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Sample Problem: Op Amps 10

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Linear Circuits 1: DC Analysis

Instituto de Tecnologia da GeórgiaInstituto de Tecnologia da Geórgia

4.5 (195 classificações) | 15K alunos inscritos

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.

Visualizar o programa do curso

Avaliações

4.5 (195 classificações)

5 stars
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4 stars
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3 stars
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2 stars
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1 star
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DM

May 09, 2018

Great course, learned a lot of things that they don't teach at my school, Thanks for the opportunity.

SA

Apr 10, 2019

its a great course to develop a greater knowledge in linear circuits and has helped me a lot.

Na lição

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

Ministrado por

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

Transcrição

The topic of this problem is Operational Amplifier Circuits and

the problem is a design problem.

Suppose we have a bio sensor with an output range

of 0.5 volts through 1 volts and it's an oxygen sensor.

And the oxygen sensor itself has a range of 1 to 250 parts per million.

So as the oxygen concentration changes from 1 to 250 parts per million,

the sensor itself puts out a signal which has a range of 0.5 to 1 volts.

But we need to do something with our signal and to boost it up and

to get over the proper range for our downstream circuitry.

So we want to design a sensor interface circuit that provides an output range

of 0 to 5 volts for this biosensor system.

The first thing we acknowledge is that we're going to use linear circuits

to solve this.

When we have linear circuits, we can use a full arm for

our expressions that is associated with Linear

circuits and with linear equations in general.

That is that the output is equal to the slope times the input plus sum intercept.

We know that we have two conditions for

our circuit based on our problem statement.

That is if we want the output of the system to be equal to 0, then

it's equal to sum m + our V sub s which is at the 0 value of output

as a V sub s value of 0.5 plus this intercept.

We can also from our problem statement utilize the fact that we

want 5 volts out when the sensor is putting out 1 volt.

So we have two equations and we have two unknowns, our two unknowns are m and b.

So if we take these two equations and we solve them And we end up with an M

which is equal to 100, and a B which is equal to -5.

So we can rewrite our equation for

the output in terms of the input as ten times V sub S- 5.

So that's our input output relationship.

This all came from the problem statement.

So what we need to do is we need to design the interface circuitry that allows

us to do this.

We can see that our output is a function of the input, it's not an inverted so

we might use a non inverting configuration for

making this relationship between the output and the input.

We have this other factor, which is a costant, and it's -5,

where the output has an inverted relationship to some input,

which gives a -5 contribution to the output voltage.

So, we might use a circuit configuration that has

some inputs for both the inverting and noninverting input of the op amp.

First of all we know that V sub s is not an inverted signal so we might try.

An input voltage which is tied to the non-inverting input of our op amp.

And we might also attach some reference voltage,

which gives us a constant value with respect to a known voltage.

At the inverting input of the op amp because we have this inverted signal that

we're also getting a contribution from.

So we'll put in a R1 here, we'll put in our feedback resistor like we have seen in

previous examples for our base circuit.

So we have an output voltage.

Which is the function of the input voltage V sub s and

our input voltage which is our reference voltage at the inverting input.

So, we can use superposition if we wanted to solve this problem with

the contribution to V out from V sub s and the contribution of V out from the Vref.

So, if we first look at the contribution from Vs,

what will we do with the reference voltage for using super position.

We know that when we are solving problems of super position.

If we are looking at the contribution from one source that we short circuit all other

voltage sources and we open circuit all other current sources.

So, if we do this, this circuit would look like what we have here except for

we'd have V ref taken out an a short circuit to ground at this point.

That circuit looks exactly like what we've done before for

the non inverting operational amplifier example.

And what we have in that

case is that V out = (1

+ R sub F / R1) V sub S.

That's what we get if we solve this circuit with V sub S in the circuit

VRef taken out replaced with the short circuit.

We get the basic configuration for

the non inverting amplifier which has this output, input relationship.

Now if we want to add the contribution from our reference voltage.

We would do the same thing as what we did just a minute ago.

We would look at V ref and we would remove all other sources.

We'd short circuit the voltage sources.

So we'd have a short circuit across this V sub s.

And we would open circuit the current sources,

which we don't have in this problem.

So if we look at that configuration where we have V ref, R1, R sub f,

and a short circuit to ground at the non-inverting input.

That is exactly what the circuit looked like when we did the example for

the inverting op-amp configuration.

And so the output that we receive from the V ref,

the contribution to the output voltage is R sub f / R 1 times Vref.

So that's what the output of this circuit would give us.

It would give us a contribution from V sub s and a contribution from Vref.

So if we compare this to what we're looking for

it fits pretty well to that scheme.

Where we have a 10, which has to be satisfied by 1 plus R sub s / R1.

And a 5, which has to be solved using R sub F / R1 times V ref.

So we have to make some decisions at this point.

We know that 1 + R sub f / R1.

That's one equation we have.

And we have another equation

where Rf / R1 x V ref is

gotta be equal to 5.

So we have to make some decisions on what values to use for

resistors since we're not given those values.

So if we choose R1 = 10k,

from our first expression,

R sub f has to be equal to 90k.

So R sub f,

Is equal to 90K.

If R sub f is equal to 90K and R1 is equal to 10K,

our second equation leads us to a value for V-Ref, and

we get a V-Ref equal to five-ninths of a volt.

So we can plug these values back into our circuit, our initial circuit, and

if we got a circuit tied to the non inverting input of our opp amp.

It's our Vs and it's a 0 to 1 volt signal.

Sorry, it's a 0.5 to 1v signal at Vs.

Then the output would be a signal which ranged from

0v to 5v, just as we had hoped with our design.

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