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### Course: Electrical engineering>Unit 3

Lesson 1: Operational amplifier

# Virtual ground - examples

Using the "virtual ground" idea, we reexamine the inverting op-amp and find a solution much quicker. We apply the virtual ground method to a unity-gain buffer (amplifier with gain = 1). Created by Willy McAllister.

## Want to join the conversation?

• What is a unity-gain buffer used for?
• Hello Josemiguel,

Short answer - a unity-gain buffer transforms a high impedance device to a low impedance device.

Long answer: Consider the audio amplifier. Here the output of a typical amplifier serves as a unity gain buffer. For sake of discussion assume the amplifier has two stages. In the first stage the voltage is increased to a level sufficient to drive the loudspeaker. Unfortunately this 1st stage has no muscle. It is unable to provide the current required to drive the loudspeaker. We now add the unity -buffer (second stage). It has no voltage gain but it is able to provide additional current.

We would say the 1st stage had a high output impedance. When we add the unity buffer we now have a low impedance output.

Regards,

APD
• What's a "hand" of volts?
• One "hand of volts" is a personal slang I use. I guess I could have said, "a handful of volts" to mean the same thing while being a bit clearer.
(1 vote)
• Can't we use virtual ground to analyze the non-inverting op-amp?
• Yes, you can use the idea of a virtual ground to figure out the non-inverting configuration of an opamp. Virtual ground works for every opamp circuit I've come across.
• The professor on op amps is brilliant, can I know his name?
• What is the use of an amplifier if it doesn't even get to amplify? I mean it can amplify around 10^5 but it can't go above the supply voltage meaning at the end it wasn't amplified?
• it amplifies the incoming signal, not the supply voltage.
• In case of buffer Vo=Vin. Sowhat is the need of Op-Amp there? Where does this buffer find an applications?
Thanks
(1 vote)
• I do not understand the last example of this video. According to you, Vout=Vin, so according to the known equation that Vout=A((V+)-(V-))=(Vin-Vout)=0, and consequently Vin=Vout=0:((. Could you explain this? Thank you.
• "Virtual ground" is a very useful approximation for figuring out how an opamp amplifier circuit works. The gain of opamps is very high, like 10k or 100k or more. When you work out the gain expressions you get gain terms like A/(A + 1). A is so big you can treat this fraction as 1. 50,000/50,001 pretty much equals 1. The overall effect is that it seems like the two input terminals have the same voltage, and you can make up an idea like "virtual ground". It's only an approximation (a very good one), so in fact the two input voltages are very slightly different, like by a microvolt or two.

When you make the virtual ground approximation the analysis of opamp circuits becomes simpler. However, once you make the assumption it is not fair to go back and use the defining equation for the opamp any more, Vo = A(v+ - v-). If you do, you confuse yourself.

Adopting the approximation is like a 1-way door you go through in your understanding of opamps. We start out teaching the hard way so you know what's going on, but then use this trick to simplify.
• The question is about using op-amp as a filter.
When it is a single power supply mode, we can prevent the negative signal from being attenuated by adding a voltage divider at the V+ input.
But won't the signal be removed by the capacitor? Since the capacitor blocks dc and allow ac to pass?
• What if ı say R2=2R1 and Vin=10sinwt. it means gain its -2 so Vo should be -20sinwt right. Also we have V+,V- and its cannot bigger than 15V ? What should ı do, what can ı say for Vo. I have one more question is there any situation for Virtual ground not equal to 0?
(1 vote)
• If you place limits on the opamp power supply to +-15V then the output cannot exceed those limits. Your output can be described as a "clipped sinewave with frequency w".

The nickname 'virtual ground' comes from analyzing this circuit with the + input of the opamp tied to ground. In this case, the - input to the opamp is "virtually" ground by reason of the feedback loop. If you make the + input another voltage the concept still works, the - input is 'virtually' the same value as the + input. We don't have another name for this case, it is still called a virtual ground when talking to another engineer.
(1 vote)
• Heres he hung right, its troubling me that we didn't addup R1 and R2 like in nodal analysis where from vo to v1 i = (vo-vin)/(R1 + R2)
(1 vote)
• The equation you have written is true. It has three variables, i, Vi, and Vo.

