# How to analyze this opamp circuit?

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?

• What are you assuming about the output impedance or resistance of the op amp? Mar 3, 2021 at 19:22
• Zero output impedance. Is it even possible to study this circuit in that condition?
– Bee
Mar 3, 2021 at 19:23
• Sure. But think about what "zero output impedance" means. Is there any limit to the current available from the op amp output? Does the output voltage change if you increase the load on the output? Mar 3, 2021 at 19:25
• You mention "V-" in the text but there is no "V-" in the circuit. Do you know how an ideal opamp behaves? What does it do? What determines the output voltage/current of the opamp? So what does that mean if R3 is present or not? Mar 3, 2021 at 19:35
• Vout doesn't change when I change 3R right?
– Bee
Mar 3, 2021 at 22:34

• zero input current
• infinite gain (V+ = V-)
• zero output resistance
• no input/output voltage restrictions

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.