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>> The first of the op-amp configurations that we're going

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to consider is known as the non-inverting amplifier.

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It gets its name from the fact that the source voltage to

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be amplified is connected to the non-inverting terminal.

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Because of that, it turns out that the output voltage will be of the same sign,

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S-I-G-N, as the input voltage.

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So, for example, if V_s is a positive voltage,

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the output voltage will also be positive.

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On the other hand, if the input voltage is negative,

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then the output voltage will be negative also.

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The sign is not inverted.

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I have here two different schematics.

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I'd encourage you to stop the video for just a second and convince

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yourself that these two schematics are identical as far as function is concerned.

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The things that you're going to be looking at are that

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the V_s is connected to the non-inverting terminal.

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Of course, the non-inverting terminal is the one that's got the positive sign on it.

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So, if this one is connected to the non-inverting,

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over here, V_s is also connected to the non-inverting.

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So, this amplifier over here is upside down from this op-amp here.

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The other thing that you look at is where does the feedback go.

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We'll talk more about feedback later,

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but suffice for now to say that feedback is the process or the act of

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taking the voltage at the output and running it back to,

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in this case, the negative or the inverting terminal.

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Again, we'll talk more about feedback later,

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but this is referred to as negative feedback.

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You can remember that because the feedback loop comes back

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to the terminal of the negative sign which is the inverting sign.

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So, in both of these feedback circuits,

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we're going from V_out through R_1.

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The node beat between R_1 and R_2 is

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then tabbed and brought back to the inverting terminal.

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Notice, over here, we have the same situation,

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coming from the output,

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going through R_1, connecting to R_2,

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and the node where R_1 and R_2 are connected is connected

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then to the inverting terminal and then brought back the ground.

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Now, let's analyze these op-amps at both of these circuits and see what it is,

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why we might draw it this way under certain circumstances

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and why we might draw it this way under other circumstances.

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Let's start here with the one on the right.

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By drawing it in this configuration,

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it makes it obvious that the output voltage goes through a voltage divider circuit,

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and only a portion of the output voltage is fed back to the inverting terminal.

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In this case here,

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V_n then,

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using our voltage divider formula,

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V_n is equal to the voltage across R_2,

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which is V_out times R_2 over R_1 plus R_2.

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Now, in order to get V_out as a function of V,

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our input, we're going to reverse the roles

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here and multiply both sides by the inverse of this.

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In solving or solving for V_out,

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we get that V_out is equal to V_n times R_1 plus R_2 over R_2.

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Now, we don't want it as a function of V_n.

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We want to know what the output is as a function of the input voltage

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V_s. We now apply one of our ideal op-amp approximations.

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That was the one that referred to as the virtual short.

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That V_p and V_n are going to be so close to each

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other that the difference V_p minus V_n is zero,

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or what we can say then is that V_n is approximately equal to V_p.

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Now, what is V_p?

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Using another op-amp approximation,

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the current going into the input terminals is zero.

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Therefore, I_p is zero.

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There will be no voltage drop across

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V_s because the current going through it is zero.

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So, V_p is in fact just our source voltage V_s.

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So, V_p equals V_s. V_n equals V_p due to the virtual short.

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We have then that V_out again replacing V_n with V_s.

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V_out is equal to V_s times R_1 plus R_2 over R_2.

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Generally speaking, we'll take this and say, well,

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note that R_2 is a denominator that's common to both of those two terms.

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So, we can then have V_out is equal to V_s times R_2 over R_2,

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that's one, plus R_1 over R_2.

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We say that the gain, G, the closed loop gain,

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the gain that we get because of this feedback circuit is equal

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to one plus R_1 plus R_2.

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We're going to refer to that as the gain of the non-inverting amplifier.

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We'll see when we get to the inverting amplifier configuration that

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its gain is just slightly different than this.

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But let's just point out now that V_out is in fact going to be

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the same sign as V_s. R_1 and R_2 are both positive quantities.

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So, the ratio of a positive quantities is positive,

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plus one is a positive number,

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times whatever the sign is on V_s gives us a V_out,

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which is the same sign as V_s. Now,

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let's look at this circuit over here.

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To analyze this, we're going to use a technique that use a node analysis,

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which is an analysis technique that we'll frequently use on op-amp circuits,

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especially as the op-amp circuit gets to be a little bit more complex.

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So, here's the deal.

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This is now V_n. We're going to write a node equation at V_n. But first,

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we're going to note that this voltage here, V_p,

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is going to equal our source voltage.

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As we saw over here, the R_s had no influence on it.

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So, I've left that out of this circuit here.

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So, V_p equals V_s. V_n equals V_p because of the virtual short.

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So, V_n then is going to equal V_s.

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Now, let's write a node equation summing the currents leaving this node.

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The current leaving this node going through R_2 to ground is V_s minus

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zero divided by R_2 plus the current going from this node in this direction.

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It's going to be the voltage across R_1, which is V_s,

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minus V_out divided by R_1,

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and then plus the current going into or leaving this node and going into the op-amp.

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But our ideal op-amp approximation tells us

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that the current going into the input terminal is zero.

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So, we could write a plus zero there.

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But let's that off and simply say that the sum of those two currents must equal zero.

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So, what that's saying is that in fact,

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just as we saw over here,

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the current is going from the output through these two resistors back.

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It's going from the output through these two resistors to ground.

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These two resistors are in series because the current going in here is zero.

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Now, let's solve this equation for V_out in terms of these paths.

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We have V_s factoring out V_s times one over R_2 plus one over

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R_1 minus V_out over R_1 is equal to zero.

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That bring this term over to the other side, and solving for V_0,

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We get then that V_0 is equal to V_s times R_1 times

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one over R_2 plus one over R_1 or distributing that

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through the R_1 over the R_1 gives you the one,

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the R_1 over R_2 gives you the other term,

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and we get the V_out is equal to

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V_s times one plus R_1 over R_2.

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So, the gain terms are the same.

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These two circuits are equivalent

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and that gives you an idea of what the non-inverting amplifier does.