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

Introduction to Electronics

326 ratings

Georgia Institute of Technology

326 ratings

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.

Â This is Dr. Robinson.

Â In this lesson we are going to design a balanced output amplifier to meet a set of

Â given specifications.

Â But let me begin by explaining to you what a balanced output amplifier is and

Â why we would want to use one.

Â Let me draw a block that represents the balanced amp.

Â Now, a balanced amplifier has a single input we'll call it, Vin and

Â produces two outputs.

Â That are related to the input.

Â One output equal to AVin, and a second output equal to minus AVin.

Â Where A is the gain of the balanced amplifier.

Â Now you can see how the outputs are related to the input and to each other.

Â The outputs are the negatives of each other.

Â Or you can think of them as two signals that are 180 degrees out of

Â phase from each other.

Â Now, why would we want to use a balanced amplifier?

Â Say for example, we have an electrical signal that represents an audio source.

Â So for example, this audio signal [NOISE] could be

Â the output of an MP3 player, possibly a phonograph.

Â And we want to transmit it through a cable or

Â a transmission line to the next stage in our audio system.

Â So for example, a preamp.

Â Pre amp.

Â Well, during this transmission.

Â Along this cable electrical noise can be introduced by external electrical

Â components.

Â So that the input of the preamp,

Â we not only have the signal that we're interested in.

Â But we have, in addition to that, this added noise component that I'm calling N.

Â So during the transmission we have distorted our signal that represents

Â the audio by this external noise.

Â But what if instead we transmit the signal using a balanced amplifier.

Â So I make the input of our balanced amplifier, the audio signal, or

Â the electrical signal that represents the audio signal.

Â We then transmit the signal using two transmission lines or two cables.

Â And we make the input stage of our preamplifier, a differential amplifier.

Â A diff amp.

Â With a gain equal to one-half.

Â Then, any noise that's introduced due to external devices.

Â The noise.

Â Is applied to both the AVin signal and the minus AVin signal equally.

Â So at this input to the diff amp, we have AVin plus the noise.

Â And at this differential amplifier input, we have minus AVin plus the noise.

Â So then at the output of the differential amplifier Vout,

Â we can write than Vout is equal to one of the inputs.

Â AVin plus the noise minus, because it's a diff amp,

Â the other input voltage, which is minus AVin plus the noise.

Â All times the gain of the diff amp, one-half.

Â So Vout is equal to one-half times

Â AVin plus AVin plus the noise minus the noise.

Â Or the output voltage is equal to AVin.

Â So this combination of a balanced output amplifier and

Â an input differential amplifier tends to cancel any common mode noise.

Â That's introduced into the signal during it's transmission.

Â Now, let's look at an a op amp circuit that can be used to implement a balanced

Â output amplifier.

Â Here is our input voltage Vin applied to the non inverting terminal of an op amp.

Â We have a feedback resistor, RF1.

Â We have another resistor, R1,

Â that's applied to the inverting terminal of a second op amp.

Â And we have another feedback resistor, RF2.

Â Here is one output of the circuit and here is the second output of the circuit.

Â Now we consider this circuit to be composed of two simpler op amp amplifiers,

Â a non-inverting amplifier and an inverting amplifier.

Â Remember, these two voltages must be equal to each other.

Â The voltage here is ground,

Â which means the voltage at this node is zero volts or ground.

Â So we can consider this portion of the circuit

Â to be a non-inverting op amp amplifier with

Â a gain of Vo1 equal to one plus RF1 over R1.

Â And we multiply that by Vin to get the output voltage Vo1.

Â Now, this voltage here must also be equal to the voltage here.

Â So the node voltage here is equal to the input voltage Vin.

Â We can consider this circuit, or this portion of this circuit,

Â to be an inverting op amp amplifier with an input voltage Vin.

Â So Vo2 is equal to negative RF2 over R1 times the Vin.

Â Now what I want to do is design this circuit so

Â that it meets a couple of conditions.

Â One, I want the magnitude.

Â The right design.

Â I want the magnitude of the gain A of this balanced output amplifier circuit

Â to be equal to 6.

Â And I want to impose a second condition, and that is that

Â Want the maximum

Â current through RF1 and

Â RF2 to be one milliamp.

Â When the magnitude of Vo1 is equal to

Â the magnitude of Vo2, is equal to 12 volts.

Â So, you can see that from these two equations we can impose

Â this gain condition.

Â We can choose RF1, RF2, and R1.

Â So that the gain here is equal to 6 and the gain here is equal to negative 6.

Â But to implement this second condition we need another equation.

Â An equation for

Â the current through RF1 and RF2 in terms of the circuit component values.

Â Now, remember there's no current into the input terminals of an op amp.

Â So the current into this inverting terminal is zero.

Â And the current into this inverting terminal is zero.

Â So any current that flows through our RF1 must also flow through R1 and

Â flow through RF2.

Â Because there is no way out at this node.

Â Because of the infinite impedance looking into the op amp here.

Â And there's no way out at this node.

Â Because of this infinite impedance.

Â So we have, flowing from here.

Â A current that flows through RF1, through R1 and through RF2.

Â So I can use Ohm's law to calculate that current.

Â The difference in voltage across this series combination of three resistors.

Â Must be equal to the current through the resistors times the resistance.

Â Or in other words, I can write that Vo1 minus Vo2, the total

Â voltage across the three resistors divided by the sum of the three resistors.

Â The series resistors are RF1 plus RF2 plus R1 must be

Â equal to the current I through the three resistors.

Â So to impose our second design condition I can

Â write that Vo1 when it is equal to 12 minus Vo2.

Â Which is equal to minus 12 divided by RF1

Â plus RF2 plus R1 must be equal to 1 milliamp.

Â Then we know that 1 plus RF1

Â over R1 must be equal to 6,

Â and we know that, minus RF2 over

Â R1 must be equal to minus 6.

Â So this equation implies that RF1

Â is equal to 5 times R1.

Â This equation implies that RF2 is equal to 6 R1.

Â So I can use these two equations to eliminate RF1 and RF2 from this equation.

Â So I can write that 24

Â divided by 5 R1 plus 6 R1,

Â plus R1 is equal to 1

Â milliamp is equal to 24

Â divided by 12 R1.

Â Which implies that R1

Â is equal to 2k ohms.

Â Then, I can use RF1 is equal to

Â 5 R1 is equal to 10k ohms.

Â And.

Â RF2 is equal to 6R1 is equal to 12k ohms.

Â So if we construct a balanced output amplifier using

Â these resistor values we'll meet our design specifications.

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