WEBVTT 00:00:00.290 --> 00:00:02.580 >> All right. Let's take a look at 00:00:02.580 --> 00:00:07.110 an example of a non-inverting amplifier with some real numbers. 00:00:07.110 --> 00:00:14.295 Let's let R_1, the feedback resistor be 10 kiloohms, 00:00:14.295 --> 00:00:20.385 and R_2 be two kiloohms. 00:00:20.385 --> 00:00:26.100 Let's calculate what the output voltage will be in terms of V_s in these ratios. 00:00:26.100 --> 00:00:28.350 What we know from our previous work, 00:00:28.350 --> 00:00:33.330 is that V_out for an non-inverting amplifier is equal 00:00:33.330 --> 00:00:38.330 to V_s times the ratio of those two resistors, 00:00:38.330 --> 00:00:45.750 R1 over R2, or 10 kiloohms divided by two kiloohms plus one. 00:00:45.750 --> 00:00:48.945 Well, 10 divided by 2 is 5 plus 1 is 6. 00:00:48.945 --> 00:00:58.290 So, V_out then is equal to six times V_s. In this case, then, 00:00:58.290 --> 00:01:01.670 we would say that the gain of the amplifier G is equal 00:01:01.670 --> 00:01:05.615 to six and know that gain is a unitless term. 00:01:05.615 --> 00:01:10.190 It's equal to the ratio of V_out divided by V_s 00:01:10.190 --> 00:01:15.200 or we say then the gain is equal to the ratio of V_out divided by V_s, 00:01:15.200 --> 00:01:19.945 the volts cancel, and gain is again a unitless term. 00:01:19.945 --> 00:01:23.300 Now, let's look at issues of 00:01:23.300 --> 00:01:27.905 saturation by just assuming that we have two voltage sources. 00:01:27.905 --> 00:01:33.020 The positive supply is going to be a plus 15 volts and let's make the negative supply, 00:01:33.020 --> 00:01:37.455 a negative 15-volt source. 00:01:37.455 --> 00:01:39.314 We now ask ourselves, 00:01:39.314 --> 00:01:42.240 what is the range on V_s, 00:01:42.240 --> 00:01:46.030 so that the amplifier will stay in it's linear range, 00:01:46.030 --> 00:01:50.360 and the output voltage will be six times whatever V_s is. 00:01:50.360 --> 00:01:52.550 Remember what the limitation is. 00:01:52.550 --> 00:01:56.900 The output voltage can't be any greater than the supply voltages. 00:01:56.900 --> 00:02:02.540 It can't be any greater than 15 volts and it can't be any less than negative 15 volts, 00:02:02.540 --> 00:02:07.550 or to put it in terms of inequalities V_out has got to be greater 00:02:07.550 --> 00:02:12.850 than negative 15 volts and less than positive 15 volts. 00:02:12.850 --> 00:02:16.035 Now, substituting V_out for six V_s, 00:02:16.035 --> 00:02:22.460 we have then six V_s must be greater than a negative 15 volts and less than 00:02:22.460 --> 00:02:30.600 a positive 15 volts dividing both this side and this side by the six, 00:02:30.600 --> 00:02:34.265 so that we get V_s by itself rather, 00:02:34.265 --> 00:02:40.460 we get V_s must be greater than a negative 15 over six and it must be 00:02:40.460 --> 00:02:46.400 less than a positive 15 over six or V_s that is constrained to 00:02:46.400 --> 00:02:54.840 being greater than a negative 2.5 volts and less than a positive 2.5 volts. 00:02:55.090 --> 00:03:03.350 Let's graph this by looking at V_out as a function of V_s. So, 00:03:03.350 --> 00:03:07.960 V_s here, V_out along here. 00:03:07.960 --> 00:03:14.895 Let's go ahead and mark our positive 15 volts there 00:03:14.895 --> 00:03:19.980 and negative 15 volts 00:03:19.980 --> 00:03:26.190 there and here's negative 2.5 volts. 00:03:26.190 --> 00:03:31.095 There's positive 2.5 volts. 00:03:31.095 --> 00:03:34.080 Just mark it up there, 00:03:34.080 --> 00:03:36.870 I mark it down here. 00:03:36.870 --> 00:03:41.920 So, for V_s less than 2.5 volts the amplifier is going to be 00:03:41.920 --> 00:03:47.330 saturated at its negative supply voltage of negative 15 volts. 00:03:47.330 --> 00:03:55.450 As V_s gets larger than negative 2.5 volts and less than positive 2.5 volts, 00:03:55.450 --> 00:03:59.420 the output is going to equal six times V_s, 00:03:59.840 --> 00:04:09.660 or a nice sort of linear, you get the idea, relationship in that linear region, 00:04:09.660 --> 00:04:11.210 hence, it's called the linear region. 00:04:11.210 --> 00:04:15.230 When V_s is greater than 2.5 volts, it saturates at the upper or 00:04:15.230 --> 00:04:20.165 the positive supply voltage and while the graph may not be beautiful, you get the idea. 00:04:20.165 --> 00:04:25.170 Saturation means that the output can't be any less than negative 15 volts, 00:04:25.170 --> 00:04:29.270 can't be any greater than positive 15 volts and in between the constraints, 00:04:29.270 --> 00:04:31.400 it's a nice linear relationship. 00:04:31.400 --> 00:04:36.770 Where then the slope of this line is 00:04:36.770 --> 00:04:47.320 15 divided by 2.5 which is six and that's just our gain term.