Electronic – Current in opamp

circuit analysisoperational-amplifier

Im trying to do a node analysis on the next opamp configuration

schematic

simulate this circuit – Schematic created using CircuitLab

But the issue is at the third node (\$Vo+\$).
What current is supposed the flow there?

\$i_{22}=\Large\frac{v_{s}-v_{1}}{22k}\$

\$i_{33}=\Large\frac{v_{1}-v_{2}}{33k}\$

\$i_{47}=\Large\frac{v_{1}}{47k}\$

\$i_{330}=\Large\frac{v_{2}-v_{3}}{330k}\$

kcl in \$1\$

\$i_{22}=i_{33}+i_{47}\$

kcl in \$2\$

\$i_{33}=i_{330}\$

kcl in \$3\$

\$i_{330}=?\$

Best Answer

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:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\text{I}_2+\text{I}_3\\ \\ 0=\text{I}_3+\text{I}_4\\ \\ \text{I}_2=\text{I}_1+\text{I}_4 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_3-\text{V}_2}{\text{R}_4} \end{cases}\tag2 $$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\ \\ 0=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}+\frac{\text{V}_3-\text{V}_2}{\text{R}_4}\\ \\ \frac{\text{V}_1}{\text{R}_2}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3-\text{V}_2}{\text{R}_4} \end{cases}\tag3 $$

Now, using an ideal opamp, we know that:

$$\text{V}_+=\text{V}_-=\text{V}_2=0\space\text{V}\tag4$$

So we can rewrite equation \$(3)\$ as follows:

$$ \begin{cases} \frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\frac{\text{V}_1-0}{\text{R}_3}\\ \\ 0=\frac{\text{V}_1-0}{\text{R}_3}+\frac{\text{V}_3-0}{\text{R}_4}\\ \\ \frac{\text{V}_1}{\text{R}_2}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3-0}{\text{R}_4} \end{cases}\tag5 $$

Now, we can solve for the transfer function:

$$\mathscr{H}:=\frac{\text{V}_3}{\text{V}_\text{i}}=-\frac{\text{R}_2\text{R}_4}{\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)}\tag6$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
V2 = 0;
FullSimplify[
 Solve[{I1 == I2 + I3, 0 == I3 + I4, I2 == I1 + I4, 
   I1 == (Vi - V1)/R1, I2 == V1/R2, I3 == (V1 - V2)/R3, 
   I4 == (V3 - V2)/R4}, {I1, I2, I3, I4, V1, V3}]]

Out[1]={{I1 -> ((R2 + R3) Vi)/(R2 R3 + R1 (R2 + R3)), 
  I2 -> (R3 Vi)/(R2 R3 + R1 (R2 + R3)), 
  I3 -> (R2 Vi)/(R2 R3 + R1 (R2 + R3)), 
  I4 -> -((R2 Vi)/(R2 R3 + R1 (R2 + R3))), 
  V1 -> (R2 R3 Vi)/(R2 R3 + R1 (R2 + R3)), 
  V3 -> -((R2 R4 Vi)/(R2 R3 + R1 (R2 + R3)))}}

My equation was also confirmed using LTspice.

Using your values we get:

$$\mathscr{H}=-\frac{1290}{281}\tag7$$

With \$\text{V}_\text{i}=1\space\text{V}\$, the output voltage is:

$$\text{V}_3=1\cdot\left(-\frac{1290}{281}\right)=-\frac{1290}{281}\approx-4.59075\space\text{V}\tag8$$

If you're interested in all the calculations:

In[2]:=Clear["Global`*"];
V2 = 0;
Vi = 1;
R1 = 22*1000;
R2 = 43*1000;
R3 = 33*1000;
R4 = 330*1000;
FullSimplify[
 Solve[{I1 == I2 + I3, 0 == I3 + I4, I2 == I1 + I4, 
   I1 == (Vi - V1)/R1, I2 == V1/R2, I3 == (V1 - V2)/R3, 
   I4 == (V3 - V2)/R4}, {I1, I2, I3, I4, V1, V3}]]

Out[2]={{I1 -> 19/772750, I2 -> 3/281000, I3 -> 43/3091000, 
  I4 -> -(43/3091000), V1 -> 129/281, V3 -> -(1290/281)}}

In[3]:=N[%2]

Out[3]={{I1 -> 0.0000245875, I2 -> 0.0000106762, I3 -> 0.0000139114, 
  I4 -> -0.0000139114, V1 -> 0.459075, V3 -> -4.59075}}
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