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## Electrical engineering

### Course: Electrical engineering > Unit 3

Lesson 1: Operational amplifier# Inverting op-amp

We analyze the inverting op-amp configuration, doing all the algebra from first principles. Created by Willy McAllister.

## Want to join the conversation?

- Why is Vout crossed out if Vout = A(v+ - v-)? Doesn't the definition of Vout make it along the same order of magnitude as A?(6 votes)
- I'm not sure if i got this right, since i'm learning this now, but from what i got from the video he is comparing:

*Vout * (A + 1)*with*Vout*

Even if you use the definition Vout = A(- v-) you would be comparing:*V- * A * (A + 1)*with*V- * A*

Which is still (A + 1) times bigger than the other.(6 votes)

- what is the difference between positive and negative feedback?(2 votes)
- In positive feedback circuits, part of the output signal is
**added**to the input signal. In negative feedback circuits, part of the output is**subtracted**from the input signal. Most feedback circuits use negative feedback, because the circuit gain stays under control. Positive feedback is less common. When someone with a microphone stands too close to the loudspeaker, you might hear a loud squeal caused by positive feedback. The amplified sound from the loudspeaker goes back into the microphone and gets amplified again, even louder.(8 votes)

- Why is VR1 = Vin - V- and why is VR2 = V- - Vout ?(2 votes)
- Looking at around4:40in the video:

Let's review the definitions of the voltages at each node. The circuit has 4 nodes: Upper left, upper middle, upper right output node, and ground node. Now identify/name the voltages at each node The upper left node is at a voltage of v_in with respect to ground. The upper middle node is at some (for now) unknown voltage we will give the name v- (suggested by its connection to the v- input of the opamp). The upper right node we will call v_out because it's connected to the output of the opamp. We don't know v_out yet either, but we will by the end of the video. The voltage of the ground node is 0 V by definion, and we label it with the triangle symbol. All these voltage names are called**node**voltages.

If we look at the two resistors we can give names to the voltages across each resistor: v_R1 and v_R2. These are called**element**voltages, and they will be useful later on in the analysis. An element voltage is the difference in the voltages of the two terminals of a resistor. We've already given node voltage names to all the points in the circuit, so let's use those names to figure out the element voltages on R1 and R2 in terms of node voltages:

For R1: v_R1 = v_in - v-

For R2: v_R2 = v- - v_out

I picked a particular orientation of the + and - signs when I defined the element voltages on the schematic. That gives rise to the particular order of the subtraction. If you flip the signs on the element voltage definition, it would flip the subtractions around the other way.(7 votes)

- could you please explain the formula for current at 3.20?

why does it involve v_in - v- ?(2 votes)- By Ohm's law the current flowing through R1 is equal to the voltage drop across R1 divided by the resistance of R1. The voltage drop is computed as the difference of electric potential between the terminals of the component. The left terminal (at a higher potential) is at Vin volts, while the right terminal (lower potential) is at V_ volts. Thus, the voltage drop equals Vin - V_.(5 votes)

- When simplifying AVin = -R1/R2vo(1 + A) - vo, why didn't we simplify (1+A) by A directly?(2 votes)
- At9:50the simplification steps are to first drop the trailing vo term, and next divide through by A. We notice that (1+A)/A is very close to unity, so it can be left out of the equation. I think you are asking if it would be ok to do the divide through by A as the first step. Yes it is. Then you simplify by noticing that vo/A is a very tiny number and can be left out. I thought my order was a little bit simpler, but your suggestion is fine, too.(3 votes)

- If I'm seeing this correct a non-inverting op-amp only effects current on the Vout terminal but an inverting op-amp effects current on both Vin and Vout.

I suppose with a robust Vin available current this wouldn't matter but with a robust Vin why bother with an op-amp at all? Without robust Vin current this Vin parasitic current could be suppressed to an insignificant level with very high value resistors but then you increase sensitivity to both external interference and the stray micro-current of a real op-amp.(1 vote)- I like your term "robust". You are correct about the difference between the inverting and non-inverting opamp configurations. The non-inverting version has very high input impedance, equal to that of the opamp chip itself.

