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
Current time:0:00Total duration:5:58

Video transcript

I want to take a look at our two op-amp circuits and make an interesting observation about how these things are behaving when they are working properly when they're hooked up right there's something these things do that is really helpful and makes life simple for us let's let the gain of our op-amp be 10 to the 3rd or 10 to the 6th really high gain a million and we're going to let the output voltage here V out let's say six volts and you remember what's not shown here in this circuit is is the is the power supplies going to both of these op amps plus or minus let's say it's plus or minus twelve volts those power supplies are implicit they're not shown in the diagram but we know they're there all right now if V out is six volts in a is ten to the sixth then what's V n VN is the difference between these two voltages here let's call this the usual thing we'll call this V plus and we'll call this V minus and we know that V n we know that V n equals V plus minus V minus and now what the question is what is V in in terms of V out well V n equals V out divided by a if we fill in the values we had it's six volts divided by 10 to the sixth or six micro volts so this is six micro volts between here and here okay so with with six volts here there are six micro volts over here okay this is a really small voltage in order for this op-amp to have an output voltage that stays between plus or minus 12 volts this voltage over here has to be really small it has to be done in the micro volts level so because I'm a practical engineer I'm just going to say this is pretty much zero volts and if I say this is zero that's pretty much the same thing as saying that V Plus approximately equals V minus so that's a little observation we're going to make right there so in this circuit when it's working right these two voltages are pretty much the same so let's take this idea of e+ pretty much equals v- and apply it to this circuit over here now this is our inverting configuration for an op-amp so this is V Plus and this is V minus in this circuit and let's do the same analysis that we did before if this is V out and if V out is 6 volts that means that V plus minus V minus divided by 10 to the 6 equals 6 micro volts and that says that this is 6 micro volts in this direction when we did this over here because the signs of the inputs are flipped this was 6 micro volts this way so again because of the enormous gain of this amplifier this is always going to be a tiny tiny number and so heck why not make it zero so if I treat this as zero what it means is I'm going to go right in here and I'm going to change this to zero volts so let's make a couple more observations okay right now it says right here V plus equals zero because it's grounded so what does that mean v minus izz well V minus is also 0 V minus is zero so that point right there is at zero volts okay so that's pretty cool so that points at zero volts now is it connected to ground it's not connected to ground but it's zero volts because of what this op-amp is doing it for us this op-amp is making sure by this feedback path that this node is always next to this node and that means it's always zero there's a really cool word that we use for this and the word is virtual you and what is the word virtual mean well virtual means that something is not there but it seems like it is you so in this case this note is not connected to ground but it seems like it is so this is referred to as a a virtual ground so these two ideas say the same thing V plus equals V minus is always the situation around the input to an op-amp when it's running properly and in the case particularly of this op-amp configuration where the plus terminal is connected to ground we say that the other terminal V minus is at a virtual ground or is a virtual ground in the next video I'm going to go back and do this inverting configuration of the op-amp I'm going to do the analysis again with this idea of a virtual ground and it's going to be really easy compared to doing all that algebra