1Copyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003EECS 42 Introduction to Digital Electronics Andrew R. NeureutherLecture #5 •Recap I vs. V•Thevenin and Norton Eq. Circuits•Load Line Graphical Solution•Nonlinear elements and powerSchwarz and Oldham 3.1-3.3http://inst.EECS.Berkeley.EDU/~ee42/Copyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003VSIEXAMPLE of I vs V GRAPHSimple Circuit, e.g. voltage source + resistor.2V2KIf two circuit elements are in series the current is the same; clearly the total voltage will be the sum of the voltages i.e. VS+ IR. We can graph this on the I-V plane. We find the I-V graph of the combination by adding the voltages VSand I R at each current I. Lets do an example for =2V, R=2KCombined 2K + 2V+-2K2V I+-VV (Volt)I24(ma)5At 2 mA2V + 4V = 6VCopyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003GRAPHICAL EQUIVALENT CIRCUITCombined CircuitVOUT (Volt)IINI24(ma)5R1R3R2ABCISSVOUTISCVOCISS= 1 mA R1= 1kΩR2 = 6 k ΩR3 = 3 kΩShort Circuit : VOUT= 0IIN= -ISS = - 1 mAIINOpen Circuit: IIN= 0Note R2||R3 = 2 kΩVOUT= ISS× R2||R3= 1 mA × 2kΩ = 2VThird PointIIN= 1 mAKCL => I2||3= ISS+ IINVOUT= I2||3× R2||R3= 2 mA × 2kΩ = 4VKey: It will always be the case that for linear circuit elements the I vs. V is a straight line.Copyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003SIMPLEST EQUIVALENT CIRCUITSRTh+−VThVOUTITheveninINVOUTRNINortonAn adequately equivalent circuit is one that has an I vs. V graph that is identical to that of the original circuit.RTH= RN= VOC / (-ISC) = 2V /(- ( -1 mA)) = 2 kΩVTH= VOC = 2VIN= -ISC = - ( -1 mA) = 1mACombined CircuitVOUT (Volt)IINI24(ma)5ISCVOCRTHis the inverse of the slopeCopyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003RTH= RNSHORTCUT METHODSR1R3R2ABCISSVOUTRTH= RN= VOC / (-ISC)VOC= ISS×R2||R3ISC= -ISSRTH= RN= (ISS×R2||R3)/(-(-ISS)) = R2||R3Look at algebraic relation for the example circuit.In General turn all of the independent sources to zero and find the remaining equivalent resistance seen looking into the terminals.Currents sources are turned to zero current (with any voltage) and voltage sources are turned to zero voltage (with any current).Copyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003Thevenin Equivalent Circuit Example+-250KS9µA1MDISCVOCISC= -9µA - 1V/1MΩ = -10 µA 1VVOC= 2V from Node Analysis Method Week 5 RTH= RN= VOC / (-ISC) = 2V/ 10 µA = 200kΩ+-2V200KSDIDVDS+ -2Copyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003• I vs. V for ideal voltage source is a vertical line at V = VSV• I vs. V for ideal current source is a horizontal line at I = ISC• I vs. V for a circuit made up of ideal independent sources and resistors is a straight line.• The simplest circuit for a straight line is an ideal voltage source and a resistor (Thevenin) or a current source and a parallel resistor (Norton)• The easiest way to find the I vs. V line is to find the intercepts where I = 0 (open circuit voltage VT) and where V = 0 (Short circuit current IN)• The short-cut for finding the (slope)-1= RT= RNis to turn off all of the independent sources to zero and find the remaining equivalent resistance between the terminals of the elements.I vs. V and Equivalent CircuitsCopyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003Example of I-V GraphsNonlinear element, e.g. lightbulbIf we think of the light bulb as a sort of resistor, then the resistance changes with current because the filament heats up. The current is reduced (sort of like the resistance increasing). “sort of” because a resistor has, by definition a linear I-V graph and R is always the same. But for a light bulb the graph kind of “rolls over”, becoming almost flat. Consider a 100 Watt bulb, which means at at the nominal line voltage of 117 V the current is 0.85A. (117 V times 0.85 A = 100W).V (Volt)I0.40.8(A)100I=0.85A at 117VI+-VCopyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003I-V Graphs as a method to solve circuitsWe can find the currents and voltages in a simple circuit graphically. For example if we apply a voltage of 2.5V to the two resistors of our earlier example: We draw the I-V of the voltage and the I-V graph of the two resistors on the same axes. Can you guess where the solution is?At the point where the voltages of the two graphs AND the currents are equal. (Because, after all, the currents are equal, as are the voltages.)1K4K2.5V+-Combined 1K + 4KI24(ma)V (Volt)52.5VSolution: I = 0.5mA, V = 2.5VI+-VThis is called the LOAD LINE method; it works for harder (non-linear) problems.Copyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003Another Example of the Load-Line MethodLets hook a 2K resistor + 2V source circuit up to an LED (light-emitting diode), which is a very nonlinear element with the IV graph shown below. Again we draw the I-V graph of the 2V/2K circuit on the same axes as the graph of the LED. Note that we have to get the sign of the voltage and current correct!! (Think of IOUTversus VOUTfor the 2V/2K circuit)I24(ma)V (Volt)5Solution: I = 0.7mA, V = 1.4VI+-V+-2V2KLEDLEDAt the point where the two graphs intersect, the voltages and the currents are equal, in other words we have the solution. IOUTVOUTCopyright 2003, Regents of University of CaliforniaLecture 5: 09/09/03 A.R. NeureutherVersion Date 09/08/03EECS 42 Intro. Digital Electronics Fall 2003The Load-Line MethodGiven a circuit containing a two-terminal non-linear element “NLE”, and some linear components.1V+-250KSNon-linear
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