The inverting version has the input connected to R1 and then to the inverting input of the opamp. We know that when this circuit is working properly the difference in voltage between the two opamp input pins (+/-) is very small. So in the circuit shown here the right side of R1 is "virtually" connected to ground.

That means v_in is required to drive R1 connected to ground, making the input impedance = R1. (Interestingly, R2 has no impact on the input impedance.)

The question of "robustness" is a matter of scale. Suppose R1 = 10kOhm and suppose the load on the output of the opamp is 100 Ohm. The opamp presents much less load on v_in than if you asked v_in to drive the ultimate 100 Ohm load. It is a matter of degree.

If you have a really feeble v_in you probably don't choose an inverting opamp as the first stage. You would prefer to place a non-inverting opamp in that position for the reason you stated.(2 votes)

- at8:40how did he take Vo out of the fraction by multiplying in R1?(1 vote)
- What is a multistage amplifier? Please explain.

Thank you(1 vote)- A multi-stage amplifier refers to a circuit with two or more amplifiers, one after the other. If you have an amplifier with a gain of 2, connected to a following amplifier with a gain of 10, the overall gain of this "multi-stage amplifier" is 2 x 10 = 20.(1 vote)

- is A is a symbol for gain? So can I say that A = -R2/R1?(1 vote)
- So the both the inverting op amp and the non inverting op amp are on the inverting terminal, but the difference is that the inverting op amp is connected to a power source?(1 vote)
- The input to the inverting configuration connects to the inverting terminal on the opamp. For the non-inverting config the input connects to the non-inverting input of the opamp (the opamp's + teminal).

A quick glance at these connections tells you if the overall circuit is inverting or non-inverting, even before you look at the resistor connections.(1 vote)

