[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.67,0:00:05.28,Default,,0000,0000,0000,,>> The first of the op-amp configurations that we're going
Dialogue: 0,0:00:05.28,0:00:08.25,Default,,0000,0000,0000,,to consider is known as the non-inverting amplifier.
Dialogue: 0,0:00:08.25,0:00:12.27,Default,,0000,0000,0000,,It gets its name from the fact that the source voltage to
Dialogue: 0,0:00:12.27,0:00:16.56,Default,,0000,0000,0000,,be amplified is connected to the non-inverting terminal.
Dialogue: 0,0:00:16.56,0:00:21.32,Default,,0000,0000,0000,,Because of that, it turns out that the output voltage will be of the same sign,
Dialogue: 0,0:00:21.32,0:00:24.08,Default,,0000,0000,0000,,S-I-G-N, as the input voltage.
Dialogue: 0,0:00:24.08,0:00:26.50,Default,,0000,0000,0000,,So, for example, if V_s is a positive voltage,
Dialogue: 0,0:00:26.50,0:00:28.76,Default,,0000,0000,0000,,the output voltage will also be positive.
Dialogue: 0,0:00:28.76,0:00:31.10,Default,,0000,0000,0000,,On the other hand, if the input voltage is negative,
Dialogue: 0,0:00:31.10,0:00:33.39,Default,,0000,0000,0000,,then the output voltage will be negative also.
Dialogue: 0,0:00:33.39,0:00:35.52,Default,,0000,0000,0000,,The sign is not inverted.
Dialogue: 0,0:00:35.52,0:00:38.03,Default,,0000,0000,0000,,I have here two different schematics.
Dialogue: 0,0:00:38.03,0:00:41.90,Default,,0000,0000,0000,,I'd encourage you to stop the video for just a second and convince
Dialogue: 0,0:00:41.90,0:00:46.62,Default,,0000,0000,0000,,yourself that these two schematics are identical as far as function is concerned.
Dialogue: 0,0:00:46.62,0:00:49.16,Default,,0000,0000,0000,,The things that you're going to be looking at are that
Dialogue: 0,0:00:49.16,0:00:52.19,Default,,0000,0000,0000,,the V_s is connected to the non-inverting terminal.
Dialogue: 0,0:00:52.19,0:00:55.36,Default,,0000,0000,0000,,Of course, the non-inverting terminal is the one that's got the positive sign on it.
Dialogue: 0,0:00:55.36,0:00:57.53,Default,,0000,0000,0000,,So, if this one is connected to the non-inverting,
Dialogue: 0,0:00:57.53,0:01:00.44,Default,,0000,0000,0000,,over here, V_s is also connected to the non-inverting.
Dialogue: 0,0:01:00.44,0:01:05.42,Default,,0000,0000,0000,,So, this amplifier over here is upside down from this op-amp here.
Dialogue: 0,0:01:05.42,0:01:09.30,Default,,0000,0000,0000,,The other thing that you look at is where does the feedback go.
Dialogue: 0,0:01:09.30,0:01:10.70,Default,,0000,0000,0000,,We'll talk more about feedback later,
Dialogue: 0,0:01:10.70,0:01:15.06,Default,,0000,0000,0000,,but suffice for now to say that feedback is the process or the act of
Dialogue: 0,0:01:15.06,0:01:21.56,Default,,0000,0000,0000,,taking the voltage at the output and running it back to,
Dialogue: 0,0:01:21.56,0:01:25.84,Default,,0000,0000,0000,,in this case, the negative or the inverting terminal.
Dialogue: 0,0:01:25.84,0:01:27.95,Default,,0000,0000,0000,,Again, we'll talk more about feedback later,
Dialogue: 0,0:01:27.95,0:01:30.86,Default,,0000,0000,0000,,but this is referred to as negative feedback.
Dialogue: 0,0:01:30.86,0:01:33.35,Default,,0000,0000,0000,,You can remember that because the feedback loop comes back
Dialogue: 0,0:01:33.35,0:01:37.38,Default,,0000,0000,0000,,to the terminal of the negative sign which is the inverting sign.
Dialogue: 0,0:01:37.38,0:01:39.80,Default,,0000,0000,0000,,So, in both of these feedback circuits,
Dialogue: 0,0:01:39.80,0:01:42.82,Default,,0000,0000,0000,,we're going from V_out through R_1.
Dialogue: 0,0:01:42.82,0:01:45.47,Default,,0000,0000,0000,,The node beat between R_1 and R_2 is
Dialogue: 0,0:01:45.47,0:01:48.44,Default,,0000,0000,0000,,then tabbed and brought back to the inverting terminal.
