Sample Problem: Op Amps 11

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來自 Georgia Institute of Technology 的課程

Linear Circuits 1: DC Analysis

94 個評分

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

94 個評分

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.

從本節課中

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

與講師見面

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 input 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 0.

And the voltages at the inverting and

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 of 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 could

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

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

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

We 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,

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

The current of 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 0.

We're going to call that equation one.

Equation two, for the node down below, sum into currents into that node.

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

A current pop-up is 0.

The curent 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 some of these 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 two and we solve equation two for V sub A.

And then we take V sub A expression and

plug into equation one, 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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