Finding closed loop gain in ideal op amp

AI Thread Summary
To find the closed-loop gain of an ideal op-amp in a negative feedback configuration, the feedback factor β must be determined using resistors R3 to R6. The load resistor RL does not affect the gain due to the ideal op-amp's zero output resistance. The overall gain can be calculated using the relationship G = α/β, where α is derived from the resistors R1 and R2. The discussion emphasizes the importance of applying the voltage divider rule and understanding the potentials at the op-amp inputs. Mastery of these concepts will enhance confidence in solving similar problems in the future.
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Hello there, this isn't necessarily a homework question. I came across this qn during revision and didn't know how to solve it. Any help would be greatyly appreciated.

The answer is 0.97

1. Homework Statement

For the ideal OP-Amp in negative feedback configuration shown in Figure Q30, calculate the closed-loop gain, AVCL=VO/VS= ________. Assume R1=10 KΩ, R2=5 KΩ, R3=2 KΩ, R4=2 KΩ, R5=5 KΩ, R6=1 KΩ, RL=10 KΩ.

Homework Equations


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The Attempt at a Solution


Im unsure how to deal with the bottom bit, the resistors R3 to R6 which is connected to the inverting input. Do i try and find the Req for that bit and then treat it as Req is the feedback resistor? How would I proceed to attempt this question?

Thank you for all your help.
 

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What you need is the feedback factor β which depends on R3...R6 only.
Forget RL because this is a load resistor only - and does not has any influence on gain (because the ideal opamp has zero output resistance - an ideal voltage source).
The feedback factor β is defined as the voltage ratio β=Vin-/Vout (Vin- is the voltage at the inverting input node).
The overall gain is according to H. Blacks formula:
G=α*Aol/(1+β*Aol)=α/[(1/Aol)+β].
For infinite open-loop gain Aol (ideal opamp) we have:
G=α/β.
(The feedforward factor is simply α=R2/(R1+R2)
 
LvW said:
What you need is the feedback factor β which depends on R3...R6 only.
Forget RL because this is a load resistor only - and does not has any influence on gain (because the ideal opamp has zero output resistance - an ideal voltage source).
The feedback factor β is defined as the voltage ratio β=Vin-/Vout (Vin- is the voltage at the inverting input node).
The overall gain is according to H. Blacks formula:
G=α*Aol/(1+β*Aol)=α/[(1/Aol)+β].
For infinite open-loop gain Aol (ideal opamp) we have:
G=α/β.
(The feedforward factor is simply α=R2/(R1+R2)

H.Blacks formula and Feedforward factor arent in my course syllabus.
How do I find the feedback factor using R3 to R6? I am aware that Vo/Vs=1/β for non inverting input
 
whatphysics said:
H.Blacks formula and Feedforward factor arent in my course syllabus.
How do I find the feedback factor using R3 to R6? I am aware that Vo/Vs=1/β for non inverting input
Yes - but the signal is applied not directly to the non-inv. input but through a diver with the factor α.
Blacks formula can be found using wikipedia.
The feedback factor contains 4 resistors only.
It shouldn`t be a problem for you to apply the voltage divider rule twice.
 
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whatphysics said:
H.Blacks formula and Feedforward factor arent in my course syllabus.
You can also approach the problem using standard circuit analysis and the simplifying properties of the ideal op-amp (you should know what these are). Start by finding the potentials at the op-amp inputs in terms of Vs. Then consider what currents can flow and where in the feedback network.
 
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Not knowing your level of expertise
remember what i told you in the EE thread?

Since they gave you values for all resistors it is just an arithmetic exercise to figure voltages at + and - inputs.
Set them equal and rearrange that equation to give Vo/Vs.

Opampforanime.jpg

A question well phrased is half answered.
"What's V@+input?" is easy, by inspection you know what it is.
Can you figure out Vintermediate and V@-input by plain old step by step simplifying that R3,4,5,6 string?

Feedback Factor and Black's formula become intuitive after you've worked a few hundred problems like this one .
I think you need to start out simpler. What before why.
When you've got confidence that your arithmetic works, you should use it to prove Black's formula.
Then you'll have confidence in that powerful tool, too.

That's how i learned, anyway.

Good luck in your studies.

old jim
 
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