1 00:00:00,420 --> 00:00:06,640 Now, what can we do in order to obtain power amplification, it is very clear that the source 2 00:00:06,640 --> 00:00:11,290 of signals that interest that cannot be the only source that is applied to the circuits, 3 00:00:11,290 --> 00:00:15,710 because then what you get out of the network will definitely be less than what you put 4 00:00:15,710 --> 00:00:22,810 in. So, there must also be an additional source of power that is you have some relevant inputs 5 00:00:22,810 --> 00:00:27,590 signals, let say v 1 or v in and there must also be an additional source of power. So, 6 00:00:27,590 --> 00:00:32,969 somehow the power must be greater than the power that you put in from the relevant signal 7 00:00:32,969 --> 00:00:39,479 source. Now, this is quite common at least even if you do not understand how things work 8 00:00:39,479 --> 00:00:44,749 you know that an amplifier works with a battery or with some source of power. So, you have 9 00:00:44,749 --> 00:00:48,989 this public system in your class room and so on, they are plugged into the mains I mean 10 00:00:48,989 --> 00:00:55,379 that is so that they can draw some power from the mains. The actual signal is let say coming 11 00:00:55,379 --> 00:01:01,780 from the microphone, so the output power that goes into the loudspeaker is more than what 12 00:01:01,780 --> 00:01:07,049 comes in from the microphone. And how does it do that it actually draw some power from 13 00:01:07,049 --> 00:01:27,090 the, and converts it into additional power in the loudspeaker. 14 00:01:27,090 --> 00:01:36,060 What it means is the following. Let me again take a two port network and let me assume 15 00:01:36,060 --> 00:01:50,810 that N is the linear two port network. Now, I said that I will have a source of signal 16 00:01:50,810 --> 00:01:57,979 and another additional source of power. So, to distinguish between the two easily, I will 17 00:01:57,979 --> 00:02:03,409 consider the signal source to be a sinusoid of some frequency and additional power source 18 00:02:03,409 --> 00:02:10,840 to be dc. Now this is the quite common case, but it does not have to be the additional 19 00:02:10,840 --> 00:02:15,530 source can also be a sinusoid but the most simple form the most commonly used form is 20 00:02:15,530 --> 00:02:22,320 where the additional source of power is dc, so that is what I am going to consider. So, 21 00:02:22,320 --> 00:02:36,420 that is by input v i is some peak value v i had cos omega t. So, by the way, this is 22 00:02:36,420 --> 00:02:57,470 not just a linear two port network, it is also passive. So, now the output is a sinusoid 23 00:02:57,470 --> 00:03:02,800 of the same frequency because it is a linear network, and for simplicity I will assume 24 00:03:02,800 --> 00:03:09,769 that it is in the same phase. It does not matter if it is or if it is not. 25 00:03:09,769 --> 00:03:18,659 Now, again I have a single independent source in the circuit. So, this we know it cannot 26 00:03:18,659 --> 00:03:29,659 give you any power gain, but I am just going to go through it. And this current that is 27 00:03:29,659 --> 00:03:47,510 going into the load will be some i hat cos omega t. We already know that defining the 28 00:03:47,510 --> 00:03:56,530 port variables like this v 1 i 1 and v 2 i 2 minus v 2 i 2 which is the output power 29 00:03:56,530 --> 00:04:05,221 is less than v 1 i 1. So, this circuit cannot give you any power gain at all. Now, what 30 00:04:05,221 --> 00:04:11,719 we can do is to add an additional source of power to this circuit somewhere, it can be 31 00:04:11,719 --> 00:04:17,320 anywhere; it does not matter. The same argument will apply, but remember this network is a 32 00:04:17,320 --> 00:04:27,050 linear passive two port network. 