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>> Now, I want to spend some time talking about operational amplifiers.

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Operational amplifiers are a device that gets used a lot in electrical circuits.

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You will see them over and over again this semester.

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They get used a lot in instrumentation systems.

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They get used a lot in control systems, etc, etc.

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They're quite often the basis for electrical circuits,

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which perform mathematical operations.

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That's why they're called operational amplifiers.

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Now, there's one point that has to be made very clear up front.

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These are not a passive device.

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So far, with the exception of our power sources,

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all of our circuit elements have been passive than resistors, essentially.

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That means that the energy delivered by the circuit to the element is non-negative.

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This element does not create power out of somewhere else and provide it to the circuit.

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Okay, it has to get any energy that it has from the circuit.

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Operational amplifiers are in active device.

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Okay. They will deliver power to your circuit.

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The way they deliver power to your circuit is because they have an external power supply.

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There's some other magical device somewhere that is feeding

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these guys power, which these guys can then provide to your circuit.

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I will tend to abbreviate operational amplifiers as op-amps.

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That's very common, primarily, just to save syllables.

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Quick overview of operational amplifiers.

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We're going to think of operational amplifiers as a device.

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It's something that performs some task.

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So, we're going to think of them as a black box.

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There is a bunch of internal circuitry in these guys.

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We won't be analyzing these on that level.

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Okay. They're going to be a black box that has

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essentially some input output characteristic.

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That's all we care about.

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One of the drawbacks of dealing with things this way is that it may

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appear as if KCL and KVL don't apply to these guys.

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That's not true.

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If you model the internal circuitry, these guys do satisfy KVL and KCL.

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It's just that there's something very complicated going on inside there.

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Number one, they've got an external power supply that's

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feeding them current or voltage or whatever,

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that we are generally not going to worry too much about when we're

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looking at the op-amp as part of an overall circuit.

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So, what we're going to end up with are several rules

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for how the op-amp is going to behave.

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Okay. Those are based on an analysis of the internal circuitry,

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but we aren't going to worry about that translation.

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We're just going to have a few rules that we're going to say

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this is the way this device behaves,

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we're going to forget about it until a 400-level class later on.

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So, we're going to use op-amps to perform operations,

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but we don't need to actually design and build the operational amplifiers

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themselves or analyze them on a detailed level at this stage in our career.

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Here's a schematic of a very common 741 operational amplifier.

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You can see that it's pretty complex.

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It has a whole bunch of bipolar junction transistors in it.

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It has a bunch of resistors. It has a couple of inputs.

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It has an output. It has a couple of external power supplies and

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it's got some stuff here that we don't even need to worry about yet.

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But we aren't going to deal with this internal view of

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the operational amplifier, we're going to treat it as a black box.

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Okay, our high level view of an operational amplifier is going to be to

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represent it just as this rightward pointing triangle.

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This device has three terminals. There are two input terminals.

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They have a positive and a negative sign associated with

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them and one output terminal here.

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V_n is the voltage applied at the inverting or negative input terminal.

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V_p is the voltage applied at the non-inverting or positive input terminal.

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The out comes at the output of the operational amplifier.

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Now, there are a number of parameters that

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this operational amplifiers operation is going to be characterized relative to.

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They're not necessarily these individual values, they're something else.

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The first of these is the difference in

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voltage between the inverting and non-inverting terminals.

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The change in voltage between V_p and V_n.

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So, Delta V_in is V_p minus V sub n. That's the voltage difference.

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Keep in mind that generally, according to our operational amplifier behavior,

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we don't care what these individual voltages are,

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we just care what the difference is between them.

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The other thing that you use to characterize

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operational amplifier behavior are the currents into the input terminals.

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We'll have some current into the positive or non-inverting terminal and

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some other current into the negative or inverting terminal.

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Okay, these parameters are what we're going to base

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our rules of operational amplifier behavior on.

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Now, I want to provide the rules by which we will

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characterize the operational amplifiers behavior.

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In order to do that, I want to give a slightly more

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complete symbol for the operational amplifier.

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I said that the operational amplifier requires an external power supply to do its job.

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The power supplies are provided here and here.

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There are actually two additional terminals

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that we need to worry about for five terminals in all.

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Generally, when I'm analyzing a circuit, I'll leave these off

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and just in the back of my mind, recognize that they're there,

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but for right now, I need to put them back in.

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Now, I said earlier, that the operation is going to be characterized by this difference in

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voltage delta V_in and the currents into the non-inverting and inverting input terminals.

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So, when we see one of these devices,

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we're going to make the following assumptions.

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These assumptions are relative to ideal operational amplifiers.

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We will assume that the current into

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the non-inverting and the inverting input terminals are both zero.

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This device is not accepting any power into these terminals.

