Understanding the Ideal Op-Amp Negative Feedback Loop

[CoolDude420]

Homework Statement

Hi,

Here is what I know about the Ideal Op-Amp.

- The open loop voltage gain is infinite
- The output voltage is given by the following $$v_o = A(v_+ - v_-)$$
- Only with a negative feedback loop is $$ v_+ = v_-$$

My query is with regards to the negative feedback loop as shown below:

My queries are as follows:

1. Clearly, $$v_o = A(v_+ - v_-)$$ should still apply and since $$ v_+ = v_-$$, shouldn't the output voltage $$ v_o = 0 $$ always?
2. Since $$v_o = A(v_+ - v_-)$$ should still apply, is A still the open-loop voltage gain which for an ideal op-amp is infinity. Thus, would the output voltage always be infinity?

Homework Equations

The Attempt at a Solution

 

[berkeman]

Post a typical opamp configuration so we can talk you through this. The output current flows back through the feedback circuit through the virtual ground node at the - input. There is a voltage drop from the output of the opamp to the - input.

 

[CoolDude420]

Post a typical opamp configuration so we can talk you through this. The output current flows back through the feedback circuit through the virtual ground node at the - input. There is a voltage drop from the output of the opamp to the - input.

Okay, I've adjusted my question. Now, after some googling I found this

from here
https://ocw.mit.edu/courses/electri...ent-spring-2006/lecture-notes/22_op_amps1.pdf

And that indicates to me that Vi = 0 which means that V+ = V- always not just for negative feedback? I'm so confused. The more I google, the more confused i get.

 

[CoolDude420]

So basically, I've managed to understand most of it I think now but I have one question. Clearly from this argument,

The differential input voltage is zero and thus inverting and non-inverting voltages should be equal. Why do people then say that only in a closed loop configuration(with negative feedback) is the differential input voltage zero?

 

[LvW]

Why do people then say that only in a closed loop configuration(with negative feedback) is the differential input voltage zero?

Because this simply is the RESULT of negative feedback - applied to an amplifier having IDEAL properties (infinite open-loop gain Aol.)
For each real amplifier (finite Aol) we have a finite Vdiff=Vout/Aol.
For a better understanding (WHY and HOW this happens) you must study in detail the various effects and the benefits of negative feedback.

 

[donpacino]

Why do people then say that only in a closed loop configuration(with negative feedback) is the differential input voltage zero?

In order for the difference to be zero the op amp gain has to be inf.
Remember that the output is always equal to the difference between the two inputs times the op amp open loop gain, like you already said.Take the attached circuit. There is a constant input voltage of V+=1.
We know that there is no current through the resistor in an ideal op amp.
So V- = Vo

We also know Vdiff = (V+) - (V-)
and Vo = Vdiff * Ao

For each gain setting there is an equilibrium state where V- and Vo must be equal to each other. It's like solving a system of equations, there is only one possible solution that exists, the voltages are driven to a set value.
Lets look at how the output (and therefore v-) changes with open loop gain.

remember v+ = 1
Ao = 1: Vo = V- = 0.5. This means the difference between V- and V+ is 0.5
Ao = 2: Vo = V- = 0.6666. This means the difference between V- and V+ is 0.3333
Ao = 10: Vo = V- = 0.909. This means the difference between V- and V+ is 0.091
Ao = 100: Vo = V- = 0.99. This means the difference between V- and V+ is 0.001
Ao = 10000: Vo = V- = 0.9999. This means the difference between V- and V+ is 0.0001

This continues as Ao increases. Once Ao gets high enough, the difference is for all practical purposes 0, which is why these op amps are designed to have very high gains, and why an ideal op amp has a gain of inf.

 

[LvW]

CoolDude - in his foregoing contribution donpacino has mentioned the term "equilibrium".
Here comes an example (step-by-step sequence) how such an equilibrium can be reached automatically:

To answer your question it helps to analyze what may happen after switch-on the power supplies +/- Vs=+/-10V. More than that, you have to apply negative feedback (resistive netork between output node and inverting input).
(The described timely sequence may be somewhat „formalistic“ - however, it helps to understand the feedback concept).
Example: Non-inverting gain stage with desired gain of "+2". That means: Feedback factor k=0.5 using a voltage divider with two equal resistors
Open-loop gain (assumption): Aol=1E4.

(1) t=0: Apply at an input voltage Vin=1V. The opamp is not yet working in its linear range (feedback not yet active due to time constants within the circuit) and the output will immediately jump to Vs=+10V.
(2) t>0: The voltage at the inverting terminal will rise to 0.5Vs=5V>Vin=1V. Hence, the voltage at this inverting terminal dominates (is larger) and the output voltage will change in the direction to minus 10V.
(3) However, on its way to -10V the ouput voltage is crossing a positive value which produces at the inverting terminal a feedback voltage of +0.99980004V .
(4) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region because the diff. voltage is Vdiff=Vin-0.99980004=1-0.99980004=0.00019996V.
As a result, the output voltage is Vout=Vdiff*Aol=0.00019996*1E4=1.9996001V.
(5) This is a stable equilibrium because: the classical feedback formula for a finite value of Aol also gives the output voltage Vout=Vin*[1E4/(1+0.5E4)]=1.9996001 V.
(6) That means: We have an equilibrium because the output voltage has a value which exactly meets the condition Vout=Vdiff*Aol.
Any larger/smaller output voltage causes a small reduction/increase for Vdiff thereby correcting this deviation from the equilibrium.
(7) In this example, the input difference voltage, of course, is NOT zero. It never will be zero - however, the diff. voltage is so small (in our case app. 0.2mV) that in can be neglected (assumed to be zero for calculations) in many cases.

