1. 1 ECE 3144 Lecture 17 Dr. Rose Q. Hu Electrical and Computer Engineering Department Mississippi State University

2. 2 Review for Chapter 3 Nodal analysis for an N-node circuit –Unknown parameters are nodal voltages –Apply KCL to each node –There are N-1 independent equations – 4 cases Loop analysis for an N-loop circuit –Unknown parameters are loop currents –Apply KVL to each loop –There are N independent equations –4 cases Ideal op-amp –For an ideal op-amp, i + = i - = 0 and v + = v -

3. 3 Ideal Op-amp example For the first op-amp v a = v b = 0 Apply the KCL at terminal a a b c d e => Apply the KCL at terminal d (1) => (2) For the second op-amp v c = v d = v o (3) There are three equations three unknowns => v o = -11.4 V

4. 4 Chapter 4: Additional analysis tools Circuit linearity Superpostition for linear circuits Thevenin’s rule and Norton rule Review for ideal op-amp Maximum power transfer Matlab and Pspice techniques

5. 5 Linear system Linear element: a passive element that has a linear voltage-current relationship- example: v(t) = Ri(t) Linear dependent source: a dependent current or voltage source whose output current or voltage is proportional only to the first power of a specified current or voltage variable in the circuit Linear circuit: a circuit composed entirely of independent sources, linear dependent sources and linear elements. All circuits we have analyzed are linear circuits For a linear system, it satisfies both additivity and homogeneity. Linear system x 1 (t)y 1 (t) Linear system x 2 (t)y 2 (t) additivity Linear system x 1 (t)+ x 2 (t)y 1 (t)+ y 2 (t) Linear system x 1 (t)y 1 (t) Linear system Ax 1 (t)Ay 1 (t) homogeneity

6. 6 Using Homogeneity to solve the circuit problem For the circuit shown, we wish to determine the output current I o. Rather than approaching the problem in the straightforward way we have introduced and calculate I 1 and I 2, then I o. We will use Homogeneity to solve the problem. Remember that Homogeneity says if the single input is scaled by A, then the output is scaled by A too. We just simply assume the output current I o =1 mA. This output current will yield a value for the input current source I. We will then find the scaling ratio of the true current value I to the calculated one. The same scaling ratio times 1 mA will yield the actual value for I o. If I o = 1mA, I 3 = 0.5 mA (based on the current divider) =>I 1 = 1.5mA R 2 = 4k+8k = 12k. (The resistance to the left side of the input source) R 1 = 2k+6k//3k = 4k (The resistance to the right side of the input source) Based on the current divider =>I 2 = 0.5 mA=> I= I 1 +I 2 = 2 mA Thus the assumption of I o = 1mA produces a source current 2 mA. Since the actual source current is 6 mA, the scaling factor is 3. Thus the actual output current is 3 mA.

7. 7 Homework for Lecture 17 Problems 4.3, 4.4 Due Feb 25