CHAPTERS 7 & 8 CHAPTERS 7 & 8 NETWORKS 1: NETWORKS 1: December 2002 – Lecture 7b ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002

networks I Today’s learning objectives – review op-amps introduce capacitance and inductance introduce first order circuits introduce concept of complete response

THE OP-AMP FUNDAMENTAL CHARACTERISTICS _+_+ INVERTING INPUT NODE NON-INVERTING INPUT NODE OUTPUT NODE i1i1 i2i2 ioio vovo v2v2 v1v1 RoRo RiRi

Op-Amp Fundamentals for KCL to apply to Op-Amps we must include all currents: i 1 + i 2 + i o + i + + i - = 0 When power supply leads are omitted from diagrams (which they most often are) KCL will not apply to the remaining 3 nodes

are Op-Amps linear elements?

yes.. and no… three conditions must be satisfied for an op-amp to be a linear element: |V o | <= V sat | i o | <= i sat Slew rate >= | dV o /dt |

Example from Text the  A741 when biased +/- 15 V has the following characteristics: v sat = 14 V i sat = 2 mA SR = 500,000 V/S So is it linear?  When R L = 20 kOhm or 2 kOhm?

Using Op-Amps Resistors in Op-Amp circuits > 5kohm Op-Amps display both linear and non- linear behavior

Remember: for Ideal Op-Amp node voltages of inputs are equal currents of input leads are zero output current is not zero

One more important Amp difference amplifier See Figure 6.5-1, page 213

Practical Op-Amps characteristic idealpracticalsample Bias current 0> nA Input resistance infinitefinite Mohm Output resistance 0> Kohm Differential gain infinitefinite V/mV Voltage saturation infinitefinite 6V-15V

What you need to know Parameters of an Ideal Op Amp Types of Amplification Gain (K) vs. Which nodes and Amps circuits are needed to achieve same How to identify which type of circuit is in use (effect) How to solve Op Amp problems

new concepts from ch. 7 energy storage in a circuit capacitors series and parallel inductors series and parallel using op amps in RC circuits

DEFINITION OF CAPACITANCE Measure of the ability of a device to store energy in the form of an electric field. CAPACITOR: IMPORTANT RELATIONSHIPS: +–+– i + + _ _

CALCULATING i c FOR A GIVEN v(t) Let v(t) across a capacitor be a ramp function. t v vv tt As Moral: You can’t change the voltage across a capacitor instantaneously.

VOLTAGE ACROSS A CAPACITOR

ENERGY STORED IN A CAPACITOR

CAPACITORS IN SERIES +–+– C1C1 C2C2 C3C3 + v 1 -+ v 2 -+ v 3 - i v KVL

Capacitors in series combine like resistors in parallel. CAPACITORS IN SERIES

CAPACITORS IN PARALLEL C1C1 C2C2 C3C3 i i1i1 i2i2 i3i3 KCL Capacitors in parallel combine like resistors in series.

HANDY CHART ELEMENT CURRENTVOLTAGE

DEFINITION OF INDUCTANCE Measure of the ability of a device to store energy in the form of a magnetic field. INDUCTOR: IMPORTANT RELATIONSHIPS: i + _ v

CALCULATING v L FOR A GIVEN i(t) Let i(t) through an inductor be a ramp function. t i ii tt As Moral: You can’t change the current through an inductor instantaneously.

CURRENT THROUGH AN INDUCTOR

ENERGY STORED IN AN INDUCTOR

INDUCTORS IN SERIES L1L1 L2L2 L3L3 + v 1 -+ v 2 -+ v 3 - i KVL Inductors in series combine like resistors in series.

INDUCTORS IN PARALLEL L1L1 L2L2 L3L3 v i1i1 i2i2 i3i3 KCL +–+–

INDUCTORS IN PARALLEL Inductors in parallel combine like resistors in parallel.

HANDY CHART ELEMENT CURRENTVOLTAGE

OP-AMP CIRCUITS WITH C & L _+_+ i1i1 i2i2 ioio vovo v2v2 v1v1 CfCf RiRi +–+– vsvs Node a

QUIZ: Find v o = f(v s ) _+_+ i1i1 i2i2 ioio vovo v2v2 v1v1 RfRf +–+– vsvs Node a LiLi

ANSWER TO QUIZ

IMPORTANT CONCEPTS FROM CH. 7 I/V Characteristics of C & L. Energy storage in C & L. Writing KCL & KVL for circuits with C & L. Solving op-amp circuits with C or L in feedback loop. Solving op-amp circuits with C or L at the input.

new concepts from ch. 8 response of first-order circuits the complete response stability of first order circuits

1st ORDER CIRCUITS WITH CONSTANT INPUT +–+– t = 0 R1R1 R2R2 R3R3 Cvsvs + v(t) -

Thevenin Equivalent at t=0 + RtRt C +–+– V oc + v(t) - KVL i(t) + -

SOLUTION OF 1st ORDER EQUATION

SOLUTION CONTINUED

WITH AN INDUCTOR +–+– t = 0 R1R1 R2R2 R3R3 Lvsvs i(t)

Norton equivalent at t=0 + RtRt I sc + v(t) - L i(t) KCL

SOLUTION

HANDY CHART ELEMENT CURRENTVOLTAGE

IMPORTANT CONCEPTS FROM CHAPTER 8 determining Initial Conditions setting up differential equations solving for v(t) or i(t)