 

[Tom.G]

For a different explanatory approach, the below is a copy of:
https://www.physicsforums.com/threads/opamps-some-basic-questions.924272/#post-5833477

Here is the image @jim hardy posted:

Let's try an approach that some people may be more comfortable with. We will be using the Inverting Amplifier circuit on the left.

The basic starting assumptions are that the op-amp has very high gain, perhaps 100 000, an input impedance high enough to be ignored, and an output impedance low enough that it too can be ignored.
It is supplied with equal positive and negative supply voltages whos common point is circuit common ("ground").

Just as an example set both R1 and R2 to 10kΩ and turn on the power. Vin is not connected to anything. Since the op-amp has gain, any voltage difference between the input pins will show up at the output.

Now assume that the op-amp is NOT ideal and has a small input offset voltage, perhaps making the Inverting input 1mV above the Non-Inverting input. With the op-amp gain being 100 000, this would try to drive the output to -100V. But keep in mind that the output is connected to Inverting input through R2. When the output gets to -1mV it has canceled the +1mV of offset at the input. That's also why the inputs are often referred to as being "at Virtual Ground", because the Inverting input is driven to match (very closely) the Non-Inverting input.

That's the long explanation of your quote in the OP: "Rule: The output always acts to make the input voltages the same." And it holds when there is a net negative feedback in the circuit, i.e when the output is fed back the Inverting input.

Now it's time to complicate matters by considering what happens when there is a voltage applied at Vin. Here it is easier to explain if we consider the currents through R1 and R2. As we saw above, due to the negative feedback via R2, the Inverting input is driven to match the Non-Inverting Input.

Lets start with the power off. Apply a voltage, let's say +1V, to Vin, which applies a voltage to Inverting input.

Now turn on the power. As described above, the feedback from the output through R2 to the Inverting input, will tend to match the voltage at the Inverting input to the voltage at the Non-Inverting input. To do this, there must be the same amount of current flowing thru R2 as there is flowing thru R1, but of the opposite polarity. You now have an Op-Amp circuit with a gain of minus one (-1). The output is at -1V.

If you want a higher gain, you increase the ratio of R2/R1. For instance for a gain of three, R2/R1 would be 3; you could change R2 to 30kΩ or change R1 to 3 333Ω. This means the output voltage must be 3 times Vin to drive a matching, compensating current thru R2.

So there is a version of OP-AMPS 101. Hope it helped.

Cheers,
Tom

 

[jim hardy]

I remember well my revulsion when first presented with operational amplifiers.
Infinite gain ?
Output with zero input ?
Sounded preposterous to this young vacuum tube audio guy.

Perhaps it's worth mentioning that the very word "Operational" is significant.
It really applies to the circuit of which the amplifier is part.
But as terminology evolved , it came to be applied to amplifiers built for service in such circuits and that's unfortunate because it confuses beginners.

An operational amplifier circuit performs some mathematical operation.
We like to start out with simple ones that just multiply by a constant, maybe even as simple as multiply by +1 or -1..
We can build such a circuit by using a high gain DC differential amplifier in a circuit.
The circuit must allow the amplifier to force its input voltages equal to one another.
It is the duty of the designer to wrap the amplifier in such a circuit. Next paragraph tells why.

Now here's the practical rub
with infinite or even very high gain our intuition says output should be free to approach infinity.
But since the amplifier has only finite supply voltage , typically + and - 15 volts or less, it can't make more output voltage than that.
Hmmm.
That means the difference between Vin+ and Vin- must be less than : ( whatever is Vsupply) / Av_o
Since Av₀ is such a big number perhaps 10^⁵ ,
the difference between Vin+ and Vin- must be a very small number, like maybe 12 /10^⁵ = 0.00012 with a 12 volt Vsupply .
Else the circuit is not performing its mathematical operation.
Because of the high gain we round off the term (difference between Vin+ and Vin-) to zero and just say 'the opamp holds its inputs equal"

That's what trips up most beginners.
The circuit only performs its math operation so long as it is operating an a range where it can hold its inputs equal(well, very nearly equal) .
Outside that range it's no longer 'operating' .

It is difficult to rein in our thinking to accept that the practical limit on output applies to the input, because physically it doesn't.
It applies only in the mind of the person analyzing the circuit - there's an implicit agreement to not overdrive the inputs to the point the amplifier can't balance them.

When your brain becomes comfortable with that train of thought you have made a good start.

old jim.