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.

Â This is Dr. Ferri.

Â In this lesson we'll go over some basic Op Amp amplifier configurations.

Â In a previous lesson we looked at Ideal Op Amp behavior and Buffer Circuits.

Â Now we want to introduce some specific types of amplifier circuits.

Â The Inverting and

Â Non-Inverting configurations, as well as the difference in summing configurations.

Â We're also going to introduce the gain of a circuit.

Â Let's look first at the non-inverting amplifier.

Â We, in order to analyze this,

Â we want to be able to use our ideal amplifier assumptions.

Â And that is, that we've got, zero amps going in to each of these terminals.

Â And a zero, voltage drop across those terminals.

Â And I show the result of this being right here.

Â This shows the output as a function of the input, now let's derive it.

Â As part of their, our derivation, we are going to be doing some KVLs.

Â The first one, will be up, around this loop right here.

Â So, going up around here and back down to, the ground.

Â Going up this way, I get minus Vn, plus the voltage drop here.

Â Well the current is zero so that means the voltage drop is zero.

Â And then plus a voltage drop here, but that's also a zero.

Â So, we're left with the only other voltage drop which would be across this resistor.

Â And I'll define this current as being i sub 3.

Â So, it's plus i sub 3, R3 equal 0.

Â And solving for i3.

Â We get that is equal to Vn, over R3.

Â Now my second KVL, I will, look across this loop.

Â Actually this node voltage right here, let me go from minus to plus right here.

Â Going from minus to plus we gain V0.

Â And also looking in this direction,

Â we can add up this voltage across these two resistors.

Â That should equal to V0.

Â So we have V0, is equal to R2 plus R3 times i3.

Â And notice that,

Â I'm using the fact that there's zero current going in this direction.

Â So i3 going down through this resistor, has to be the same i3

Â as going through this resistor, since no current is going in this direction.

Â So, if I were to plug in this expression, N for i3.

Â I gain this expression out here.

Â So V0 is equal to this quantity times VN.

Â I'm going to define a quantity like that as being G.

Â In other words the gain.

Â The gain here is equal to R2 plus R3 over R3.

Â So it's whatever multiplies the input gives me the output.

Â And because my circuit has only resistors in it, my gain is a static gain.

Â Let's look at a particular example here.

Â Now in this example, I've given numerical values of 200 to R2 and R3.

Â Well, remember our gain was equal to R2 plus R3 over R3.

Â In this case if I plugged in the values I would get 400 over 200,

Â would be a gain of two.

Â Since the gain is equal to one the input is amplified.

Â And remember what our gain equation is, the output is equal to G times the input.

Â So in this case the gain is two.

Â So for example, if, V in is equal

Â to 5 volts, then V out is equal to 10 volts.

Â So the gain is amp, the output is amplified.

Â The input is amplified to give me the output.

Â Now let's look at a different configuration, the inverting amplifier.

Â We're going to use the same ideal op-amp assumptions,

Â which again zero amps, zero amps into those two terminals,

Â and the voltage drop is zero volts.

Â So, I'm going to do the same type of derivation as I did before,

Â which is to look at KVLs around different loops.

Â So, I'm going to start with this loop right here.

Â Again, we're going to be using KVLs.

Â So my first loop, I've got, minus V in plus R1 times,

Â and I'll call this current i1,

Â because the current goes through here and up through this way.

Â So I've got zero current here, so

Â all the current through V1 are, through R1 has to go through RF.

Â So I've got R1 times i1.

Â In other words, i1 is

Â equal to V in, over R1.

Â Now I want to do a second KVL,

Â and I'm going to do that KVL up and around this way.

Â And remember that this being a no voltage, it's as equivalent to having that

Â voltage here, with it being connected to ground this way.

Â So then I can actually do my KVL completely around this way.

Â So going around this leap I'm going to get a minus

Â V in plus R1 i1 plus Rf i1 plus V0 equal to 0.

Â We've already established that these first two terms are going to cancel out.

Â So these two terms cancel out.

Â And what I'm left is this part here.

Â So I can solve for i1.

Â Well actually, what I want to do is plug in for i1 right here.

Â So I have Rf over R1 times V in is equal to V0,

Â and actually there's a minus sign here.

Â So this gives me the equation,

Â the input output equation, for an inverting amplifier.

Â And I highlighted it right here.

Â So this term right here, minus Rf over R1 is equal to the gain.

Â Now notice in this case the gain could be greater than one or

Â less than one depending on this ratio.

Â How I select those resistors, if I want to amplify, or

Â if I want to make the output smaller, or what we say attenuate.

Â Lets look at a specific example with numbers in here.

Â Suppose R1 is equal to 1,000, Rf is equal to 2,000.

Â So our gain is equal to minus Rf over R1, and

Â in this particular case that would be equal to minus two.

Â So V out is equal to minus 2 times V in.

Â That's my equation.

Â That's how the input relates to the output.

Â And in this case, the input is if the magnitude let's say,

Â if the magnitude of G is greater than one the input is amplified.

Â If the magnitude is less than one the input is attenuated.

Â So the magnitude of V out is made larger than that of V in, so

Â we call it an, an amplified, a, an amplified signal.

Â So in this case if V in is equal to ten volts,

Â V out is equal to minus 20 volts.

Â Now, the term inverting corresponds to

Â the fact that we've got a negative sign right here.

