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

317 ratings

Georgia Institute of Technology

317 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 solve for the transfer function or

Â the output voltage versus input voltage relationship for

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

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

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

Â of an op-amp.

Â The non-inverting terminal is grounded.

Â Here is a feedback resistor, R2.

Â The output here is connected through a resistor R4

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

Â This is the output voltage of the circuit.

Â This is one of the input voltages.

Â Now we have a second input to the circuit,

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

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

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

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

Â this more complicated circuit.

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

Â identify it as an op-amp inverting amplifier.

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

Â As a summing circuit or an op-amp summer.

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

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

Â Now this technique of identifying subcircuits within more complicated

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

Â because we can use the known results for

Â the subcircuits to speed up our overall analysis.

Â So for example, the inverting amplifier.

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

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

Â voltage times minus R2, the feedback resistor over R1.

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

Â Now let's look at the summing circuit alone and

Â analyze its output voltage versus input voltages.

Â So let me redraw the summing circuit, like this.

Â Here is our resister R3 with our input voltage V2.

Â Here is the resister R4 with input voltage V1.

Â They're connected together and

Â connected to the inverting terminal of the op-amp and

Â I can draw the feedback resistor R5 output voltage and

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

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

Â the output voltage of Vout for the summing circuit.

Â Now remember, when we use superposition,

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

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

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

Â each source individually,

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

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

Â V2 on and Vx source off.

Â Now Vx is a voltage source.

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

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

Â Here's our resistor R3.

Â Here's our resistor R4 with Vx now grounded.

Â Here is V2.

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

Â R5, Vout and I want to solve for

Â a Vout in terms of V2.

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

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

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

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

Â voltage at the inverting terminal.

Â This voltage is ground, this voltage is also ground.

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

Â So the voltage difference across R4 is equal to 0.

Â So the current through R4 is equal to 0.

Â Another way to see that is you could actually write

Â the Ohm's Law equation, V equals IR.

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

Â 0 minus 0.

Â So that IR must be equal to 0.

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

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

Â So let me redraw the circuit one more time.

Â Here is V2.

Â Here is a resistor R3.

Â R 4 is an open circuit.

Â Or in other words is just left out.

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

Â So you can see that what we have here is another

Â inverting amplifier configuration with

Â Vout equal to V2 times minus R5 over R3.

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

Â Now we go back to the original circuit and

Â we turn Vx on and turn V2 off.

Â So Vx on,

Â V2 is off.

Â So we're going to get a similar configuration.

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

Â Makes this R3.

Â Here is Vx and that is connected

Â to the op-amp, like this.

Â Here is R5 and here is Vout.

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

Â because there's no current through it.

Â So again, redraw the circuit with this being Vx.

Â Here is the resistor R4 minus,

Â plus feedback resistor R5 and

Â here is Vout.

Â And again, by inspection,

Â we know the result that Vout is equal

Â to Vx times minus R5 over R4.

Â So we obtain these two results.

Â This one and this one using super position.

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

Â output voltages.

Â So, let me write Vout for

Â the Summer is equal to V2

Â time minus R5 over R3 minus

Â Vx times R5 over R4.

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

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

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

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

Â V1 over minus R2 over R1 for Vx.

Â So, overall.

Â We have Vout for the entire circuit

Â is equal to V2 times minus R5 over

Â R3 minus R5 over R4 times Vx,

Â but we know that Vx is equal to V1 times

Â minus R2 over R1, like that.

Â So this is a solution to the problem.

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

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

Â then combining the two in this way.

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

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

Â making some assumptions about the resistor values in the circuit.

Â So for example, if we let the resistor

Â R2 equal R1 and R4 equal R3,

Â then we can rewrite the output voltage

Â expression as Vout is equal to V2

Â times a minus R 5 over R3 minus R5

Â over now R3 times V1 times a minus 1.

Â Or we can write the Vout equals,

Â I'll factor out the R5 over

Â R3 times V1 minus V2.

Â The formula for a true differential amplifier.

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