[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:02.50,Default,,0000,0000,0000,,>> To wrap up our discussion on Op-amps,
Dialogue: 0,0:00:02.50,0:00:05.53,Default,,0000,0000,0000,,let's see if we can understand how this negative feedback loop here
Dialogue: 0,0:00:05.53,0:00:08.32,Default,,0000,0000,0000,,constraints the Op-amp output,
Dialogue: 0,0:00:08.32,0:00:11.97,Default,,0000,0000,0000,,so that our ideal Op-amp approximations work.
Dialogue: 0,0:00:11.97,0:00:14.55,Default,,0000,0000,0000,,We've pointed out that an Op-amp is designed as
Dialogue: 0,0:00:14.55,0:00:18.89,Default,,0000,0000,0000,,a high-gain device such that V_out is equal to some A or A,
Dialogue: 0,0:00:18.89,0:00:20.56,Default,,0000,0000,0000,,some really large number,
Dialogue: 0,0:00:20.56,0:00:27.90,Default,,0000,0000,0000,,times the difference between V_p and V_n. A is on the order of 10_5,
Dialogue: 0,0:00:27.90,0:00:31.00,Default,,0000,0000,0000,,say 50,000, 100,000, or something like that.
Dialogue: 0,0:00:31.00,0:00:35.98,Default,,0000,0000,0000,,But our source, our power supply voltages,
Dialogue: 0,0:00:35.98,0:00:40.48,Default,,0000,0000,0000,,are going to be significantly smaller than 10_5.
Dialogue: 0,0:00:40.48,0:00:42.24,Default,,0000,0000,0000,,Which means that, say,
Dialogue: 0,0:00:42.24,0:00:43.47,Default,,0000,0000,0000,,let's just put some numbers on here,
Dialogue: 0,0:00:43.47,0:00:47.37,Default,,0000,0000,0000,,let's say that this is a plus or minus 15-volt source.
Dialogue: 0,0:00:47.71,0:00:51.50,Default,,0000,0000,0000,,That means that the output voltage is constrained to
Dialogue: 0,0:00:51.50,0:00:54.96,Default,,0000,0000,0000,,being within the range of plus or minus 15 volts.
Dialogue: 0,0:00:54.96,0:00:56.44,Default,,0000,0000,0000,,If that's the case,
Dialogue: 0,0:00:56.44,0:01:00.30,Default,,0000,0000,0000,,then this difference V sub p minus V sub n is going to have to be very small,
Dialogue: 0,0:01:00.30,0:01:04.20,Default,,0000,0000,0000,,or V_p minus V_n is going to
Dialogue: 0,0:01:04.20,0:01:10.31,Default,,0000,0000,0000,,equal V_out divided by A. V_out can't be any larger than 15 volts.
Dialogue: 0,0:01:10.31,0:01:16.01,Default,,0000,0000,0000,,So, 15 volts divided by 10 to the fifth means that V_p minus V_n is relatively small.
Dialogue: 0,0:01:16.01,0:01:18.58,Default,,0000,0000,0000,,That's our approximation, that's our assumption
Dialogue: 0,0:01:18.58,0:01:23.90,Default,,0000,0000,0000,,that we used in deriving our ideal Op-amp approximations,
Dialogue: 0,0:01:23.90,0:01:26.16,Default,,0000,0000,0000,,and the gains for these different amplifiers.
Dialogue: 0,0:01:26.16,0:01:30.29,Default,,0000,0000,0000,,So, what is it about this feedback, this R_1,
Dialogue: 0,0:01:30.29,0:01:37.28,Default,,0000,0000,0000,,R_2 voltage divider, that keeps V_p minus V_n so small?
Dialogue: 0,0:01:37.28,0:01:39.18,Default,,0000,0000,0000,,Let's see if we can understand that.
Dialogue: 0,0:01:39.18,0:01:40.97,Default,,0000,0000,0000,,To do that, let's just pick a few more numbers.
Dialogue: 0,0:01:40.97,0:01:45.85,Default,,0000,0000,0000,,Let's say that R_1 is equal to four kiloohms,
Dialogue: 0,0:01:45.85,0:01:50.43,Default,,0000,0000,0000,,and R_2 is equal to one kiloohm.
Dialogue: 0,0:01:50.43,0:01:55.85,Default,,0000,0000,0000,,So, with that then V sub n is just a voltage divided portion of V_out.
