1Lecture #2Nonlinear Modeling• Nonlinear problems are difficult to solve• The diode is a nonlinear device• Picewise linear models can simplifying the solution of non linear circuits problems.Lecture #2The purpose of modeling• Nonlinear problems are much more difficult than linear ones. These problems could be impossible to solve manually and could require huge amount of time if solved on a computer.• One possible solution of the above mentioned problem is to approximate the nonlinear relationship with a model that has a linear relationship.• The trust of nonlinear modeling is direct towards this end.• The modeling not only simplifies the solution, it also allows the designer to understand how the circuit behaves. Modeling often increases the conceptual understanding of the circuit operation.Lecture #2Demonstration of the difficulty in solving nonlinear problems• Consider a linear circuit (a) and nonlinear circuit (b). Determine I and Vout?• Solution: In circuit (a):By simple ohm’s law we can find current I as,I = V1/R1 + R2 = 6 /(200 + 300) = 0.012AThe output voltage is then Vout= IR2 = 0.012 X 300 = 3.6VCircuit (b):The current I can be determine by using diode equation as I = Is(eqV/kT-1) = 10-10(eVout/0.026 - 1)There is no close form solution of the above equation.Lecture #2Demonstration of the difficulty in solving nonlinear problems (cont.)In order to determine Voutwe have to solve another equation which can be written as by Kirchhoff’s law,V1 = IR1 + Vout⇒V1= 200 X 10-8(eVout/0.026-1) + VoutAgain, there is no close form solution of the above equation.Perhaps the quickest method for solving this problems is a trial and error iterative method. If we guess many time, finally we will be able to show that, when Vout= 0.505215 ~ 0.5V, the right side of the above equation is 5.99V, which is essentially equal to the value of the left side of the equation.Finally ,I =0.02747≅0.027A.2Lecture #2Possible model of the problem(constant voltage drop model)• One possible model for the forward biasdiode is a simple 0.6V voltage source.• When this model replaces the diode, the circuit appear as shown in the figure and is very easy to analyze. • For this circuit the current is calculated to be•I=(V1-0.6)/200 = 0.027A•And the Vout= 0.6V• These values compare well to the results calculated from the exact equations, but much easier to obtained.• The above example demonstrate that how model simplifies the solution.Lecture #2A load line approach• An alternate and more traditional graphical method to analyze a circuit containing a nonlinear element is that of using a load line.• The load line can yield accurate results and used extensively in the evaluation of the electronic circuits.Lecture #2Load line analysis• In this approach the series circuit shown here can be split into a non linear element and the remaining external circuit.Load line equation: Vab=V1-IR1 Lecture #2Ideal diode modelOpen circuitOffOnShort circuitID = (6-0)/200=0.03AVOUT=0V3Lecture #2 Lecture #2Lecture #2 Lecture #24Lecture #2 Lecture