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Current time:0:00Total duration:12:32

okay now we're going to work on our first op-amp circuit here's what the circuit is going to look like watch where I put 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 VN that's our input signal and over on the output will have V out and it's hooked up this way there's a resistor another resistor to ground and this goes back to the inverting input and we're going to look at this circuit and see what it does now we know that connected up here the power supplies hooked up to these points here and the ground symbol is zero volts and we want we want to analyze this circuit and what do we know about this we know that V out equals some gain will write the gain right there a big big number times V minus sorry v+ minus v- 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 is describe what's going on inside this triangle symbol in more detail by building a circuit model alright and the 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 haven't 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 is this 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 and drawn a circuit diagram that represents our voltage expression for our circuit now specifically over here we draw on an open circuit on v+ + v - so we know that those currents are zero so this model this this circuit sketch represents our two properties of our of our op-amp so I'm going to take a second here and I'm going to draw our 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 VI 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 n that's what we're shooting for so let's see if we can do that let's give our resistor some names let's call this r1 and r2 our favorite names always and now everything is labeled oh and we can label this point here and this point we can call V - V - so that's our two unknowns our unknowns are V naught 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 I know that V out equals a times V Plus - V - all right that's that's what 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 - 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 a voltage on the top and ground on the bottom this is the expression for the voltage at the mid point okay 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 get enough room okay 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 right so far so good let's keep going let's keep working on this V naught equals a times V plus minus a V naught r2 over r1 plus r2 all right so now I'm going to gather all the V naught terms over on the left hand side let's try that so that gives me V naught plus a V naught 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 naught V naught is 1 plus a r2 over r1 plus r2 and that equals AV in all right 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 naught on this side and V in on the other side this makes more room I can do that I can say V naught equals a V n divided by this big ol expression 1 plus a r2 over r1 plus r2 all right so that's our answer that's the answer that's V out equals some fun action of V in now I want to make it a really important observation here we're going to 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 now 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 1/2 then this total thing is one half of 10 to the 6 or 1/2 a million and that's huge compared to 1 so I can pretty safely ignore the 1 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 naught equals V n times what times r1 plus r2 divided by r2 so our amplifier our feedback circuit came down to V out is V n 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 could chosen the gain of our circuit based on the components that we pick 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 it's that's a good assumption it is really big so this expression came out with a positive sign right all the ours or 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 R 1 plus R 2 R plus R over R is equal to 2 so the gain is 2 times V n so just to do a quick sketch just to remind ourselves what this looks like this was V n and we had what out here we had a resistor and a resistor to 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 a non-inverting op-amp circuit looks like and it's going to be one of the familiar patterns that you'll see over and over again as you read schematics and you design your own circuits