Download Simplifying Complex Circuits with Nonlinear Modeling - Prof. Aurangzeb Khan and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! 1 Lecture #2 Nonlinear 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 #2 The 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 #2 Demonstration 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) = 0.012A The output voltage is then Vout = IR2 = 0.012 X 300 = 3.6V Circuit (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 #2 Demonstration of the difficulty in solving nonlinear problems (cont.) In order to determine Vout we 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) + Vout Again, 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. 2 Lecture #2 Possible model of the problem (constant voltage drop model) • One possible model for the forward bias diode 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 #2 A 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 #2 Load 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 #2 Ideal diode model Open circuit OffOn Short circuit ID = (6-0)/200=0.03A VOUT=0V