1
00:00:00,000 --> 00:00:04,755
>> Now, let's do an example with our non-inverting operational amplifier.

2
00:00:04,755 --> 00:00:06,960
So, here comes an example.

3
00:00:06,960 --> 00:00:09,750
Let's start with our non-ideal model.

4
00:00:09,750 --> 00:00:12,120
This is the non-ideal model and let's connect it

5
00:00:12,120 --> 00:00:15,195
up in the non-inverting amplifier case that I showed you.

6
00:00:15,195 --> 00:00:18,120
So, we want to define our gain.

7
00:00:18,120 --> 00:00:20,340
This way we want to find the circuit gain of

8
00:00:20,340 --> 00:00:25,305
this amplifier circuit and the gain is defined as the output divided by the input.

9
00:00:25,305 --> 00:00:28,020
So, let's use some,

10
00:00:28,020 --> 00:00:29,340
the node-voltage method.

11
00:00:29,340 --> 00:00:32,100
Let's first consider this node A.

12
00:00:32,100 --> 00:00:35,505
So, let's consider our currents that are going out this way.

13
00:00:35,505 --> 00:00:38,520
We know sum of those must be equal to zero.

14
00:00:38,520 --> 00:00:40,950
So, what voltage do we have here?

15
00:00:40,950 --> 00:00:43,745
V out minus what voltages over here.

16
00:00:43,745 --> 00:00:51,560
That is A times Vp minus Vn and divide that by the resistor R0.

17
00:00:51,560 --> 00:00:54,545
Now, let's consider this value right here,

18
00:00:54,545 --> 00:00:59,530
V out minus this voltage is Vn.

19
00:00:59,530 --> 00:01:04,250
We're going to divide that by R1 and that is going to be equal to zero.

20
00:01:04,250 --> 00:01:08,165
Now, let's do our node at this point.

21
00:01:08,165 --> 00:01:18,244
We can say Vn minus V out divided by R1 plus Vn minus zero divided by R2,

22
00:01:18,244 --> 00:01:25,520
plus right here, Vn minus going all the way,

23
00:01:25,520 --> 00:01:28,440
keep going right there.

24
00:01:28,440 --> 00:01:34,315
Vn minus Vp divided by Ri and this all equals zero.

25
00:01:34,315 --> 00:01:36,005
Those are the two values.

26
00:01:36,005 --> 00:01:38,900
These are the two equations and we'd call them the A equation and

27
00:01:38,900 --> 00:01:42,320
the B equation that define the two nodes that we have.

28
00:01:42,320 --> 00:01:44,060
Let's also take a look at the circuit,

29
00:01:44,060 --> 00:01:45,680
see if there's anything else that we know.

30
00:01:45,680 --> 00:01:48,875
Because at the moment, we have three unknowns and only two equations.

31
00:01:48,875 --> 00:01:51,560
We don't know V out, Vn or Vp.

32
00:01:51,560 --> 00:01:52,925
But if you look right here,

33
00:01:52,925 --> 00:01:57,695
you can see that Vp is equal to Vs. Because of the way the circuit is hooked up.

34
00:01:57,695 --> 00:02:01,290
So, let's just substitute that back in for

35
00:02:01,290 --> 00:02:06,140
Vp in each of those cases right there to make these equations a little bit simple.

36
00:02:06,140 --> 00:02:07,970
Now we have two unknowns,

37
00:02:07,970 --> 00:02:12,870
Vn and V out and we have two equations to solve them.

38
00:02:13,150 --> 00:02:20,255
So, in order to write those as a matrix equation and apply Cramer's rule,

39
00:02:20,255 --> 00:02:22,310
I've given you a little example here.

40
00:02:22,310 --> 00:02:24,470
You might want to just stop the video for a minute and

41
00:02:24,470 --> 00:02:26,720
take a look at the [inaudible] this example.

42
00:02:26,720 --> 00:02:29,630
I took the two equations right here and I converted

43
00:02:29,630 --> 00:02:32,630
them into matrix form for V out and Vn.

44
00:02:32,630 --> 00:02:34,205
Then I applied Cramer's rule.

45
00:02:34,205 --> 00:02:37,100
You may have to look that up online if you don't remember it from math.

46
00:02:37,100 --> 00:02:41,120
Let's stop the video right now and do a little bit of practice on

47
00:02:41,120 --> 00:02:46,910
this particular problem and see that you can get the answer for the gain of this Op Amp.

