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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}}\$:

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

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.

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\$.

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

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

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}}\$:

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

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.

Temporary Summary

I do NOT want to write any more until I see some commentary from you. You've been silent, so far. So I'm going to hold short at this point and wait.

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\$.

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

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