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[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:05.60,0:00:10.29,Default,,0000,0000,0000,,>> Now, I want to spend some time talking about operational amplifiers.
Dialogue: 0,0:00:10.29,0:00:15.64,Default,,0000,0000,0000,,Operational amplifiers are a device that gets used a lot in electrical circuits.
Dialogue: 0,0:00:15.64,0:00:18.46,Default,,0000,0000,0000,,You will see them over and over again this semester.
Dialogue: 0,0:00:18.46,0:00:20.67,Default,,0000,0000,0000,,They get used a lot in instrumentation systems.
Dialogue: 0,0:00:20.67,0:00:24.03,Default,,0000,0000,0000,,They get used a lot in control systems, etc, etc.
Dialogue: 0,0:00:24.03,0:00:27.45,Default,,0000,0000,0000,,They're quite often the basis for electrical circuits,
Dialogue: 0,0:00:27.45,0:00:30.21,Default,,0000,0000,0000,,which perform mathematical operations.
Dialogue: 0,0:00:30.21,0:00:33.45,Default,,0000,0000,0000,,That's why they're called operational amplifiers.
Dialogue: 0,0:00:33.45,0:00:37.60,Default,,0000,0000,0000,,Now, there's one point that has to be made very clear up front.
Dialogue: 0,0:00:37.60,0:00:40.06,Default,,0000,0000,0000,,These are not a passive device.
Dialogue: 0,0:00:40.06,0:00:43.37,Default,,0000,0000,0000,,So far, with the exception of our power sources,
Dialogue: 0,0:00:43.37,0:00:48.44,Default,,0000,0000,0000,,all of our circuit elements have been passive than resistors, essentially.
Dialogue: 0,0:00:48.44,0:00:55.12,Default,,0000,0000,0000,,That means that the energy delivered by the circuit to the element is non-negative.
Dialogue: 0,0:00:55.12,0:01:01.40,Default,,0000,0000,0000,,This element does not create power out of somewhere else and provide it to the circuit.
Dialogue: 0,0:01:01.40,0:01:06.18,Default,,0000,0000,0000,,Okay, it has to get any energy that it has from the circuit.
Dialogue: 0,0:01:06.52,0:01:11.26,Default,,0000,0000,0000,,Operational amplifiers are in active device.
Dialogue: 0,0:01:11.26,0:01:15.69,Default,,0000,0000,0000,,Okay. They will deliver power to your circuit.
Dialogue: 0,0:01:16.21,0:01:23.63,Default,,0000,0000,0000,,The way they deliver power to your circuit is because they have an external power supply.
Dialogue: 0,0:01:23.63,0:01:27.56,Default,,0000,0000,0000,,There's some other magical device somewhere that is feeding
Dialogue: 0,0:01:27.56,0:01:31.68,Default,,0000,0000,0000,,these guys power, which these guys can then provide to your circuit.
Dialogue: 0,0:01:31.68,0:01:37.42,Default,,0000,0000,0000,,I will tend to abbreviate operational amplifiers as op-amps.
Dialogue: 0,0:01:37.42,0:01:41.61,Default,,0000,0000,0000,,That's very common, primarily, just to save syllables.
Dialogue: 0,0:01:41.86,0:01:45.22,Default,,0000,0000,0000,,Quick overview of operational amplifiers.
Dialogue: 0,0:01:45.22,0:01:48.40,Default,,0000,0000,0000,,We're going to think of operational amplifiers as a device.
Dialogue: 0,0:01:48.40,0:01:51.84,Default,,0000,0000,0000,,It's something that performs some task.
Dialogue: 0,0:01:51.84,0:01:55.67,Default,,0000,0000,0000,,So, we're going to think of them as a black box.
Dialogue: 0,0:01:55.67,0:01:58.86,Default,,0000,0000,0000,,There is a bunch of internal circuitry in these guys.
Dialogue: 0,0:01:58.86,0:02:02.39,Default,,0000,0000,0000,,We won't be analyzing these on that level.
Dialogue: 0,0:02:02.39,0:02:04.34,Default,,0000,0000,0000,,Okay. They're going to be a black box that has
Dialogue: 0,0:02:04.34,0:02:06.66,Default,,0000,0000,0000,,essentially some input output characteristic.
Dialogue: 0,0:02:06.66,0:02:08.82,Default,,0000,0000,0000,,That's all we care about.
Dialogue: 0,0:02:08.82,0:02:13.67,Default,,0000,0000,0000,,One of the drawbacks of dealing with things this way is that it may
Dialogue: 0,0:02:13.67,0:02:18.23,Default,,0000,0000,0000,,appear as if KCL and KVL don't apply to these guys.
Dialogue: 0,0:02:18.23,0:02:19.74,Default,,0000,0000,0000,,That's not true.
