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

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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.

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

Â the output voltage of an op-amp circuit.

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

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

Â That's a two op-amp circuit.

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

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

Â So the voltage at this node,

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

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

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

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

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

Â the output voltage of this op-amp.

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

Â So, I say that V01 is equal to

Â V plus at the non-inverting terminal

Â plus I times R3 is equal to Vin

Â plus Vin over R4 times R3.

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

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

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

Â the current through this resistor,

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

Â Which implies that V01 over R2 is

Â equal to negative Vout over R1 or

Â V01 is equal to negative R2 over R1

Â times the output voltage, Vout.

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

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

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

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

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

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

Â So, I can write that Vout over

Â Vin is equal to negative R1 over

Â R2 times 1 plus R3 over R4.

Â The answer.

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

Â simplify our analysis.

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

Â so we know the relationship between V01 and Vout.

Â VO1 is equal to negative R2 over R1 time Vout.

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

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

Â plus R4.

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

Â the voltage here, which is equal to Vin.

Â So we can write by inspection that

Â Vin is equal to Vout times negative

Â R2 over R1 times R4 over R3 plus R4.

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

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

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

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

Â Then we can write that V0 is equal to or

Â V0 over Vin is equal to negative

Â R1 over R2 times 1 plus R3 over R4.

Â The same answer we obtained previously.

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

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

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

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

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

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

Â which may appear initially to be positive feedback.

Â But in this path between the output voltage and

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

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

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

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

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