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So, any power that comes into the output comes from the power

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supplies that you've connected up here and here.

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We're also going to assume that the difference in voltage

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between the positive and negative terminals has to be zero.

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Therefore, V_p is going to be equal to V_n.

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One other thing has to be true for operational amplifiers, this output voltage.

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The amount of output voltage you can get here is constrained by these two power supplies.

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This voltage has to be greater than the negative power supply

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and less than the positive power supply,

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and those are strict inequalities.

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Generally, most operational amplifiers you can only get to within

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a volt or two of your power supply voltages.

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Okay, a few points about operational amplifier behavior.

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The output current is generally not known,

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you cannot make any assumptions relative to the output current.

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Right? It's provided by the external power supplies.

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That's where people sometimes early on get themselves into trouble.

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They say, "Okay, there's no current into the input terminals,

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but I've got some current out of the output terminals.

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KCL doesn't apply." Well, it does.

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The current coming out of the operational amplifier is

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coming from the external power supplies,

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you know nothing about those.

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You don't know anything about the output current unless you

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analyze the circuit to determine what that is.

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In general, when I'm analyzing an operational amplifier,

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I will start out by applying KCL at the input nodes.

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Okay. It doesn't always cure all of your problems,

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but it's generally a good place to start.

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The operation of the operational amplifiers is generally based on Delta V_in.

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If I look at this amplifier as an input-output relationship,

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what it looks like is that I have some Delta V_in,

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I multiply that by some large number K.

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That gives me V_out.

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Now, for an ideal operational amplifier,

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we assume that K goes to infinity.

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Okay. For non-ideal or realistic operational amplifiers,

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K is on the order of millions.

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We'll assume it's infinite.

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Also, I mentioned earlier the output voltage is limited by the external power supplies.

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Okay, your output voltage must be lower than the positive power supply.

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It must be higher than the negative power supply.

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These two things in conjunction with one another lead

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us to the conclusion that Delta V_in has to be equal to zero,

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because if V_out is infinity times Delta V_in and V_out must be finite.

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Right, we can only apply so much voltage at the power supply terminals.

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In order to make this a finite number,

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if this is infinite, this guy has to be zero.

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Okay, let's do an example of analyzing an operational amplifier based circuit.

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Here is my op-amp. It has two inputs and one output, okay?

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It also has some other voltage supplies and some resistors hanging around here,

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and what I haven't shown are the external power supplies.

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I generally won't show those. Generally, when I start out analyzing one of these,

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the first thing I'm going to do is employ my op-amp rules, okay?

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There is no voltage difference between the inverting and non-inverting terminals.

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So this delta V is zero. I've tied the non-inverting terminal to ground.

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So, this voltage is zero volts. That means that since I

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can't have a voltage difference between here and here,

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this voltage is zero volts.

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Likewise, the current here into the non-inverting terminal is zero,

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the current here into the inverting terminal is also zero.

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Now, I've labeled everything that I know about this operational amplifier.

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I can go ahead and analyze it to determine V_out, okay?

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As usual, I'll start out applying KCL at an input node.

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KCL at an input node is kind of a good idea,

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because you already know something's there.

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You know there's no current into the op-amp itself.

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So if I call this node A and do KCL at A,

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the current through R_in is this voltage,

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V_in minus this voltage which the op amp is constraining to be zero volts over RN.

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So V_in minus zero over R_in,

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this current into this node is equal to this current out of the node,

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because there's no current flow through this leg, here.

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This current is this voltage minus this voltage over R_f.

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So that's equal to zero minus V_out over R_f.

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V_in, as my input voltage,

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I don't know what it is,

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but I have to be told it before I can determine a number for V_out.

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So V_out, let's multiply this by R_f,

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is R_f over Rn times V_in taking this negative sign over here.

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V_out is equal to minus R_f over R_in times V_in.

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So whatever you give me for V_in,

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I'm going to multiply that by a number,

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take the negative of that,

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and this op-amp will give you that as V_out.

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This has a particular name.

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This is an inverting voltage amplifier, okay?

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It's amplifying voltage.

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You give it a voltage in,

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it gives you a voltage out.

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It inverts that.

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The voltage out you get as the negative of the voltage in.

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It is also amplifying that,

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according to whatever you choose for R_f and R_in.

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If R_f is 10 ohms and R_in is one ohms,

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then this is going to be 10 and the output voltage is

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going to be negative 10 times whatever the input voltage is.

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Let's analyze another operational amplifier circuit.

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I want to find V_out with this operational amplifier based circuit.

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Notice, again, that I have my three terminal device.

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It has some non-inverting and inverting terminal in and output terminal.

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I'm not showing my power supplies,

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but if I wired this circuit up in the lab,

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I would need to provide power to it, okay?

0:14:55.150,0:14:57.565
So let's find V_out as a function of V_in.

