How does the feedback work in an inverting op amp amplifier?

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Thank you for your recommendation, Steve.

I think as a beginner in electronics as I am, I must first understand something basic. I belive there are more nad more interesting things waiting there for me to discover.
 
Heidi said:
Because the circuit only involves resistors, a DC voltage source and a dependent source, I think we don't have to consider the frequency influences for now.
Click to expand...
Here is the simplest way I can explain why frequency is important. Take this circuit you showed ...




Now swap the inverting input (-) and noninverting input (+) to the opamp. Now do the DC analysis. Would you not get a similar answer with the terminals swapped? Now try to build the circuit that way and see if it works. The circuit would be unstable and the output voltage would latch to the rail. Can you understand why with a DC analysis only? I think you will need to think about what happens in time to see why.
 

Two things for you to consider:

1. When using a real opamp, the input current at either the inverting or non-inverting inputs is zero (or small enough that it can be neglected. That means the current through the input resistor is equal to the current in the feedback resistor.

2. Since the gain of the opamp is almost infinite (200K is pretty close to ∞), the voltage at the non-inverting input = the voltage at the inverting input = zero.
 
OK, but give me some time. I'll have to study more op-amp chapters in my text.
 
Here is the same opamp circuit with 3 different DC input voltages and the resulting DC output voltages.
 

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OK Heidi

You are a breath of fresh air here. You don't waste Members time. And you have the likes of the very best helping you (not me). The absolute attention of Pro's.

You are an exceptional Student

Everyone wants to give you the ability to fly (succeed).
 
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For a real opamp in post #22, A(Vp-Vn)=Vo where A is the open-loop gain which is close to be infinite, Vp the non-inverting voltage which is zero, Vn the inverting input voltage.

Or

Vp-Vn = Vo/A = 0, so Vp=Vn. Correct?
 
Heidi said:
For a real opamp in post #22, A(Vp-Vn)=Vo where A is the open-loop gain which is close to be infinite, Vp the non-inverting voltage which is zero, Vn the inverting input voltage.

Or

Vp-Vn = Vo/A = 0, so Vp=Vn. Correct?
Click to expand...

Actually, (Vp-Vn)*A = Vo

but Vp=0

so (0-Vn)*A = Vo, or Vn=-Vo/A


Now go back and study post #2 in this thread.
 
MikeMl said:
Actually, (Vp-Vn)*A = Vo

but Vp=0

so (0-Vn)*A = Vo, or Vn=-Vo/A


Now go back and study post #2 in this thread.
Click to expand...
For your circuit in post #2, when OLG>1k, the circuit gets similar to a opamp, the current between two input terminals becoming neglectable, Vo/Vin=Rf/Ri , even with Ri=1k. What can it tell us about the inverting feedback? Could you please tell me specifically?

EDIT: should be Vo/Vin=Rf/Rs
 
audioguru said:
Here is the same opamp circuit with 3 different DC input voltages and the resulting DC output voltages.
Click to expand...
I can solve for all three responses because they (my teachers and my texts) tell me how to. But I still do not know where the feedback comes from.

Please let's go back to the circuit in Fig.1 in post #12. Specifically, I think the "feedback" goes on Node n is 2*Vd*(Rp/(Rf+Rp)) which is a negative value and Rp=Rs*Ri/(Rs+Ri).

Do you agree?
 
The feedback comes from the output of the opamp and develops the same amount of current in the feedback resistor as the input voltage creates in the input resistor.
Since the two resistors connect to the inverting input then if the input goes positive the output goes negative.
 
The feedback comes from the output of the opamp and develops the same amount of current in the feedback resistor as the input voltage creates in the input resistor.
Since the two resistors connect to the inverting input then if the input goes positive the output goes negative.

I cannot answer your question about the circuit in post #12 because it does not have an opamp with an extremely high voltage gain.
 

Well you can call it what you want, but it is the combination of the feedback voltage and the input voltage, which is simply the inverting terminal voltage.
If there is zero offset voltage the voltage will eventually be equal to the non inverting terminal voltage. This is never the case with a real op amp but it can be very close to zero, like 2mv, or even 20uv.


