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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Из курса от партнера 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.

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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 branch currents in a differential amplifier circuit.

Here is a schematic of the circuit we're going to analyze and

I have labeled the currents that we're going to solve for

as we precede through the example, I1 through I7 plus the load current IL.

Now let's begin by solving for the most apparent branch current in the circuit and

those are I3 and I4.

You can see that I3 and

I4 are the current into the input terminals of an ideal op-amp.

So we know immediately that I3 must equal I4, must equal 0 amps.

Now let's find the current I1.

We know that I3 is equal to 0, so the current that flows through the 4k resistor

must also flow through the 16k ohm resistor.

So we know that we have a total of ten volts from ground to this node or

a total of ten volts across the series combination of the 16k resistor and

the 4k resistor, so I can use Ohm's law to solve for a I1.

I1 is equal to 10 volts, the voltage across

the series combination of the 16k and

the 4k resistor, which is equal to 10 over

20 milliamps is equal to 0.5 milliamps.

Now again, because I3 is equal to 0,

we know that the magnitude of I1 must equal the magnitude of I6.

Because they're the same current, but I6 is defined in this direction.

So we can write that I6 is equal to negative

I1 is equal to negative 0.5 milliamps.

Let's now find the current I2 and we'll begin by determining the voltage

here at the inverting terminal of the off amp.

We know we have an op-amp circuit with negative feedback, so

the non-inverting voltage here must equal the inverting terminal voltage here and

we can solve for the voltage here by voltage division.

So let me write that V plus the non-inverting voltage must be equal

to the inverting terminal voltage is equal to 10 volts times this voltage divider.

So 16k over 4k plus 16k is

equal to 10 times 16 over 20 or

80% of 10 or 8 volts.

Now we used voltage division here, another way to solve for

the non-inverting terminal voltage would be just to multiple the 16k

ohm resistor by the current through it, which is a 0.5 milliamp.

So 16k times a 0.5 miliamp would be equal to 8 volts.

So we now know the voltage here at the inverting terminal is eight volts and

we know the voltage here is twelve volts.

So we can use Ohm's law to solve for the current through this 4k ohm resistor.

So, I can write that I2 is equal to 12

volts minus 8 volts divided by 4k is equal

to 4 volts over 4k is equal to 1 milliamp.

Now, I4 is equal to 0, so the current that flows through the 4k

ohm resistor must also flow through the 16k ohm resistor,

because there's no way out at this node.

So, I2 must be equal to I5.

So, I2 is equal to I5 is equal to 1 milliamp.

Now, I want to solve for the load current IL, but

to do that, I'm first going to solve for the voltage Vout.

Because I can divide Vout by 2k ohms to get IL.

And I'm going to solve for Vout by beginning with

this known voltage here at the inverting terminal, eight volts.

And then I'm going to subtract from that node voltage,

the drop across this 16k ohm resistor.

So this voltage minus this voltage drop will give us the voltage at this node,

which is equal to the output voltage.

So, I can write that Vout is equal to 8 volts.

The voltage at the inverting terminal minus 16k ohms times the current

through the 16k ohms resistor, which is I5, which is one milliamp.

1 milliamp is equal to 8 minus 16

volts is equal to negative 8 volts.

So we can now solve for

the load current IL using Ohm's Law as the output voltage divided by 2k ohms.

So IL is equal to negative 8, the output voltage

divided by 2k is equal to negative 4 milliamps.

Now the low current is defined in this direction, but this negative sign

indicates that the actual load current is flowing in the opposite direction.

So, if we have negative 4 milliamps flowing in this direction,

then we can think of that a 4 miliamps flowing in this direction.

So we have four miliamps into this nod from this direction,

we know we have one miliamps flowing into the nod from this direction.

So the total current in this branch,

flowing into the output of the op-amp is 4 plus 1 or 5 miliamps.

Now we can see that I7 is defined in the other direction, so

I7 must be equal to negative 5 milliamps or we can solve for I7 using an equation.

We can write that I7 is equal to IL minus I5

is equal to negative 4 minus 1 is equal

to negative 5 milliamps of current.

Now, in solving for the output voltage in this problem,

I used this known node voltage and the drop across this resistor, but

another way to do it is to use the known result for

the gain of a differential amplifier if we recognize that this is a diff-amp.

These two resistors are equal and these two resistors are equal.

So we can write that the gain of this diff-amp.

Let me write it here.

Vout is equal to 16 divided by 4.

16, the value of these two resistors and 4,

the value of these two resistors times the difference of the input voltages.

The ten volt source is applied to the non-inverting terminal and

the twelve volt source is applied to the inverting terminal.

So we write it as 10 minus 12 is equal to 4 times

a negative 2 is equal to a negative 8 volts.

The same answer we obtained before.

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