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schematic

simulate this circuit – Schematic created using CircuitLab

I've tried solving for Vo in two ways, one is I took V+ by voltage divider getting V+ = V1(R2/R1+R2) = 0.05V. Since V+ = V-, it follows that V- = 0.05V. But I tried simulating it in the circuit simulator applet but it shows there that its voltage should be 4.99V so i stopped solving there.

Second solution I tried is using nodal analysis. I get (V+-V3)/R2 + (V+ - V1)/R1 = 0, V3/R3 + (V3-V+/R2) + (V3-V-)/R4 = 0, getting V3 to be 0.05V but again, the applet says that V3 is 9.88V.

Im not sure where I went wrong, maybe its the voltage divider or the way i used nodal analysis. Circuit simulator applet link if you want to try out for yourselves: Falstad.

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  • \$\begingroup\$ You can remove R3 from your calculations as it is driven by the op-amp output. \$\endgroup\$ – Transistor Apr 9 at 6:31
  • \$\begingroup\$ V+ still turns 0.05V even using nodal analysis, does that mean i simulated it wrong? \$\endgroup\$ – Aeden Schmidt Apr 9 at 6:35
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I get exactly 9.90V for an ideal op-amp. There is net negative feedback because R4 < R2 so it is stable.

Just equate the voltages at the two inputs and solve for V3.

What you call V+ is not just dependent on V1, it also depends on V3. V- also depends on V3, so V3 appears on both sides of the above-mentioned equation. Give it a try.

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  • \$\begingroup\$ I got the same for the ideal opamp output: (.198/2)/(1-1.98/2). \$\endgroup\$ – jonk Apr 9 at 7:04
  • \$\begingroup\$ I just cant get it, I tried (V1-V+)/R1 = (V+ - V3)/R2 and (V+-V3)/R2 = (V3-V-)/R4 and still got it wrong, is my equation wrong? \$\endgroup\$ – Aeden Schmidt Apr 9 at 7:13
  • \$\begingroup\$ @AedenSchmidt V+ is (V3*100k+.1 V*100k)/200k. It's a simple divider between two sources. I think you got that wrong, already. \$\endgroup\$ – jonk Apr 9 at 7:29
  • \$\begingroup\$ @jonk may i ask where did V3*100k come from? \$\endgroup\$ – Aeden Schmidt Apr 9 at 7:36
  • \$\begingroup\$ @AedenSchmidt V3 is the opamp output. You have two voltage sources and two resistors on the left side. You multiply one source by the resistance of the other source and add the product of the other source and its opposing resistance and then divide all that by the sum of the two resistances. Basic divider formula. \$\endgroup\$ – jonk Apr 9 at 7:40
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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:

enter image description here

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

$$ \begin{cases} \text{I}_2=\text{I}_2\\ \\ 0=\text{I}_4+\text{I}_5\\ \\ \text{I}_\text{o}+\text{I}_1+\text{I}_3=\text{I}_2+\text{I}_4 \end{cases}\tag1 $$

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

$$ \begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_3}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_3}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_2}{\text{R}_5} \end{cases}\tag2 $$

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

$$ \begin{cases} \frac{\text{V}_1-\text{V}_3}{\text{R}_2}=\frac{\text{V}_1-\text{V}_3}{\text{R}_2}\\ \\ 0=\frac{\text{V}_2-\text{V}_3}{\text{R}_4}+\frac{\text{V}_2}{\text{R}_5}\\ \\ \text{I}_\text{o}+\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3}{\text{R}_3}=\frac{\text{V}_1-\text{V}_3}{\text{R}_2}+\frac{\text{V}_2-\text{V}_3}{\text{R}_4} \end{cases}\tag3 $$

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:

$$ \begin{cases} \frac{\text{V}_x-\text{V}_3}{\text{R}_2}=\frac{\text{V}_x-\text{V}_3}{\text{R}_2}\\ \\ 0=\frac{\text{V}_x-\text{V}_3}{\text{R}_4}+\frac{\text{V}_x}{\text{R}_5}\\ \\ \text{I}_\text{o}+\frac{\text{V}_\text{i}-\text{V}_x}{\text{R}_1}+\frac{\text{V}_3}{\text{R}_3}=\frac{\text{V}_x-\text{V}_3}{\text{R}_2}+\frac{\text{V}_x-\text{V}_3}{\text{R}_4} \end{cases}\tag5 $$