Dialogue: 0,0:02:19.74,0:02:25.52,Default,,0000,0000,0000,,If you model the internal circuitry, these guys do satisfy KVL and KCL.
Dialogue: 0,0:02:25.52,0:02:29.58,Default,,0000,0000,0000,,It's just that there's something very complicated going on inside there.
Dialogue: 0,0:02:29.58,0:02:32.18,Default,,0000,0000,0000,,Number one, they've got an external power supply that's
Dialogue: 0,0:02:32.18,0:02:35.18,Default,,0000,0000,0000,,feeding them current or voltage or whatever,
Dialogue: 0,0:02:35.18,0:02:38.36,Default,,0000,0000,0000,,that we are generally not going to worry too much about when we're
Dialogue: 0,0:02:38.36,0:02:41.97,Default,,0000,0000,0000,,looking at the op-amp as part of an overall circuit.
Dialogue: 0,0:02:41.97,0:02:46.25,Default,,0000,0000,0000,,So, what we're going to end up with are several rules
Dialogue: 0,0:02:46.25,0:02:48.94,Default,,0000,0000,0000,,for how the op-amp is going to behave.
Dialogue: 0,0:02:48.94,0:02:53.70,Default,,0000,0000,0000,,Okay. Those are based on an analysis of the internal circuitry,
Dialogue: 0,0:02:53.70,0:02:56.27,Default,,0000,0000,0000,,but we aren't going to worry about that translation.
Dialogue: 0,0:02:56.27,0:02:58.10,Default,,0000,0000,0000,,We're just going to have a few rules that we're going to say
Dialogue: 0,0:02:58.10,0:03:00.29,Default,,0000,0000,0000,,this is the way this device behaves,
Dialogue: 0,0:03:00.29,0:03:04.74,Default,,0000,0000,0000,,we're going to forget about it until a 400-level class later on.
Dialogue: 0,0:03:04.74,0:03:09.20,Default,,0000,0000,0000,,So, we're going to use op-amps to perform operations,
Dialogue: 0,0:03:09.20,0:03:13.37,Default,,0000,0000,0000,,but we don't need to actually design and build the operational amplifiers
Dialogue: 0,0:03:13.37,0:03:19.05,Default,,0000,0000,0000,,themselves or analyze them on a detailed level at this stage in our career.
Dialogue: 0,0:03:19.37,0:03:26.46,Default,,0000,0000,0000,,Here's a schematic of a very common 741 operational amplifier.
Dialogue: 0,0:03:26.46,0:03:28.37,Default,,0000,0000,0000,,You can see that it's pretty complex.
Dialogue: 0,0:03:28.37,0:03:31.90,Default,,0000,0000,0000,,It has a whole bunch of bipolar junction transistors in it.
Dialogue: 0,0:03:31.90,0:03:36.41,Default,,0000,0000,0000,,It has a bunch of resistors. It has a couple of inputs.
Dialogue: 0,0:03:36.41,0:03:40.85,Default,,0000,0000,0000,,It has an output. It has a couple of external power supplies and
Dialogue: 0,0:03:40.85,0:03:44.40,Default,,0000,0000,0000,,it's got some stuff here that we don't even need to worry about yet.
Dialogue: 0,0:03:44.40,0:03:47.81,Default,,0000,0000,0000,,But we aren't going to deal with this internal view of
Dialogue: 0,0:03:47.81,0:03:52.32,Default,,0000,0000,0000,,the operational amplifier, we're going to treat it as a black box.
Dialogue: 0,0:03:52.36,0:03:57.20,Default,,0000,0000,0000,,Okay, our high level view of an operational amplifier is going to be to
Dialogue: 0,0:03:57.20,0:04:01.57,Default,,0000,0000,0000,,represent it just as this rightward pointing triangle.
Dialogue: 0,0:04:01.57,0:04:07.28,Default,,0000,0000,0000,,This device has three terminals. There are two input terminals.
Dialogue: 0,0:04:07.28,0:04:10.19,Default,,0000,0000,0000,,They have a positive and a negative sign associated with
Dialogue: 0,0:04:10.19,0:04:13.78,Default,,0000,0000,0000,,them and one output terminal here.
Dialogue: 0,0:04:13.78,0:04:20.15,Default,,0000,0000,0000,,V_n is the voltage applied at the inverting or negative input terminal.
Dialogue: 0,0:04:20.15,0:04:26.52,Default,,0000,0000,0000,,V_p is the voltage applied at the non-inverting or positive input terminal.
Dialogue: 0,0:04:26.52,0:04:32.02,Default,,0000,0000,0000,,The out comes at the output of the operational amplifier.
Dialogue: 0,0:04:32.02,0:04:34.73,Default,,0000,0000,0000,,Now, there are a number of parameters that
Dialogue: 0,0:04:34.73,0:04:39.90,Default,,0000,0000,0000,,this operational amplifiers operation is going to be characterized relative to.
