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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From the course by Georgia Institute of Technology

电子学基础

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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.

From the lesson

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, we are going to solve for the transfer function or

the output voltage versus input voltage relationship for

a circuit known as a two op-amp diff-amp or two op-amp differential amplifier.

Let me begin by drawing the circuit schematic for the two op-amp, diff-amp.

Here we have an input resistor R1 connected to the inverting terminal

of an op-amp.

The non-inverting terminal is grounded.

Here is a feedback resistor, R2.

The output here is connected through a resistor R4

to the inverting terminal of a second op-amp that has a feedback resistor R5.

This is the output voltage of the circuit.

This is one of the input voltages.

Now we have a second input to the circuit,

which I'll call V2 that is connected through a resistor R3

to the inverting terminal of the second op-amp, like this.

So this circuit, a two op-amp has two inputs and single output.

Now, I want to begin our analysis of this circuit by identifying subcircuits within

this more complicated circuit.

So for example, we can look at this portion of the circuit and

identify it as an op-amp inverting amplifier.

And we can identify this circuit or this portion of the overall circuit.

As a summing circuit or an op-amp summer.

A two input summer where one of the inputs is V2 and

let me label the second input, this no voltage as Vx.

Now this technique of identifying subcircuits within more complicated

circuits can greatly simplify the analysis of the more complicated circuit,

because we can use the known results for

the subcircuits to speed up our overall analysis.

So for example, the inverting amplifier.

We know that the output voltage is related to the input voltage for

this inverting amp by Vx, the output voltage is equal to the input

voltage times minus R2, the feedback resistor over R1.

So no analysis was required, we just used our known result to relate V1 to Vx.

Now let's look at the summing circuit alone and

analyze its output voltage versus input voltages.

So let me redraw the summing circuit, like this.

Here is our resister R3 with our input voltage V2.

Here is the resister R4 with input voltage V1.

They're connected together and

connected to the inverting terminal of the op-amp and

I can draw the feedback resistor R5 output voltage and

this should be Vx, the Vx input is applied to R4.

So what I want to is use superposition of V2 and Vx to solve for

the output voltage of Vout for the summing circuit.

Now remember, when we use superposition,

we turn one of the input sources on with all of the other sources off and solve for

the output voltage, then we repeat that for every other input voltage source.

Then once we've determined the contribution to the output voltage for

each source individually,

we add all the contributions together to determine the total output voltage.

So, I'm going to begin by turning the V2 source on.

V2 on and Vx source off.

Now Vx is a voltage source.

When we turn a voltage source off, its voltage becomes zero volts or ground.

So, I can, for this condition, rewrite the circuit, like this.

Here's our resistor R3.

Here's our resistor R4 with Vx now grounded.

Here is V2.

Then I connect the rest of the circuit, like this.

R5, Vout and I want to solve for

a Vout in terms of V2.

Now the first thing to notice here in the circuit is that R4 has no effect

on the circuit and the reason for that is the voltage on this side of R4 is equal

to the voltage on this side of R4, so no current flows through R4.

This is an ideal op-amp, so the voltage at the non-inverting terminal is equal to

voltage at the inverting terminal.

This voltage is ground, this voltage is also ground.

So we have ground on this side, ground on this side.

So the voltage difference across R4 is equal to 0.

So the current through R4 is equal to 0.

Another way to see that is you could actually write

the Ohm's Law equation, V equals IR.

In this case, V, the voltage across R4 is equal to 0.

0 minus 0.

So that IR must be equal to 0.

R is a non-zero quantity, so the current I must be equal to 0.

So we can replace the resistor R4 by an open circuit.

So let me redraw the circuit one more time.

Here is V2.

Here is a resistor R3.

R 4 is an open circuit.

Or in other words is just left out.

Ground the non-inverting terminal and here is the feedback resistor R5, Vout.

So you can see that what we have here is another

inverting amplifier configuration with

Vout equal to V2 times minus R5 over R3.

For the case where Vx is off and V2 is on.

Now we go back to the original circuit and

we turn Vx on and turn V2 off.

So Vx on,

V2 is off.

So we're going to get a similar configuration.

We have two resistors, like this with Vx on, which makes this R4.

Makes this R3.

Here is Vx and that is connected

to the op-amp, like this.

Here is R5 and here is Vout.

And again, for the same reasons as before, our three can be neglected,

because there's no current through it.

So again, redraw the circuit with this being Vx.

Here is the resistor R4 minus,

plus feedback resistor R5 and

here is Vout.

And again, by inspection,

we know the result that Vout is equal

to Vx times minus R5 over R4.

So we obtain these two results.

This one and this one using super position.

And then the total output voltage of the summing circuit is the sum of these two

output voltages.

So, let me write Vout for

the Summer is equal to V2

time minus R5 over R3 minus

Vx times R5 over R4.

Let's go back and look at the original circuit.

Here remember, we had this inverting amplifier connected between V1 and VX,

so VX and V1 were related by this inverting amplifier game formula.

So, I'm going to replace in our expression below,

V1 over minus R2 over R1 for Vx.

So, overall.

We have Vout for the entire circuit

is equal to V2 times minus R5 over

R3 minus R5 over R4 times Vx,

but we know that Vx is equal to V1 times

minus R2 over R1, like that.

So this is a solution to the problem.

You can see that we obtain the output voltage by multiplying the input

voltage V2 by one gain and the input voltage V1 by another gain and

then combining the two in this way.

Now we can have this circuit implement a true diff-amp in that it's output voltage

is equal to a gain times the difference of the two input voltages by

making some assumptions about the resistor values in the circuit.

So for example, if we let the resistor

R2 equal R1 and R4 equal R3,

then we can rewrite the output voltage

expression as Vout is equal to V2

times a minus R 5 over R3 minus R5

over now R3 times V1 times a minus 1.

Or we can write the Vout equals,

I'll factor out the R5 over

R3 times V1 minus V2.

The formula for a true differential amplifier.

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