Now, we can solve for the transfer function:

$$\mathscr{H}:=\frac{\text{V}_x}{\text{V}_\text{i}}=\frac{\text{R}_2\text{R}_5}{\text{R}_2\text{R}_5-\text{R}_1\text{R}_4}\tag6$$

Where I used the following Mathematica-code:

In[1]:=Clear["Global`*"];
V1 = Vx;
V2 = Vx;
FullSimplify[
 Solve[{I1 == I2, 0 == I4 + I5, I1 + Io + I3 == I2 + I4, 
   I1 == (Vi - V1)/R1, I2 == (V1 - V3)/R2, I3 == V3/R3, 
   I4 == (V2 - V3)/R4, I5 == V2/R5}, {Io, I1, I2, I3, I4, I5, Vx, 
   V3}]]

Out[1]={{Io -> (R2 (R3 + R4 + R5) Vi)/(R3 (R1 R4 - R2 R5)), 
  I1 -> (R4 Vi)/(R1 R4 - R2 R5), I2 -> (R4 Vi)/(R1 R4 - R2 R5), 
  I3 -> (R2 (R4 + R5) Vi)/(R3 (-R1 R4 + R2 R5)), 
  I4 -> (R2 Vi)/(R1 R4 - R2 R5), I5 -> (R2 Vi)/(-R1 R4 + R2 R5), 
  Vx -> (R2 R5 Vi)/(-R1 R4 + R2 R5), 
  V3 -> (R2 (R4 + R5) Vi)/(-R1 R4 + R2 R5)}}

My equation was also confirmed using LTspice.


Now, using \$\text{R}:=\text{R}_1=\text{R}_2=\text{R}_3=\text{R}_5\$, we can simplify the transfer function as follows:

$$\mathscr{H}=\frac{\text{R}}{\text{R}-\text{R}_4}\tag7$$

So, we get:

$$\text{V}_x=\frac{100\cdot10^3}{100\cdot10^3-98\cdot10^3}\cdot100\cdot10^{-3}=5\space\text{V}\tag8$$

Running the code again with your resistor values, we get:

In[2]:=Clear["Global`*"];
V1 = Vx;
V2 = Vx;
R1 = R;
R2 = R;
R3 = R;
R5 = R;
FullSimplify[
 Solve[{I1 == I2, 0 == I4 + I5, I1 + Io + I3 == I2 + I4, 
   I1 == (Vi - V1)/R1, I2 == (V1 - V3)/R2, I3 == V3/R3, 
   I4 == (V2 - V3)/R4, I5 == V2/R5}, {Io, I1, I2, I3, I4, I5, Vx, 
   V3}]]

Out[2]={{Io -> -(((2 R + R4) Vi)/(R (R - R4))), I1 -> (R4 Vi)/(R (-R + R4)), 
  I2 -> (R4 Vi)/(R (-R + R4)), I3 -> ((R + R4) Vi)/(R (R - R4)), 
  I4 -> -(Vi/(R - R4)), I5 -> Vi/(R - R4), Vx -> (R Vi)/(R - R4), 
  V3 -> ((R + R4) Vi)/(R - R4)}}

In[3]:=Clear["Global`*"];
V1 = Vx;
V2 = Vx;
Vi = 100*10^(-3);
R1 = 100*1000;
R2 = 100*1000;
R3 = 100*1000;
R4 = 98*1000;
R5 = 100*1000;
FullSimplify[
 Solve[{I1 == I2, 0 == I4 + I5, I1 + Io + I3 == I2 + I4, 
   I1 == (Vi - V1)/R1, I2 == (V1 - V3)/R2, I3 == V3/R3, 
   I4 == (V2 - V3)/R4, I5 == V2/R5}, {Io, I1, I2, I3, I4, I5, Vx, 
   V3}]]

Out[3]={{Io -> -(149/1000000), I1 -> -(49/1000000), I2 -> -(49/1000000), 
  I3 -> 99/1000000, I4 -> -(1/20000), I5 -> 1/20000, Vx -> 5, 
  V3 -> 99/10}}

In[4]:=N[%3]

Out[4]={{Io -> -0.000149, I1 -> -0.000049, I2 -> -0.000049, I3 -> 0.000099, 
  I4 -> -0.00005, I5 -> 0.00005, Vx -> 5., V3 -> 9.9}}
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