Dialogue: 0,0:01:55.85,0:02:02.13,Default,,0000,0000,0000,,So, V sub n is equal to V_out times 1 divided by 1 plus 4, that's 5.
Dialogue: 0,0:02:02.13,0:02:06.46,Default,,0000,0000,0000,,So, V sub n is equal to one-fifth of V_out.
Dialogue: 0,0:02:06.68,0:02:11.12,Default,,0000,0000,0000,,When analyzing loops, you make some assumptions and
Dialogue: 0,0:02:11.12,0:02:15.16,Default,,0000,0000,0000,,then see what this circuit does to the assumptions that you've made.
Dialogue: 0,0:02:15.16,0:02:19.15,Default,,0000,0000,0000,,So, let's assume that V sub p has been 0,
Dialogue: 0,0:02:19.15,0:02:22.40,Default,,0000,0000,0000,,which means V_out would be 0,
Dialogue: 0,0:02:22.40,0:02:24.23,Default,,0000,0000,0000,,and V sub n would be 0.
Dialogue: 0,0:02:24.23,0:02:26.54,Default,,0000,0000,0000,,So, V sub p has been 0,
Dialogue: 0,0:02:26.54,0:02:29.93,Default,,0000,0000,0000,,and it is transitioning to V sub p equals 2.
Dialogue: 0,0:02:29.93,0:02:31.88,Default,,0000,0000,0000,,So, let's just say that it just switched,
Dialogue: 0,0:02:31.88,0:02:36.66,Default,,0000,0000,0000,,and V sub p now equals 2 volts. All right.
Dialogue: 0,0:02:36.66,0:02:40.55,Default,,0000,0000,0000,,An Op-amp doesn't change the output voltage instantaneously.
Dialogue: 0,0:02:40.55,0:02:42.65,Default,,0000,0000,0000,,It's limited by something called the slew rate,
Dialogue: 0,0:02:42.65,0:02:43.88,Default,,0000,0000,0000,,and it doesn't really matter what that is,
Dialogue: 0,0:02:43.88,0:02:45.50,Default,,0000,0000,0000,,you'll learn more about that in later.
Dialogue: 0,0:02:45.50,0:02:47.94,Default,,0000,0000,0000,,In later classes, we'll learn more about Op-amps.
Dialogue: 0,0:02:47.94,0:02:51.62,Default,,0000,0000,0000,,The supply is to say that at an instantaneous change here, the input,
Dialogue: 0,0:02:51.62,0:02:59.18,Default,,0000,0000,0000,,will have some delay before its output is singing or the changes in to the output.
Dialogue: 0,0:02:59.18,0:03:04.05,Default,,0000,0000,0000,,In fact, what happens is V_out starts to grow,
Dialogue: 0,0:03:04.05,0:03:05.82,Default,,0000,0000,0000,,and under these circumstances,
Dialogue: 0,0:03:05.82,0:03:08.60,Default,,0000,0000,0000,,it would be heading towards A,
Dialogue: 0,0:03:08.60,0:03:12.30,Default,,0000,0000,0000,,this big number, times V sub p which is 2,
Dialogue: 0,0:03:12.30,0:03:16.22,Default,,0000,0000,0000,,minus V sub n. So, initially,
Dialogue: 0,0:03:16.22,0:03:20.64,Default,,0000,0000,0000,,this V sub p minus V sub n is going to be about 2, it's right down here.
Dialogue: 0,0:03:20.64,0:03:23.33,Default,,0000,0000,0000,,V_out would be heading towards,
Dialogue: 0,0:03:23.33,0:03:24.70,Default,,0000,0000,0000,,is not going to get there.
Dialogue: 0,0:03:24.70,0:03:26.42,Default,,0000,0000,0000,,This feedback to work it's going to constrain it,
Dialogue: 0,0:03:26.42,0:03:28.54,Default,,0000,0000,0000,,but if it was just open circuit,
Dialogue: 0,0:03:28.54,0:03:31.20,Default,,0000,0000,0000,,V_out would be heading towards A,
Dialogue: 0,0:03:31.20,0:03:33.52,Default,,0000,0000,0000,,which in our case is 10 to the fifth,
Dialogue: 0,0:03:33.52,0:03:35.54,Default,,0000,0000,0000,,times V sub p which is 2,
Dialogue: 0,0:03:35.54,0:03:40.59,Default,,0000,0000,0000,,minus V sub n. All right.