Dialogue: 0,0:01:48.44,0:01:50.92,Default,,0000,0000,0000,,Notice, over here, we have the same situation,
Dialogue: 0,0:01:50.92,0:01:52.20,Default,,0000,0000,0000,,coming from the output,
Dialogue: 0,0:01:52.20,0:01:55.58,Default,,0000,0000,0000,,going through R_1, connecting to R_2,
Dialogue: 0,0:01:55.58,0:01:59.97,Default,,0000,0000,0000,,and the node where R_1 and R_2 are connected is connected
Dialogue: 0,0:01:59.97,0:02:04.94,Default,,0000,0000,0000,,then to the inverting terminal and then brought back the ground.
Dialogue: 0,0:02:04.94,0:02:11.39,Default,,0000,0000,0000,,Now, let's analyze these op-amps at both of these circuits and see what it is,
Dialogue: 0,0:02:11.39,0:02:13.91,Default,,0000,0000,0000,,why we might draw it this way under certain circumstances
Dialogue: 0,0:02:13.91,0:02:16.79,Default,,0000,0000,0000,,and why we might draw it this way under other circumstances.
Dialogue: 0,0:02:16.79,0:02:18.95,Default,,0000,0000,0000,,Let's start here with the one on the right.
Dialogue: 0,0:02:18.95,0:02:21.32,Default,,0000,0000,0000,,By drawing it in this configuration,
Dialogue: 0,0:02:21.32,0:02:27.83,Default,,0000,0000,0000,,it makes it obvious that the output voltage goes through a voltage divider circuit,
Dialogue: 0,0:02:27.83,0:02:34.26,Default,,0000,0000,0000,,and only a portion of the output voltage is fed back to the inverting terminal.
Dialogue: 0,0:02:34.70,0:02:36.72,Default,,0000,0000,0000,,In this case here,
Dialogue: 0,0:02:36.72,0:02:38.92,Default,,0000,0000,0000,,V_n then,
Dialogue: 0,0:02:39.01,0:02:41.63,Default,,0000,0000,0000,,using our voltage divider formula,
Dialogue: 0,0:02:41.63,0:02:45.25,Default,,0000,0000,0000,,V_n is equal to the voltage across R_2,
Dialogue: 0,0:02:45.25,0:02:53.43,Default,,0000,0000,0000,,which is V_out times R_2 over R_1 plus R_2.
Dialogue: 0,0:02:53.43,0:02:57.44,Default,,0000,0000,0000,,Now, in order to get V_out as a function of V,
Dialogue: 0,0:02:57.44,0:03:01.13,Default,,0000,0000,0000,,our input, we're going to reverse the roles
Dialogue: 0,0:03:01.13,0:03:04.73,Default,,0000,0000,0000,,here and multiply both sides by the inverse of this.
Dialogue: 0,0:03:04.73,0:03:06.59,Default,,0000,0000,0000,,In solving or solving for V_out,
Dialogue: 0,0:03:06.59,0:03:16.22,Default,,0000,0000,0000,,we get that V_out is equal to V_n times R_1 plus R_2 over R_2.
Dialogue: 0,0:03:16.22,0:03:18.32,Default,,0000,0000,0000,,Now, we don't want it as a function of V_n.
Dialogue: 0,0:03:18.32,0:03:21.80,Default,,0000,0000,0000,,We want to know what the output is as a function of the input voltage
Dialogue: 0,0:03:21.80,0:03:26.82,Default,,0000,0000,0000,,V_s. We now apply one of our ideal op-amp approximations.
Dialogue: 0,0:03:26.82,0:03:29.95,Default,,0000,0000,0000,,That was the one that referred to as the virtual short.
Dialogue: 0,0:03:29.95,0:03:35.00,Default,,0000,0000,0000,,That V_p and V_n are going to be so close to each
Dialogue: 0,0:03:35.00,0:03:40.84,Default,,0000,0000,0000,,other that the difference V_p minus V_n is zero,
Dialogue: 0,0:03:40.84,0:03:45.80,Default,,0000,0000,0000,,or what we can say then is that V_n is approximately equal to V_p.
Dialogue: 0,0:03:45.80,0:03:49.18,Default,,0000,0000,0000,,Now, what is V_p?
Dialogue: 0,0:03:49.18,0:03:51.54,Default,,0000,0000,0000,,Using another op-amp approximation,
Dialogue: 0,0:03:51.54,0:03:55.58,Default,,0000,0000,0000,,the current going into the input terminals is zero.
