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Sample Problem: Op Amps 1

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Linear Circuits 1: DC Analysis

Instituto de Tecnología de Georgia

4.5 (198 calificaciones) | 16K estudiantes 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.

Revisiones

4.5 (198 calificaciones)

5 stars
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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.

YZ

Jan 22, 2018

Excellent course where I get to learn a lot of things that is left from my AP Physics C E&M Course.

De la lección

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

Impartido por:

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

Transcripción

The topic of this problem is operational amplifier circuits.

And the problem is a design problem.

Suppose we have a biosensor with an output range of 0.5 volts through 1.0 volts,

and it's an oxygen sensor and

the oxygen sensor itself has a range of 1.0 to 250.0 parts per million.

So as the oxygen concentration changes from 1.0 to 250.0 parts per million,

the sensor itself puts out a signal which has a range of 0.5 to 1.0 volts.

But we need to do something with that signal 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 0.0 to 5.0 volts for this biosensor system.

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

to solve this.

So when we have linear circuits we can use a form

for our expressions that is associated with

plenty of circuits, and with linear equations in general.

That is that the output is equal to the slope times the input plus some 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 zero, then

it's equal to sum m plus our v sub s which is at the zero 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 1.0 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, then we end up for

the m which is equal to 10, and a b which is equal to minus 5.

So we can rewrite our equation for the output in terms of

the input as 10Vs- 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 an interface circuitry that allows us to

do this.

We can see that our output is a function of the input.

It's not 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 constant and it's minus 5,

where the output has an inverted relationship to some input,

which gives a minus 5 contribution to the output voltage.

So we might use a circuit configuration that has.

Some inputs for both the inverting and non-inverting 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 at the op-amp,

because we have this inverted signal that we're also getting a contribution from.

So we'll put in an R1 here, we'll put an R feedback resistor

like we have seen in previous examples for our base circuit.

So we have an output voltage.

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

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

So we can use super position if we wanted to solve this problem for

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

So if we first look at the contribution from Vs,

what would we do with the reference voltage for using super position?

We know that when we're solving problems with super position,

if we're looking at the contribution for one source, that we short circuit

all other voltage sources and we open circuit all of the current sources.

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

we'd have Vref taken out, and a short circuit to ground at this point.

That circuit looks exactly like what we have done before for

the non-inverting operational amplifier example.

And what we have in that

case is that Vo = (1

+ Rf/R1) x Vs.

That's what we get if we solve this circuit with Vs in the circuit,

Vref taken out, replaced with a 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 Vref, and

we would remove all other sources, we'd short circuit the voltage sources.

So we'd have a short circuit across this Vs, 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 Vref, R1, Rf 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 received from Vref, the contribution

to the output voltage is Rf/R1 x Vref.

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

It would give us contribution from this Vs, 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 + Rs/R1, and

a 5 which has to be solved using

Rf/R1 x Vref.

So we have to make some decisions at this point.

We know that 1 + Rf/R1 = 10.

That's one equation we have, and

we have another equation where Rf or

V feedback, sorry, divided by R1

times Vref has got to 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

Rf has to be equal to 90k.

So Rf.

Is equal to 90k.

If Rf = 90k, and R1 = 10k,

our second equation leads us to a value for Vref.

And we get a Vref = 5/9 of a volt.

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

and if we have a circuit tied to the non-inverting input of our op-amp,

it's our Vs and it's a 0.0 to 1.0 volt signal.

Sorry, it's a 0.5 to 1.0 volt signal at Vs.

Then the output would be a signal which will range from 0.0 volts to 5.0 volts.

Just as we had hoped with our design.

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