Dialogue: 0,0:04:39.90,0:04:44.33,Default,,0000,0000,0000,,They're not necessarily these individual values, they're something else.
Dialogue: 0,0:04:44.33,0:04:47.21,Default,,0000,0000,0000,,The first of these is the difference in
Dialogue: 0,0:04:47.21,0:04:50.60,Default,,0000,0000,0000,,voltage between the inverting and non-inverting terminals.
Dialogue: 0,0:04:50.60,0:04:54.06,Default,,0000,0000,0000,,The change in voltage between V_p and V_n.
Dialogue: 0,0:04:54.06,0:05:02.56,Default,,0000,0000,0000,,So, Delta V_in is V_p minus V sub n. That's the voltage difference.
Dialogue: 0,0:05:02.56,0:05:06.95,Default,,0000,0000,0000,,Keep in mind that generally, according to our operational amplifier behavior,
Dialogue: 0,0:05:06.95,0:05:09.65,Default,,0000,0000,0000,,we don't care what these individual voltages are,
Dialogue: 0,0:05:09.65,0:05:13.10,Default,,0000,0000,0000,,we just care what the difference is between them.
Dialogue: 0,0:05:13.10,0:05:15.80,Default,,0000,0000,0000,,The other thing that you use to characterize
Dialogue: 0,0:05:15.80,0:05:20.65,Default,,0000,0000,0000,,operational amplifier behavior are the currents into the input terminals.
Dialogue: 0,0:05:20.65,0:05:25.86,Default,,0000,0000,0000,,We'll have some current into the positive or non-inverting terminal and
Dialogue: 0,0:05:25.86,0:05:29.68,Default,,0000,0000,0000,,some other current into the negative or inverting terminal.
Dialogue: 0,0:05:29.68,0:05:32.39,Default,,0000,0000,0000,,Okay, these parameters are what we're going to base
Dialogue: 0,0:05:32.39,0:05:36.23,Default,,0000,0000,0000,,our rules of operational amplifier behavior on.
Dialogue: 0,0:05:36.23,0:05:39.92,Default,,0000,0000,0000,,Now, I want to provide the rules by which we will
Dialogue: 0,0:05:39.92,0:05:43.10,Default,,0000,0000,0000,,characterize the operational amplifiers behavior.
Dialogue: 0,0:05:43.10,0:05:45.74,Default,,0000,0000,0000,,In order to do that, I want to give a slightly more
Dialogue: 0,0:05:45.74,0:05:49.30,Default,,0000,0000,0000,,complete symbol for the operational amplifier.
Dialogue: 0,0:05:49.30,0:05:55.19,Default,,0000,0000,0000,,I said that the operational amplifier requires an external power supply to do its job.
Dialogue: 0,0:05:55.19,0:05:57.89,Default,,0000,0000,0000,,The power supplies are provided here and here.
Dialogue: 0,0:05:57.89,0:06:00.86,Default,,0000,0000,0000,,There are actually two additional terminals
Dialogue: 0,0:06:00.86,0:06:04.30,Default,,0000,0000,0000,,that we need to worry about for five terminals in all.
Dialogue: 0,0:06:04.30,0:06:08.48,Default,,0000,0000,0000,,Generally, when I'm analyzing a circuit, I'll leave these off
Dialogue: 0,0:06:08.48,0:06:12.08,Default,,0000,0000,0000,,and just in the back of my mind, recognize that they're there,
Dialogue: 0,0:06:12.08,0:06:14.88,Default,,0000,0000,0000,,but for right now, I need to put them back in.
Dialogue: 0,0:06:14.88,0:06:19.61,Default,,0000,0000,0000,,Now, I said earlier, that the operation is going to be characterized by this difference in
Dialogue: 0,0:06:19.61,0:06:27.23,Default,,0000,0000,0000,,voltage delta V_in and the currents into the non-inverting and inverting input terminals.
Dialogue: 0,0:06:27.23,0:06:30.38,Default,,0000,0000,0000,,So, when we see one of these devices,
Dialogue: 0,0:06:30.38,0:06:32.62,Default,,0000,0000,0000,,we're going to make the following assumptions.
Dialogue: 0,0:06:32.62,0:06:37.82,Default,,0000,0000,0000,,These assumptions are relative to ideal operational amplifiers.
Dialogue: 0,0:06:37.82,0:06:42.11,Default,,0000,0000,0000,,We will assume that the current into
Dialogue: 0,0:06:42.11,0:06:47.27,Default,,0000,0000,0000,,the non-inverting and the inverting input terminals are both zero.
Dialogue: 0,0:06:47.27,0:06:52.08,Default,,0000,0000,0000,,This device is not accepting any power into these terminals.
Dialogue: 0,0:06:52.08,0:06:57.08,Default,,0000,0000,0000,,So, any power that comes into the output comes from the power
Dialogue: 0,0:06:57.08,0:07:00.56,Default,,0000,0000,0000,,supplies that you've connected up here and here.
