This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

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Del curso dictado por Georgia Institute of Technology

Introduction to Electronics

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This course introduces students to the basic components of electronics: diodes, transistors, and op amps. It covers the basic operation and some common applications.

De la lección

Op Amps Part 1

Learning Objectives: 1. Develop an understanding of the operational amplifier and its applications. 2. Develop an ability to analyze op amp circuits.

- Dr. Bonnie H. FerriProfessor

Electrical and Computer Engineering - Dr. Robert Allen Robinson, Jr.Academic Professional

School of Electrical and Computer Engineering

Welcome back to Electronics. This is Dr. Robinson.

In this lesson, I'm want to work an op-amp example problem where we solve for

the output voltage of an op-amp circuit.

Here's the schematic of the circuit, we're going to analyze.

Here's the input voltage, here's the output voltage of the circuit.

That's a two op-amp circuit.

Let's begin by noting that the voltage at the inverted terminal

of this op-amp is equal to the input voltage.

So the voltage at this node,

because of the ideal op-amp must also be equal to Vin.

We can calculate the current I through this R4 resistor as Vin divided by R4.

So, I is equal to Vin divided by R4 is equal V plus,

the voltage at the non-inverting terminal divided by R4.

Now to calculate the voltage at this node, let me label it V01,

the output voltage of this op-amp.

We'll start with this node voltage and add the IR drop across R3.

So, I say that V01 is equal to

V plus at the non-inverting terminal

plus I times R3 is equal to Vin

plus Vin over R4 times R3.

Now let's introduce Vout, the voltage we were trying to solve for

into our set of equations by writing a node equation at this node.

So, I can write that V01 minus 0 over R2,

the current through this resistor,

plus the 0 or Vout minus 0 over R1 is equal to 0.

Which implies that V01 over R2 is

equal to negative Vout over R1 or

V01 is equal to negative R2 over R1

times the output voltage, Vout.

Now we know that V01 is equal to Vin plus Vin times R3 over R4.

So, I'm going to make that substitution into this equation.

So, I can write that Vin plus Vin times R3 over

R4 is equal to negative R2 over R1 times the output voltage, Vout.

Now, on this side of this equation, I can factor Vin out, bring it to

this side to solve for the ratio would be Vout to Vin or the gain of the circuit.

So, I can write that Vout over

Vin is equal to negative R1 over

R2 times 1 plus R3 over R4.

The answer.

Now, let's rework this problem in another way where we use known results to

simplify our analysis.

Now we recognize that this portion of the circuit is an inverting op-amp amplifier,

so we know the relationship between V01 and Vout.

VO1 is equal to negative R2 over R1 time Vout.

Then we recognize this portion of the circuit as a two resistor voltage divider,

where the output voltage here is equal to the input voltage times R4 over R3

plus R4.

And because of this ideal op-amp, we know that the voltage here must be equal to

the voltage here, which is equal to Vin.

So we can write by inspection that

Vin is equal to Vout times negative

R2 over R1 times R4 over R3 plus R4.

Where again, Vout times negative R2 over R1 is equal to VO1 and

VO1 is the input to the voltage divider with a gain of R4 over R3 plus R4.

So V01 is this portion, we multiply by the voltage divider to get the voltage here,

which is equal to Vin, because of this idea op-amp.

Then we can write that V0 is equal to or

V0 over Vin is equal to negative

R1 over R2 times 1 plus R3 over R4.

The same answer we obtained previously.

Now let's look at something to note about this circuit.

It may appear at first, that this circuit does not have negative feedback and

because of that, we cannot consider the voltage at the inverting terminal

to be equal to the voltage at the non-inverting terminal.

You can see that there's no path from the output voltage to the inverting terminal.

In fact, there's a path from the output voltage to the non-inverting terminal,

which may appear initially to be positive feedback.

But in this path between the output voltage and

the non-inverting terminal is an inverting op-amp that introduces a negative sign.

So this path from output to non-inverting terminal is actually a negative feedback

path and because of that, the voltage at the inverting terminal is

equal to the voltage at the non-inverting terminal.

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