Sample Problem: Op Amps 4

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Linear Circuits 1: DC Analysis

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Georgia Institute of Technology

Linear Circuits 1: DC Analysis

135 classificações

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.

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

Conheça os instrutores

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 a design problem.

Suppose we have a biosensor 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 ppm.

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

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

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

to get it over the proper range for our downstream circuitry.

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

range of 0 to 5 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 form for

our expression that associate with linear circuits and

with linear equations in general.

That is that the output equal to 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 0,

as you put the sum m plus 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 five volts out when the sensor is putting out one 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 a 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 10 times V sub s minus 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

make 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 inverted as signal, so we might try an amp of the voltage which is tied to

the non-inverting input of our op-amp.

And we want also attach some reference voltage which gives us

a cost in value with respect to unknown voltage.

At the inverting input of the op-amp as we have this inverted signal that we're

also getting a contribution from.

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

previous examples for our base as 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 voltage at the inverting input.

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

the contribution to V out for V sub S and the contribution of V out from VRef.

So if we first look at the contribution for Vs,

what will we do with the reference voltage for easy superposition.

We know that when we're solving problems with superposition,

if we're 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 will look like what we have here,

except for we have VRef taken out and a short circuit to ground at this point.

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

the non-inverting operation amplifier example.

And what we have in that case

is that V out is equal to 1 plus

R sub F over R1 times 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 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 V sub S and

we would open circuit the current source which we don't have in this problem.

So if we look at that configuration where we have VRef, 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 received from VRef,

the contribution to the output voltage is R sub F divided by R one,

times VRef.

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

It will give us 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 set as 5 by 1 plus R sub s divided by R1.

And a 5 which has to be solved using

R sub F divided by R 1 times VRef.

So we had to make some decisions at this point, we know that 1 plus R sub

F over R 1 is gotta be equal to 10.

That's one equation we have and we have another equation where

Rf of the feed back side

divided by R 1 times VRef

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 equal to 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 R sub 1 is equal to 10K,

our second equation leads us to a value for VRef.

And we've got VRef,

equal to five-ninths of a volt.

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

if we had the circuit tied to the non-inverting input of our op-amp.

It's our V sub S and it's a 0 to 1 volts signal, so

it's a 0.5 to 1 volt signal at V sub S.

Then the output be a signal which range from 0

volts to 5 volts just as we had hope with our design.

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