It doesnt 'really' happen this way, it takes time, that's why i suggested in my last post that if you need the time response then we have to start to include the time response of the op amp itself. A quick idea what actually happens is the input 'sees' a different voltage at t1 from what it was at t0, and that gets amplified, but the amplifier output can not get from what it was before to what it should be now in zero time, so it ramps up. As it ramps up, more and more feedback is applied to node 'n' and that little by little sums with the input voltage and eventually reaches equilibrium. Depending on some other time factors, it may reach that point smoothly or overshoot first. If the response is fairly damped, it will smoothly increase little by little. Because the input will now be constantly changing it is possible that the ramp slope decreases as it reaches the stable state, or it ramps up a lot then slows down later.
 
Thank you everyone, I appreciate all your help here. I think of another way that might help visualize how feedback is working, or exactly how feedback can be calculated. How about illustrating with a simplest circuit containing only resistors, caps, power sources or transistors (but I have very little knowledge on transistors), that it can still involve the feedback concept. Would it be a challenge?
 
An opamp is simply an amplifier with extremely high DC voltage gain. Most opamps have very low input current and some opamps have no input current. An opamp has many parts inside, look at the datasheet of an opamp to see.
The voltage gain is so high that if the output went a little too high then the negative feedback resistor causes the voltage at the inverting input to also go a little too high then the inverting voltage gain forces the output voltage to drop so that the (-) input voltage is the same as the (+) reference input voltage which is 0V.
 
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Hi again,

I think your example with the voltage controlled voltage source was a good one because one view of the op amp is as a voltage controlled voltage source. As others have suggested though i think you would do well to make the following changes and calculations:
1. Make the gain at least 100, maybe 1000 to start.
2. Make the input impedance high so that only the input and feedback resistors matter.
3. After the above changes, calculate the output voltage as well as the voltage at node 'n'.
4. Increase the gain by a factor of 10, calculate the output voltage and voltage at node 'n' again, compare to those from #3 above.

If you would like to investigate how the input offset of the op amp affects the output, add a small battery of perhaps 0.002v in series with the non inverting terminal. Do the calculations once with the positive terminal of this battery connected to the non inverting terminal, then again after you flip the battery polarity. This will show what happens with a typically small but non zero input offset voltage.

As another experiment, if you'd like to see the op amp in a different light then do the calculations with infinitely high input resistance again and non inverting terminal connected to ground, and this time view the op amp as a current operated device where the basic operation is defined by the current flowing AROUND the op amp. The current goes into the input (left side of input resistor) and that same current flows through the feedback resistor and into the output.
This shows that the same current that flows into the input also flows though the feedback resistance. The input current is Vin/Rin and the feedback current is Vout/Rfb, and the op amp (in the linear inverting mode) forces the relationship:
Vin/Rin=-Vout/Rfb
Looking at the current flow alone however makes it appear that the current Vin/Rin flows though the input resistor then through the feedback resistor and into the output terminal. So it flows from left to right looking at the schematic.
In this view the op amp is trying to put a current through the feedback resistor that is equal to the input current Vin/Rin.

Op amps typically have high gain because that makes the output set point more accurate when the op amp is used as an error amplifier, and that is what it really is used for most of the time. It amplifies the error between the input and output (times a ratio of resistances) and tries to make up for the difference by changing the output to compensate. That's the whole idea in a nut shell. Without the high gain it's just a simple voltage controlled voltage source, but with high gain it starts to look like what we call an "op amp".
 
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I'll do some calculation to see if I can make your statement more concrete.

Using the same circuit.

By superposition, the voltage Vn at Node n consists of two components Vn1 and Vn2 where

Vn1=Vin*Rp1/(Rs+Rp1) , Rp1=Ri*Rf/(Ri+Rf)

Vn2=A*Vd*Rp2/(Rf+Rp2) , Rp2=Ri*Rs/(Ri+Rs), and

Vd= -(Vn1+Vn2)= -Vn

Solving for Vn and Vo, I get

Vn=(Vin*Rp1/(Rs+Rp1))/(1+A*Rp2/(Rf+Rp2))

Vo=A*Vd= -A*Vn

Now if A is close to infinity, Vo/Vin= -(Rp1/(Rs+Rp1))/(Rp2/(Rf+Rp2)),

if Ri is close to infinity, Rp1=Rf, Rp2=Rs.

So if both A and Ri are close to infinity, Vo/Vin= -Rf/Rs.

Therefore, if the gain A and Ri approach ∞, the circuit above has the same property as an ideal op-amp does.

This is not surprising because when A and Ri are close to ∞, the circuit itself is a model of an ideal op-amp.

What about if A and Ri are not that large? Can it still show some negative feedback?

Let me do some experiments ... but sorry, now it's almost 3 am in the morning in Taipei, I need to get some sleep. I'll be back later.
 
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