Chap.2-handout

Chap.2-handout - 8/27/2010 1 1 Operational Amplifiers 1 1-...

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Unformatted text preview: 8/27/2010 1 1 Operational Amplifiers 1 1- Covered are Sections 2.1-2.4 and 2.7 - Skipped are Sections 2.5, 2.6 and 2.8 Chapter expectations Use opamp to learn and apply the amplifier concepts: gain, io resistances, circuit model, and applications in inverting, non-inverting, difference amps and others. Apply opamp properties to the analysis and design of opamp circuits 2 In this chapter, we will study a most typical voltage amplifier the opamp Opamp possesses characteristics that are very close to an ideal voltage amplifier: infinite Ri, zero Ro, and infinite gain. In terms of the amplifier model, how is the ideal voltage amplifier defined??? An ideal Volt.Amp has R i =( ) R o =( ) A vo = ( ) 8/27/2010 2 3 We will see in the opamp and its circuits, That all the voltage AMPLIFIER CONCEPTS are realized, which we learned in Chapter One. 4 The Circuit Symbol: What is opamp? An amplifier built as an integrated circuit and possesses almost ideal voltage amp property! Terminals: 1=Inverting input 2=Noninverting input 3=Output showing DC supplies 4, 5=DC supply terminals if more than one opamp in the package, then the DC supply pins are shared by all opamps. 8/27/2010 3 5 14-pin 5-pin 8-pin How many opamps in each chip ?? Hint : DC supplies lines are shared by different opamps IC Opamp Packages 6 Ideal Opamp Property : (1) R i = (2) R o = 0 (3) A cm = 0 [ Common-mode input: V cm = (v 1 +v 2 ) ] (4) A = [ Differential-mode input: V d = v 2-v 1 ] (5) Infinite Bandwidth Equivalent circuit: showing properties (1)-(4) Common-mode & Differential-mode Gain: V o = A cm V cm + A V d Meaning of Ideal Property : (1) Input terminals take in 0 Amp (2) v o =A(v 2-v 1 ) indept. of R L (3) (4) V d =0 v 1 = v 2 ( virtual short-circuit ) (5) 6 V o = A cm V cm + A V d ideal voltage amp 8/27/2010 4 7 V o = A cm V cm + A V d A cm = 0 we want to eliminate all common-mode (noise) A = all important signal is in the differential-mode 8 Example: TV signal flowing in Coaxial Cable center conductor to one input ( + ) shield to the other input (- ) V d = V center V shield = V signal The noise V noise is the same in the center conductor and the shield V center = V noise + V signal V shield = V noise V center-V shield = V signal Why do you want to handle the signals differentially??? 8/27/2010 5 Note: When there are two inputs, v 1 and v 2 , they can be represented in terms of Common-mode and Differential-mode...
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Chap.2-handout - 8/27/2010 1 1 Operational Amplifiers 1 1-...

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