Dialogue: 0,0:07:00.56,0:07:05.00,Default,,0000,0000,0000,,We're also going to assume that the difference in voltage
Dialogue: 0,0:07:05.00,0:07:10.50,Default,,0000,0000,0000,,between the positive and negative terminals has to be zero.
Dialogue: 0,0:07:10.60,0:07:23.88,Default,,0000,0000,0000,,Therefore, V_p is going to be equal to V_n.
Dialogue: 0,0:07:23.88,0:07:30.100,Default,,0000,0000,0000,,One other thing has to be true for operational amplifiers, this output voltage.
Dialogue: 0,0:07:30.100,0:07:39.72,Default,,0000,0000,0000,,The amount of output voltage you can get here is constrained by these two power supplies.
Dialogue: 0,0:07:39.72,0:07:44.81,Default,,0000,0000,0000,,This voltage has to be greater than the negative power supply
Dialogue: 0,0:07:44.81,0:07:48.52,Default,,0000,0000,0000,,and less than the positive power supply,
Dialogue: 0,0:07:48.52,0:07:50.76,Default,,0000,0000,0000,,and those are strict inequalities.
Dialogue: 0,0:07:50.76,0:07:54.28,Default,,0000,0000,0000,,Generally, most operational amplifiers you can only get to within
Dialogue: 0,0:07:54.28,0:07:59.37,Default,,0000,0000,0000,,a volt or two of your power supply voltages.
Dialogue: 0,0:07:59.66,0:08:03.96,Default,,0000,0000,0000,,Okay, a few points about operational amplifier behavior.
Dialogue: 0,0:08:03.96,0:08:07.46,Default,,0000,0000,0000,,The output current is generally not known,
Dialogue: 0,0:08:07.46,0:08:10.94,Default,,0000,0000,0000,,you cannot make any assumptions relative to the output current.
Dialogue: 0,0:08:10.94,0:08:14.62,Default,,0000,0000,0000,,Right? It's provided by the external power supplies.
Dialogue: 0,0:08:14.62,0:08:17.60,Default,,0000,0000,0000,,That's where people sometimes early on get themselves into trouble.
Dialogue: 0,0:08:17.60,0:08:20.40,Default,,0000,0000,0000,,They say, "Okay, there's no current into the input terminals,
Dialogue: 0,0:08:20.40,0:08:22.73,Default,,0000,0000,0000,,but I've got some current out of the output terminals.
Dialogue: 0,0:08:22.73,0:08:25.38,Default,,0000,0000,0000,,KCL doesn't apply." Well, it does.
Dialogue: 0,0:08:25.38,0:08:27.83,Default,,0000,0000,0000,,The current coming out of the operational amplifier is
Dialogue: 0,0:08:27.83,0:08:29.93,Default,,0000,0000,0000,,coming from the external power supplies,
Dialogue: 0,0:08:29.93,0:08:32.00,Default,,0000,0000,0000,,you know nothing about those.
Dialogue: 0,0:08:32.00,0:08:35.09,Default,,0000,0000,0000,,You don't know anything about the output current unless you
Dialogue: 0,0:08:35.09,0:08:38.81,Default,,0000,0000,0000,,analyze the circuit to determine what that is.
Dialogue: 0,0:08:38.81,0:08:42.77,Default,,0000,0000,0000,,In general, when I'm analyzing an operational amplifier,
Dialogue: 0,0:08:42.77,0:08:47.80,Default,,0000,0000,0000,,I will start out by applying KCL at the input nodes.
Dialogue: 0,0:08:47.80,0:08:52.26,Default,,0000,0000,0000,,Okay. It doesn't always cure all of your problems,
Dialogue: 0,0:08:52.26,0:08:55.73,Default,,0000,0000,0000,,but it's generally a good place to start.
Dialogue: 0,0:08:55.73,0:09:03.06,Default,,0000,0000,0000,,The operation of the operational amplifiers is generally based on Delta V_in.
Dialogue: 0,0:09:03.06,0:09:09.04,Default,,0000,0000,0000,,If I look at this amplifier as an input-output relationship,
Dialogue: 0,0:09:09.04,0:09:14.15,Default,,0000,0000,0000,,what it looks like is that I have some Delta V_in,
Dialogue: 0,0:09:14.15,0:09:19.50,Default,,0000,0000,0000,,I multiply that by some large number K.
Dialogue: 0,0:09:19.50,0:09:28.23,Default,,0000,0000,0000,,That gives me V_out.
Dialogue: 0,0:09:28.23,0:09:32.16,Default,,0000,0000,0000,,Now, for an ideal operational amplifier,
Dialogue: 0,0:09:35.81,0:09:40.10,Default,,0000,0000,0000,,we assume that K goes to infinity.
