0:00:00.000,0:00:03.975 >> Hello, this is Dr. Cynthia Furse at the University of Utah, 0:00:03.975,0:00:08.080 and today we're going to talk about designing op-amp systems. 0:00:08.180,0:00:12.105 Electrical engineering is about what can you do to a voltage. 0:00:12.105,0:00:14.550 A lot of times we like to add voltages together, 0:00:14.550,0:00:16.079 we like to add constants to them, 0:00:16.079,0:00:18.405 we like to multiply them by various values, 0:00:18.405,0:00:20.580 in order to make the circuit do what we want. 0:00:20.580,0:00:22.770 Let's suppose that we want our output to be 0:00:22.770,0:00:26.895 a linear combination of a constant and our input voltage like this. 0:00:26.895,0:00:30.764 This math might look like 2 times v_2 0:00:30.764,0:00:36.435 plus minus 2 times some voltage, something like this. 0:00:36.435,0:00:39.415 So, how do we design that system? 0:00:39.415,0:00:41.280 Whenever we are designing systems, 0:00:41.280,0:00:44.105 we like to break them down into individual components. 0:00:44.105,0:00:46.190 For example, we wouldn't really want to have to 0:00:46.190,0:00:48.230 consider the power plant, the breaker box, 0:00:48.230,0:00:50.450 and all the outlets in your house individually 0:00:50.450,0:00:52.955 every time we wanted to design a lamp or a fan. 0:00:52.955,0:00:56.510 We'll take that entire distribution system and we'll model that as 0:00:56.510,0:01:00.345 a single source voltage with its source resistance, 0:01:00.345,0:01:03.440 and then we might consider the lamp and the fan in 0:01:03.440,0:01:07.415 parallel like so and we design each of them independently. 0:01:07.415,0:01:11.894 In order to design individual blocks like this independently, 0:01:11.894,0:01:13.725 there's a very important concept, 0:01:13.725,0:01:17.225 that concept is input and output resistances. 0:01:17.225,0:01:19.030 Let's suppose that we have a circuit, 0:01:19.030,0:01:20.965 any circuit, that's this black box. 0:01:20.965,0:01:22.840 If we look in, 0:01:22.840,0:01:26.825 that's going to give us the input resistance and if we look into the output side, 0:01:26.825,0:01:29.045 that's going to give us the output resistance. 0:01:29.045,0:01:32.540 A lot of times in the book and elsewhere, you will see Z. 0:01:32.540,0:01:35.505 That is impedance as opposed to resistance, 0:01:35.505,0:01:36.960 it's a complex resistance, 0:01:36.960,0:01:41.015 so just consider that to be the same as resistance for this case. 0:01:41.015,0:01:43.190 If we are looking in to either side, 0:01:43.190,0:01:44.660 that's the same thing as using 0:01:44.660,0:01:49.045 the Thevenin resistances and you calculate it in the same way. 0:01:49.045,0:01:54.305 Now, let's consider the basics of input and output resistance. 0:01:54.305,0:01:58.895 Here is an example of an equivalent circuit that we would be very likely to build, 0:01:58.895,0:02:01.940 we would have a source right here with its input equivalent, 0:02:01.940,0:02:06.095 connected onto some amplifier circuit driving some load. 0:02:06.095,0:02:08.660 If we wanted to figure out how these we're working, 0:02:08.660,0:02:12.545 we would consider the input and output impedances of each of our blocks. 0:02:12.545,0:02:15.140 Here for example is our input circuit. 0:02:15.140,0:02:16.760 Now even though it's called an input circuit, 0:02:16.760,0:02:19.300 you notice it doesn't really have an input resistance, 0:02:19.300,0:02:21.110 it only has an output resistance. 0:02:21.110,0:02:23.105 The output resistance looking in, 0:02:23.105,0:02:24.530 resistance would be here. 0:02:24.530,0:02:29.650 We would short out our voltage source and the only resistance there would be Zs. 0:02:29.650,0:02:32.220 Now, let's consider the amplifier circuit. 0:02:32.220,0:02:33.929 Looking into the amplifier, 0:02:33.929,0:02:39.880 remember the fact that when we have a op-amp, 0:02:39.880,0:02:44.235 the input resistance is approximately infinity, it's very high. 0:02:44.235,0:02:46.215 So, when we look into an op-amp, 0:02:46.215,0:02:49.500 Zin is equal to infinity. 0:02:49.500,0:02:51.300 Let's consider the Zout, 0:02:51.300,0:02:53.185 remember that when we have an op-amp, 0:02:53.185,0:02:55.670 we have an output resistance Rout, 0:02:55.670,0:02:58.895 so our Zout is approximately zero. 0:02:58.895,0:03:00.980 Then let's look into our load, 0:03:00.980,0:03:04.600 Zin, is ZL like this. 0:03:04.600,0:03:08.870 So, we now have looked at our input and output resistances or 0:03:08.870,0:03:12.530 impedances for each one of the elements in our circuit, 0:03:12.530,0:03:14.780 and input impedance is looking into the input 0:03:14.780,0:03:18.300 and an output impedance is looking in to the output. 