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# Inverting op-amp

## Video transcript

now I come to another configuration for an op-amp and it's partially drawn here and I'm going to talk about this as I draw the rest of this circuit in so this is going to be made from a resistor configuration that looks like this will 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 now this is upside down from what we've done 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 VN and in particular what we want to do is find an expression for V out as a function of V n alright and in this video I'm going to do it the hard way and what the hard way means is we're going to do all the algebra to do this and then you know in the next video I'm going to 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+ minus v- this is V Plus this is V minus usually the expression here is v+ minus v- and since V Plus is zero we're just going to put in minus V minus and this is equivalent to to saying that V minus equals minus V out over a so what else can we write for this circuit okay let's let's look at these resistors we have let's call this plus and minus V r1 and we'll call this one plus minus v r2 so there's a current flowing here and that will call I so I'm just going to use Ohm's law on r1 here and write an expression for I and I equals V R one 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 n minus V minus over r1 all right so that's the current going through this this guy here and now I'm going to use something special I'm going to 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 can take advantage of that what that means is that I flows in r2 so let me write an expression for I based on what I find over here based on r2 and I can write I equals let's do it it's it's V R 2 over R 2 and i can write V r2 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 alright so now we're going to set these two equal to each other now we're going to make these two equal to each other let me go over here and do that VN minus V minus over R 1 that equals this term here which is V minus minus V naught V out rather over r2 so what I'm going for here is I have how many variables do we have here we have V out we have V N and we have V minus and what I want is just V out and V in so I'm going to try to eliminate V minus and the way I'm going to do that is this expression over here we're going to take advantage of this this statement right here to replace V minus with the - V out over a so I'll do that right here so let me rewrite this it's going to be V n 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 going to do next next next I'm going to multiply both sides by a just to get a out of the out of the bottom there so now I get a times V n plus V out over r1 equals minus V out - a V out yeah over r2 there's a lot of algebra here but trust me it's going to simplify down here in just a minute all right so now I'm going to break this up into separate terms so I can handle them separately a V n over R 1 plus V out over R 1 equals minus V out over R 2 minus AV out over R 2 let's change color so we don't get bored okay next what I'm going to do is start to gather the V out terms on one side and the V n terms on the other side so that means that this V out term here is going to go to the other side so we get a the N 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 okay haven't made any sign errors yet and now let me clear the r1 we'll multiply both sides by r1 AV n equals r1 over r2 minus V not minus r1 over r2 a V not minus V not he had the are ones cancel on that last term all right AV n equals out of here I can factor this term here minus r1 over r2 times V not I can factor out of that out of here and here so I can do minus r1 over r2 V naught times 1 plus a minus V not so let's take a look at this expression and use our judgment to decide what to do next now because a is so huge that means that this first term is going to be gigantic compared to this V naught term here you know V naught is some value like 5 volts or minus 5 volts or something like that and a times this is something like a hundred thousand or two hundred thousand something like that it Dwarfs it Dwarfs this V nought so I'm going to ignore this for now I'm just going to cross that out and we'll move forward without without that little V knot on the end of the expression now this is after we've we've left that out now we have V in on this side and I'm going to take a over to the other side one plus a over a and this is a point where we get to use our judgment again again a is a huge number you know like a million and so a plus one is a million and one that fraction is really really close to one so I'm going to ignore it I'm going to 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 going to spin this around and we get V out equals minus r2 over r1 times VN 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 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 okay so this pattern with the resistor going over the top to the - 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 going to show you a really easy way to short-circuit all this and be able to do this analysis really quickly and that's called the virtual ground