48
00:02:47,630 --> 00:02:50,660
Okay, I'm hoping that you spent a little bit of time

49
00:02:50,660 --> 00:02:53,120
with that example and did the non-ideal case.

50
00:02:53,120 --> 00:02:59,180
That's the last time you're going to have to do the non-ideal amplifier in this class.

51
00:02:59,180 --> 00:03:04,790
Now, let's rewrite our node a equation V out minus A times V

52
00:03:04,790 --> 00:03:12,980
n minus V p divided by R out plus V out minus Vn divided by R1 equals zero.

53
00:03:12,980 --> 00:03:16,510
Here's the KCL node that I had at node B.

54
00:03:16,510 --> 00:03:20,855
Now, let's apply our ideal Op Amp equations.

55
00:03:20,855 --> 00:03:23,585
Remember we know that Vs is equal to Vp.

56
00:03:23,585 --> 00:03:27,305
But we also know because its ideal that Vp equals Vn.

57
00:03:27,305 --> 00:03:29,949
So, if Vn equals Vp,

58
00:03:29,949 --> 00:03:31,440
then this is zero.

59
00:03:31,440 --> 00:03:35,100
Vn is Vs, Vn is Vs,

60
00:03:35,100 --> 00:03:40,230
Vn is Vs, and Vn minus Vp is equal to zero.

61
00:03:40,230 --> 00:03:43,200
Great, our A equation and our B equation,

62
00:03:43,200 --> 00:03:44,945
we've applied that simplification.

63
00:03:44,945 --> 00:03:46,885
Do we need to apply anything for the currents?

64
00:03:46,885 --> 00:03:49,095
No, so we don't need to use that information.

65
00:03:49,095 --> 00:03:50,600
But keep your eye on this one.

66
00:03:50,600 --> 00:03:53,995
Remember that the output resistance is zero. We're going to need that.

67
00:03:53,995 --> 00:03:58,220
Right here this equation gives us the idea that we could solve for V out

68
00:03:58,220 --> 00:04:02,510
very easily because the only values we have are V out and Vs.

69
00:04:02,510 --> 00:04:05,270
But remember that this value is zero.

70
00:04:05,270 --> 00:04:07,220
So, it's like we're dividing by zero.

71
00:04:07,220 --> 00:04:09,650
This is not a good equation for us to use.

72
00:04:09,650 --> 00:04:11,930
Let's use the B equation instead.

73
00:04:11,930 --> 00:04:13,370
So, for the V equation,

74
00:04:13,370 --> 00:04:18,709
I want to solve for V out and I can say right here there's my V out.

75
00:04:18,709 --> 00:04:25,995
So, I can say that V out is equal to Vs times one divided by R2,

76
00:04:25,995 --> 00:04:31,415
plus one divided by R1 and then I have a V out over R1 on the other side.

77
00:04:31,415 --> 00:04:36,200
So, if I solve this, I can say that V out over Vs and happens to be

78
00:04:36,200 --> 00:04:43,760
my gain is going to multiply this thing out R1 plus R2 over R1 R2.

79
00:04:43,760 --> 00:04:46,340
Then I'm going to bring this R1 over to the other side,

80
00:04:46,340 --> 00:04:48,100
see this R1 came over here.

81
00:04:48,100 --> 00:04:52,160
So, now my R1 cancel out and my gain which is my output

82
00:04:52,160 --> 00:04:56,210
divided by my input is R1 plus R2 over R2.

83
00:04:56,210 --> 00:05:00,320
Awesome. If we actually calculate that for the numbers that we have here,

84
00:05:00,320 --> 00:05:01,520
the value is five.

85
00:05:01,520 --> 00:05:06,860
If we had calculated it using the non-ideal case, it was 4.99975.

86
00:05:06,860 --> 00:05:11,330
So, this is a very good approximation for an ideal Op Amp.

87
00:05:11,330 --> 00:05:15,240
This is how we are going to do our problems in the future.

88
00:05:15,740 --> 00:05:19,775
So remember, that the circuit gain depends on a circuit.

89
00:05:19,775 --> 00:05:21,665
Here's the gain we just derived it.

90
00:05:21,665 --> 00:05:23,915
Remember saturation is going to take over too.

91
00:05:23,915 --> 00:05:26,015
We can't make this gain just as large as we want.

92
00:05:26,015 --> 00:05:30,940
It's controlled by the power supplies that we put on this Op Amp.

93
00:05:30,940 --> 00:05:34,450
Again, back to our Wild Horse Corral.