1. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 1 LECTURE 14 Today: Summary of Linear and Nonlinear Rules Load-line graphical method for circuits with nonlinear elements Three terminal devices, solution methods and power Unsolicited Advice: Make learning your priority

2. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 2 Art, Michelle and Rob (Be Like Michelle) Reduce or eliminate: Over-focusing on policy and grades –Better off devoting energy towards understanding Perfectionism –Don’t expect perfection of yourself or your instructors Competition –Cooperation, helping others benefits you too Big don’ts for success at school and at work

3. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 3 Circuit Laws and Tricks Linear Circuits: KCL, KVL Nodal Analysis Thévenin and Norton equivalents Nonlinear Circuits: KCL, KVL Nodal Analysis Special Graphical Technique for Circuits containing linear and nonlinear elements (Load-line Method): Combine all linear parts into a simple Thévenin equivalent Circuit attached to the nonlinear element. Plot the I-V Characteristics of NLE and Thévenin circuit on same axes, recognizing the intersection as the solution.

4. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 4 Example of Load-Line Method We have a circuit containing a two-terminal non-linear element “NLE”, and some linear components. 1V +-+- 250K S Non- linear element 9A9A 1M D +-+- 2V 200K S NLENLE D Then define I and V at the NLE terminals (typically associated signs) First replace the entire linear part of the circuit by its Thevenin equivalent. IDID V DS +-+-

5. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 5 Example of Load-Line method (con’t) And have this connected to a linear (Thévenin) circuit +-+- 2V 200K The solution ! Given the graphical properties of two terminal non-linear circuit (i.e. the graph of a two terminal device) V DS IDID  A) 10 (V) 1 2 IDID D NLENLE S Whose I-V can also be graphed on the same axes (“load line”) Application of KCL, KVL gives circuit solution +-+- 2V 200K S NLENLE D IDID V DS +-+-

6. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 6 Load-Line method But if we use equations instead of graphs, it could be accurate The method is graphical, and therefore approximate It can also be use to find solutions to circuits with three terminal nonlinear devices (like transistors) +-+- 2V 200K The solution ! V DS IDID  A) 10 (V) 1 2 IDID D S Application of KCL, KVL gives circuit solution

7. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 7 Three-Terminal Devices We can set one variable at each terminal and the I-V behavior of the device will determine the value of the two unknowns. For example apply a V GS and V DS, then measure I G and I D. 3-Terminal Device IDID D G S With three terminals we have four independent variables (two voltages and two currents) Similarly we could apply a fixed V GS and I G and plot the two-terminal characteristic I D versus V DS.

8. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 8 V DS IDID  A) 10 (V) 1 2 Three-Terminal Parametric Graphs Concept of 3-Terminal Parametric Graphs: We set a voltage (or current) at one set of terminals (here we will apply a fixed V GS, IG=0) and conceptually draw a box around the device with only two terminals emerging so we can again plot the two-terminal characteristic (here I D versus V DS ). 3-Terminal Device IDID D G S V GS +-+- V GS = 3 V GS = 2 V GS = 1 But we can do this for a variety of values of V GS with the result that we get a family of curves.

9. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 9 Graphical Solutions for 3-Terminal Devices Now draw I D vs V DS for the 2V - 200K  Thevenin source. First select V GS (e.g. 2V) and draw I D vs V DS for the 3-Terminal device. V DS IDID  A) 10 (V) 1 2 V GS = 3 V GS = 2 V GS = 1 IDID G +-+- +-+- V 2V D 200K S V DS IDID  A) 10 (V) 1 2 The only point on the I vs V plane which obeys KCL and KVL is I D = 5  A at V DS = 1V. The solution ! We can only find a solution for one input (V GS ) at at time:

10. EECS 40 Fall 2002 Copyright Regents of University of California S. Ross and W. G. Oldham 10 Three-Terminal Devices – Power Flow We can set any two variables and the I-V behavior of the device will determine the value of the other two. 3-Terminal Device IDID D G S With three terminals we have two independent measures of power (e.g. one at GS and one at DS). Defining all currents as inward, the power dissipated is simply the algebraic sum of V GS I G and V DS I D, i. e., P (in) = V GS I G + V DS I D. See the textbook, p. 117 for more generality. IGIG