Dialogue: 0,0:03:40.59,0:03:44.54,Default,,0000,0000,0000,,Again, it can't get anywhere close to that because our power supplies are limited,
Dialogue: 0,0:03:44.54,0:03:46.76,Default,,0000,0000,0000,,but it would saturate at plus or minus 15,
Dialogue: 0,0:03:46.76,0:03:49.16,Default,,0000,0000,0000,,in this case it's saturated at 15 volts.
Dialogue: 0,0:03:49.16,0:03:56.01,Default,,0000,0000,0000,,But look what happens, as V_out starts to grow, V sub n,
Dialogue: 0,0:03:56.01,0:03:58.41,Default,,0000,0000,0000,,which is equal to one-fifth of V_out,
Dialogue: 0,0:03:58.41,0:04:01.59,Default,,0000,0000,0000,,let's write that here one-fifth V_out,
Dialogue: 0,0:04:01.59,0:04:05.18,Default,,0000,0000,0000,,V sub n now starts to grow also.
Dialogue: 0,0:04:05.18,0:04:07.80,Default,,0000,0000,0000,,As V sub n starts to grow,
Dialogue: 0,0:04:07.80,0:04:13.31,Default,,0000,0000,0000,,this difference here V_p minus V sub n starts to get smaller.
Dialogue: 0,0:04:13.31,0:04:14.93,Default,,0000,0000,0000,,Let's write that down here.
Dialogue: 0,0:04:14.93,0:04:17.40,Default,,0000,0000,0000,,As V_out starts to grow,
Dialogue: 0,0:04:17.40,0:04:24.43,Default,,0000,0000,0000,,V sub n which is equal to one-fifth V_out starts to increase also,
Dialogue: 0,0:04:24.43,0:04:25.100,Default,,0000,0000,0000,,and thus the difference,
Dialogue: 0,0:04:25.100,0:04:29.34,Default,,0000,0000,0000,,V_p minus V sub n,
Dialogue: 0,0:04:29.34,0:04:31.51,Default,,0000,0000,0000,,starts to get smaller.
Dialogue: 0,0:04:31.51,0:04:33.62,Default,,0000,0000,0000,,V_out continues to grow,
Dialogue: 0,0:04:33.62,0:04:39.28,Default,,0000,0000,0000,,it's still a number times a relatively small number times a big number.
Dialogue: 0,0:04:39.28,0:04:41.39,Default,,0000,0000,0000,,So, V_out is going to continue to grow,
Dialogue: 0,0:04:41.39,0:04:48.22,Default,,0000,0000,0000,,and it will continue to grow until it stabilizes at a voltage where in fact V_out
Dialogue: 0,0:04:48.22,0:04:55.71,Default,,0000,0000,0000,,is equal to A times V sub p minus V_n,
Dialogue: 0,0:04:55.71,0:05:02.97,Default,,0000,0000,0000,,but V_n is minus one-fifth V_out.
Dialogue: 0,0:05:02.97,0:05:06.99,Default,,0000,0000,0000,,Let's say that again. V_out will continue to grow,
Dialogue: 0,0:05:06.99,0:05:10.22,Default,,0000,0000,0000,,until it gets to the point where V_out is equal to
Dialogue: 0,0:05:10.22,0:05:14.78,Default,,0000,0000,0000,,A times the difference V_p minus V sub n at V sub n is one-fifth V_out.
Dialogue: 0,0:05:14.78,0:05:24.48,Default,,0000,0000,0000,,This point then becomes the stabilized value of V_out.
Dialogue: 0,0:05:24.74,0:05:28.68,Default,,0000,0000,0000,,Okay. Let's solve this for V_out.
Dialogue: 0,0:05:28.68,0:05:35.34,Default,,0000,0000,0000,,While distributing A, we've got V_out is equal
Dialogue: 0,0:05:35.34,0:05:42.10,Default,,0000,0000,0000,,to AV sub p minus one-fifth AV_out.
Dialogue: 0,0:05:42.10,0:05:44.98,Default,,0000,0000,0000,,Bring the one-fifth AV_out to the other side.
Dialogue: 0,0:05:44.98,0:05:51.57,Default,,0000,0000,0000,,We've got a V_out plus one-fifth AV_out is equal to AV sub p,
Dialogue: 0,0:05:51.57,0:05:59.10,Default,,0000,0000,0000,,factor out the V_out and you're going to get V_out times 1 plus one-fifth A,
Dialogue: 0,0:05:59.10,0:06:01.10,Default,,0000,0000,0000,,that's a V_out there,
Dialogue: 0,0:06:01.10,0:06:08.85,Default,,0000,0000,0000,,one-fifth A is equal to A times V sub p. So, V_out then stabilizes,
Dialogue: 0,0:06:08.85,0:06:18.60,Default,,0000,0000,0000,,or the final value of V_out is equal to AV_p divided by one plus one-fifth of A.
