOR AN AMPLIFIER AMPLIFIER SYMBOL Differential Amplifier V+ V + A V0 A(V V ) AMPLIFIER MODEL Circuit Model in linear region V0 Ri + V1 + AV1 + V0 V0 depends only on input (V+ V-) See the utility of this: this Model when used correctly mimics the behavior of an amplifier but omits the complication of the many many transistors and other components. 12 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross The 4 Basic Linear Dependent Sources Constant of proportionality Parameter being sensed Output Voltage-controlled voltage source … V = Av Vcd Current-controlled voltage source … V = Rm Ic Current-controlled current source … I = Ai Ic Voltage-controlled current source … I = Gm Vcd + Av Vcd - + - Rm Ic Ai Ic Gm Vcd 13 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross NODAL ANALYSIS WITH DEPENDENT SOURCES Example circuit: Voltage controlled voltage source in a branch R5 R1 R3 Va Vb Vc R2 + VAA + AvVc R4 R6 ISS Write down node equations for nodes a, b, and c. (Note that the voltage at the bottom of R2 is “known” so current flowing down from node a is (Va AvVc)/R2.) Va VAA Va A v Vc Va Vb 0 R1 R2 R3 Vc Vb Vc Vb Va Vb Vb Vc ISS 0 R5 R6 R3 R4 R5 CONCLUSION: Standard nodal analysis works 14 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross ANOTHER EXAMPLE OF NODAL ANALYSIS WITH DEPENDENT SOURCES R1 I Va I2 R3 R2 R4 + - Rm I2 I2 Standard technique, except an additional equation is needed if the dependent variable is an unknown current as here. Dependent voltage sources also have I = Va / R2 + (Va - Rm I2)/ R3 and I2 = Va / R2 unknown current—so no KCL at attached nodes. Supernode I = Va (1/R2 + 1/R3 - Rm /R2 R3 ) Solving: around floating dependent voltage So Va = I R2 R3 /(R2 + R3 - Rm ) sources! 15 EECS 40 Fall 2000 Lecture 9-10 W. G. Oldham and S. Ross FINDING VT, RT BY ANALYSIS 1 Calculate VOC. VT = VOC 2a) If have only resistors and independent voltage/current sources, turn off independent voltage/current sources and simplify remaining resistors 2b) If only dependent sources and resistors present, you will need to apply test voltage, calculate i and use RT v TEST i as in previous slide 2c) If both independent and dependent sources (with resistors) present, you may find ISC and let RT=VOC/ISC, or turn v TEST off independent sources and use RT i 16