Your goal is to write a gain equation with the ratio of Vo/Vi on one side and whatever else on the other side. You have to figure out how to eliminate the i term and replace it with just voltage and/or resistance terms. What do you think you might do next?
(1 vote)

## Video transcript

- [Voiceover] Alright, so now I'm gonna do the analysis of this op-amp configuration again, and I'm gonna do it using the idea of virtual ground. And the idea of a virtual ground actually, it makes really short work of analyzing a circuit like this. To review the virtual ground idea, it says if this voltage here, V out, is in a reasonable range between the power supplies of its op-amp, and remember the power supplies are not shown here but they're connected up to the op-amp, so this is a number that's sort of like one hand of volts, or two hands of volts, between zero and plus or minus 10 volts, something like that. And the gain of this op-amp is huge, it's just gigantic. It's up in the 100s of thousands or millions. And so, whenever there's a normal, finite size voltage here, it means the voltage here is about zero, it's very close to zero, it's down in the very small microvolts. And, the idea of a virtual ground, says that if this node here is at zero volts, which we've drawn it grounded, so it's at zero volts, that means that this minus terminal is gonna be also held at zero volts, very close to zero volts. And that makes this virtually a ground, or a virtual ground. Another way to write this is to say that the V plus is approximately equal to V minus. Now, my professor who taught me this, had a little symbol for it, what he did was he drew it like this, so you can sketch that on your schematics, just a little symbol there, and that symbol will help remind you that those two nodes are at the same voltage. Okay, so let's do the analysis again, we had, this was R1, and this is R2, and we have a current flowing here, called i. And we can write an expression for i, so the voltage on this side... The voltage on this side of the resistor is V in, and the voltage on this side of the resistor is, it's zero volts, it's zero volts, because we have a virtual ground here, so we know the voltage on both sides. So I can say right away, i equals the voltage difference V in minus zero, so just V in, divided by R1. So we have an expression for i. Now, as you remember, the thing we also know about op-amps is that this current is zero here, there's no current going into the input of an ideal op-amp. A real op-amp there'll be a really tiny current there. For our purposes right now we can treat this as zero current going in here, so what does that mean? That means that i goes through R2. So let's write an expression for i going through R2, and we need to know the voltage on each side of R2, well this side, this is V naught, or V out. This side is V out, and this side, again, we get to use the zero, so i through R2 equals, let's get the sign right, so the current's going in this way, so this is the positive, this is the relatively positive side and this is the negative side, so i is zero minus V out, divided by R2. Now, let's set these two currents equal to each other, so that means that V in over R1 equals minus V out over R2, and what we want is an expression that tells us what V out is in terms of V in, so V out equals V in times, let's see, R2 goes on top, R1 goes on the bottom, and there's a minus sign. And we did it, and that's the expression for V out in terms of V in, for the inverting op-amp configuration. Now, by using the virtual ground idea, the analysis became really simple, it's basically one two three steps to get to the answer. Compare that to the algebra that we did in the previous video when we did this from first principles. So that's an example of applying the virtual ground idea to analyze an op-amp circuit. I'll do one more real quick. So here's a different op-amp configuration, we haven't seen this one before. First thing to notice, first thing to notice is now plus is on top. Always take a peek at your op-amp to see which terminal is on top. We have V in here, as usual, and here's V out, and again, what we wanna do is find out what's V out in terms of V in. Now this circuit has no resistors in it, okay, that's alright, we'll figure out what's going on here. Okay, let's apply the idea of a virtual ground, so I'm gonna make my little virtual ground symbol here, like that, that reminds me that these two terminals are the same voltage if we're doing the right thing, so let's figure out what V out is. V out equals this terminal here, and this one is pretty much equal to this one, and what's this one? This is V out, sorry V in. So, boom. We just did it, in one step, that's the answer. This configuration is called the buffer. Or it's also called a unity-gain buffer. This is just a long way to say the gain is one. So that's an example of a unity-gain buffer, and we use the virtual ground idea to analyze it almost instantly.