[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:04.41,Default,,0000,0000,0000,,>> The next of the common op-amp configurations
Dialogue: 0,0:00:04.41,0:00:07.46,Default,,0000,0000,0000,,that we're going to consider is known as the inverting amplifier.
Dialogue: 0,0:00:07.46,0:00:11.10,Default,,0000,0000,0000,,It takes its name from the fact that the source voltage that's going to be
Dialogue: 0,0:00:11.10,0:00:16.66,Default,,0000,0000,0000,,amplified is connected to the inverting terminal.
Dialogue: 0,0:00:16.66,0:00:19.02,Default,,0000,0000,0000,,So, with that then,
Dialogue: 0,0:00:19.02,0:00:24.24,Default,,0000,0000,0000,,we might have rightfully assume that because of the name inverting amplifier,
Dialogue: 0,0:00:24.24,0:00:31.54,Default,,0000,0000,0000,,that the output voltage is going to be the opposite in sign of the source voltage.
Dialogue: 0,0:00:31.54,0:00:35.49,Default,,0000,0000,0000,,So, let's go ahead and analyze this as in the same manner that we have
Dialogue: 0,0:00:35.49,0:00:42.48,Default,,0000,0000,0000,,the other op-amp circuits by writing a node equation at the inverting op-amp terminal.
Dialogue: 0,0:00:42.48,0:00:46.58,Default,,0000,0000,0000,,Before we do that we're going to remember that due to the virtual short,
Dialogue: 0,0:00:46.58,0:00:49.60,Default,,0000,0000,0000,,the voltage at the inverting terminal V sub n is
Dialogue: 0,0:00:49.60,0:00:52.76,Default,,0000,0000,0000,,equal to the voltage at the non-inverting terminal,
Dialogue: 0,0:00:52.76,0:00:54.65,Default,,0000,0000,0000,,and in this case, V sub p
Dialogue: 0,0:00:54.65,0:00:58.54,Default,,0000,0000,0000,,the voltage at the non-inverting terminal has been tied to ground.
Dialogue: 0,0:00:58.54,0:01:01.62,Default,,0000,0000,0000,,So, V sub n is going to equal zero also.
Dialogue: 0,0:01:01.62,0:01:05.48,Default,,0000,0000,0000,,Lets go ahead and write the equation leaving V sub n in place so that we can see what's
Dialogue: 0,0:01:05.48,0:01:09.56,Default,,0000,0000,0000,,happening and we'll go back and replace V sub n with zero.
Dialogue: 0,0:01:09.56,0:01:13.61,Default,,0000,0000,0000,,So, starting here, adding the currents leaving
Dialogue: 0,0:01:13.61,0:01:17.84,Default,,0000,0000,0000,,this node starting with the current going in that direction we have,
Dialogue: 0,0:01:17.84,0:01:24.78,Default,,0000,0000,0000,,V sub n minus V sub s divided by R sub s,
Dialogue: 0,0:01:24.78,0:01:29.80,Default,,0000,0000,0000,,plus the current leaving the node going in that direction is going to be
Dialogue: 0,0:01:29.80,0:01:35.88,Default,,0000,0000,0000,,V sub n minus V out divided by R sub f,
Dialogue: 0,0:01:35.88,0:01:41.27,Default,,0000,0000,0000,,plus the current entering the inverting terminal of the op-amp which of course is zero.
Dialogue: 0,0:01:41.27,0:01:42.48,Default,,0000,0000,0000,,So, there's nothing right there,
Dialogue: 0,0:01:42.48,0:01:46.86,Default,,0000,0000,0000,,thus the sum of those two terms must equals zero.
Dialogue: 0,0:01:46.86,0:01:50.76,Default,,0000,0000,0000,,Now, let's replace V sub n with zero in both places.
Dialogue: 0,0:01:50.76,0:01:54.92,Default,,0000,0000,0000,,That's zero. That then is zero and we're left with
Dialogue: 0,0:01:54.92,0:02:00.87,Default,,0000,0000,0000,,negative V sub s over R sub s minus V out over R sub f equals 0.
Dialogue: 0,0:02:00.87,0:02:04.19,Default,,0000,0000,0000,,So, let's just take this negative V out over R sub f to the other side of
Dialogue: 0,0:02:04.19,0:02:10.74,Default,,0000,0000,0000,,the equation as a positive V out over R sub f.