33 00:04:27,050 --> 00:04:40,540 Now just for illustration sake let me assume that the additional source of power is something 34 00:04:40,540 --> 00:04:54,770 connected in series with the input source. So, this is the input signal source, and the 35 00:04:54,770 --> 00:05:20,160 additional source of power let say is another source that is connected in series, 36 00:05:20,160 --> 00:05:44,280 and let say it has the value of v 1 0. So, now we will have some v o, and some load current; 37 00:05:44,280 --> 00:06:00,950 and this will have a component due to v i, and also a component due to v 1 0. All I am 38 00:06:00,950 --> 00:06:07,240 saying is that both of these sources will have some effect on the output voltage, so 39 00:06:07,240 --> 00:06:15,840 that is a very general statement very obvious. Now, the question is if the power in the load, 40 00:06:15,840 --> 00:06:20,940 due to the input source v i, because this is the signal source, this is the source that 41 00:06:20,940 --> 00:06:32,680 has useful information. So, as I said I call this v i hat cos omega t. So, the component 42 00:06:32,680 --> 00:06:38,740 due to v i in the load, if it has more power than what is being fed from the left side, 43 00:06:38,740 --> 00:06:46,160 what is being delivered by the input source. And the answer is it turns out that it cannot 44 00:06:46,160 --> 00:06:57,420 and the answer is very simple, because this is a linear network superposition applies. 45 00:06:57,420 --> 00:07:23,240 What does it mean, let me copy this over, it means that the solution to this can be 46 00:07:23,240 --> 00:07:49,010 obtained by superposition. And what are the two conditions, two conditions are one in 47 00:07:49,010 --> 00:08:27,830 which only v i is present and one in which only v 1 0 is present. This is 48 00:08:27,830 --> 00:08:35,219 the load, so the solution to this can be obtained as the superposition of this where we have 49 00:08:35,219 --> 00:08:43,250 only v i applied to the network, and this where we have only v 1 0 applied to the network, 50 00:08:43,250 --> 00:08:51,149 this is because of linearity. Now, clearly the output voltage in this case, let say v 51 00:08:51,149 --> 00:09:01,240 o 1 and the load current i L 1 are only due to the signal v i. And the output voltage 52 00:09:01,240 --> 00:09:19,060 in this case, let say v o 2 and i L 2 are only due to the dc source v 1 0. So if you 53 00:09:19,060 --> 00:09:25,149 compute the power, that is contained only in the signal component; like I said v i is 54 00:09:25,149 --> 00:09:36,699 v i hat cos omega t both i L 1 and v o 1 will be sinusoids, and their product is the power. 55 00:09:36,699 --> 00:09:46,019 In this case, we already know that minus v 2 i 2 is less than or equal to v 1 i 1, so 56 00:09:46,019 --> 00:09:52,800 the output power due to the signal component is less than or equal to the input power due 57 00:09:52,800 --> 00:09:59,279 to the signal component. So, what it means is the following. If you 58 00:09:59,279 --> 00:10:04,100 have a linear network, even if you use an additional source of power you cannot have 59 00:10:04,100 --> 00:10:09,600 an output power in the signal component that is more than the input power in the signal 60 00:10:09,600 --> 00:10:15,639 component. You surely have an additional output power, because the voltage across this for 61 00:10:15,639 --> 00:10:21,709 instance if you consider this load as a resistor, the voltage across this will have component 62 00:10:21,709 --> 00:10:28,079 from the input source as well as the dc. So, there will be additional power in the load 63 00:10:28,079 --> 00:10:34,779 for sure, but that would not get translated to power in the signal component, because 64 00:10:34,779 --> 00:10:40,699 by linearity you can analyze the circuit with the signal and the dc separately. And we will 65 00:10:40,699 --> 00:10:46,019 assume that the signal is sinusoid, so the signal component of the power will only come 66 00:10:46,019 --> 00:10:52,199 from this one and the dc component of the load power will only come from that one. So, 67 00:10:52,199 --> 00:10:56,499 even if you add an additional source, you are not going to get a higher output power 68 00:10:56,499 --> 00:11:00,319 at the signal component; you will surely get higher power in the load, but that would not 69 00:11:00,319 --> 00:11:05,910 be signal component. Now, it is a very obvious and you can do the 70 00:11:05,910 --> 00:11:11,209 analysis yourself, this is not the only way to provide an additional dc source. You can 71 00:11:11,209 --> 00:11:16,699 be provided using a third port, but by linearity exactly the same thing will happen. You can 72 00:11:16,699 --> 00:11:24,180 always decompose the linear network into a different circuit in which only one is source 73 00:11:24,180 --> 00:11:27,680 activated at a time. So, you can activate only the signal source at a time, you can 74 00:11:27,680 --> 00:11:33,370 activate the dc source at a time; and from this, you will conclude that the power in 75 00:11:33,370 --> 00:11:38,690 