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The first thing I want to do is apply my op-amp rules.

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This voltage source is insisting that

0:15:07.645,0:15:13.495
the voltage at the non-inverting terminal is going to be set to be V_in.

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The op-amp itself is insisting that

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there is no voltage difference between these two terminals,

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therefore I must have voltage V_in at this terminal.

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Therefore, this voltage here is V_in.

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Now, I have no current into these terminals.

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Notice, very importantly, this voltage source is not providing any power, okay?

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The current out of the voltage source is zero.

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It's not providing any power in order to create this output voltage.

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Any power in this output signal is coming from the external power supplies.

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So these currents are zero.

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I know something about that current.

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Now I'm going to my old fallback standard position.

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I'm going to do KCL at the input nodes.

0:16:06.865,0:16:11.515
Let me call this node A. I'll apply a KCL there.

0:16:11.515,0:16:18.085
Let me say that this current through the resistor Rf is I_f,

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and this current through R_1 is I_1,

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and those are going to be my positive directions.

0:16:24.475,0:16:31.330
So KCL, at A, tells me that the current going into node A

0:16:31.330,0:16:33.760
is equal to the current coming out of node A,

0:16:33.760,0:16:37.915
so I_f is equal to I_1.

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This current is zero, right?

0:16:42.760,0:16:45.970
I don't need to list it in my KCL.

0:16:45.970,0:16:51.190
Now, I_f is this voltage minus this voltage, over R_f.

0:16:51.190,0:17:01.450
So, V_out minus V_in over R_f is equal to the current going through here,

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which is just V_in minus,

0:17:03.775,0:17:05.290
I'm going to take this as my reference,

0:17:05.290,0:17:07.819
it's going to be zero volts.

0:17:08.579,0:17:15.040
V_in minus zero over R_1.

0:17:15.040,0:17:21.040
So V_out over R_f is equal

0:17:21.040,0:17:27.880
to V_in over R_f taking this term over to the other side,

0:17:27.880,0:17:32.635
plus V_in over R_1.

0:17:32.635,0:17:39.805
Grouping terms, V_out is equal to R_f

0:17:39.805,0:17:48.100
times one over R_f plus one over R_1 times V_in.

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This becomes a one plus R_f over R_1.

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So therefore, V_out is equal

0:18:00.100,0:18:06.740
to one plus R_f over R_1 times V_in.

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This device takes a voltage V_in,

0:18:16.015,0:18:19.375
multiplies it by a number,

0:18:19.375,0:18:22.225
and actually a positive number.

0:18:22.225,0:18:24.250
In order to get V_out,

0:18:24.250,0:18:33.440
this is a non-inverting voltage amplifier.

0:18:36.990,0:18:41.830
Okay, you're still amplifying voltage by taking your input voltage,

0:18:41.830,0:18:44.635
multiplying it by a number to get the output voltage,

0:18:44.635,0:18:48.010
but you're not changing the sign, it not-inverting.

0:18:48.010,0:18:54.415
Now one quick thing I want to point out about both this example and the previous one.

0:18:54.415,0:18:57.820
Both of these had a resistor which was

0:18:57.820,0:19:01.510
feeding back from the output to one of the input terminals.

0:19:01.510,0:19:04.720
That is very typical of op-amp based circuits.

0:19:04.720,0:19:07.315
It's generally called the feedback loop.

0:19:07.315,0:19:14.425
Almost invariably you will feedback from the output to the inverting input terminal.

0:19:14.425,0:19:19.145
That's necessary in order to keep this entire device stable.

0:19:19.145,0:19:22.020
If I feedback to the positive terminal,

0:19:22.020,0:19:25.485
generally this device will do what is called going unstable,

0:19:25.485,0:19:28.470
the output will try to go to infinity.

0:19:28.470,0:19:32.190
So what you'll have happen is that your output voltage will either

0:19:32.190,0:19:36.130
go to the positive or the negative voltage rail, and stay there.

0:19:36.130,0:19:37.540
It can't get to infinity,

0:19:37.540,0:19:40.420
but it's going to go as high as it can go,

0:19:40.420,0:19:42.800
and it's not going to come back.

0:19:42.990,0:19:46.015
Okay, this concludes Lecture 12.

0:19:46.015,0:19:50.185
Next lecture, we'll do some more work with operational amplifiers.

0:19:50.185,0:19:52.810
We'll look at their operation in a little bit more depth,

0:19:52.810,0:19:58.240
we'll talk more about the voltage rails applied by the power supplies,

0:19:58.240,0:20:02.515
and we'll start looking at them as dependent sources.

0:20:02.515,0:20:06.340
Okay, later on in your schooling career you'll see a lot of dependent sources.

0:20:06.340,0:20:11.090
We'll just start kind of looking at things in that way in this class.