0:00:00.290,0:00:02.580
>> All right. Let's take a look at

0:00:02.580,0:00:07.110
an example of a non-inverting amplifier with some real numbers.

0:00:07.110,0:00:14.295
Let's let R_1, the feedback resistor be 10 kiloohms,

0:00:14.295,0:00:20.385
and R_2 be two kiloohms.

0:00:20.385,0:00:26.100
Let's calculate what the output voltage will be in terms of V_s in these ratios.

0:00:26.100,0:00:28.350
What we know from our previous work,

0:00:28.350,0:00:33.330
is that V_out for an non-inverting amplifier is equal

0:00:33.330,0:00:38.330
to V_s times the ratio of those two resistors,

0:00:38.330,0:00:45.750
R1 over R2, or 10 kiloohms divided by two kiloohms plus one.

0:00:45.750,0:00:48.945
Well, 10 divided by 2 is 5 plus 1 is 6.

0:00:48.945,0:00:58.290
So, V_out then is equal to six times V_s. In this case, then,

0:00:58.290,0:01:01.670
we would say that the gain of the amplifier G is equal

0:01:01.670,0:01:05.615
to six and know that gain is a unitless term.

0:01:05.615,0:01:10.190
It's equal to the ratio of V_out divided by V_s

0:01:10.190,0:01:15.200
or we say then the gain is equal to the ratio of V_out divided by V_s,

0:01:15.200,0:01:19.945
the volts cancel, and gain is again a unitless term.

0:01:19.945,0:01:23.300
Now, let's look at issues of

0:01:23.300,0:01:27.905
saturation by just assuming that we have two voltage sources.

0:01:27.905,0:01:33.020
The positive supply is going to be a plus 15 volts and let's make the negative supply,

0:01:33.020,0:01:37.455
a negative 15-volt source.

0:01:37.455,0:01:39.314
We now ask ourselves,

0:01:39.314,0:01:42.240
what is the range on V_s,

0:01:42.240,0:01:46.030
so that the amplifier will stay in it's linear range,

0:01:46.030,0:01:50.360
and the output voltage will be six times whatever V_s is.

0:01:50.360,0:01:52.550
Remember what the limitation is.

0:01:52.550,0:01:56.900
The output voltage can't be any greater than the supply voltages.

0:01:56.900,0:02:02.540
It can't be any greater than 15 volts and it can't be any less than negative 15 volts,

0:02:02.540,0:02:07.550
or to put it in terms of inequalities V_out has got to be greater

0:02:07.550,0:02:12.850
than negative 15 volts and less than positive 15 volts.

0:02:12.850,0:02:16.035
Now, substituting V_out for six V_s,

0:02:16.035,0:02:22.460
we have then six V_s must be greater than a negative 15 volts and less than

0:02:22.460,0:02:30.600
a positive 15 volts dividing both this side and this side by the six,

0:02:30.600,0:02:34.265
so that we get V_s by itself rather,

0:02:34.265,0:02:40.460
we get V_s must be greater than a negative 15 over six and it must be

0:02:40.460,0:02:46.400
less than a positive 15 over six or V_s that is constrained to

0:02:46.400,0:02:54.840
being greater than a negative 2.5 volts and less than a positive 2.5 volts.

0:02:55.090,0:03:03.350
Let's graph this by looking at V_out as a function of V_s. So,

0:03:03.350,0:03:07.960
V_s here, V_out along here.

0:03:07.960,0:03:14.895
Let's go ahead and mark our positive 15 volts there

0:03:14.895,0:03:19.980
and negative 15 volts

0:03:19.980,0:03:26.190
there and here's negative 2.5 volts.

0:03:26.190,0:03:31.095
There's positive 2.5 volts.

0:03:31.095,0:03:34.080
Just mark it up there,

0:03:34.080,0:03:36.870
I mark it down here.

0:03:36.870,0:03:41.920
So, for V_s less than 2.5 volts the amplifier is going to be

0:03:41.920,0:03:47.330
saturated at its negative supply voltage of negative 15 volts.

0:03:47.330,0:03:55.450
As V_s gets larger than negative 2.5 volts and less than positive 2.5 volts,

0:03:55.450,0:03:59.420
the output is going to equal six times V_s,

0:03:59.840,0:04:09.660
or a nice sort of linear, you get the idea, relationship in that linear region,

0:04:09.660,0:04:11.210
hence, it's called the linear region.

0:04:11.210,0:04:15.230
When V_s is greater than 2.5 volts, it saturates at the upper or

0:04:15.230,0:04:20.165
the positive supply voltage and while the graph may not be beautiful, you get the idea.

0:04:20.165,0:04:25.170
Saturation means that the output can't be any less than negative 15 volts,

0:04:25.170,0:04:29.270
can't be any greater than positive 15 volts and in between the constraints,

0:04:29.270,0:04:31.400
it's a nice linear relationship.

0:04:31.400,0:04:36.770
Where then the slope of this line is

0:04:36.770,0:04:47.320
15 divided by 2.5 which is six and that's just our gain term.