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I am asking this question to understand this.

enter image description here

It states that V-/Vout is equal R1/(R1+R2)

Here :

enter image description here

V- is equal 0V. Because V+ is grounded.

So V-/Vout = R1/(R1+R2) => V- = Vout*R1/(R1+R2). But V- is equal 0V. So why this is not working ? Because either Vout is 0V or R1/(R1+R2) is 0. But I know both of them aren't equal 0.

Source : http://www.ecircuitcenter.com/Circuits/op_bandwidth1/op_bandwidth1.htm

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    \$\begingroup\$ Different question, different answear. I'm now aiming for V-/Vout. Where V- is equal 0V. Closed loop gain is equal Vout/Vin. Like here : electrical4u.com/… \$\endgroup\$
    – user331990
    Aug 6 at 12:54
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    \$\begingroup\$ in the previous one I asked for the equation I think. But now i want to understand how it works for V-/Vout. I've send the source where I've got it. I've also gave some equations I belive it is. \$\endgroup\$
    – user331990
    Aug 6 at 12:56
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    \$\begingroup\$ Read up on AN31, do some hand calculations and simulations and see it it becomes more clear: ti.com/lit/an/snla140d/snla140d.pdf \$\endgroup\$
    – winny
    Aug 6 at 13:00

2 Answers 2

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The first equation in your answer is correct! It defines the amount of feedback (beta). However - this is only one half of the truth, which means: This is only one part of the voltage which exist as V- at the inverting opamp input.

You must apply superposition when finding the correct value of this voltage V- because there are two voltage sources contributing to V- :

  1. Feedback part Vr=(beta)=Vout[R1/(R1+R2)], and

  2. Forward part Vf=Vin[R2/(R1+R2)].

Only now you are allowed to set V-=Vr+Vf=0 because both parts contribute to V- :

From this, we get:

Vout[R1/(R1+R2)]=-Vin[R2/(R1+R2)] and:

Vout/Vin=-R2/R1.

Comment 1: Because you were asking about the difference inverting/nin-inverting:

In the NON-inverting case, we have only one source (Vout) which determines the voltage at the inverting termnal (V-) because the input is connected directly to the other input termnal of the opamp.

Comment 2: The transfer function for a system with feedback is

Vout/Vin=Hf[Aol/(1+Aol*Hr)]=Hf/[(1/Aol)+Hr]. For Aol approaching infinite this reduces to

Vout/Vin=Hf/Hr.

Here Hf is a (forward) factor which contains the signal reduction in case the input signal is not connected directly at the opamp input. This is the case for the inverting configuration (EDIT: Hr also contains sign information when effectiv at the inverting input). Therefore: Hf=-R2/(R1+R2)

For a non-inverting circuit the input is normally directly conncted to the pos. input without any signal reduction - hence, Hf=1.

Therefore (non-inverting): Vout/Vin=1/Hr=1/beta.

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  • \$\begingroup\$ @user331990 - Hi, (a) This ongoing series of comments was becoming excessive and I see you have now asked your follow-up question here. Therefore you must stop asking in comments here: (a) because new questions should not be asked in comments, and (b) to avoid duplication with your new question. || I will also move these comments to chat, to avoid future readers being put-off from the answer, by the long series of comments. \$\endgroup\$
    – SamGibson
    Aug 7 at 15:00
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – SamGibson
    Aug 7 at 15:01
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Your circuit diagram is of an inverting amplifier.

However, the equation you have in your question is for a non-inverting amplifier and hence V- has some value other than 0 volts when a real signal is present.

For an inverting amplifier, \$\beta\$ is the same value i.e. R1/(R1 + R2) as explained previously: -

What is the feedback factor in the inverting Op Amp?

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  • \$\begingroup\$ Ok so as I understand V-/Vout is for non-inverting. So can you also tell me what does it say this B value ? does it say what is the voltage on R1 ? Or what exactly ? \$\endgroup\$
    – user331990
    Aug 6 at 13:09
  • \$\begingroup\$ But on this site you also mentioned : ecircuitcenter.com/Circuits/op_bandwidth1/op_bandwidth1.htm In the section inverting author also mentiones that it works with V-/Vout. But maybe my english is pretty bad. Correct me if you can please. \$\endgroup\$
    – user331990
    Aug 6 at 13:12
  • \$\begingroup\$ The bottom line is that it hasn't really got anything to do with input voltages; it's all about how much feedback is permitted by the two resistors and, remember that feedback factor is just a man-made definition. I never use it because I find it to be pointless but, in my knowledge base, I know what it is defined as @user331990 \$\endgroup\$
    – Andy aka
    Aug 6 at 13:16
  • \$\begingroup\$ And the thing is I need this feedback factor, to identify which feedback is stronger when I have 1 Op Amp and 2 feedbacks. The thing is that I read a lot about it on the internet. And I am confused because I know that V- is equal 0V. But also I know that there is that universal equation for feedback factor which was V-/Vout. But when I was calculating this V- = 0V made me go crazy. Like in the site you send me that this V-/Vout works for inverting and non-inverting. But also you said it works for inverting. And I don't know what is true and what not. \$\endgroup\$
    – user331990
    Aug 6 at 13:20
  • \$\begingroup\$ Feedback factor applies exclusively to negative feedback systems and mechanisms as far as I'm aware. I suggest that your solution might well be more easily solved by using a simulator. Or, going back to a link in a comment on your very first question, this tells you the answer I think you need: electronics.stackexchange.com/questions/112472/… \$\endgroup\$
    – Andy aka
    Aug 6 at 13:25

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