# Why does this circuit have unity gain??

Discussion in 'Homework Help' started by daviddeakin, Oct 8, 2010.

1. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
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(This isn't actually homework, but it seemed like the right forum.

Taking a perfectly ordinary inverting opamp circuit, using equal value resistors, I know the gain will be very close to unity. But why?

The universal feedback equation says:
Closed loop gain = A / (1 + AB)
Where A is the open loop gain and B is the feedback fraction.

But in the circuit above it seems to me that the two resistors form a divider with a B of 0.5, so half of the output is fed back and the gain should be roughly 2, no unity! What am I missing?

Apr 5, 2008
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3. ### Wendy Moderator

Mar 24, 2008
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Actually the gain isn't unity, it is -1. It is an inverting amplifier.

The key is the - input is a virtual ground. With the circuit powered up, if you probe the -input with an ohm meter to ground, it will read very close to zero ohms.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The undeclared assumption for the given circuit is that the open loop gain is infinite.

In reality one can write the overall gain as

$A_v=-\frac{K_2 A_{ol}}{(1+K_1 A_{ol})}$

where

$A_{ol}$

is the amplifier open loop gain.

and

$K_1=\frac{R_i}{(R_i+R_f)}$

$K_2=\frac{R_f}{(R_i+R_f)}$

Av=-Rf/Ri as the open loop gain tends to infinity.

5. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
Ok, something is starting to sink in. With that extra K2 in the numerator, I can see where A=Rf/Ri comes from. I just have to visualise where that K2 comes from in the first place. Evidently the presence of the feedback network inhibits the available open loop gain, even before you consider the feedback itself. Hmm...

6. ### Wendy Moderator

Mar 24, 2008
20,772
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You want to try an simple intellectual exercise, try the following conditions using your schematic. Input is -1V, 0V, and 1V. Remember, the inputs of the op amp (+ input and - input) will be the same voltages due to the negative feedback and the resistors are the same values. What will the output voltage be to make this so?

7. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
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I understand that, but it's a circular argument because you are starting with the assumption that NFB will always force the inputs to achieve the same voltages. That only works if the open-loop gain is extremely high. My question relates more directly to the universal feedback equation- I want to visualize where that K2 in he numerator comes from.

8. ### Georacer Moderator

Nov 25, 2009
5,151
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Use the followin set of equations to get the formula:
$\left{ \begin{array}{l} V_{out}=A_{ol} \cdot (V_+-V_{\small{-}}) \\
I \cdot R_i =V_i - V_{\small{-}} \\
I \cdot R_f = V_{\small{-}} - V_o \\
V_+ = 0 \\ \end{array} \right$

and solve for the transfer function of Vi to Vo.

I did so and found R1-R2 to the denominator of the fraction, opposed to t_n_k who found R1+R2. It's 3 o'clock here and I 'm a couple of beer cans too many for clear thought, so I keep my reservations.

9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
All op amp circuit have equal inputs, if they don't then the output is saturated one direction or another. If you ever troubleshoot something like this that basic fact (not assumption) will come in handier than formulas.

10. ### Ghar Active Member

Mar 8, 2010
655
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He is clearly aware of that idea, the problem is that it is an assumption, you just provided an example where they won't be equal.
To actually understand the mechanics of feedback this is an important discussion.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The K2 term is derived here ...

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12. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The reality is that the open loop gain is extremely high with due consideration given to the gain-bandwidth product constraint.

100dB DC gain with 1MHz (unity gain) bandwidth is respectable for a voltage driven operational amplifier.

Last edited: Oct 10, 2010
13. ### daviddeakin Thread Starter Active Member

Aug 6, 2009
207
27
And suddenly the light dawns- amazing!

I now see how it operates! Rf and Ri can be considered as two superimposed potential dividers (K1 and K2) working in 'opposite directions'. K1 works to attenuate the input voltage and K2 works to attenuate the output voltage (effectively reducing the open loop gain by a factor of K2)!
(That description probably made no sense, but it works visually in my head! )

14. ### Georacer Moderator

Nov 25, 2009
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I did the caclulations this morning and the result agrees with t_n_k. I made a mistake moving a sign last night.

I wouldn't (and didn't) use superposition for that proof, because it just didn't "feel" right for me, but since it works (me and t_n_k have the same result) it's fine with me.

Studying the finite OpAmp gain can be useful in very high frequency circuitry analysis.

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Your disquiet probably stems from a misinformed (& widely taught) belief that superposition cannot be applied to circuits with dependent / controlled sources.

The Electrician corrected this misconception on my part some time back on the AAC forum.

See this paper by a notable advocate [Marshall Leach] of the contrary viewpoint.

http://users.ece.gatech.edu/mleach/papers/superpos.pdf

Last edited: Oct 10, 2010
16. ### Georacer Moderator

Nov 25, 2009
5,151
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I had tracked that particular thread very closely, and had actually saved the link to that pdf. I know that superposition can be applied at circuits with dependent sources. It's just that sometimes I 'm not sure of the selection of equations. Be it habit or university policy, I find myself usually going back to the old Kirchoff's laws to get my set of equations. I guess it has to do with the line of the university that promotes linear equation sets, due to their application to informatics and an inclination of my studies towards robotics and control systems, rather than pure electronics.

In this particular case, I found it a bit "weird" to use two voltage dividers with different refference nodes. I also found it strange that the two voltages had opposite influence on the V- and yet they were added to get the result.