Dialogue: 0,0:09:40.10,0:09:43.73,Default,,0000,0000,0000,,Okay. For non-ideal or realistic operational amplifiers,
Dialogue: 0,0:09:43.73,0:09:45.64,Default,,0000,0000,0000,,K is on the order of millions.
Dialogue: 0,0:09:45.64,0:09:48.05,Default,,0000,0000,0000,,We'll assume it's infinite.
Dialogue: 0,0:09:48.26,0:09:55.19,Default,,0000,0000,0000,,Also, I mentioned earlier the output voltage is limited by the external power supplies.
Dialogue: 0,0:09:55.19,0:10:00.74,Default,,0000,0000,0000,,Okay, your output voltage must be lower than the positive power supply.
Dialogue: 0,0:10:00.74,0:10:04.62,Default,,0000,0000,0000,,It must be higher than the negative power supply.
Dialogue: 0,0:10:04.62,0:10:08.90,Default,,0000,0000,0000,,These two things in conjunction with one another lead
Dialogue: 0,0:10:08.90,0:10:13.22,Default,,0000,0000,0000,,us to the conclusion that Delta V_in has to be equal to zero,
Dialogue: 0,0:10:13.22,0:10:20.53,Default,,0000,0000,0000,,because if V_out is infinity times Delta V_in and V_out must be finite.
Dialogue: 0,0:10:24.69,0:10:30.31,Default,,0000,0000,0000,,Right, we can only apply so much voltage at the power supply terminals.
Dialogue: 0,0:10:30.31,0:10:32.78,Default,,0000,0000,0000,,In order to make this a finite number,
Dialogue: 0,0:10:32.78,0:10:37.01,Default,,0000,0000,0000,,if this is infinite, this guy has to be zero.
Dialogue: 0,0:10:39.90,0:10:47.32,Default,,0000,0000,0000,,Okay, let's do an example of analyzing an operational amplifier based circuit.
Dialogue: 0,0:10:47.32,0:10:52.96,Default,,0000,0000,0000,,Here is my op-amp. It has two inputs and one output, okay?
Dialogue: 0,0:10:52.96,0:10:57.76,Default,,0000,0000,0000,,It also has some other voltage supplies and some resistors hanging around here,
Dialogue: 0,0:10:57.76,0:11:01.26,Default,,0000,0000,0000,,and what I haven't shown are the external power supplies.
Dialogue: 0,0:11:01.26,0:11:07.09,Default,,0000,0000,0000,,I generally won't show those. Generally, when I start out analyzing one of these,
Dialogue: 0,0:11:07.09,0:11:12.78,Default,,0000,0000,0000,,the first thing I'm going to do is employ my op-amp rules, okay?
Dialogue: 0,0:11:12.78,0:11:17.59,Default,,0000,0000,0000,,There is no voltage difference between the inverting and non-inverting terminals.
Dialogue: 0,0:11:17.59,0:11:23.47,Default,,0000,0000,0000,,So this delta V is zero. I've tied the non-inverting terminal to ground.
Dialogue: 0,0:11:23.47,0:11:29.50,Default,,0000,0000,0000,,So, this voltage is zero volts. That means that since I
Dialogue: 0,0:11:29.50,0:11:32.52,Default,,0000,0000,0000,,can't have a voltage difference between here and here,
Dialogue: 0,0:11:32.52,0:11:35.81,Default,,0000,0000,0000,,this voltage is zero volts.
Dialogue: 0,0:11:37.11,0:11:43.69,Default,,0000,0000,0000,,Likewise, the current here into the non-inverting terminal is zero,
Dialogue: 0,0:11:43.69,0:11:48.76,Default,,0000,0000,0000,,the current here into the inverting terminal is also zero.
Dialogue: 0,0:11:48.76,0:11:53.44,Default,,0000,0000,0000,,Now, I've labeled everything that I know about this operational amplifier.
Dialogue: 0,0:11:53.44,0:11:59.59,Default,,0000,0000,0000,,I can go ahead and analyze it to determine V_out, okay?
Dialogue: 0,0:11:59.59,0:12:04.42,Default,,0000,0000,0000,,As usual, I'll start out applying KCL at an input node.
Dialogue: 0,0:12:04.42,0:12:07.00,Default,,0000,0000,0000,,KCL at an input node is kind of a good idea,
Dialogue: 0,0:12:07.00,0:12:08.56,Default,,0000,0000,0000,,because you already know something's there.
Dialogue: 0,0:12:08.56,0:12:11.64,Default,,0000,0000,0000,,You know there's no current into the op-amp itself.
Dialogue: 0,0:12:11.64,0:12:18.47,Default,,0000,0000,0000,,So if I call this node A and do KCL at A,
Dialogue: 0,0:12:19.38,0:12:24.07,Default,,0000,0000,0000,,the current through R_in is this voltage,
Dialogue: 0,0:12:24.07,0:12:31.74,Default,,0000,0000,0000,,V_in minus this voltage which the op amp is constraining to be zero volts over RN.
