If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Non-inverting op-amp

An op-amp can be used to amplify a voltage. The gain is determined by the ratio of two resistors. Created by Willy McAllister.

Want to join the conversation?

  • blobby green style avatar for user Atharva
    It the current is zero, what is the point of the circuit?
    also isnt it impossible to have 0 current when there is a voltage.
    And what does a -negative voltage even mean?
    (17 votes)
    Default Khan Academy avatar avatar for user
    • piceratops ultimate style avatar for user Nathan Joseph
      Good questions. For your first question, the current is not literally zero, but it is so close to zero that the concepts and equations work well while assuming zero current.

      Your point about the impossibility of zero current when there is a voltage is addressed by considering the power outlets in your house. The outlets have voltage, but no current will flow until there is a complete circuit, e.g. plugging in an electric razor or hair dryer.

      Your last question, about what does a negative voltage mean, requires that one way or another, you become comfortable with the convention that any voltage at a lower potential then a defined reference is negative when compared to that reference, and any voltage higher than the defined reference is positive when compared to that reference. For illustration, picture a volt meter measuring a DC power supply, when measuring the output of the power supply with the positive lead of the meter on the positive terminal of the power supply, and the negative lead of the meter on the negative terminal of the power supply, the meter will read positive voltage. If the leads are swapped so that the positive lead of the meter is on the negative terminal of the power supply and the negative lead of the meter is on the positive terminal of the power supply, the meter will read negative voltage. This is because the meter has a fixed reference, namely the negative lead. So, when the negative lead is placed on the positive (high) terminal of the power supply, the meter sees negative voltage at the negative terminal of the power supply through the positive lead. Please believe me when I say that everyone was confused by negative voltages and understanding measuring with a reference in mind in the beginning of their learning about voltages. Do not stop trying to grasp the concept, it will become more clear as you continue to learn.
      (50 votes)
  • blobby green style avatar for user kunal.anilkumar
    At , how can you say V+ is Vin ? Wasn't Vin=V+-V-?
    (9 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user APDahlen
      Hello Kunal,

      Assuming the gain is infinite then feedback will make V+ = V-.

      This is a good assumption as long as the signal has a low frequency e.g. audio. Things start to fall apart when the signals get higher.

      Let us know when you reach these more interesting levels with op-amp.

      Quick question for you - have you had the chance to build and experiment with an op-amp?

      Regards,

      APD
      (4 votes)
  • blobby green style avatar for user Atharva
    Why cant the inverting input (v-) simply be connected to v_out?
    Why do we need tgose extra resistors?
    (5 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user APDahlen
      Hi Atharva,

      Those extra resistors are used to set the gain of the op-amp stage. Without them the non-inverting amp has a gain of 1. While this is useful in certain applications there are many where you want more gain.

      For example, if you had a microphone you may want a gain of 100 or even more.

      Regards,

      APD
      (8 votes)
  • blobby green style avatar for user abhishek baijal
    There were two gains referred to, the A and the R2/(R2+R2). Could you please tell me more about them? I understand A is large and a property of the particular op-amp whereas R1 and R2 are decided by us and that which we actually use? So the resisters decide the 'gain' in the voltage? Then whats the working of A?If so could please elaborate on the difference between their workings? Is it that A does not matter as long as it's very large?
    Thank you for the video, Khan Academy has never let me down and I am immensely indebted to you. All of you provide a immeasurable service to the word!
    (5 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      This is a good question. You have it all right. Big A is the gain of the opamp itself. The gain of the whole circuit (opamp plus resistors) is the resistor ratio expression. The resistor ratio answer is valid as long as Big A is huge. A's job is to be huge enough to make it disappear from the gain expression. The value of A does not matter as long as it is very large. It is large enough when we decide that A and (A+1) are so close together we can cancel them in a fraction.

      You can explore this more by using specific values for A. Go back to the equation just before I decided to ignore the "1" right at in the video. Plug in real A values and get a "feeling" for how appropriate the cancellation is. Try A = 5 and A = 100 and A = 10,000. How much of a mistake do you make when you ignore the "1"?
      (5 votes)
  • blobby green style avatar for user festavarian2
    In the previous video you said Vin=A(Vplus-Vminus). In this video you say Vin = Vplus. I suppose this would be true if Vminus was ZERO, but here, Minus is NOT zero. Why did you label Vplus as Vin to begin with? You keep using Vin and Vplus interchangeably. Please explain
    (3 votes)
    Default Khan Academy avatar avatar for user
    • old spice man green style avatar for user Willy McAllister
      The previous video "What is an operational amplifier" gave the general equation for what an opamp does,

      V_out = A(v+ - v-)

