So with the following circuit (currently studying for an exam):
I am able to get V- just through the voltage divider composed of 2R and R at the non inverting terminal. However, once there, I am unsure of how to get Vout just because of the 3R resistor going from Vout and ground. Can I treat the feedback loop resistor, R2, and the 3R resistor, as a voltage divider that gives me Vout in terms of V-? If so, then what about the current that exits the opamp?
Premises about the op. amp.:
First part is a simple voltage divider:
$$V_+ = V_{in}\frac{R}{3R} = \frac{V_{in}}{3}$$
The other is a also a voltage divider with equal resistors:
$$ V_- = \frac{V_{out}}{2}$$
Since V+ = V-:
$$ V_{out} = \frac{2 V_{in}}{3} $$
First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).
Well, we are trying to analyze the following opamp-circuit:
simulate this circuit – Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$ \begin{cases} \text{I}_1=\text{I}_2\\ \\ 0=\text{I}_3+\text{I}_4\\ \\ \text{I}_5=\text{I}_4+\text{I}_6\\ \\ \text{I}_1+\text{I}_6=\text{I}_2+\text{I}_3+\text{I}_5 \end{cases}\tag1 $$
When we use and apply Ohm's law, we can write the following set of equations:
$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_3}{\text{R}_5} \end{cases}\tag2 $$
Substitute \$(2)\$ into \$(1)\$, in order to get:
$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}\\ \\ 0=\frac{\text{V}_2}{\text{R}_3}+\frac{\text{V}_2-\text{V}_3}{\text{R}_4}\\ \\ \frac{\text{V}_3}{\text{R}_5}=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}+\text{I}_6\\ \\ \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\text{I}_6=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_2}{\text{R}_3}+\frac{\text{V}_3}{\text{R}_5} \end{cases}\tag3 $$
Now, using an ideal opamp, we know that:
$$\text{V}_x:=\text{V}_+=\text{V}_-=\text{V}_1=\text{V}_2\tag4$$
So we can rewrite equation \$(3)\$ as follows:
$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_x}{\text{R}_1}=\frac{\text{V}_x}{\text{R}_2}\\ \\ 0=\frac{\text{V}_x}{\text{R}_3}+\frac{\text{V}_x-\text{V}_3}{\text{R}_4}\\ \\ \frac{\text{V}_3}{\text{R}_5}=\frac{\text{V}_x-\text{V}_3}{\text{R}_4}+\text{I}_6\\ \\ \frac{\text{V}_\text{i}-\text{V}_x}{\text{R}_1}+\text{I}_6=\frac{\text{V}_x}{\text{R}_2}+\frac{\text{V}_x}{\text{R}_3}+\frac{\text{V}_3}{\text{R}_5} \end{cases}\tag5 $$
Now, we can solve for the transfer function:
$$\mathcal{H}:=\frac{\text{V}_3}{\text{V}_\text{i}}=\frac{\text{R}_2\left(\text{R}_3+\text{R}_4\right)}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)}\tag6$$
Where I used the following Mathematica-code:
In[1]:=Clear["Global`*"];
V1 = Vx;
V2 = Vx;
FullSimplify[
Solve[{I1 == I2, 0 == I3 + I4, I5 == I4 + I6,
I1 + I6 == I2 + I3 + I5, I1 == (Vi - V1)/R1, I2 == V1/R2,
I3 == V2/R3, I4 == (V2 - V3)/R4, I5 == V3/R5}, {I1, I2, I3, I4, I5,
I6, Vx, V3}]]
Out[1]={{I1 -> Vi/(R1 + R2), I2 -> Vi/(R1 + R2),
I3 -> (R2 Vi)/((R1 + R2) R3), I4 -> -((R2 Vi)/((R1 + R2) R3)),
I5 -> (R2 (R3 + R4) Vi)/((R1 + R2) R3 R5),
I6 -> (R2 (R3 + R4 + R5) Vi)/((R1 + R2) R3 R5),
Vx -> (R2 Vi)/(R1 + R2), V3 -> (R2 (R3 + R4) Vi)/((R1 + R2) R3)}}
My equation was also confirmed using LTspice.
Now, using \$\text{R}_x:=\text{R}_3=\text{R}_4\$, \$\text{R}_1=2\text{R}\$ and \$\text{R}_2=\text{R}\$ we can simplify the transfer function as follows:
$$\mathcal{H}=\frac{\text{R}\left(\text{R}_x+\text{R}_x\right)}{\text{R}_x\left(2\text{R}+\text{R}\right)}=\frac{\text{R}\left(2\text{R}_x\right)}{\text{R}_x\left(3\text{R}\right)}=\frac{2\text{R}\text{R}_x}{3\text{R}\text{R}_x}=\frac{2}{3}\tag7$$
Running the code again with your resistor values, we get:
In[2]:=Clear["Global`*"];
V1 = Vx;
V2 = Vx;
R1 = 2 R;
R2 = R;
R3 = Rx;
R4 = Rx;
R5 = 3 R;
FullSimplify[
Solve[{I1 == I2, 0 == I3 + I4, I5 == I4 + I6,
I1 + I6 == I2 + I3 + I5, I1 == (Vi - V1)/R1, I2 == V1/R2,
I3 == V2/R3, I4 == (V2 - V3)/R4, I5 == V3/R5}, {I1, I2, I3, I4, I5,
I6, Vx, V3}]]
Out[2]={{I1 -> Vi/(3 R), I2 -> Vi/(3 R), I3 -> Vi/(3 Rx), I4 -> -(Vi/(3 Rx)),
I5 -> (2 Vi)/(9 R), I6 -> 1/9 (2/R + 3/Rx) Vi, Vx -> Vi/3,
V3 -> (2 Vi)/3}}
Simply put, this is a scaled non-inverting amplifier. It consists of two cascaded devices - voltage divider and non-inverting amplifier. This connection makes some sense in the circuit of a simple op-amp differential amplifier where it is used to equalize the gain at both inputs.
If you want to say it in a more intriguing way, two voltage dividers - "straight" (2R and R) and "inverted" (R2, R2 and the op-amp) are cascaded. So, the total ratio is R/(2R + R) x (R2 + R2)/R2 = 2/3... and accordingly, the output voltage is Vin.2/3.
The resistor 3R serves as a load consuming a current IL = Vout/3R... but it does not affect the output voltage.