Dialogue: 0,0:06:18.60,0:06:22.10,Default,,0000,0000,0000,,Let's go ahead and plug in our numbers and see what V_out then would be.
Dialogue: 0,0:06:22.10,0:06:25.61,Default,,0000,0000,0000,,So, V_out would equal A,
Dialogue: 0,0:06:25.61,0:06:28.10,Default,,0000,0000,0000,,which is 10 to the fifth,
Dialogue: 0,0:06:28.10,0:06:34.05,Default,,0000,0000,0000,,times V sub p which we just chose to be 2 divided
Dialogue: 0,0:06:34.05,0:06:41.22,Default,,0000,0000,0000,,by 1 plus one-fifth of 10 to the fifth.
Dialogue: 0,0:06:41.22,0:06:50.21,Default,,0000,0000,0000,,Plug that into a calculator and you get that V_out is equal to 9.9995 volts.
Dialogue: 0,0:06:50.21,0:07:00.15,Default,,0000,0000,0000,,V_n, the sub-divided version of that would just be one-fifth of 9.9995 volts,
Dialogue: 0,0:07:00.15,0:07:11.55,Default,,0000,0000,0000,,which is equal to 1.9999,
Dialogue: 0,0:07:11.55,0:07:19.84,Default,,0000,0000,0000,,and thus V sub p minus V sub n is equal to two minus 1.9999,
Dialogue: 0,0:07:19.84,0:07:28.23,Default,,0000,0000,0000,,or V sub p minus V sub n is equal to 0.0001 volts,
Dialogue: 0,0:07:28.23,0:07:30.38,Default,,0000,0000,0000,,or one-tenth of a millivolt.
Dialogue: 0,0:07:30.38,0:07:36.42,Default,,0000,0000,0000,,When you compare that difference, that 0.0001 volts,
Dialogue: 0,0:07:36.42,0:07:39.62,Default,,0000,0000,0000,,the voltage here to here,
Dialogue: 0,0:07:39.62,0:07:45.10,Default,,0000,0000,0000,,we see that's one-ten thousandth of a fold.
Dialogue: 0,0:07:45.10,0:07:48.92,Default,,0000,0000,0000,,Our approximation again was that this voltage wasn't exactly 0,
Dialogue: 0,0:07:48.92,0:07:52.73,Default,,0000,0000,0000,,but it was so small compared to V sub p which is 2,
Dialogue: 0,0:07:52.73,0:07:57.05,Default,,0000,0000,0000,,or V sub n which turns out to be 1.9999 volts,
Dialogue: 0,0:07:57.05,0:08:01.76,Default,,0000,0000,0000,,or V_out which turned out to be 9.9995 volts.
Dialogue: 0,0:08:01.76,0:08:04.07,Default,,0000,0000,0000,,That this approximate or that
Dialogue: 0,0:08:04.07,0:08:07.55,Default,,0000,0000,0000,,this voltage across here was so small that we can neglect it.
Dialogue: 0,0:08:07.55,0:08:11.87,Default,,0000,0000,0000,,Thus, our approximation V sub p being approximately equal to
Dialogue: 0,0:08:11.87,0:08:16.69,Default,,0000,0000,0000,,V sub n is true under linear conditions,
Dialogue: 0,0:08:16.69,0:08:22.73,Default,,0000,0000,0000,,and this feedback, this voltage divider circuit right here,
Dialogue: 0,0:08:22.73,0:08:27.60,Default,,0000,0000,0000,,constrains the value on V sub n,
Dialogue: 0,0:08:27.60,0:08:29.74,Default,,0000,0000,0000,,thus constrain the value on V_out,
Dialogue: 0,0:08:29.74,0:08:36.70,Default,,0000,0000,0000,,so that this approximation holds as long as we don't allow V sub s to get too big.
Dialogue: 0,0:08:36.70,0:08:41.36,Default,,0000,0000,0000,,Under those circumstances this non-inverting amplifier then would have a gain of
Dialogue: 0,0:08:41.36,0:08:49.50,Default,,0000,0000,0000,,1 plus 4 over 1 or a gain of 5. Pretty close.