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>> This amplifier circuit is referred to as a difference amplifier.

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As the name suggests, we're going to see that at the output,

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we're going to find the difference or the subtraction of V2 and V1 or V2 minus V1.

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Before we get started on this circuit,

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and before you analyze this circuit,

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let's just make a few observations.

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First of all, in the circuit that we've analyzed involving Op-amps until now,

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the sources have been on either the inverting terminal,

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or the non-inverting terminal.

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This configuration has sources V1 connected to the inverting terminal and V2

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connected to the non-inverting terminal albeit through

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this resistive network before it gets to the non-inverting terminal.

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That points out the fact that VP,

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the voltage at the inverting terminal is not exactly V2.

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So, what effect is this resistive network have?

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Well, we know that the current going into the input of the Op-amp is zero.

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Therefore, any current coming from V2 going through R3 will also go through R4.

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R3 and R4 are in series with each other,

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we have a voltage divider taking place here.

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Thus, the voltage at the non-inverting terminal is the voltage across to R4.

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It's not exactly V2 but a subdivided portion of V2.

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Lets just go ahead and make that observation,

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we'll do it up here because we need to keep room,

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this is a fairly long derivation.

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Let's just say that V sub p,

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the voltage of the non-inverting terminal is going to equal

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V2 times R4 over R3 plus R4.

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So, V sub p is just a little less than V2.

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Now keeping in mind, we have the virtual short between the terminals,

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that then means that V sub n is equal to V sub p which is equal to that.

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This should give us a little bit of flavor of what's going on here.

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We know that sources connected to

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the non-inverting terminal in our non-inverting amplifier configuration had a gain

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that is slightly larger than

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the gain experienced by sources connected to the inverting terminal.

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The non-inverting gain was one plus R2 over R1,

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whereas the inverting gain was negative of course,

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just R2 over R1.

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So, this voltage divider circuit is intended to

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reduce V2 down just a little bit so that at the output,

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V2 and V1 will experience the same gain,

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and this circuit then we'll implement a true subtraction.

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All right. So, with that,

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let's go ahead and start our analysis of this by doing

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exactly what we've done in the other circuits involving amplifiers,

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and that is write a node equation summing the currents leaving the inverting terminal.

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Again, we'll do it in terms of V sub n until we get things simplify then will substitute

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V sub n for V sub p which is going to be V2 times R4 over R3 plus R4.

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So, KCL at the node.

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We got the current leaving the node going this way.

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That's going to be V sub n minus V1 divided by R1,

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plus the current going this way through the feedback loop is going to be,

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and we're adding that in,

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V sub n minus Vout divided by R2.

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There's no current going into the inverting terminal,

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so we only have those two terms that must add to give us zero.

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Now, combining terms and factoring out the V sub n,

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then we've got V sub n times one over R1 plus one over R2,

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and then we have minus V1 over R1.

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We still have this minus Vout over R2,

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lets take it to the other side as a positive, Vout over R2.

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Now, let's get a common denominator here.

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So, we have V sub n times R1 plus R2 over R1 times R2,

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minus V1 over R1 equals Vout over R2.

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Now, we're going to multiply both sides of the equation by R2,

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to solve for Vout and we'll reverse the order of the equation.

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We're left with then Vout is equal to,

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multiplying both sides by R2,

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we have an R2 here that's kind of cancel that R2.

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So, we're left with then V sub n times R1 plus R2 over R1,

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minus V1 times R2 over R1.

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Okay, so we've got the two terms.

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We have the inverting term here,

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V1 times R2 over R1.

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We have V sub n,

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now we need to replace V sub n with V2 over

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R4 over R3 plus R4 so that we can see the dependency of the output on V2.

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Let's do that on up here.

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We've got the Vout is equal to V sub n,

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then V sub n is equal to V2 times R4 over R3 plus R4.

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So that's V sub n times R1 plus R2 over R1.

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That's our first term, minus V1 times R2 over R1.

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Thus, we see that we have different gain terms.

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We've got this multiplying effect here,

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minus V1 times R2 over R1.

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We've got a couple more steps to take until we get there,

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but let's be explicit and show all of our work here.

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First of all, let's notice that this term here R1 plus R2 over R1,

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R1 is a common denominator to both of them.

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That then gives us,

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if we identify that as such,

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let's just write this then as R4 over R3 plus R4,

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times R1 over R1.

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That's just one plus R2 over R1,

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minus V1 times R2 over R1.

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We see that this term right here really is just

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that the gain term for the non-inverting amplifier,

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there's the gain term for the inverting amplifier.

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Here's this voltage divider term which is taking V2 and subdividing it just a little bit.

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Now, what we want to do is determine the relationships between R1, R2, R3,

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and R4 that will make it so that this gain term right here,

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multiplying V2, is the same as the gain term

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multiplying V1 with the obvious exception that V1 it's got the minus sign on it.

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All right. In order to do that,

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we're going to factor in R3 out of the denominator of this term right here.

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So, we have then equals

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V2 factoring in R3 out and I'm also going to bring the R4 out in front.

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So, we'll have, let's see,

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R4 over R3 times one

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over factor in the R3 that leaves

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as a one plus factoring in R4 out there it leads me to R3,

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refactoring R3 out of this term here leaves me

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an R3 in the denominator or I have a R4 over

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R3 times one plus R2 over

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R1 minus V1 times R2 over R1.

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Kind of a mess but we've got this V2 times R4 over R3 then

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multiplying this one divided by one plus R4 over R3.

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Then multiplying this one plus R2 over R1.

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Just a second, I'm going to ask you to stop the video.

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Let me observe first of all,

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that this gain term here will equal this gain term here,

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if the ratio of R4 over R3 equals the ratio of R2 over R1.

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If the ratio R4 over R3 equals R2 over R1,

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then this term here in the denominator is the same as

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this term here in the numerator and they will cancel.

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We're left with simply then Vout is equal

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to V2 times again R4 over R3 equals R2 over R1.

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So, let's replace it here also, R2 over R1,

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minus V1 times R2 over R1.

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Now, if we factor out this R2 over R1,

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we get then and this is where we've been trying to get this whole way.

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Vout then is equal to R2 over R1 times V2 minus V1.

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From here, you can see then why this is called a difference amplifier.

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The output voltage is equal to a scaled version of

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this source voltage minus this source voltage.

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The usefulness of this amplifier configuration can't be overstated.

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What this allows us to do and it's sometimes

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also referred to as a differential amplifier.

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It allows or this circuit will sample voltages at

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two different points in a circuit and subtract one from the other.

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In control circuits, this comparison is used to or they'll

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wraps circuitry around it to drive that difference to zero.

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This type of an amplifier becomes the core of

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feedback systems or control systems where

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one voltage is meant to control another voltage.

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It also performs the obvious operation of taking one source and

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subtracting it from another or adding in depending upon the size of the V1 and V2.

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So, if I would say for now this is a very useful amplifier configuration

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and you'll see more of it as you go on with

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this course and then also into your linear electronics classes.