Find Vo of an op-amp circuit

Question

^{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.

Answer 1

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.

Answer 2

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:

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}}