Dialogue: 0,0:03:55.58,0:03:58.64,Default,,0000,0000,0000,,Therefore, I_p is zero.
Dialogue: 0,0:03:58.64,0:04:00.71,Default,,0000,0000,0000,,There will be no voltage drop across
Dialogue: 0,0:04:00.71,0:04:03.28,Default,,0000,0000,0000,,V_s because the current going through it is zero.
Dialogue: 0,0:04:03.28,0:04:07.91,Default,,0000,0000,0000,,So, V_p is in fact just our source voltage V_s.
Dialogue: 0,0:04:07.91,0:04:15.06,Default,,0000,0000,0000,,So, V_p equals V_s. V_n equals V_p due to the virtual short.
Dialogue: 0,0:04:15.06,0:04:21.82,Default,,0000,0000,0000,,We have then that V_out again replacing V_n with V_s.
Dialogue: 0,0:04:21.82,0:04:31.22,Default,,0000,0000,0000,,V_out is equal to V_s times R_1 plus R_2 over R_2.
Dialogue: 0,0:04:31.22,0:04:33.96,Default,,0000,0000,0000,,Generally speaking, we'll take this and say, well,
Dialogue: 0,0:04:33.96,0:04:38.27,Default,,0000,0000,0000,,note that R_2 is a denominator that's common to both of those two terms.
Dialogue: 0,0:04:38.27,0:04:46.11,Default,,0000,0000,0000,,So, we can then have V_out is equal to V_s times R_2 over R_2,
Dialogue: 0,0:04:46.11,0:04:53.35,Default,,0000,0000,0000,,that's one, plus R_1 over R_2.
Dialogue: 0,0:04:53.35,0:04:57.57,Default,,0000,0000,0000,,We say that the gain, G, the closed loop gain,
Dialogue: 0,0:04:57.57,0:05:01.64,Default,,0000,0000,0000,,the gain that we get because of this feedback circuit is equal
Dialogue: 0,0:05:01.64,0:05:08.62,Default,,0000,0000,0000,,to one plus R_1 plus R_2.
Dialogue: 0,0:05:08.62,0:05:13.94,Default,,0000,0000,0000,,We're going to refer to that as the gain of the non-inverting amplifier.
Dialogue: 0,0:05:13.94,0:05:16.69,Default,,0000,0000,0000,,We'll see when we get to the inverting amplifier configuration that
Dialogue: 0,0:05:16.69,0:05:20.54,Default,,0000,0000,0000,,its gain is just slightly different than this.
Dialogue: 0,0:05:20.54,0:05:24.91,Default,,0000,0000,0000,,But let's just point out now that V_out is in fact going to be
Dialogue: 0,0:05:24.91,0:05:29.44,Default,,0000,0000,0000,,the same sign as V_s. R_1 and R_2 are both positive quantities.
Dialogue: 0,0:05:29.44,0:05:32.97,Default,,0000,0000,0000,,So, the ratio of a positive quantities is positive,
Dialogue: 0,0:05:32.97,0:05:34.67,Default,,0000,0000,0000,,plus one is a positive number,
Dialogue: 0,0:05:34.67,0:05:38.52,Default,,0000,0000,0000,,times whatever the sign is on V_s gives us a V_out,
Dialogue: 0,0:05:38.52,0:05:41.78,Default,,0000,0000,0000,,which is the same sign as V_s. Now,
Dialogue: 0,0:05:41.78,0:05:44.38,Default,,0000,0000,0000,,let's look at this circuit over here.
Dialogue: 0,0:05:44.38,0:05:50.19,Default,,0000,0000,0000,,To analyze this, we're going to use a technique that use a node analysis,
Dialogue: 0,0:05:50.19,0:05:56.66,Default,,0000,0000,0000,,which is an analysis technique that we'll frequently use on op-amp circuits,
Dialogue: 0,0:05:56.66,0:06:02.00,Default,,0000,0000,0000,,especially as the op-amp circuit gets to be a little bit more complex.
Dialogue: 0,0:06:02.00,0:06:04.45,Default,,0000,0000,0000,,So, here's the deal.
Dialogue: 0,0:06:04.45,0:06:13.77,Default,,0000,0000,0000,,This is now V_n. We're going to write a node equation at V_n. But first,
Dialogue: 0,0:06:13.77,0:06:18.63,Default,,0000,0000,0000,,we're going to note that this voltage here, V_p,
Dialogue: 0,0:06:18.63,0:06:21.64,Default,,0000,0000,0000,,is going to equal our source voltage.