the output at the signal component will be no different from if you have only the signal 76 00:11:38,690 --> 00:11:43,209 input. So, the whole idea was if you have only the signal input, the output power will 77 00:11:43,209 --> 00:11:47,100 be smaller than the input power. So we thought we will add an additional source of power, 78 00:11:47,100 --> 00:11:50,569 let say we stick a battery somewhere, but that is not going to help if the circuit is 79 00:11:50,569 --> 00:11:56,790 linear, because once the circuit is linear then whether you have two sources together 80 00:11:56,790 --> 00:12:08,449 or acting separately, the effect of each source will remain the same, that is what is meant 81 00:12:08,449 --> 00:12:11,339 by superposition. 82 00:12:11,339 --> 00:12:23,720 So, to conclude passive devices means P out less than or equal to P in, so which means 83 00:12:23,720 --> 00:12:44,329 no power amplification if the signal source is the only power source that is if you have 84 00:12:44,329 --> 00:12:50,779 box with an input and an output, and the input consist only of the signal source, the output 85 00:12:50,779 --> 00:12:54,970 power has got to be less than what is being delivered by the signal source. So, the next 86 00:12:54,970 --> 00:13:10,139 idea is, an additional power source. Now this you know by experience, you know that all 87 00:13:10,139 --> 00:13:14,730 your electronic gadget work with the battery. Why is that because you need amplification 88 00:13:14,730 --> 00:13:20,000 and you need an extra source of power, so that when you get a small signal, you can 89 00:13:20,000 --> 00:13:25,029 augmented to with the power from the battery, so that is the general idea. Although you 90 00:13:25,029 --> 00:13:39,430 do not know exactly how it is work yet, the idea behind this is it can convert some power 91 00:13:39,430 --> 00:13:55,980 from this into the signal component, but this also does not work if you have a linear device 92 00:13:55,980 --> 00:14:10,180 and of course all devices are passive, then the output power in this signal component 93 00:14:10,180 --> 00:14:16,930 is going to be less than the input power in the signal component. 94 00:14:16,930 --> 00:14:26,589 This is because with ah linear network, when you have two sources that is the signal source 95 00:14:26,589 --> 00:14:31,939 and the additional power source, there is no effect on the output because of the signal 96 00:14:31,939 --> 00:14:36,259 source just because you connected an additional power source. This is very easily seen when 97 00:14:36,259 --> 00:14:43,119 you decompose it into superposition with each source acting by itself. So, whether the additional 98 00:14:43,119 --> 00:15:01,100 sources is there or not, the output because of the signal source will be exactly the same. 99 00:15:01,100 --> 00:15:18,689 This will be independent of the additional source. So, you cannot use a linear device 100 00:15:18,689 --> 00:15:19,740 for amplification. 101 00:15:19,740 --> 00:15:36,759 So, the very important conclusion is that we need nonlinear devices. Now, some of you 102 00:15:36,759 --> 00:15:42,600 may have already been expose to the transistor or other amplifying devices, and you know 103 00:15:42,600 --> 00:15:49,220 that they are nonlinear. It is not arbitrary, it is in fact something that we need if you 104 00:15:49,220 --> 00:15:53,040 have only linear devices, you could never get any power amplification. It is only with 105 00:15:53,040 --> 00:15:57,839 nonlinear devices that you need power amplification. Now, this also gives you very good motivation 106 00:15:57,839 --> 00:16:05,019 for studying analysis methods of nonlinear circuits, and understands them properly and 107 00:16:05,019 --> 00:16:09,139 so on. So, those things we will see in the following lessons that gives you motivation 108 00:16:09,139 --> 00:16:13,490 to study nonlinear devices and method to analyze the nonlinear circuits and so on. Although 109 00:16:13,490 --> 00:16:18,779 they are more complicated than linear circuits, because without nonlinear devices, we cannot 110 00:16:18,779 --> 00:16:23,290 have any amplifiers; and without amplifiers, we cannot have any of the wonders of the information 111 00:16:23,290 --> 00:16:27,649 is because it is with amplifiers that you can build all the wonderful analog and digital 112 00:16:27,649 --> 00:16:30,309 circuitry that are all around us.