0:03:18.320,0:03:20.700 Now, let's go back to our circuit, 0:03:20.700,0:03:22.850 let's take a look again at the circuit and 0:03:22.850,0:03:25.655 decide what our input and output resistances are. 0:03:25.655,0:03:27.755 Just like in my previous case, 0:03:27.755,0:03:31.990 the output resistance of the source block is simply Zs. 0:03:31.990,0:03:35.110 Now, let's look at this fan block right here. 0:03:35.110,0:03:38.630 The input resistance right there would be 0:03:38.630,0:03:42.520 Rfan and the output resistance would also be Rfan. 0:03:42.520,0:03:45.420 Now, here's our last load right here, that's the lamp, 0:03:45.420,0:03:50.070 and looking into the lamp that gives me an input resistance of Rlamp. 0:03:50.600,0:03:53.660 Okay. Now, let's consider how we connect 0:03:53.660,0:03:56.405 circuits that have different input and output resistances. 0:03:56.405,0:03:58.670 If we wanted to connect circuit number one, 0:03:58.670,0:04:02.090 which has its input and output resistances right here 0:04:02.090,0:04:06.700 and we wanted to consider circuit two with its input and output resistances, 0:04:06.700,0:04:09.320 let's see what would happen if we hook them together. 0:04:09.320,0:04:13.215 Here's an example where I'll just be connecting a source impedance to a load. 0:04:13.215,0:04:14.605 So, if I look in here, 0:04:14.605,0:04:16.950 the output resistance is Zs, 0:04:16.950,0:04:18.285 and if I look in here, 0:04:18.285,0:04:20.760 the input resistance is ZL. 0:04:20.760,0:04:24.055 Now, imagine what would happen if I hook them together. 0:04:24.055,0:04:28.345 The output voltage that I might want like here would be Vout1, 0:04:28.345,0:04:33.025 and let's suppose that I wanted to drive this circuit with the source Vs, 0:04:33.025,0:04:34.920 and I'm going to drive this, 0:04:34.920,0:04:37.100 maybe a mixer input impedance or something, 0:04:37.100,0:04:39.630 and I connect them up like this. 0:04:39.820,0:04:43.820 What does that give me? That is a voltage divider, 0:04:43.820,0:04:45.905 we know something about voltage dividers. 0:04:45.905,0:04:51.840 We know that Vout1 is equal to 0:04:51.840,0:04:58.860 V_s times ZL over ZL plus Zs because that's the voltage divider. 0:04:58.860,0:05:04.765 Now, if ZL is small compared to Zs, 0:05:04.765,0:05:07.790 we're not going to get the voltage that we wanted at all. 0:05:07.790,0:05:09.290 The only time that we are going to get 0:05:09.290,0:05:11.615 the voltage that we want it to be deriving it with, 0:05:11.615,0:05:15.220 is if ZL is very large compared to the Zs. 0:05:15.220,0:05:24.430 So, ZL much greater than Zs will give us the result that Vout1 is equal to Vs. 0:05:25.400,0:05:29.070 So, this is a really important feature. 0:05:29.070,0:05:34.395 When we are designing circuits and I showed it graphically here, 0:05:34.395,0:05:39.680 if the input resistance of the second circuit is very large, 0:05:39.680,0:05:42.440 Zin2 is much greater than Zin1, 0:05:42.440,0:05:46.290 then we can consider these two circuits to be independent. 0:05:46.290,0:05:47.930 We can design them separately. 0:05:47.930,0:05:51.950 Any other case we can't do that and we'd have to analyze the entire circuit together. 0:05:51.950,0:05:57.660 So, we like it very much if the input impedance of a circuit is very high. 0:05:58.510,0:06:01.485 Now, in the event that doesn't happen, 0:06:01.485,0:06:04.505 what we're going to do is put something called a buffer in the circuit. 0:06:04.505,0:06:07.340 A buffer multiplies the incoming voltage by one, 0:06:07.340,0:06:11.870 but it has this magic thing that the input impedance of the buffer is always large, 0:06:11.870,0:06:14.885 and the output impedance of the buffer is always small. 0:06:14.885,0:06:17.310 That makes us so that we can always put a buffer in here, 0:06:17.310,0:06:19.670 we can always design our block separately. 0:06:19.670,0:06:21.890 So, the buffer allows us to design 0:06:21.890,0:06:26.500 our input equivalent circuit separate from our load equivalent circuit. 0:06:26.500,0:06:31.380 Now, another word that we often use for this is loading. 0:06:32.260,0:06:40.605 So, loading occurs when ZL is approximately equal to or less than Zs. 0:06:40.605,0:06:44.490 In this case, loading happens to make it so that 0:06:44.490,0:06:51.395 the output voltage that we want isn't the same as the output voltage that we input, 0:06:51.395,0:06:54.480 so loading is a bad thing. 0:06:54.890,0:06:59.620 Now, let's review again connecting input and output resistances. 0:06:59.620,0:07:04.930 If I have a large input resistance and I connect it to a small output resistance, 0:07:04.930,0:07:07.210 I can design my circuit without a buffer, 0:07:07.210,0:07:13.065 I can individually designed circuit one and circuit two as if they were not interrelated, 0:07:13.065,0:07:16.390 any other time I have to analyze the entire circuit. 0:07:16.390,0:07:21.205 In the event that I didn't have Zin2 much greater than Zout2, 0:07:21.205,0:07:24.270 what I would do is put a buffer in that case, 0:07:24.270,0:07:26.560 and that is going to make it so that I always have 0:07:26.560,0:07:30.010 a large input impedance and a small output impedance. 0:07:30.010,0:07:33.400 So, these are our two design criteria that allow us 0:07:33.400,0:07:37.820 to design individual elements of a more complex circuit.