Sample Problem: Op Amps 2

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En provenance du cours de Georgia Institute of Technology

Linear Circuits 1: DC Analysis

106 notes

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

106 notes

This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

À partir de la leçon

Module 2

The module describes some basic principles used in circuit analysis.

Sample Problem: DC Op Amp 19:29

Sample Problem: DC Op Amp 25:33

Sample Problem: DC Op Amp 34:23

Sample Problem: DC Op Amp 46:22

Sample Problem: DC Op Amp 55:33

Sample Problem: Current Division 13:57

Sample Problem: Current Division 25:20

Sample Problem: Parallel and Series Resistors 12:24

Sample Problem: Parallel and Series Resistors 28:07

Sample Problem: KVL6:05

Sample Problem: KVL 25:48

Sample Problem: KVL 32:28

Sample Problem: KCL 12:12

Sample Problem: KCL 24:43

Sample Problem: KCL 32:51

Sample Problem: KCL 41:26

Sample Problem: KCL 52:10

Sample Problem: Op Amps 19:17

Sample Problem: Op Amps 25:29

Sample Problem: Op Amps 34:59

Sample Problem: Nodes, Branches, Paths, and Loops3:19

Sample Problem: Summing Op Amp6:05

Sample Problem: Differential Amp6:47

Sample Problem: Inverting Op Amp5:09

Sample Problem: Non-inverting Op Amp7:54

Sample Problem: Max Power (Depend Sources)4:06

Sample Problem: Dependent Sources 15:10

Sample Problem: Dependent Sources 24:48

Sample Problem: Series/Parallel (Independ Sources) 12:59

Sample Problem: Series/Parallel (Independ Sources) 27:22

Sample Problem: Series/Parallel (Independ Sources) 33:00

Sample Problem: Series/Parallel (Independ Sources) 44:35

Sample Problem: Op Amps 49:10

Sample Problem: Op Amps 59:29

Sample Problem: Op Amps 65:33

Sample Problem: Op Amps 74:23

Sample Problem: Op Amps 86:22

Sample Problem: Op Amps 95:33

Sample Problem: Op Amps 109:17

Sample Problem: Op Amps 115:29

Sample Problem: Op Amps 124:59

Rencontrer les enseignants

Dr. Bonnie H. Ferri
Professor
Electrical and Computer Engineering
Dr. Joyelle Harris
Director of the Engineering for Social Innovation Center
School of Electrical and Computer Engineering
Dr Mary Ann Weitnauer

The topic of this problem is Operational Amplifier Circuits.

And the problem is to determine V out in terms of the input voltages.

In this circuit we have two op amps, an upper and lower op amp.

We also have input voltages V sub 1 associated with

the non-inverting input of the upper op amp, and

V sub 2 associated with the non-inverting amp of the lower op amp.

We have this resistor network at the output of the op amp where V out is

measured across that series of resistors and parallel combinations of resistors.

So we're going to use our properties of the ideal op amp to solve the circuit.

Namely that the currents into the op amp are equal to zero and the voltages

at the inverting and the non-inverting input are equal to one another.

So if we use that property, then we know that V sub 1,

which is tied to the non-inverting input of the upper op amp, is the same voltage

down below at the inverting input and we have the same voltage out

at this node at the output of our op amp circuit.

So that node is at V sub 1.

Similarly, if we look at the lower circuit or the lower op amp,

which is part of our circuit, we see that the voltage at the node

between R2 and R1 is equal to V sub 2 using the same approach.

So now we have our input

voltages conveyed over to the output side of our circuit.

Now if we use Kirchhoff's Current Law to sum the currents into the node where

the voltage is V1 and also into the node where the voltage is V2, we

can find some equations which relate those input voltages to the output voltage.

Let's start this upper node where the voltage is V1.

And so we are going to sum the currents into that node.

We have the current down through R2.

We have the current out of the inverting input of the op amp,

we have the current up through R1 and we have a current up through R sub G.

There's four currents.

So let's start with the current through R2 at the top coming down.

That current is going to be V out, minus V1, divided by R2.

The current out of the op amp is 0,

there's no current flowing out of the op amp.

The current through R1 is going to be V sub A,

minus V1, divided by R1,

plus the current through R sub G which

is V2 minus V1, divided by R sub G.

So there's our four currents and they're equal to zero.

We're going to call that equation 1.

Equation 2 for the node down below summing the currents into that node.

First of all, the current up through R2 is going to be 0 minus V sub 2,

divided by R2.

The current out of the op amp is 0.

The current down through R sub G is

going to be V1 minus V2 over R sub G.

And then we have the current through R sub 1 from node A down to

the node where the value of the voltage is V2.

So it's going to be VA minus V2, divided by R1.

That's our last current, and the sum of those currents is equal to 0.

So we have two equations which have the input voltages V1 and V2.

The output voltage V sub out, and this V sub A, for node A in our circuit.

So if we take for example equation 2, and we solve equation 2 for V sub A.

And then we take V sub A expression and

plug into equation 1, then we'd have an equation

that related V out, and V1, and V2.

So if we indeed do that, then we end up with a V out which is equal to V1

minus V2, so it's a difference amplifier.

Times this factor which is a function of the resistors that are in our circuit.

The R1 and R2 values and the R sub G values as well.

So it's a difference amplifier where the voltage is equal

to the difference in the input values.

For the voltages times this factor, which is a function of the resistors.

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