Virtual ground

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 (10^3) or (10^6), really high, gained a million. We're going to let the output voltage here, (V_{out}), let's say 6 volts. And you remember what's not shown here in this circuit is the power supply going to both of these op-amps plus or minus. Let's say it's plus or minus 12 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 6 volts and (a) is (10^6), then what's (V_n)? (V_n) is the difference between these two voltages here. Let's call this the usual thing; we'll call this (V_{+}) and we'll call this (V_{-}). And we know that (V_n) equals (V_{+} - V_{-}).

Now, what the question is: what is (V_n) in terms of (V_{out})? Well, (V_n = \frac{V_{out}}{a}). If we fill in the values we had: it's 6 volts divided by (10^6) or 6 microvolts. So this is 6 microvolts between here and here. Okay, so with 6 volts here, there's 6 microvolts over here.

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 down to the microvolts level. So, because I'm a practical engineer, I'm just going to say this is pretty much zero volts. If I say this is zero, that's pretty much the same thing as saying that (V_{+} \approx V_{-}).

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 (V_{+} \approx V_{-}) and apply it to this circuit over here. Now this is our inverting configuration for an op-amp. So this is (V_{+}) and this is (V_{-}) in this circuit.

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 (\frac{V_{+} - V_{-}}{10^6} = 6 \text{ microvolts}). That says that this is 6 microvolts in this direction. When we did this over here because the signs of the inputs are flipped, this was 6 microvolts this way.

So again because of the enormous gain of this amplifier, this is always going to be a tiny, tiny number. So heck, why not make it zero? 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_{+} = 0) because it's grounded. So what does that mean (V_{-}) is? Well, (V_{-}) is also zero. (V_{-}) is zero, so that point right there is at 0 volts.

Okay, so that's pretty cool. So that point is at 0 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 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." What does the word virtual mean? Well, virtual means that something is not there, but it seems like it is. So, in this case, this node is not connected to ground, but it seems like it is. So this is referred to as a virtual ground.

These two ideas say the same thing: (V_{+} = V_{-}) is always the situation around the input to an op-amp when it's running properly. In the case particularly of this op-amp configuration, where the plus terminal is connected to ground, we say that the other terminal (V_{-}) 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.