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### Course: Electrical engineering>Unit 3

Lesson 1: Operational amplifier

# Virtual ground

The idea of a "virtual ground" makes some simplifying assumptions, giving us an elegant and simple way to look at op-amp circuits. Created by Willy McAllister.

## Want to join the conversation?

• at , what does he mean by the output voltage will be between +/- 12 volts , shouldn't it be 6 V ?
• Hello Balraj,

The +/- 12 volts is an assumption. Here Willy assumed the op amp is powered by a dual power supply. The upper "rail" is 12 VDC and the lower "power rail" is -12 VDC. The op-amp can operate from many voltage (with stipulation) but +/- 12 is a common value seen in the literature.

The output of the op-amp may take on any value between the rails. In this example the output will be between +/- 12 volts.

The 6 VDC is the calculated voltage for a given assumption.

Regards,

APD
• The op-amp is connected to a power supply. How can I know the exact rating (voltage and power?) that the power supply should have in order to operate the op-amp?
• Hello Feneva,

The best place to look is the data sheet for the op amp. This document will give you the design maximum power supply voltage - never go higher than this. Most manufactures will also present a typical power supply voltage.

For traditional op-amps you will see +and - power supplies. Today you will also find many single supply op amps.

Regards,

APD
• Everybody is getting confused because he explains that what we see in the real world is not exactly the same as the way we make calculations.
He constantly switches between the two without making it clear.

He should make a few videos about Ideal OP-AMPS without ever mentioning real world op amps, and after people understand that part he can go back to the real world explain how it really works.
That way people would understand why those problems were solved that way in the previous videos and appreciate it.

This way is just CONFUSING.
You go back and forth between the two.

And I see in every video people have the exact same question.
If Vin=v+ - v-
And Vout=A*Vin
Why isn't Vout=0 since Vin=0 since v+=v-
THE ANSWER IS YOU ARE USING ONE FORMULA FROM THE REAL WORLD AND ONE FORMULA THAT WE USE IN THE BOOK WORLD TO MAKE CALCULATIONS.
• Sir, where is v+ = v- applicable and where it isn't ?
• Good question. The answer may sound odd.

For any opamp circuit that does something useful, the assumption v+ = v- has to hold. That's because if it doesn't hold, the circuit is not working as intended and in fact is pretty much useless.

If the assumption is not true, and v+ - v- is some non-tiny voltage, the output of the opamp, A(v+ - v-), tries to become some huge number, like 100,000 volts. No opamp can actually do that. Instead, the output voltage becomes "pinned" to either the positive or negative power supply voltage. That is a useless circuit.

If you build an opamp circuit and you see the output voltage equals one of the power supplies, that's an indication something is wrong with the feedback connection.
• This may be exactly what Eldi is talking about below, but at about he says that Vin=V+-V-; yet, regarding the very same circuit, he says at in the non-inverting op-amp video that Vin=V+. So... which is it? Similarly, he says that Vin=Vout/A, but I thought we proved that Vin=Vout*(R2/(R1+R2))....
• Bob - I think your question is different than Eldi's question.
What you've noticed is inconsistency in how I used the variable name v_in various opamp videos. I inadvertently laid a trap for you to fall into. I apologize for that.
Let me try to explain the confusion.

In the introductory video "What is an operational amplifier?" at , the variable v_in refers to the voltage difference between the two input pins of the opamp,
v_in = v+ - v-

In the next video, “Non-inverting opamp”, at I change how I use the variable v_in. It refers to the name of the input signal coming from the voltage source on the left. The quantity ( v+ - v- ) comes up at , but it does not get its own special variable name.

In the next two videos on Feedback and Inverting Opamp v_in is used in the same way, as the name of the input signal.

The blunder comes in this Virtual Ground video. v_in goes back to being the name of
( v+ - v- ). All the info in the Virtual Ground video is correct, but I can totally see where the confusion comes from with v_in getting used differently in the sequence of videos.

Here’s what I would ask you to do… Please watch the first part of the Virtual Ground video again, and when I write v_in (two times) put your thumb over the top of it (two times). Ignore when I speak its name. Just focus on ( v+ - v- ). Does the virtual ground concept make more sense that way?

I may attempt to repair this video, but it will be hard to get released, since my fellowship at KA wrapped up in 2016. I’m continuing my teaching on my own web site, spinningnumbers.org. I'll post the repaired video there.

Thank you for pointing this out. - Willy
• if Vin is zero how is it possible for there to be an output voltage, considering Vout=AVin .
(1 vote)
• Don't forget that opamps operate with both a positive and negative power supply. The output of the opamp can swing down below ground, all the way down to near the negative supply. An output value of 0 volts is just an ordinary value near the middle of the full output range.
• If the op amp saturates v+ is not equal to v-. why?
(1 vote)
• An opamp circuit has this aspect that "it works when it works". What I mean is that is that all these circuit theories like "virtual ground" work only if all assumptions are true. One of the assumptions is that the output of the opamp is capable of achieving a voltage that allows the virtual ground idea to work. If the output of the opamp is in saturation (stuck at either the high or low power supply) then the assumption is not true and all bets are off. The two inputs, v+ and v- will not be identical.
• awesome teaching skills.just one thing how does the power supply determines the range of Vo and ive seen many take I1 equal to I2 saying that input resistance is very high so no current flows between V+ and V- and all current from R1 i.e I1 goest to R2 i.e I2
(1 vote)
• The power supply voltages of an opamp limit on how high or low the output voltage, Vo, can go. If the power supplies are +12V and -12V, that's the total possible range of Vo. Inside the opamp there are two transistors connected to the Vo pin. One transistor pulls Vo up, the other pulls Vo down. The up-pulling transistor's other terminal is connected to +12, making +12 the highest possible voltage it can cause on Vo.
• At how are we getting Vin = Vo/Va in terms of Vout? Saying in terms of Vout makes me think were solving for Vout, such as Vout = Vin/Va.. but where are we getting this equation in the first place?
(1 vote)
• This equation comes from the definition of the opamp symbol.
The opamp amplifies the difference between its two input pins.F
Give the voltage difference a name, like v+ - v- is called Vin. Then the opamp function is Vo = A Vin. (This is probably the equation you were expecting). All I did was solve for Vin. Vin = Vo/A.