[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.29,0:00:02.58,Default,,0000,0000,0000,,>> All right. Let's take a look at
Dialogue: 0,0:00:02.58,0:00:07.11,Default,,0000,0000,0000,,an example of a non-inverting amplifier with some real numbers.
Dialogue: 0,0:00:07.11,0:00:14.30,Default,,0000,0000,0000,,Let's let R_1, the feedback resistor be 10 kiloohms,
Dialogue: 0,0:00:14.30,0:00:20.38,Default,,0000,0000,0000,,and R_2 be two kiloohms.
Dialogue: 0,0:00:20.38,0:00:26.10,Default,,0000,0000,0000,,Let's calculate what the output voltage will be in terms of V_s in these ratios.
Dialogue: 0,0:00:26.10,0:00:28.35,Default,,0000,0000,0000,,What we know from our previous work,
Dialogue: 0,0:00:28.35,0:00:33.33,Default,,0000,0000,0000,,is that V_out for an non-inverting amplifier is equal
Dialogue: 0,0:00:33.33,0:00:38.33,Default,,0000,0000,0000,,to V_s times the ratio of those two resistors,
Dialogue: 0,0:00:38.33,0:00:45.75,Default,,0000,0000,0000,,R1 over R2, or 10 kiloohms divided by two kiloohms plus one.
Dialogue: 0,0:00:45.75,0:00:48.94,Default,,0000,0000,0000,,Well, 10 divided by 2 is 5 plus 1 is 6.
Dialogue: 0,0:00:48.94,0:00:58.29,Default,,0000,0000,0000,,So, V_out then is equal to six times V_s. In this case, then,
Dialogue: 0,0:00:58.29,0:01:01.67,Default,,0000,0000,0000,,we would say that the gain of the amplifier G is equal
Dialogue: 0,0:01:01.67,0:01:05.62,Default,,0000,0000,0000,,to six and know that gain is a unitless term.
Dialogue: 0,0:01:05.62,0:01:10.19,Default,,0000,0000,0000,,It's equal to the ratio of V_out divided by V_s
Dialogue: 0,0:01:10.19,0:01:15.20,Default,,0000,0000,0000,,or we say then the gain is equal to the ratio of V_out divided by V_s,
Dialogue: 0,0:01:15.20,0:01:19.94,Default,,0000,0000,0000,,the volts cancel, and gain is again a unitless term.
Dialogue: 0,0:01:19.94,0:01:23.30,Default,,0000,0000,0000,,Now, let's look at issues of
Dialogue: 0,0:01:23.30,0:01:27.90,Default,,0000,0000,0000,,saturation by just assuming that we have two voltage sources.
Dialogue: 0,0:01:27.90,0:01:33.02,Default,,0000,0000,0000,,The positive supply is going to be a plus 15 volts and let's make the negative supply,
Dialogue: 0,0:01:33.02,0:01:37.46,Default,,0000,0000,0000,,a negative 15-volt source.
Dialogue: 0,0:01:37.46,0:01:39.31,Default,,0000,0000,0000,,We now ask ourselves,
Dialogue: 0,0:01:39.31,0:01:42.24,Default,,0000,0000,0000,,what is the range on V_s,
Dialogue: 0,0:01:42.24,0:01:46.03,Default,,0000,0000,0000,,so that the amplifier will stay in it's linear range,
Dialogue: 0,0:01:46.03,0:01:50.36,Default,,0000,0000,0000,,and the output voltage will be six times whatever V_s is.
Dialogue: 0,0:01:50.36,0:01:52.55,Default,,0000,0000,0000,,Remember what the limitation is.
Dialogue: 0,0:01:52.55,0:01:56.90,Default,,0000,0000,0000,,The output voltage can't be any greater than the supply voltages.
Dialogue: 0,0:01:56.90,0:02:02.54,Default,,0000,0000,0000,,It can't be any greater than 15 volts and it can't be any less than negative 15 volts,
Dialogue: 0,0:02:02.54,0:02:07.55,Default,,0000,0000,0000,,or to put it in terms of inequalities V_out has got to be greater
Dialogue: 0,0:02:07.55,0:02:12.85,Default,,0000,0000,0000,,than negative 15 volts and less than positive 15 volts.
