Sample Problem: Differential Amp

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En provenance du cours de Georgia Institute of Technology

Linear Circuits 1: DC Analysis

89 notes

Georgia Institute of Technology

Linear Circuits 1: DC Analysis

89 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 amplifiers.

And we're going to work with a circuit that is a differential amplifier.

The problem is to find the output voltage, V sub out, in terms of the input voltages.

So we look at our circuit that's given,

it's an operational amplifier circuit with four resistors and two input voltages.

One input voltage is V sub 1 which is associated with the inverting input of

the op amp and a second voltage source which is associated with the non inverting

input of the op amp.

We also have resistors R1, R2, R3, and R4.

And V out is measured at the output of the op amp.

So to solve this problem, we have to first understand the basic properties of an op

amp that are important for linear circuits.

First of all, we have the symbol for

the op amp as its inverting input, it's non-inverting input.

It has currents associated with both of these inputs, and

it has voltages also associated with these two inputs.

The properties of the ideal op amp that we use in linear circuits are that

the currents flowing into the op amp are equal to zero,

i- = i+ = 0, and the voltages are equal across,

whether it's the inverting or non-inverting input of the op amp.

So these are the properties that we use along with the other tools that we've

already used in previous problems to solve these op amp circuits.

So in general we may use these properties of op amp along with nodal analysis or

in some circuits mesh analysis to solve for

things like the output voltage in terms of the input voltage.

So, let's see how we solve this problem.

The first thing that we notice is that if we're trying to solve this circuit,

we need to somehow relate the input voltages to the output voltage.

And so, we can do that by using the ideal properties of the op amp and

one of our previous tools that we've learned, in voltage division.

Across R3 and R4 allows us to find

the voltage at the non-inverting input of the op-amp.

So our voltage at the non-inverting input of the op-amp

using voltage division is equal to V sub 2 times R sub 4 divided by R3 plus R4.

So that's our voltage at this point in a circuit.

So we know that this voltage is equal to the voltage at the inverting input

which is what we've called node one here.

So V plus is equal to V minus and in terms of the input voltage V sub

2 this is the value for V plus and V minus.

Now we can use Kirchhoff's current law to sum the currents into node 1 to

find the relationship between the output voltage and the input voltages V1 and V2.

So we're going to go to Node 1 and we're going to use Kirchhoff's current law

We're going to sum the currents into node 1.

So, the first current is the current through R1 is going to be equal

to the voltage on one side of the resistor minus that on the other

divided by the resistor value.

So, it's going to be V1 minus V

minus divided by R1 and we know what V minus is.

It's the terms of V sub 2.

And so if we sum the second current through the feedback resistor R sub 2,

it's going to be V out minus V sub minus divided by R2.

That's going to be plus V out minus V minus divided by R2.

And our third current is the current that is flowing out of

the op amp back into that node 1.

And we know that value is 0.

I'm going to write it in, but It's one of our basic

characteristics of op amps that we want to make sure that we understand.

So now we can write V out in terms of V sub 1 and

V sub 2, which is part of our V minus term.

And so if we do that, then we end up with a V out.

Which is equal to a rather long expression,

R2 divided by R1 times 1 plus R1 divided by R2.

Plus R4, or times R4, divided by R3, plus R4,

times V2, minus R2, over R1,

times V1, which is V2,

minus R2, over R1, V1.

With these problem,

with this differential amplifier circuits like what was drawing here.

Many times what people do is all set R1 equal to R3.

And R2 equal to R4.

That our equation which relates V out to V1 and

V2 reduces to V out is equal to R2 divided by R1,

V2 minus V1.

And so what we end up with is an output voltage which is

the amplification of the difference between the two input voltages.

So we have an input voltage signal from V2 and from V1.

And if you look back the circuit, v2 is tied to the non inverting terminal.

It's not inverted.

And V1 is tied to the inverting terminal and is inverted.

And we end up with this amplification factor that we can control by

setting our values for R1, R2, R3 and R4 in our circuit.

So V out becomes a factor

based on R1 and R2 values.

And also the difference in the signals between the two input voltages.

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