1 00:00:00,000 --> 00:00:02,505 >> To wrap up our discussion on Op-amps, 2 00:00:02,505 --> 00:00:05,533 let's see if we can understand how this negative feedback loop here 3 00:00:05,533 --> 00:00:08,325 constraints the Op-amp output, 4 00:00:08,325 --> 00:00:11,970 so that our ideal Op-amp approximations work. 5 00:00:11,970 --> 00:00:14,550 We've pointed out that an Op-amp is designed as 6 00:00:14,550 --> 00:00:18,890 a high-gain device such that V_out is equal to some A or A, 7 00:00:18,890 --> 00:00:20,560 some really large number, 8 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, 9 00:00:27,900 --> 00:00:31,005 say 50,000, 100,000, or something like that. 10 00:00:31,005 --> 00:00:35,985 But our source, our power supply voltages, 11 00:00:35,985 --> 00:00:40,480 are going to be significantly smaller than 10_5. 12 00:00:40,480 --> 00:00:42,240 Which means that, say, 13 00:00:42,240 --> 00:00:43,470 let's just put some numbers on here, 14 00:00:43,470 --> 00:00:47,370 let's say that this is a plus or minus 15-volt source. 15 00:00:47,710 --> 00:00:51,500 That means that the output voltage is constrained to 16 00:00:51,500 --> 00:00:54,965 being within the range of plus or minus 15 volts. 17 00:00:54,965 --> 00:00:56,445 If that's the case, 18 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, 19 00:01:00,305 --> 00:01:04,205 or V_p minus V_n is going to 20 00:01:04,205 --> 00:01:10,310 equal V_out divided by A. V_out can't be any larger than 15 volts. 21 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. 22 00:01:16,010 --> 00:01:18,575 That's our approximation, that's our assumption 23 00:01:18,575 --> 00:01:23,905 that we used in deriving our ideal Op-amp approximations, 24 00:01:23,905 --> 00:01:26,165 and the gains for these different amplifiers. 25 00:01:26,165 --> 00:01:30,290 So, what is it about this feedback, this R_1, 26 00:01:30,290 --> 00:01:37,275 R_2 voltage divider, that keeps V_p minus V_n so small? 27 00:01:37,275 --> 00:01:39,185 Let's see if we can understand that. 28 00:01:39,185 --> 00:01:40,970 To do that, let's just pick a few more numbers. 29 00:01:40,970 --> 00:01:45,850 Let's say that R_1 is equal to four kiloohms, 30 00:01:45,850 --> 00:01:50,430 and R_2 is equal to one kiloohm. 31 00:01:50,430 --> 00:01:55,850 So, with that then V sub n is just a voltage divided portion of V_out. 32 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. 33 00:02:02,130 --> 00:02:06,460 So, V sub n is equal to one-fifth of V_out. 34 00:02:06,680 --> 00:02:11,120 When analyzing loops, you make some assumptions and 35 00:02:11,120 --> 00:02:15,155 then see what this circuit does to the assumptions that you've made. 36 00:02:15,155 --> 00:02:19,150 So, let's assume that V sub p has been 0, 37 00:02:19,150 --> 00:02:22,400 which means V_out would be 0, 38 00:02:22,400 --> 00:02:24,230 and V sub n would be 0. 39 00:02:24,230 --> 00:02:26,540 So, V sub p has been 0, 40 00:02:26,540 --> 00:02:29,930 and it is transitioning to V sub p equals 2. 41 00:02:29,930 --> 00:02:31,880 So, let's just say that it just switched, 42 00:02:31,880 --> 00:02:36,660 and V sub p now equals 2 volts. All right. 43 00:02:36,660 --> 00:02:40,550 An Op-amp doesn't change the output voltage instantaneously. 44 00:02:40,550 --> 00:02:42,650 It's limited by something called the slew rate, 45 00:02:42,650 --> 00:02:43,880 and it doesn't really matter what that is, 46 00:02:43,880 --> 00:02:45,500 you'll learn more about that in later. 47 00:02:45,500 --> 00:02:47,945 In later classes, we'll learn more about Op-amps. 48 00:02:47,945 --> 00:02:51,615 The supply is to say that at an instantaneous change here, the input, 49 00:02:51,615 --> 00:02:59,175 will have some delay before its output is singing or the changes in to the output. 50 00:02:59,175 --> 00:03:04,050 In fact, what happens is V_out starts to grow, 51 00:03:04,050 --> 00:03:05,820 and under these circumstances, 52 00:03:05,820 --> 00:03:08,595 it would be heading towards A, 53 00:03:08,595 --> 00:03:12,300 this big number, times V sub p which is 2, 54 00:03:12,300 --> 00:03:16,215 minus V sub n. So, initially, 55 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. 56 00:03:20,645 --> 00:03:23,330 V_out would be heading towards, 57 00:03:23,330 --> 00:03:24,700 is not going to get there. 58 00:03:24,700 --> 00:03:26,420 This feedback to work it's going to constrain it, 59 00:03:26,420 --> 00:03:28,540 but if it was just open circuit, 60 00:03:28,540 --> 00:03:31,205 V_out would be heading towards A, 61 00:03:31,205 --> 00:03:33,515 which in our case is 10 to the fifth, 62 00:03:33,515 --> 00:03:35,540 times V sub p which is 2, 63 00:03:35,540 --> 00:03:40,590 minus V sub n. All right. 