      In this video the circuit shown has a signal named v_in coming from the input voltage source. It is connected to the v+ input of the opamp symbol. So the signal name v_in is interchangeable with the opamp pin v+.
      (3 votes)
  • hopper jumping style avatar for user Yuya Fujikawa
    At , Why can't the voltage divider equation be Vo = V in*(R2)/R1+R2 ? Why does R2 go on the numerator and not R1?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • male robot johnny style avatar for user rpiitcbme
    At , it is stated that v+ = v_in. But isn't v_in = v+ - v-? How are these two statements consistent with each other?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      In this video, v_in is the name of the voltage created by the voltage source on the left side of the circuit. It is connected directly to the v+ input of the opamp. In other opamp videos, the same variable name might be assigned a different meaning. like v_in = v+ - v- as you suggest. There are no standard rules or conventions for which signal gets to be called v_in. It all depends on the person who draws the schematic.
      (3 votes)
  • blobby green style avatar for user Julio Rentas
    @, (R1 + R2) / R2 is not 2 if R1 is not close to being equal to R2; right?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user APDahlen
      Hi Julio,

      Correct, the op-amp circuit's gain is determined by the resistors as :

      Vout = [(R1 + R2) / R2] x Vin

      There is nothing special about the gain of 2. It simply occurs when R1 = R2:

      (A + B)/B = 2 when A = B

      I'm not sure if I answered your question. Please leave a comment below if you would like to continue the conversation.

      Regards,

      APD
      (2 votes)
  • marcimus pink style avatar for user Vanessa Avila
    Is the non-inverting amplifier ALWAYS designed to have two resistors stacked on one another while connected to Vout? And how do you know the size of the resistors to use for this?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      Vanessa - Good question. The non-inverting opamp configuration pretty much always looks like the one in this video, with a feedback resistor connected between the output pin and the - input, and another resistor to ground. (Another way to do non-inverting is called an "instrumentation amplifier". That's a pretty fancy circuit.)

      Designing resistor values is a balance. Small resistors mean the opamp output will have to drive a lot of current all the time (lots of power wasted). Really big resistors (1M and above) mean the current in the resistor stack is nice and low, but you have to be wary of the tiny (but never 0) current that actually flows into the negative input pin. Even a tiny input current like 1 uA produces a significant voltage drop across a 1 MOhm resistor. That gives you errors in the gain you expect. So a good balanced approach is to select resistors in the range of 10K to 100K ohms. You can find the input current for any opamp on its data sheet.
      (2 votes)
  • blobby green style avatar for user swaraj890
    does negative voltage mean it is drawing voltage from amp.
    (2 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      To figure this out, sketch the circuit and include two batteries attached to the power supply pins of the opamp. Then assume the output voltage is some value like +1V. Trace how the current flows all the way around in a loop, starting from one of the batteries. Then set the output voltage to -1V and do it again.

      When the output voltage, Vo, is positive, current flows from the positive power supply, through the opamp to its output pin, through the resistors attached to the output pin, into the ground node, and back to the positive supply.

      When the output voltage, Vo, is negative, current flows from the ground node, up through the load resistor(s), into the opamp's output pin, then out of the opamp's negative supply pin, and back into the negative supply.
      (2 votes)

