WEBVTT 00:00:00.000 --> 00:00:02.505 >> To wrap up our discussion on Op-amps, 00:00:02.505 --> 00:00:05.533 let's see if we can understand how this negative feedback loop here 00:00:05.533 --> 00:00:08.325 constraints the Op-amp output, 00:00:08.325 --> 00:00:11.970 so that our ideal Op-amp approximations work. 00:00:11.970 --> 00:00:14.550 We've pointed out that an Op-amp is designed as 00:00:14.550 --> 00:00:18.890 a high-gain device such that V_out is equal to some A or A, 00:00:18.890 --> 00:00:20.560 some really large number, 00:00:20.560 --> 00:00:27.900 times the difference between V_p and V_n. A is on the order of 10_5, 00:00:27.900 --> 00:00:31.005 say 50,000, 100,000, or something like that. 00:00:31.005 --> 00:00:35.985 But our source, our power supply voltages, 00:00:35.985 --> 00:00:40.480 are going to be significantly smaller than 10_5. 00:00:40.480 --> 00:00:42.240 Which means that, say, 00:00:42.240 --> 00:00:43.470 let's just put some numbers on here, 00:00:43.470 --> 00:00:47.370 let's say that this is a plus or minus 15-volt source. 00:00:47.710 --> 00:00:51.500 That means that the output voltage is constrained to 00:00:51.500 --> 00:00:54.965 being within the range of plus or minus 15 volts. 00:00:54.965 --> 00:00:56.445 If that's the case, 00:00:56.445 --> 00:01:00.305 then this difference V sub p minus V sub n is going to have to be very small, 00:01:00.305 --> 00:01:04.205 or V_p minus V_n is going to 00:01:04.205 --> 00:01:10.310 equal V_out divided by A. V_out can't be any larger than 15 volts. 00:01:10.310 --> 00:01:16.010 So, 15 volts divided by 10 to the fifth means that V_p minus V_n is relatively small. 00:01:16.010 --> 00:01:18.575 That's our approximation, that's our assumption 00:01:18.575 --> 00:01:23.905 that we used in deriving our ideal Op-amp approximations, 00:01:23.905 --> 00:01:26.165 and the gains for these different amplifiers. 00:01:26.165 --> 00:01:30.290 So, what is it about this feedback, this R_1, 00:01:30.290 --> 00:01:37.275 R_2 voltage divider, that keeps V_p minus V_n so small? 00:01:37.275 --> 00:01:39.185 Let's see if we can understand that. 00:01:39.185 --> 00:01:40.970 To do that, let's just pick a few more numbers. 00:01:40.970 --> 00:01:45.850 Let's say that R_1 is equal to four kiloohms, 00:01:45.850 --> 00:01:50.430 and R_2 is equal to one kiloohm. 00:01:50.430 --> 00:01:55.850 So, with that then V sub n is just a voltage divided portion of V_out. 00:01:55.850 --> 00:02:02.130 So, V sub n is equal to V_out times 1 divided by 1 plus 4, that's 5. 00:02:02.130 --> 00:02:06.460 So, V sub n is equal to one-fifth of V_out. 00:02:06.680 --> 00:02:11.120 When analyzing loops, you make some assumptions and 00:02:11.120 --> 00:02:15.155 then see what this circuit does to the assumptions that you've made. 00:02:15.155 --> 00:02:19.150 So, let's assume that V sub p has been 0, 00:02:19.150 --> 00:02:22.400 which means V_out would be 0, 00:02:22.400 --> 00:02:24.230 and V sub n would be 0. 00:02:24.230 --> 00:02:26.540 So, V sub p has been 0, 00:02:26.540 --> 00:02:29.930 and it is transitioning to V sub p equals 2. 00:02:29.930 --> 00:02:31.880 So, let's just say that it just switched, 00:02:31.880 --> 00:02:36.660 and V sub p now equals 2 volts. All right. 00:02:36.660 --> 00:02:40.550 An Op-amp doesn't change the output voltage instantaneously. 00:02:40.550 --> 00:02:42.650 It's limited by something called the slew rate, 00:02:42.650 --> 00:02:43.880 and it doesn't really matter what that is, 00:02:43.880 --> 00:02:45.500 you'll learn more about that in later. 00:02:45.500 --> 00:02:47.945 In later classes, we'll learn more about Op-amps. 00:02:47.945 --> 00:02:51.615 The supply is to say that at an instantaneous change here, the input, 00:02:51.615 --> 00:02:59.175 will have some delay before its output is singing or the changes in to the output. 00:02:59.175 --> 00:03:04.050 In fact, what happens is V_out starts to grow, 00:03:04.050 --> 00:03:05.820 and under these circumstances, 00:03:05.820 --> 00:03:08.595 it would be heading towards A, 00:03:08.595 --> 00:03:12.300 this big number, times V sub p which is 2, 00:03:12.300 --> 00:03:16.215 minus V sub n. So, initially, 00:03:16.215 --> 00:03:20.645 this V sub p minus V sub n is going to be about 2, it's right down here. 00:03:20.645 --> 00:03:23.330 V_out would be heading towards, 00:03:23.330 --> 00:03:24.700 is not going to get there. 00:03:24.700 --> 00:03:26.420 This feedback to work it's going to constrain it, 00:03:26.420 --> 00:03:28.540 but if it was just open circuit, 00:03:28.540 --> 00:03:31.205 V_out would be heading towards A, 00:03:31.205 --> 00:03:33.515 which in our case is 10 to the fifth, 00:03:33.515 --> 00:03:35.540 times V sub p which is 2, 00:03:35.540 --> 00:03:40.590 minus V sub n. All right. 