Dialogue: 0,0:02:10.74,0:02:14.63,Default,,0000,0000,0000,,It's a simple task now to solve for V out as
Dialogue: 0,0:02:14.63,0:02:19.84,Default,,0000,0000,0000,,R sub f. Multiplying both sides of the equation by R sub f we get then that
Dialogue: 0,0:02:19.84,0:02:24.88,Default,,0000,0000,0000,,V out is equal to negative V sub s
Dialogue: 0,0:02:24.88,0:02:31.24,Default,,0000,0000,0000,,times R sub f over R sub s. As we anticipated,
Dialogue: 0,0:02:31.24,0:02:35.50,Default,,0000,0000,0000,,the sign on the output is going to be opposite sign of the source voltage,
Dialogue: 0,0:02:35.50,0:02:39.05,Default,,0000,0000,0000,,and we then can note that the gain for
Dialogue: 0,0:02:39.05,0:02:45.34,Default,,0000,0000,0000,,the inverting op-amp is equal to negative R sub f over R sub
Dialogue: 0,0:02:45.34,0:02:51.23,Default,,0000,0000,0000,,s. It's good to compare this gain term with
Dialogue: 0,0:02:51.23,0:02:54.56,Default,,0000,0000,0000,,a gain term that we derived in
Dialogue: 0,0:02:54.56,0:02:58.01,Default,,0000,0000,0000,,the non-inverting amplifier and go back and look at your notes.
Dialogue: 0,0:02:58.01,0:03:04.47,Default,,0000,0000,0000,,But, you may recall that the gain for the non-inverting amplifier was equal to
Dialogue: 0,0:03:04.47,0:03:12.42,Default,,0000,0000,0000,,one plus R sub f over r sub s. This is supposed to be R sub s there.
Dialogue: 0,0:03:12.42,0:03:15.00,Default,,0000,0000,0000,,When you compare these two terms,
Dialogue: 0,0:03:15.00,0:03:16.43,Default,,0000,0000,0000,,you'll notice they're very similar.
Dialogue: 0,0:03:16.43,0:03:22.16,Default,,0000,0000,0000,,They both have the ratio R sub f over R sub s. But in the non-inverting case,
Dialogue: 0,0:03:22.16,0:03:24.46,Default,,0000,0000,0000,,there's also one added to it.
Dialogue: 0,0:03:24.46,0:03:28.34,Default,,0000,0000,0000,,So, for the same circuit, the gain,
Dialogue: 0,0:03:28.34,0:03:31.66,Default,,0000,0000,0000,,if the source was on the non-inverting terminal,
Dialogue: 0,0:03:31.66,0:03:34.46,Default,,0000,0000,0000,,would be one larger than the gain
Dialogue: 0,0:03:34.46,0:03:38.62,Default,,0000,0000,0000,,experienced when the source is on the inverting terminal.
Dialogue: 0,0:03:38.62,0:03:40.92,Default,,0000,0000,0000,,Now, what does that inversion mean?
Dialogue: 0,0:03:40.92,0:03:44.64,Default,,0000,0000,0000,,Well, let's just take a couple of examples here real fast.
Dialogue: 0,0:03:44.64,0:03:52.36,Default,,0000,0000,0000,,Let's let R sub f equal 2 kilo-ohms,
Dialogue: 0,0:03:52.36,0:03:56.88,Default,,0000,0000,0000,,and we'll let R sub s equal 1 kilo ohm.
Dialogue: 0,0:03:56.88,0:04:02.19,Default,,0000,0000,0000,,So, that the ratio R sub f over R sub s is equal to 2,
Dialogue: 0,0:04:02.19,0:04:04.44,Default,,0000,0000,0000,,and for this first instance,
Dialogue: 0,0:04:04.44,0:04:05.55,Default,,0000,0000,0000,,for the first example,
Dialogue: 0,0:04:05.55,0:04:10.72,Default,,0000,0000,0000,,let's just assume that V sub s equals say 5 volts.