Â We negate the voltage.

Â So that's when we say it's an inverting amplifier.

Â Now let's look at a circuit that's a little bit more complicated,

Â because I've got two inputs to it.

Â And in this particular circuit we're going to be

Â finding that the output is a difference between the two inputs.

Â We show the result right here, so we see we're, we're taking the difference between

Â V2 and V1 and multiplying it by a gain to give me the output.

Â Let's show the derivation here.

Â Now this derivation is a little bit easier if we use super position.

Â Super position is a method that we use for linear circuits.

Â And this amplifier, we're looking at the linear range so we can use super position.

Â In super position we look at the output due to

Â the individual inputs applied separately.

Â So in one case I would zero out one input, and calculate the output due to,

Â the other input.

Â And then I would reverse it.

Â I would say well,

Â okay let's, zero out this input, and look at the output due to, that input.

Â And then I'd sum them together.

Â So let's try super position.

Â So in the first case, we'll let V2 equal to 0 and we are going to solve for V0.

Â So, let me actually redraw this circuit here.

Â So, I have still got V1, and R1.

Â I am redrawing this circuit because we can make an assumption.

Â We can simplify it a bit.

Â Now this part right here, if V2 is going to be zeroed out.

Â So, that's actually the same as setting that to ground.

Â Then that means that there's zero current that flows through R1 or Rf.

Â And with zero current there, then this right here is equivalent to ground.

Â I can just connect that to ground, because it's got the same potential as ground.

Â So in that case, this looks exactly like an inverting amplifier.

Â And with the inverting amplifier, G is equal to minus Rf over R1.

Â So V out, due to just V1, is equal

Â to minus Rf, over R1 times V1.

Â So now we're going to, to finish super position.

Â We're going to have to do the opposite which is set V1 equal to 0 and

Â solve for the output due to V2.

Â So, taking a look at this, again, I want to set V1 equal to 0, so

Â let me zero it out.

Â Set that equal to zero.

Â If I zero it out, then that's equivalent to setting that to ground.

Â Now in this particular case, it's a little bit easier if I define a voltage here,

Â call it V sub b, and a voltage here call it V sub a.

Â Since this is V sub b right here, then, and

Â this is ground, this node voltage actually is V sub b.

Â And this node voltage right here, since this is ground, is V sub a.

Â Going through the derivation, where I let V sub 1 equal to 0.

Â We show that V sub a has to equal to V sub b,

Â because there's zero potential, difference between this.

Â It's an ideal amp op amp.

Â Now, if I want to find what Va is, I just have to look at this right here.

Â And note that this is a complete loop.

Â So I can use the voltage divider law.

Â [NOISE] And the voltage divider law says that V sub

Â a is equal to the ratio of Rf over the sum of the two

Â resistors in the loop, R1 plus RF times V2.

Â I can do the same thing across this loop right here, so going from V0,

Â being treated as if it was a source here, but this potential right

Â here is divided across these two resistors depending on their ratio.

Â So, Vb is equal to R1, that resistor,

Â over the sum of the two resistors, R1 plus RF times V0.

Â Now, I can equate V2 or V0 and Vb and

Â if do that I notice they have got the same denominators.

Â So, just simplifying that and, after I equate them,

Â I get Rf V2 is equal to R1 V0.

Â So solving for V0, I get Rf over R1, times V2.

Â Now, putting them together, the last step.

Â Maybe I'll put this as three.

Â The last step in super position is the total value of V0, and

Â that's the result due to V1 by itself, plus the result due to V2 by itself.

Â So in that case we got V0 is equal to minus Rf,

Â R1, V1 that was our first thing that we solve for.

Â Plus what we just solve for, Rf over R1 times V2.

Â And that gives us this relationship here,

Â where this now is the gain of the different circuit.

Â The last configuration I wanted to look at is a summing amplifier.

Â Now I'm not going to go through the derivation of that.

Â We use the same methods that we've used before.

Â We use the ideal op amp assumptions, and we can also use super position here.

Â In this particular case, if R1 is not equal to R2 then, we get this result here.

Â We define a gain, G1, and a gain, G2, like this.

Â So, V0 is equal to G1 times V1, plus G2 times V2.

Â This is what we might call a weighted sum, where we weight them differently.

Â Now if R1 is equal to R2, then this result would actually be equal to

Â V0 is equal to minus Rf over

Â R1 times V1 plus V2.

Â So in that particular case we have one gain G, times the sum of the two things.

Â If I wanted, I can actually add in another resistor here, and

Â I can actually keep going like this adding in lots of resistors, and

Â different voltages, and call this V3.

Â And if I did that, I would have V0 is equal to

Â minus Rf over R1, if it's the same thing.

Â If R1, R2, and R3 are the same, then I'd have V1 plus V2 plus V3.

Â That's if R1 equal to R2 equal to R3, and

Â this case to get this equation we had made the assumption that R1 is equal to R2.

Â So it just sums the, the different voltages together.

Â So in summary, one of the most important things we've talked about was the gain.

Â The gain is what relates the input voltage, to the output voltage.

Â We've been calling it G.

Â In all these cases, G is the constant, and

Â it's dependent on the values of the resistors that we've chosen.

Â We've defined four different amplifier circuit configurations.

Â The not inverting, inverting, difference and summing amplifiers.

Â In the next lesson, we will cover differentiators and

Â integrators, thank you.

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