Dialogue: 0,0:12:31.74,0:12:36.32,Default,,0000,0000,0000,,So V_in minus zero over R_in,
Dialogue: 0,0:12:36.32,0:12:41.11,Default,,0000,0000,0000,,this current into this node is equal to this current out of the node,
Dialogue: 0,0:12:41.11,0:12:44.20,Default,,0000,0000,0000,,because there's no current flow through this leg, here.
Dialogue: 0,0:12:44.20,0:12:49.93,Default,,0000,0000,0000,,This current is this voltage minus this voltage over R_f.
Dialogue: 0,0:12:49.93,0:12:58.64,Default,,0000,0000,0000,,So that's equal to zero minus V_out over R_f.
Dialogue: 0,0:12:58.64,0:13:01.12,Default,,0000,0000,0000,,V_in, as my input voltage,
Dialogue: 0,0:13:01.12,0:13:02.14,Default,,0000,0000,0000,,I don't know what it is,
Dialogue: 0,0:13:02.14,0:13:05.83,Default,,0000,0000,0000,,but I have to be told it before I can determine a number for V_out.
Dialogue: 0,0:13:05.83,0:13:12.86,Default,,0000,0000,0000,,So V_out, let's multiply this by R_f,
Dialogue: 0,0:13:12.86,0:13:21.46,Default,,0000,0000,0000,,is R_f over Rn times V_in taking this negative sign over here.
Dialogue: 0,0:13:21.46,0:13:28.27,Default,,0000,0000,0000,,V_out is equal to minus R_f over R_in times V_in.
Dialogue: 0,0:13:28.27,0:13:31.66,Default,,0000,0000,0000,,So whatever you give me for V_in,
Dialogue: 0,0:13:31.66,0:13:35.16,Default,,0000,0000,0000,,I'm going to multiply that by a number,
Dialogue: 0,0:13:35.16,0:13:37.33,Default,,0000,0000,0000,,take the negative of that,
Dialogue: 0,0:13:37.33,0:13:41.69,Default,,0000,0000,0000,,and this op-amp will give you that as V_out.
Dialogue: 0,0:13:42.57,0:13:46.45,Default,,0000,0000,0000,,This has a particular name.
Dialogue: 0,0:13:46.45,0:14:02.26,Default,,0000,0000,0000,,This is an inverting voltage amplifier, okay?
Dialogue: 0,0:14:02.26,0:14:04.58,Default,,0000,0000,0000,,It's amplifying voltage.
Dialogue: 0,0:14:04.58,0:14:05.90,Default,,0000,0000,0000,,You give it a voltage in,
Dialogue: 0,0:14:05.90,0:14:07.34,Default,,0000,0000,0000,,it gives you a voltage out.
Dialogue: 0,0:14:07.34,0:14:08.60,Default,,0000,0000,0000,,It inverts that.
Dialogue: 0,0:14:08.60,0:14:12.24,Default,,0000,0000,0000,,The voltage out you get as the negative of the voltage in.
Dialogue: 0,0:14:12.24,0:14:14.44,Default,,0000,0000,0000,,It is also amplifying that,
Dialogue: 0,0:14:14.44,0:14:18.42,Default,,0000,0000,0000,,according to whatever you choose for R_f and R_in.
Dialogue: 0,0:14:18.42,0:14:21.79,Default,,0000,0000,0000,,If R_f is 10 ohms and R_in is one ohms,
Dialogue: 0,0:14:21.79,0:14:24.91,Default,,0000,0000,0000,,then this is going to be 10 and the output voltage is
Dialogue: 0,0:14:24.91,0:14:29.03,Default,,0000,0000,0000,,going to be negative 10 times whatever the input voltage is.
Dialogue: 0,0:14:29.22,0:14:33.13,Default,,0000,0000,0000,,Let's analyze another operational amplifier circuit.
Dialogue: 0,0:14:33.13,0:14:38.38,Default,,0000,0000,0000,,I want to find V_out with this operational amplifier based circuit.
Dialogue: 0,0:14:38.38,0:14:41.89,Default,,0000,0000,0000,,Notice, again, that I have my three terminal device.
Dialogue: 0,0:14:41.89,0:14:47.47,Default,,0000,0000,0000,,It has some non-inverting and inverting terminal in and output terminal.
Dialogue: 0,0:14:47.47,0:14:49.69,Default,,0000,0000,0000,,I'm not showing my power supplies,
Dialogue: 0,0:14:49.69,0:14:51.80,Default,,0000,0000,0000,,but if I wired this circuit up in the lab,
Dialogue: 0,0:14:51.80,0:14:55.15,Default,,0000,0000,0000,,I would need to provide power to it, okay?