Dialogue: 0,0:06:21.64,0:06:25.58,Default,,0000,0000,0000,,As we saw over here, the R_s had no influence on it.
Dialogue: 0,0:06:25.58,0:06:28.32,Default,,0000,0000,0000,,So, I've left that out of this circuit here.
Dialogue: 0,0:06:28.32,0:06:36.47,Default,,0000,0000,0000,,So, V_p equals V_s. V_n equals V_p because of the virtual short.
Dialogue: 0,0:06:36.47,0:06:40.20,Default,,0000,0000,0000,,So, V_n then is going to equal V_s.
Dialogue: 0,0:06:40.20,0:06:50.12,Default,,0000,0000,0000,,Now, let's write a node equation summing the currents leaving this node.
Dialogue: 0,0:06:50.12,0:06:57.57,Default,,0000,0000,0000,,The current leaving this node going through R_2 to ground is V_s minus
Dialogue: 0,0:06:57.57,0:07:04.69,Default,,0000,0000,0000,,zero divided by R_2 plus the current going from this node in this direction.
Dialogue: 0,0:07:04.69,0:07:07.73,Default,,0000,0000,0000,,It's going to be the voltage across R_1, which is V_s,
Dialogue: 0,0:07:07.73,0:07:16.27,Default,,0000,0000,0000,,minus V_out divided by R_1,
Dialogue: 0,0:07:16.27,0:07:23.40,Default,,0000,0000,0000,,and then plus the current going into or leaving this node and going into the op-amp.
Dialogue: 0,0:07:23.40,0:07:26.60,Default,,0000,0000,0000,,But our ideal op-amp approximation tells us
Dialogue: 0,0:07:26.60,0:07:29.30,Default,,0000,0000,0000,,that the current going into the input terminal is zero.
Dialogue: 0,0:07:29.30,0:07:31.96,Default,,0000,0000,0000,,So, we could write a plus zero there.
Dialogue: 0,0:07:31.96,0:07:37.86,Default,,0000,0000,0000,,But let's that off and simply say that the sum of those two currents must equal zero.
Dialogue: 0,0:07:37.86,0:07:40.22,Default,,0000,0000,0000,,So, what that's saying is that in fact,
Dialogue: 0,0:07:40.22,0:07:42.06,Default,,0000,0000,0000,,just as we saw over here,
Dialogue: 0,0:07:42.06,0:07:45.92,Default,,0000,0000,0000,,the current is going from the output through these two resistors back.
Dialogue: 0,0:07:45.92,0:07:52.52,Default,,0000,0000,0000,,It's going from the output through these two resistors to ground.
Dialogue: 0,0:07:52.52,0:07:58.36,Default,,0000,0000,0000,,These two resistors are in series because the current going in here is zero.
Dialogue: 0,0:07:58.36,0:08:03.40,Default,,0000,0000,0000,,Now, let's solve this equation for V_out in terms of these paths.
Dialogue: 0,0:08:03.40,0:08:11.91,Default,,0000,0000,0000,,We have V_s factoring out V_s times one over R_2 plus one over
Dialogue: 0,0:08:11.91,0:08:20.95,Default,,0000,0000,0000,,R_1 minus V_out over R_1 is equal to zero.
Dialogue: 0,0:08:22.49,0:08:26.52,Default,,0000,0000,0000,,That bring this term over to the other side, and solving for V_0,
Dialogue: 0,0:08:26.52,0:08:36.79,Default,,0000,0000,0000,,We get then that V_0 is equal to V_s times R_1 times
Dialogue: 0,0:08:36.79,0:08:44.81,Default,,0000,0000,0000,,one over R_2 plus one over R_1 or distributing that
Dialogue: 0,0:08:44.81,0:08:47.32,Default,,0000,0000,0000,,through the R_1 over the R_1 gives you the one,
Dialogue: 0,0:08:47.32,0:08:49.19,Default,,0000,0000,0000,,the R_1 over R_2 gives you the other term,
Dialogue: 0,0:08:49.19,0:08:50.98,Default,,0000,0000,0000,,and we get the V_out is equal to
Dialogue: 0,0:08:50.98,0:08:59.100,Default,,0000,0000,0000,,V_s times one plus R_1 over R_2.
Dialogue: 0,0:08:59.100,0:09:02.88,Default,,0000,0000,0000,,So, the gain terms are the same.
Dialogue: 0,0:09:02.88,0:09:05.20,Default,,0000,0000,0000,,These two circuits are equivalent
Dialogue: 0,0:09:05.20,0:09:09.87,Default,,0000,0000,0000,,and that gives you an idea of what the non-inverting amplifier does.