Dialogue: 0,0:04:10.72,0:04:14.12,Default,,0000,0000,0000,,Then our output voltage would be V out
Dialogue: 0,0:04:14.12,0:04:19.30,Default,,0000,0000,0000,,would equal negative V sub s
Dialogue: 0,0:04:19.30,0:04:26.30,Default,,0000,0000,0000,,times 2 or negative 10 volts.
Dialogue: 0,0:04:26.30,0:04:30.76,Default,,0000,0000,0000,,So, DC values are just inverted or just have an opposite sign.
Dialogue: 0,0:04:30.76,0:04:33.17,Default,,0000,0000,0000,,What about time-varying voltages?
Dialogue: 0,0:04:33.17,0:04:39.69,Default,,0000,0000,0000,,What if our V source was equal to
Dialogue: 0,0:04:39.69,0:04:49.10,Default,,0000,0000,0000,,say 5 cosine of omega t. What would the output voltage be?
Dialogue: 0,0:04:49.10,0:04:58.90,Default,,0000,0000,0000,,Well, V out would be negative 2 times 5 times the cosine of omega t,
Dialogue: 0,0:04:58.90,0:05:05.19,Default,,0000,0000,0000,,or that's equal to negative 10 cosine of omega t. Well,
Dialogue: 0,0:05:05.19,0:05:08.13,Default,,0000,0000,0000,,what does that sine, what's the amplitude of the output?
Dialogue: 0,0:05:08.13,0:05:11.48,Default,,0000,0000,0000,,Where the amplitude is going to be twice as large as the amplitude of the input,
Dialogue: 0,0:05:11.48,0:05:15.72,Default,,0000,0000,0000,,but it's also going to have a sign reversal. What does that mean?
Dialogue: 0,0:05:15.72,0:05:20.35,Default,,0000,0000,0000,,Well, if this was our original,
Dialogue: 0,0:05:22.67,0:05:26.20,Default,,0000,0000,0000,,where this was five,
Dialogue: 0,0:05:26.72,0:05:30.68,Default,,0000,0000,0000,,the output is going to have twice the amplitude.
Dialogue: 0,0:05:30.68,0:05:35.16,Default,,0000,0000,0000,,Let's just say that that's twice the amplitude.
Dialogue: 0,0:05:35.16,0:05:37.88,Default,,0000,0000,0000,,But it's also going to have a sign reversal.
Dialogue: 0,0:05:37.88,0:05:40.62,Default,,0000,0000,0000,,That means wherever the input was positive,
Dialogue: 0,0:05:40.62,0:05:48.04,Default,,0000,0000,0000,,the output is going to be negative and wherever the input with negative,
Dialogue: 0,0:05:48.04,0:05:54.24,Default,,0000,0000,0000,,the output will be positive and so on.
Dialogue: 0,0:05:54.24,0:05:56.81,Default,,0000,0000,0000,,In other words, we say also,
Dialogue: 0,0:05:56.81,0:05:59.51,Default,,0000,0000,0000,,or another way of putting it is that the output is
Dialogue: 0,0:05:59.51,0:06:03.74,Default,,0000,0000,0000,,180 degrees out of phase with the input and
Dialogue: 0,0:06:03.74,0:06:11.88,Default,,0000,0000,0000,,that's what the minus sign in the inverting op-amp gain term does to us.
Dialogue: 0,0:06:11.88,0:06:15.89,Default,,0000,0000,0000,,Putting it that way makes it sound a little ominous.
Dialogue: 0,0:06:15.89,0:06:19.49,Default,,0000,0000,0000,,What it does to us, now it's not doing anything to us,
Dialogue: 0,0:06:19.49,0:06:21.71,Default,,0000,0000,0000,,it's not hurting us and generally speaking,
Dialogue: 0,0:06:21.71,0:06:25.68,Default,,0000,0000,0000,,that sign inversion doesn't really impact anything.
Dialogue: 0,0:06:25.68,0:06:26.86,Default,,0000,0000,0000,,It just sits there,
Dialogue: 0,0:06:26.86,0:06:29.87,Default,,0000,0000,0000,,and when you build these kinds of circuits up in
Dialogue: 0,0:06:29.87,0:06:32.81,Default,,0000,0000,0000,,the laboratory and look at them on the oscilloscope,
Dialogue: 0,0:06:32.81,0:06:36.30,Default,,0000,0000,0000,,you'll note that there's a 180 degree phase shift.