Dialogue: 0,0:14:55.15,0:14:57.56,Default,,0000,0000,0000,,So let's find V_out as a function of V_in.
Dialogue: 0,0:14:57.56,0:15:02.38,Default,,0000,0000,0000,,The first thing I want to do is apply my op-amp rules.
Dialogue: 0,0:15:02.38,0:15:07.64,Default,,0000,0000,0000,,This voltage source is insisting that
Dialogue: 0,0:15:07.64,0:15:13.50,Default,,0000,0000,0000,,the voltage at the non-inverting terminal is going to be set to be V_in.
Dialogue: 0,0:15:13.50,0:15:17.11,Default,,0000,0000,0000,,The op-amp itself is insisting that
Dialogue: 0,0:15:17.11,0:15:20.30,Default,,0000,0000,0000,,there is no voltage difference between these two terminals,
Dialogue: 0,0:15:20.30,0:15:24.55,Default,,0000,0000,0000,,therefore I must have voltage V_in at this terminal.
Dialogue: 0,0:15:24.55,0:15:29.99,Default,,0000,0000,0000,,Therefore, this voltage here is V_in.
Dialogue: 0,0:15:30.18,0:15:36.22,Default,,0000,0000,0000,,Now, I have no current into these terminals.
Dialogue: 0,0:15:36.22,0:15:44.06,Default,,0000,0000,0000,,Notice, very importantly, this voltage source is not providing any power, okay?
Dialogue: 0,0:15:44.06,0:15:46.23,Default,,0000,0000,0000,,The current out of the voltage source is zero.
Dialogue: 0,0:15:46.23,0:15:50.16,Default,,0000,0000,0000,,It's not providing any power in order to create this output voltage.
Dialogue: 0,0:15:50.16,0:15:55.36,Default,,0000,0000,0000,,Any power in this output signal is coming from the external power supplies.
Dialogue: 0,0:15:55.36,0:15:57.38,Default,,0000,0000,0000,,So these currents are zero.
Dialogue: 0,0:15:57.38,0:15:59.42,Default,,0000,0000,0000,,I know something about that current.
Dialogue: 0,0:15:59.42,0:16:04.06,Default,,0000,0000,0000,,Now I'm going to my old fallback standard position.
Dialogue: 0,0:16:04.06,0:16:06.86,Default,,0000,0000,0000,,I'm going to do KCL at the input nodes.
Dialogue: 0,0:16:06.86,0:16:11.52,Default,,0000,0000,0000,,Let me call this node A. I'll apply a KCL there.
Dialogue: 0,0:16:11.52,0:16:18.08,Default,,0000,0000,0000,,Let me say that this current through the resistor Rf is I_f,
Dialogue: 0,0:16:18.08,0:16:21.52,Default,,0000,0000,0000,,and this current through R_1 is I_1,
Dialogue: 0,0:16:21.52,0:16:24.48,Default,,0000,0000,0000,,and those are going to be my positive directions.
Dialogue: 0,0:16:24.48,0:16:31.33,Default,,0000,0000,0000,,So KCL, at A, tells me that the current going into node A
Dialogue: 0,0:16:31.33,0:16:33.76,Default,,0000,0000,0000,,is equal to the current coming out of node A,
Dialogue: 0,0:16:33.76,0:16:37.92,Default,,0000,0000,0000,,so I_f is equal to I_1.
Dialogue: 0,0:16:37.92,0:16:42.76,Default,,0000,0000,0000,,This current is zero, right?
Dialogue: 0,0:16:42.76,0:16:45.97,Default,,0000,0000,0000,,I don't need to list it in my KCL.
Dialogue: 0,0:16:45.97,0:16:51.19,Default,,0000,0000,0000,,Now, I_f is this voltage minus this voltage, over R_f.
Dialogue: 0,0:16:51.19,0:17:01.45,Default,,0000,0000,0000,,So, V_out minus V_in over R_f is equal to the current going through here,
Dialogue: 0,0:17:01.45,0:17:03.78,Default,,0000,0000,0000,,which is just V_in minus,
Dialogue: 0,0:17:03.78,0:17:05.29,Default,,0000,0000,0000,,I'm going to take this as my reference,
Dialogue: 0,0:17:05.29,0:17:07.82,Default,,0000,0000,0000,,it's going to be zero volts.
Dialogue: 0,0:17:08.58,0:17:15.04,Default,,0000,0000,0000,,V_in minus zero over R_1.
Dialogue: 0,0:17:15.04,0:17:21.04,Default,,0000,0000,0000,,So V_out over R_f is equal
Dialogue: 0,0:17:21.04,0:17:27.88,Default,,0000,0000,0000,,to V_in over R_f taking this term over to the other side,
Dialogue: 0,0:17:27.88,0:17:32.64,Default,,0000,0000,0000,,plus V_in over R_1.