64 00:03:40,590 --> 00:03:44,540 Again, it can't get anywhere close to that because our power supplies are limited, 65 00:03:44,540 --> 00:03:46,760 but it would saturate at plus or minus 15, 66 00:03:46,760 --> 00:03:49,160 in this case it's saturated at 15 volts. 67 00:03:49,160 --> 00:03:56,010 But look what happens, as V_out starts to grow, V sub n, 68 00:03:56,010 --> 00:03:58,410 which is equal to one-fifth of V_out, 69 00:03:58,410 --> 00:04:01,590 let's write that here one-fifth V_out, 70 00:04:01,590 --> 00:04:05,175 V sub n now starts to grow also. 71 00:04:05,175 --> 00:04:07,800 As V sub n starts to grow, 72 00:04:07,800 --> 00:04:13,310 this difference here V_p minus V sub n starts to get smaller. 73 00:04:13,310 --> 00:04:14,930 Let's write that down here. 74 00:04:14,930 --> 00:04:17,404 As V_out starts to grow, 75 00:04:17,404 --> 00:04:24,434 V sub n which is equal to one-fifth V_out starts to increase also, 76 00:04:24,434 --> 00:04:25,995 and thus the difference, 77 00:04:25,995 --> 00:04:29,340 V_p minus V sub n, 78 00:04:29,340 --> 00:04:31,510 starts to get smaller. 79 00:04:31,510 --> 00:04:33,620 V_out continues to grow, 80 00:04:33,620 --> 00:04:39,275 it's still a number times a relatively small number times a big number. 81 00:04:39,275 --> 00:04:41,390 So, V_out is going to continue to grow, 82 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 83 00:04:48,225 --> 00:04:55,710 is equal to A times V sub p minus V_n, 84 00:04:55,710 --> 00:05:02,970 but V_n is minus one-fifth V_out. 85 00:05:02,970 --> 00:05:06,990 Let's say that again. V_out will continue to grow, 86 00:05:06,990 --> 00:05:10,220 until it gets to the point where V_out is equal to 87 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. 88 00:05:14,780 --> 00:05:24,480 This point then becomes the stabilized value of V_out. 89 00:05:24,740 --> 00:05:28,680 Okay. Let's solve this for V_out. 90 00:05:28,680 --> 00:05:35,340 While distributing A, we've got V_out is equal 91 00:05:35,340 --> 00:05:42,105 to AV sub p minus one-fifth AV_out. 92 00:05:42,105 --> 00:05:44,975 Bring the one-fifth AV_out to the other side. 93 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, 94 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, 95 00:05:59,100 --> 00:06:01,095 that's a V_out there, 96 00:06:01,095 --> 00:06:08,850 one-fifth A is equal to A times V sub p. So, V_out then stabilizes, 97 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. 98 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. 99 00:06:22,100 --> 00:06:25,610 So, V_out would equal A, 100 00:06:25,610 --> 00:06:28,105 which is 10 to the fifth, 101 00:06:28,105 --> 00:06:34,050 times V sub p which we just chose to be 2 divided 102 00:06:34,050 --> 00:06:41,215 by 1 plus one-fifth of 10 to the fifth. 103 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. 104 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, 105 00:07:00,150 --> 00:07:11,550 which is equal to 1.9999, 106 00:07:11,550 --> 00:07:19,845 and thus V sub p minus V sub n is equal to two minus 1.9999, 107 00:07:19,845 --> 00:07:28,230 or V sub p minus V sub n is equal to 0.0001 volts, 108 00:07:28,230 --> 00:07:30,375 or one-tenth of a millivolt. 109 00:07:30,375 --> 00:07:36,419 When you compare that difference, that 0.0001 volts, 110 00:07:36,419 --> 00:07:39,615 the voltage here to here, 111 00:07:39,615 --> 00:07:45,100 we see that's one-ten thousandth of a fold. 112 00:07:45,100 --> 00:07:48,920 Our approximation again was that this voltage wasn't exactly 0, 113 00:07:48,920 --> 00:07:52,730 but it was so small compared to V sub p which is 2, 114 00:07:52,730 --> 00:07:57,050 or V sub n which turns out to be 1.9999 volts, 115 00:07:57,050 --> 00:08:01,755 or V_out which turned out to be 9.9995 volts. 116 00:08:01,755 --> 00:08:04,070 That this approximate or that 117 00:08:04,070 --> 00:08:07,550 this voltage across here was so small that we can neglect it. 118 00:08:07,550 --> 00:08:11,870 Thus, our approximation V sub p being approximately equal to 119 00:08:11,870 --> 00:08:16,694 V sub n is true under linear conditions, 120 00:08:16,694 --> 00:08:22,730 and this feedback, this voltage divider circuit right here, 121 00:08:22,730 --> 00:08:27,600 constrains the value on V sub n, 122 00:08:27,600 --> 00:08:29,745 thus constrain the value on V_out, 123 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. 124 00:08:36,695 --> 00:08:41,360 Under those circumstances this non-inverting amplifier then would have a gain of 125 00:08:41,360 --> 00:08:49,500 1 plus 4 over 1 or a gain of 5. Pretty close.