Dialogue: 0,0:02:12.85,0:02:16.04,Default,,0000,0000,0000,,Now, substituting V_out for six V_s,
Dialogue: 0,0:02:16.04,0:02:22.46,Default,,0000,0000,0000,,we have then six V_s must be greater than a negative 15 volts and less than
Dialogue: 0,0:02:22.46,0:02:30.60,Default,,0000,0000,0000,,a positive 15 volts dividing both this side and this side by the six,
Dialogue: 0,0:02:30.60,0:02:34.26,Default,,0000,0000,0000,,so that we get V_s by itself rather,
Dialogue: 0,0:02:34.26,0:02:40.46,Default,,0000,0000,0000,,we get V_s must be greater than a negative 15 over six and it must be
Dialogue: 0,0:02:40.46,0:02:46.40,Default,,0000,0000,0000,,less than a positive 15 over six or V_s that is constrained to
Dialogue: 0,0:02:46.40,0:02:54.84,Default,,0000,0000,0000,,being greater than a negative 2.5 volts and less than a positive 2.5 volts.
Dialogue: 0,0:02:55.09,0:03:03.35,Default,,0000,0000,0000,,Let's graph this by looking at V_out as a function of V_s. So,
Dialogue: 0,0:03:03.35,0:03:07.96,Default,,0000,0000,0000,,V_s here, V_out along here.
Dialogue: 0,0:03:07.96,0:03:14.90,Default,,0000,0000,0000,,Let's go ahead and mark our positive 15 volts there
Dialogue: 0,0:03:14.90,0:03:19.98,Default,,0000,0000,0000,,and negative 15 volts
Dialogue: 0,0:03:19.98,0:03:26.19,Default,,0000,0000,0000,,there and here's negative 2.5 volts.
Dialogue: 0,0:03:26.19,0:03:31.10,Default,,0000,0000,0000,,There's positive 2.5 volts.
Dialogue: 0,0:03:31.10,0:03:34.08,Default,,0000,0000,0000,,Just mark it up there,
Dialogue: 0,0:03:34.08,0:03:36.87,Default,,0000,0000,0000,,I mark it down here.
Dialogue: 0,0:03:36.87,0:03:41.92,Default,,0000,0000,0000,,So, for V_s less than 2.5 volts the amplifier is going to be
Dialogue: 0,0:03:41.92,0:03:47.33,Default,,0000,0000,0000,,saturated at its negative supply voltage of negative 15 volts.
Dialogue: 0,0:03:47.33,0:03:55.45,Default,,0000,0000,0000,,As V_s gets larger than negative 2.5 volts and less than positive 2.5 volts,
Dialogue: 0,0:03:55.45,0:03:59.42,Default,,0000,0000,0000,,the output is going to equal six times V_s,
Dialogue: 0,0:03:59.84,0:04:09.66,Default,,0000,0000,0000,,or a nice sort of linear, you get the idea, relationship in that linear region,
Dialogue: 0,0:04:09.66,0:04:11.21,Default,,0000,0000,0000,,hence, it's called the linear region.
Dialogue: 0,0:04:11.21,0:04:15.23,Default,,0000,0000,0000,,When V_s is greater than 2.5 volts, it saturates at the upper or
Dialogue: 0,0:04:15.23,0:04:20.16,Default,,0000,0000,0000,,the positive supply voltage and while the graph may not be beautiful, you get the idea.
Dialogue: 0,0:04:20.16,0:04:25.17,Default,,0000,0000,0000,,Saturation means that the output can't be any less than negative 15 volts,
Dialogue: 0,0:04:25.17,0:04:29.27,Default,,0000,0000,0000,,can't be any greater than positive 15 volts and in between the constraints,
Dialogue: 0,0:04:29.27,0:04:31.40,Default,,0000,0000,0000,,it's a nice linear relationship.
Dialogue: 0,0:04:31.40,0:04:36.77,Default,,0000,0000,0000,,Where then the slope of this line is
Dialogue: 0,0:04:36.77,0:04:47.32,Default,,0000,0000,0000,,15 divided by 2.5 which is six and that's just our gain term.