Video transcript

- [Voiceover] Okay, now we're going to work on our first Op-amp circuit. Here's what the circuit's going to look like. Watch where it puts the plus sign is on the top on this one. And we're going to have a voltage source over here. This will be plus or minus V in, that's our input signal. And over on the output, we'll have V out, and it's hooked up this way. The resistor, another resistor, to ground, and this goes back to the inverting input. Now we're going to look at this circuit and see what it does. Now we know that connected up here the power supply's hooked up to these points here, and the ground symbol is zero volts. And we want to analyze this circuit. And what do we know about this? We know that V out equals some gain, I'll write the gain right there. A big, big number times V minus, sorry V plus minus V minus, and let's label that. V plus is this point right here, and V minus is this point right here. And we also know that the currents, let's call them i plus and i minus, equals zero, and that's the currents going in here. This is i minus here, and that's i plus, and we know those are both zero. So now what I want to do it describe what's going on inside this triangle symbol in more detail by building a circuit model. Alright, and a circuit model for an amplifier looks like this. We have V minus here, V plus here, so this is V in, and over on this side we have an, here's a new symbol that you haven't seen before. It's usually drawn as a diamond shape, and this is a voltage source, but it's a special kind of voltage source. It's called a voltage-dependent voltage source. And it's the same as a regular ideal voltage source except for one thing, it says that the V, in this case V out, equals gain times V plus minus V minus. So the voltage here depends on the voltage somewhere else, and that's what makes it a voltage-dependent, that's what that means. So, we've just taken our gain expression here, added, drawn circuit diagram that represents our voltage expression for our circuit. Now, specifically over here we've drawn an open circuit on V plus, and V minus so we know that those currents are zero. So this model, this circuit sketch represents our two properties of our Op-amp. So I'm going to take a second here and I'm going to draw the rest of our circuit surrounding this model, but I need a little bit more space. So let's put in the rest of our circuit here. We had our voltage source, connected to V plus, and that's V in, and over here we had V out. Let's check, V out was connected to two resistors, and the bottom is connected to ground, and this was connected there. So what our goal is right now, we want to find V out as a function of V in. That's what we're shooting for. So let's see if we can do that. Let's give our resistors some names. Let's call this R1, and R2, our favorite names always, and now everything is labeled. Now and we can label this point here, and this point we can call V minus, V minus. So that's our two unknowns. Our unknowns are V not, V out, and V minus, so let's see if we can find them. So what I'm going to do is just start writing some expressions for things that I know are true. For example, I know that V out equals A times V plus minus V minus. Alright, that's what this Op-amp is telling us is true. Now what else do I know? Let's look at this resistor chain here. This resistor chain actually looks a lot like a voltage divider, and it's actually a very good voltage divider. Remember we said this current here, what is this current here? It's zero. I can use the voltage divider expression that I know. In that case, I know that V minus, this is the voltage divider equation, equals V out times what? Times the bottom resistor remember this? R2 over R1 plus R2, so the voltage divider expression says that when you have a stack of resistors like this, with the voltage on the top and ground on the bottom, this is the expression for the voltage at the midpoint. Kay, so what I'm going to do next is I'm going to take this expression and stuff it right in there. Let's do that. See if we got enough room, okay now let's go over here. Now I can say that V out equals A times V plus minus V out times R2 over R1 plus R2, alright so far so good. Let's keep going, let's keep working on this. V not equals A times V plus minus A V not, R2 over R1 plus R2. Alright, so now I'm going to gather all the V not terms over on the left hand side. Let's try that. So that gives me, V not plus A V not, times R2 over R1 plus R2 and that equals A times V plus, and actually I can change that now V plus is what? V plus is V in. Okay let's keep going I can factor out the V not. V not is one plus A R2 over R1 plus R2 and that equals AV in. Alright so we're getting close, and our original goal, we want to find V out in terms of V in. So I'm going to take this whole expression here and divide it over to the other side, so then I have just V not on this side, and V in on the other side. Make some more room. I can do that, I can say V not equals A V in divided by this big old expression, one plus A R2 over R1 plus R2. Alright so that's our answer. That's the answer. That's V out equals some function of V in. Now I want to make a really important observation here. This is going to be a real cool simplification. Okay, so this is the point where Op-amp theory gets really cool. Watch what happens here. We know that A is a giant number. A is something like 10 to the fifth, or 10 to the sixth, and it's whatever we have here, if our resistors are sort of normal-sized resistors we know that a giant number times a normal number is still going to be a very big number compared to one. So this one is almost insignificant in this expression down here, so what I'm going to do, bear with me, I'm going to cross it out. I'm going to say no, I don't need that anymore. So if this, if this number here, if A is a million, 10 to the sixth, and this expression here is something like one half then this total thing is one half 10 to the sixth or a half a million, and that's huge compared to one. So I can pretty safely ignore the one, it's very, very small. Now when I do that, well look what happens next, now I have A top and bottom in the expression, and I can cancel that too. So the A goes away, now this is pretty astonishing. We have this amplifier circuit and all of a sudden I have an expression here where A doesn't appear, the gain does not appear, and what does this turn into? This is called V not equals V in, times what? Times R1 plus R2, divided by R2. So our amplifier, our feedback circuit came down to V out is V in multiplied by the ratio of the resistors that we added to the circuit. This is one of the really cool properties of using Op-amps in circuits, really high-gain amplifiers. What we've done is we have chosen the gain of our circuit based on the components that we picked to add to the amplifier. It's not determined by the gain of the amplifier as long as the amplifier gain is really, really big. And for Op-amps, that's a good assumption, it is really big. So this expression came out with a positive sign, right? All the R's are positive values, so this is referred to as a non-inverting Op-amp circuit amplifier. So just to do a quick example, if R1 and R2 are the same, then we end up with an expression that looks like this V out equals R1 plus R2, R plus R over R is equal to two so the gain is two times V in. So just to do a quick sketch just to remind ourselves what this looks like, this was V in, and we had what out here? We had a resistor, we had a resistor to the ground, and this is V out. So this is the configuration of a non-inverting amplifier built with an Op-amp, the two resistors in this voltage divider string connected to the negative input. So that's what non-inverting Op-amp circuit looks like, and it's going to be one of the familiar patterns that you see over and over again as you read schematics and you design your own circuits.