00:03:40.590 --> 00:03:44.540 Again, it can't get anywhere close to that because our power supplies are limited, 00:03:44.540 --> 00:03:46.760 but it would saturate at plus or minus 15, 00:03:46.760 --> 00:03:49.160 in this case it's saturated at 15 volts. 00:03:49.160 --> 00:03:56.010 But look what happens, as V_out starts to grow, V sub n, 00:03:56.010 --> 00:03:58.410 which is equal to one-fifth of V_out, 00:03:58.410 --> 00:04:01.590 let's write that here one-fifth V_out, 00:04:01.590 --> 00:04:05.175 V sub n now starts to grow also. 00:04:05.175 --> 00:04:07.800 As V sub n starts to grow, 00:04:07.800 --> 00:04:13.310 this difference here V_p minus V sub n starts to get smaller. 00:04:13.310 --> 00:04:14.930 Let's write that down here. 00:04:14.930 --> 00:04:17.404 As V_out starts to grow, 00:04:17.404 --> 00:04:24.434 V sub n which is equal to one-fifth V_out starts to increase also, 00:04:24.434 --> 00:04:25.995 and thus the difference, 00:04:25.995 --> 00:04:29.340 V_p minus V sub n, 00:04:29.340 --> 00:04:31.510 starts to get smaller. 00:04:31.510 --> 00:04:33.620 V_out continues to grow, 00:04:33.620 --> 00:04:39.275 it's still a number times a relatively small number times a big number. 00:04:39.275 --> 00:04:41.390 So, V_out is going to continue to grow, 00:04:41.390 --> 00:04:48.225 and it will continue to grow until it stabilizes at a voltage where in fact V_out 00:04:48.225 --> 00:04:55.710 is equal to A times V sub p minus V_n, 00:04:55.710 --> 00:05:02.970 but V_n is minus one-fifth V_out. 00:05:02.970 --> 00:05:06.990 Let's say that again. V_out will continue to grow, 00:05:06.990 --> 00:05:10.220 until it gets to the point where V_out is equal to 00:05:10.220 --> 00:05:14.780 A times the difference V_p minus V sub n at V sub n is one-fifth V_out. 00:05:14.780 --> 00:05:24.480 This point then becomes the stabilized value of V_out. 00:05:24.740 --> 00:05:28.680 Okay. Let's solve this for V_out. 00:05:28.680 --> 00:05:35.340 While distributing A, we've got V_out is equal 00:05:35.340 --> 00:05:42.105 to AV sub p minus one-fifth AV_out. 00:05:42.105 --> 00:05:44.975 Bring the one-fifth AV_out to the other side. 00:05:44.975 --> 00:05:51.570 We've got a V_out plus one-fifth AV_out is equal to AV sub p, 00:05:51.570 --> 00:05:59.100 factor out the V_out and you're going to get V_out times 1 plus one-fifth A, 00:05:59.100 --> 00:06:01.095 that's a V_out there, 00:06:01.095 --> 00:06:08.850 one-fifth A is equal to A times V sub p. So, V_out then stabilizes, 00:06:08.850 --> 00:06:18.595 or the final value of V_out is equal to AV_p divided by one plus one-fifth of A. 00:06:18.595 --> 00:06:22.100 Let's go ahead and plug in our numbers and see what V_out then would be. 00:06:22.100 --> 00:06:25.610 So, V_out would equal A, 00:06:25.610 --> 00:06:28.105 which is 10 to the fifth, 00:06:28.105 --> 00:06:34.050 times V sub p which we just chose to be 2 divided 00:06:34.050 --> 00:06:41.215 by 1 plus one-fifth of 10 to the fifth. 00:06:41.215 --> 00:06:50.210 Plug that into a calculator and you get that V_out is equal to 9.9995 volts. 00:06:50.210 --> 00:07:00.150 V_n, the sub-divided version of that would just be one-fifth of 9.9995 volts, 00:07:00.150 --> 00:07:11.550 which is equal to 1.9999, 00:07:11.550 --> 00:07:19.845 and thus V sub p minus V sub n is equal to two minus 1.9999, 00:07:19.845 --> 00:07:28.230 or V sub p minus V sub n is equal to 0.0001 volts, 00:07:28.230 --> 00:07:30.375 or one-tenth of a millivolt. 00:07:30.375 --> 00:07:36.419 When you compare that difference, that 0.0001 volts, 00:07:36.419 --> 00:07:39.615 the voltage here to here, 00:07:39.615 --> 00:07:45.100 we see that's one-ten thousandth of a fold. 00:07:45.100 --> 00:07:48.920 Our approximation again was that this voltage wasn't exactly 0, 00:07:48.920 --> 00:07:52.730 but it was so small compared to V sub p which is 2, 00:07:52.730 --> 00:07:57.050 or V sub n which turns out to be 1.9999 volts, 00:07:57.050 --> 00:08:01.755 or V_out which turned out to be 9.9995 volts. 00:08:01.755 --> 00:08:04.070 That this approximate or that 00:08:04.070 --> 00:08:07.550 this voltage across here was so small that we can neglect it. 00:08:07.550 --> 00:08:11.870 Thus, our approximation V sub p being approximately equal to 00:08:11.870 --> 00:08:16.694 V sub n is true under linear conditions, 00:08:16.694 --> 00:08:22.730 and this feedback, this voltage divider circuit right here, 00:08:22.730 --> 00:08:27.600 constrains the value on V sub n, 00:08:27.600 --> 00:08:29.745 thus constrain the value on V_out, 00:08:29.745 --> 00:08:36.695 so that this approximation holds as long as we don't allow V sub s to get too big. 00:08:36.695 --> 00:08:41.360 Under those circumstances this non-inverting amplifier then would have a gain of 00:08:41.360 --> 00:08:49.500 1 plus 4 over 1 or a gain of 5. Pretty close.