## Video transcript

- [Voiceover] Now I come
to another configuration for an op-amp. And it's partially drawn here. And I'm gonna talk about this as I draw the rest of this circuit in. So this is gonna be made
from a resistor configuration that looks like this. We'll have a resistor on the top. And this will be v-out, as we did before. And now we have a connection like this and the connection here to ground. This terminal is the minus terminal; and this terminal is
the positive terminal. This is upside down from
what we've done so far. But now pay particular attention here, this one has the minus on top. So now we have R1 and R2. And this is v-in. In particular, what we wanna
do is find an expression for v-out as a function of v-in. All right. And in this video, I'm
gonna do it the hard way. And what the hard way
means is we're gonna do all the algebra to do this. And then in the next video I'm
gonna show you the easy way. The easy way is really fun to learn, and it really helps to see
it the hard way one time just so you appreciate the easy way. The other thing we get to do in this video is we'll do the algebra and
we'll see how this gain, we take advantage of this
gain to make some assumptions. OK, so let's go after this. Let's develop an expression
for v-out in terms of v-in. First, let's write some things
we know about v-out, OK? We know that v-out equals A times... Now, it's usually v-plus minus v-minus. This is v-plus. This is v-minus. Usually the expression here
is v-plus minus v-minus. And since v-plus is zero, we're just gonna put in minus v-minus. This is equivalent to saying that v-minus equals minus v-out over A. So what else can we
write for this circuit? OK, let's look at these resistors. Let's call this plus and minus vR1; and we'll call this one plus minus vR2. So there's a current flowing
here, and that we'll call I. So I'm just gonna use Ohm's Law on R1 here and write an expression for I. I equals vR1 over R1. Right? Another way I can write that. What's this voltage here? This is v-minus. So I can write this in terms
of v-minus, and that equals v-in minus v-minus over R1. That's the current going
through this guy here. Now I'm gonna use something special. I'm gonna use something special that I know about this amplifier. What I know about an op-amp
is that this current here is equal to zero. There's no current that
flows into an ideal op-amp. So I could take advantage of that. What that means is that I flows in R2. So let me write and expression for I based on what I find
over here, based on R2. I can write I equals, let's do it, it's vR2 over R2. And I can write vR2 as: v-minus minus v-out over R2. All right, so I took
advantage of the zero current flowing in here to write
an expression for current going all the way through. All right. So now we're gonna set these
two equal to each other. Now we're gonna make these
two equal to each other. Let me go over here and do that. V-in minus v-minus over R1. That equals this term here, which is v-minus minus v-not, v-out rather, over R2. How many variables do we have here? We have v-out, we have
v-in, and we have v-minus. And what I want is just v-out and v-in, so I'm gonna try to eliminate v-minus; and the way I'm gonna do that
is this expression over here. We're gonna take advantage
of this statement right here to replace minus v-out over A. So I'll do that right here. So let me rewrite this. It's gonna be v-in minus minus v-out over A, so I get to make this a plus, and this becomes v-out over A all divided by R1. And that equals... V-minus is minus v-out over A minus v-out over R2. All right, let's roll down a little bit, get some room, and we'll keep going. What am I gonna do next? Next, I'm gonna multiply both sides by A, just to get A out of the bottom there. So now I get A times v-in plus v-out over R1 equals minus v-out minus Av-out over R2. There's a lot of algebra
here, but trust me, it's gonna simplify down
here in just a minute. All right, so now I'm gonna break this up into separate terms so I
can handle them separately. Av-in over R1 plus v-out over R1 equals minus v-out over R2 minus Av-out over R2. Let's change colors so we don't get bored. Next, what I'm gonna do is start to gather the v-out terms on one side and the v-in terms on the other side. So that means that this v-out term here is gonna go to the other side. Av-in over R1 equals, let's do minus v-out over R2 minus Av-out over R2. And this term comes over as minus v-out over R1. Haven't made any sign errors yet. Now let me clear the R1. We'll multiply both sides by R1. Av-in equals R1 over R2 minus v-not minus R1 over R2 Av-not minus v-not. Yeah, the R1s cancel on that last term. All right. Av-in equals. Out of here I can factor this term here. Minus R1 over R2 times v-not, I can factor that out of here and here. So I can do minus R1 over R2 v-not times one plus A minus v-not. So let's take a look at this expression and use our judgement to
decide what to do next. Now, because A is so huge, that means that this first
term is gonna be gigantic compared to this v-not term here. V-not is some value like five volts or minus five volts or
something like that. And A times this is something like 100,000 or 200,000, something like that. It dwarfs this v-not, so I'm
gonna ignore this for now. I'm just gonna cross that out, and we'll move forward
without that little v-not on the end of the expression. This is after we've left that out. Now we have v-in on this side. And I'm gonna take A
over to the other side. One plus A over A. And this is a point where we
get to use our judgement again. Again, A is a huge number, like a million; and so A plus one is a million and one. That fraction is really
really close to one, so I'm gonna ignore it; I'm
gonna just say it's one. So we'll send this one to one. And let me roll up a little bit more, just to have a little bit more room. Now what we have is what? V-in equals minus R1 over R2 times v-out. And I want the expression
just in terms of v-out, so I'm gonna spin this around, and we'll get v-out equals minus R2 over R1 times v-in. So this is what our
op-amp is doing for us. It basically says v-out is the ratio of two resistors times v-in. The gain of the overall circuit is determined by the ratio
of those two resistors. And very importantly, there's
a minus sign in front of it. The minus sign came all the way through. Again, let me sketch
the circuit real quick. We had the minus sign on top. There's v-out. And there was R1 and R2. And the positive input,
the non-inverting input, was connected to ground. OK? So this pattern with the resistor going over the top to the minus, this is called a inverting op-amp. And this is a really familiar
pattern in op-amp circuits. You can see these on schematics, and you'll be designing these on your own. So this was quite a bit of algebra it took to get down to this point, and in the next video I'm gonna show you a really easy way to shortcircuit all this and be able to do this
analysis really quickly; and that's called the virtual ground.