1. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 1 Linear Circuits -Special Properties Circuits consisting only of linear elements are linear circuits. There are simple “equivalent circuits” for “one-port” linear circuits.

2. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 2 TWO-TERMINAL LINEAR RESISTIVE NETWORKS (“One Port” Circuit) Interconnection of two-terminal linear resistive elements with only two “accessible” terminals  + a b

3. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 3 I-V CHARACTERISTICS OF LINEAR TWO-TERMINAL NETWORKS i +  v +  5V Apply v, measure i, or vice versa 5.5 1 v(V) -.5 i(mA) Associated (i defined in) 5.5 1 v(V) -.5 i(mA) Unassociated (i defined out) 5K If V = 2.5V If R = 2.5K Associated i +  v +  5V Apply v, measure i, or vice versa Unassociated

4. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 4 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 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 V T is called the Thévenin equivalent voltage The resistance R T is called the Thévenin equivalent resistance THUS

5. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 5 FINDING V T, R T BY MEASUREMENT +  VTVT RTRT +  V OC 1) V T is the open-circuit voltage V OC (i.e., i = 0) 2a) R T is the ratio of V OC to i SC, the short-circuit current i SC 2b) If V T = 0, you need to apply test voltage, then i V R TEST T  +  V TEST i Note direction of i !

6. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 6 FINDING V T, R T BY ANALYSIS 1) Calculate V OC. V T = V OC 2a) If have only resistors and independent voltage/current sources, turn off independent voltage/current sources and simplify remaining resistors OR, if V OC IS NONZERO, Find I SC, then R T =V OC /I SC Finish this later when we look at dependent sources…

7. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 7 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)

8. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 8 NORTON EQUIVALENT CIRCUIT Corollary to Thévenin: R N is found the same way as for Thévenin equivalent R N I N

9. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 9 EXAMPLE B So, V AB =  3 + 6 = 3V = V OC I SC = 6V/75K+2V/25K = 0.16 mA Find the Thévenin and Norton equivalents of: A 25 K   + 2 V  + 6 V 75 K  (4 V rise across 25K + 75K)  3 V across 75K, 1 V across 25k, + at bottom. and equivalent to 18.8 K 0.16 mA B A Norton Find V AB = V OC from voltage divider. Left to right: 18.8K 3V B A equivalent to Thévenin +

10. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 10 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 V 1 and manually adjust a voltage source V s to be 2 times this value. This constitutes a voltage-dependent voltage source. V1V1 + - Circuit A V s =2V 1 + - 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.

11. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 11 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. 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. + - c d V cd + - V = A v x V cd Here the voltage V is proportional to the voltage across the element c-d.

12. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 12 EXAMPLE OF THE USE OF DEPENDENT SOURCE IN THE MODEL FOR AN AMPLIFIER V 0 depends only on input (V +  V - ) +  A V+V+ VV V0V0 Differential Amplifier AMPLIFIER SYMBOL +  +  V0V0 AV 1 +  V1V1 RiRi Circuit Model in linear region AMPLIFIER MODEL 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.

13. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 13 G m V cd Voltage-controlled current source … I = G m V cd A i I c Current-controlled current source … I = A i I c +-+- R m I c Current-controlled voltage source … V = R m I c The 4 Basic Linear Dependent Sources Voltage-controlled voltage source … V = A v V cd A v V cd +-+- Parameter being sensedConstant of proportionality Output

14. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 14 NODAL ANALYSIS WITH DEPENDENT SOURCES Example circuit: Voltage controlled voltage source in a branch Write down node equations for nodes a, b, and c. (Note that the voltage at the bottom of R 2 is “known” so current flowing down from node a is (V a  A v V c )/R 2.) CONCLUSION: Standard nodal analysis works

15. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 15 ANOTHER EXAMPLE OF NODAL ANALYSIS WITH DEPENDENT SOURCES Standard technique, except an additional equation is needed if the dependent variable is an unknown current as here. I2I2 I = V a / R 2 + (V a - R m I 2 )/ R 3 and I 2 = V a / R 2 Solving: I = V a (1/ R 2 + 1/ R 3 - R m / R 2 R 3 ) So V a = I R 2 R 3 /(R 2 + R 3 - R m ) R 4 I 2 R 2 I R 3 R 1 V a + - R m I 2 Dependent voltage sources also have unknown current—so no KCL at attached nodes. Supernode around floating dependent voltage sources!

16. W. G. Oldham and S. RossEECS 40 Fall 2000 Lecture 9-10 16 FINDING V T, R T BY ANALYSIS 1 Calculate V OC. V T = V OC 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 as in previous slide i v R TEST T  2c) If both independent and dependent sources (with resistors) present, you may find I SC and let R T =V OC /I SC, or turn off independent sources and use i v R TEST T 