Dialogue: 0,0:17:32.64,0:17:39.80,Default,,0000,0000,0000,,Grouping terms, V_out is equal to R_f
Dialogue: 0,0:17:39.80,0:17:48.10,Default,,0000,0000,0000,,times one over R_f plus one over R_1 times V_in.
Dialogue: 0,0:17:48.10,0:17:52.10,Default,,0000,0000,0000,,This becomes a one plus R_f over R_1.
Dialogue: 0,0:17:52.10,0:18:00.10,Default,,0000,0000,0000,,So therefore, V_out is equal
Dialogue: 0,0:18:00.10,0:18:06.74,Default,,0000,0000,0000,,to one plus R_f over R_1 times V_in.
Dialogue: 0,0:18:10.86,0:18:16.02,Default,,0000,0000,0000,,This device takes a voltage V_in,
Dialogue: 0,0:18:16.02,0:18:19.38,Default,,0000,0000,0000,,multiplies it by a number,
Dialogue: 0,0:18:19.38,0:18:22.22,Default,,0000,0000,0000,,and actually a positive number.
Dialogue: 0,0:18:22.22,0:18:24.25,Default,,0000,0000,0000,,In order to get V_out,
Dialogue: 0,0:18:24.25,0:18:33.44,Default,,0000,0000,0000,,this is a non-inverting voltage amplifier.
Dialogue: 0,0:18:36.99,0:18:41.83,Default,,0000,0000,0000,,Okay, you're still amplifying voltage by taking your input voltage,
Dialogue: 0,0:18:41.83,0:18:44.64,Default,,0000,0000,0000,,multiplying it by a number to get the output voltage,
Dialogue: 0,0:18:44.64,0:18:48.01,Default,,0000,0000,0000,,but you're not changing the sign, it not-inverting.
Dialogue: 0,0:18:48.01,0:18:54.42,Default,,0000,0000,0000,,Now one quick thing I want to point out about both this example and the previous one.
Dialogue: 0,0:18:54.42,0:18:57.82,Default,,0000,0000,0000,,Both of these had a resistor which was
Dialogue: 0,0:18:57.82,0:19:01.51,Default,,0000,0000,0000,,feeding back from the output to one of the input terminals.
Dialogue: 0,0:19:01.51,0:19:04.72,Default,,0000,0000,0000,,That is very typical of op-amp based circuits.
Dialogue: 0,0:19:04.72,0:19:07.32,Default,,0000,0000,0000,,It's generally called the feedback loop.
Dialogue: 0,0:19:07.32,0:19:14.42,Default,,0000,0000,0000,,Almost invariably you will feedback from the output to the inverting input terminal.
Dialogue: 0,0:19:14.42,0:19:19.14,Default,,0000,0000,0000,,That's necessary in order to keep this entire device stable.
Dialogue: 0,0:19:19.14,0:19:22.02,Default,,0000,0000,0000,,If I feedback to the positive terminal,
Dialogue: 0,0:19:22.02,0:19:25.48,Default,,0000,0000,0000,,generally this device will do what is called going unstable,
Dialogue: 0,0:19:25.48,0:19:28.47,Default,,0000,0000,0000,,the output will try to go to infinity.
Dialogue: 0,0:19:28.47,0:19:32.19,Default,,0000,0000,0000,,So what you'll have happen is that your output voltage will either
Dialogue: 0,0:19:32.19,0:19:36.13,Default,,0000,0000,0000,,go to the positive or the negative voltage rail, and stay there.
Dialogue: 0,0:19:36.13,0:19:37.54,Default,,0000,0000,0000,,It can't get to infinity,
Dialogue: 0,0:19:37.54,0:19:40.42,Default,,0000,0000,0000,,but it's going to go as high as it can go,
Dialogue: 0,0:19:40.42,0:19:42.80,Default,,0000,0000,0000,,and it's not going to come back.
Dialogue: 0,0:19:42.99,0:19:46.02,Default,,0000,0000,0000,,Okay, this concludes Lecture 12.
Dialogue: 0,0:19:46.02,0:19:50.18,Default,,0000,0000,0000,,Next lecture, we'll do some more work with operational amplifiers.
Dialogue: 0,0:19:50.18,0:19:52.81,Default,,0000,0000,0000,,We'll look at their operation in a little bit more depth,
Dialogue: 0,0:19:52.81,0:19:58.24,Default,,0000,0000,0000,,we'll talk more about the voltage rails applied by the power supplies,
Dialogue: 0,0:19:58.24,0:20:02.52,Default,,0000,0000,0000,,and we'll start looking at them as dependent sources.
Dialogue: 0,0:20:02.52,0:20:06.34,Default,,0000,0000,0000,,Okay, later on in your schooling career you'll see a lot of dependent sources.
Dialogue: 0,0:20:06.34,0:20:11.09,Default,,0000,0000,0000,,We'll just start kind of looking at things in that way in this class.