How to find the gain of an inverting amplifier in terms of feedback?

Question

I have been struggling to answer this question from The Art of Electronics where it asks me to find the closed loop gain of a non-inverting amplifier in terms of feedback and says that the derivation is straight-forward while it seems like anything but that to me.

I can understand that (Vin-B*Vin) is equal to the voltage V, but I don't understand how the (1 + AB) part of the equation is derived.

Figure 2.89. Input and output impedances for […] inverting amplifier.

It is a straightforward exercise to derive an expression for the closed-loop voltage gain of the inverting amplifier with finite loop gain. The answer is $$G = −A(1 − B)/(1 + AB)$$

Answer 1

I looked at this circuit quickly by applying superposition and my results match simulation results quite well, involving the input and output resistances:

If you now neglect the contribution of \$r_i\$ and \$r_o\$ (infinite input resistance and zeroed output resistance), then the formula simplifies and matches the elegant factorization proposed by the book. It is easy to do when you leave paralleled terms in the || form without expanding them. I leave it to you to rearrange the formula the right way.

Answer 2

I once had a go at answering that challenge from H&H. This is what I came up with ...

Answer 3

I love the fact that you got a few answers that developed from theory to quantitative results. Nice.

Let me start just a little differently.

The factor, \$B\$

If you look at the output of the opamp (ignore, for now, the amplifier's \$r_o\$, which is likely small by comparison) and if you treat the input source as a low-impedance (zero, really) voltage source, then changes at \$V_{_\text{OUT}}\$ are divided down by the resistor divider pair, \$R_1\$ and \$R_2\$, such that the feedback into the input node (shared by both resistors) is the divided part: \$V_{_\text{OUT}}\cdot\frac{R_1}{R_1+R_2}\$. So:

$$B=\frac{R_1}{R_1+R_2}$$

is just the NFB factor -- the amount of \$V_{_\text{OUT}}\$ that is fed back to the input.

Now, this isn't strictly true, because there is also \$r_i\$. And I completely discounted that. So the simplification of \$B=\frac{R_1}{R_1+R_2}\$ only applies if you can discount things and say that \$r_o \ll R_2\$ and \$r_i\gg R_1\$, so that both \$r_o\$ and \$r_i\$ can be neglected.

The output impedance, \$Z_{_\text{OUT}}\$

Now, looking into \$V_{_\text{OUT}}\$ to see what impedance might be seen, it would seem to be mostly related to \$r_o\$. It's true that the sum, \$R_1+R_2\$, is also seen in parallel. So keep that in mind. But let's just focus on \$r_o\$ for now. The schematic looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

(\$A_{v_o}\$ is the open-loop voltage gain of the opamp.)

If we know the current as a function of \$V_{_\text{OUT}}\$ then we can divide \$V_{_\text{OUT}}\$ by that current to find the impedance. This is:

$$\begin{align*} Z_{_\text{OUT}}&=\frac{V_{_\text{OUT}}}{\left[\frac{V_{_\text{OUT}}-\,\left(-A_{v_o}\,\cdot\,V_{_\text{OUT}}\,\cdot\, \frac{R_1}{R_1+R_2}\right)}{r_o}\right]} \\\\ &=\frac{1}{\left[\frac{1-\,\left( A_{v_o}\,\cdot\,\frac{R_1}{R_1+R_2}\right)}{r_o}\right]} \\\\ &=\frac{r_o}{1+\,\left( A_{v_o}\,\cdot\,\frac{R_1}{R_1+R_2}\right)} \\\\ &=\frac{r_o}{1+\,A_{v_o}\,\cdot\,B} \end{align*}$$

So, it's not too difficult to see their simplified view of the output impedance.

The reality will be a little more complicated because we neglected \$R_1+R_2\$, which sits in parallel with the above. But it is a reasonable simplification for many circumstances. That said, you do have to always keep in mind the assumptions that were made in arriving there. If those assumptions are no longer true, the above simplification fails.

Hopefully, you follow this much.

Summary

Rmano nicely gave you the nice picture I was going to draw about the closed loop situation! Nice.

All that's left is for you to recognize that \$\frac{R_2}{R_1+R_2}=1-\frac{R_1}{R_1+R_2}=1-B\$.

I'd planned on dropping in this diagram:

Then you can solve this for \$V_{_\text{OUT}}\$:

$$\left[V_{_\text{IN}}\cdot -\left(1-B\right)-V_{_\text{OUT}}\cdot B\right]\cdot A_v=V_{_\text{OUT}}$$

Do note that none of this takes into account \$r_i\$ or \$r_o\$. They are assumed to be negligible for the above purposes. Yes, your diagram includes them. But the simplified equations provided by the authors ignore them.

Answer 4

You can see it using blocks in quite a visual way.

Your output it

$$v_o = A_v(v⁺ - v⁻) = A_v v_i$$

and by superposition

$$ v_i = - v_o\frac{R_1}{R_1+R_2} - v_s \frac{R_2}{R_1+R_2} $$

which graphically is:

Now you see that the feedback to the amplifier is really $$B=\frac{R_1}{R_1+R_2}$$

This kind of analysis is quite intuitive, I think, and shows easily why, for example, when applying the Gain Bandwidth Product method the inverting and non-inverting amplifiers have the same high cut frequency.

Now, the feedback block is the classical and simpler one, and can be obtained by simple algebraic derivation as $$ \frac{A_v}{1+BA_v} $$ which, when the loop gain is big, tends to 1/B. So when this happens $$ \frac{v_o}{v_s}=-\frac{R_2}{R_1+R_2}\frac{1}{B} = -\frac{R_2}{R_1}$$ which is the expected value.

Answer 5

if v=0 Iin=0 and R1, R2 share the same current.

This is because the output pulls in the oppositive polarity of Vin to maintain that v=0.

What does that tell you about the voltage drop across each R (V=IR) and thus the ratio of each is now inverted V gain or Av= - ?

When A = 1e4 to 1e10 how much difference does 1+ make to Zo when Ro is often 220 ohms with >10 % tolerance. So 1+ is ignored. and Zout= Ro/AB.
But if A=10 the then might be important.