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For the op amp below

  • Open-loop gain is \$A=2\times10^5\$
  • Input resistance is \$R_i=2\,M\Omega\$
  • Output resistance is \$R_o=50\,\Omega\$

I am asked to calculate the close-loop gain \$Vo/Vs\$ and find \$i_o\$ when \$V_S=1\$ .

schematic diagram

(Schematic diagram above from "Fundamentals of Electric Circuits" by Charles Alexander and Matthew Sadiku.)

Then I've redrawn the circuit and defined the currents as shown in the figure.

schematic

simulate this circuit – Schematic created using CircuitLab

Then according to those currents above I've applied KCL and also I've got these equations below.

\$I = I_1+I_2\qquad I_2=I_3+I_4\$ $$I_4=\frac{V_1}{5\,k\Omega}$$ $$I_3=\frac{V_1-V_o}{40\,k\Omega}$$ $$I_3+I_1=\frac{V_o}{20\,k\Omega}$$ $$A \times V_d=2\times10^5\times 2\,M\Omega\times I$$ $$I=\frac{V_S-V_1}{2\,M\Omega}$$

Then I've also written some equalities according to KVL in the closed loops but I was never able to find a relation between \$V_S\$ and \$V_o\$ or just ended up crosschecking the same equations. It all got tangled up.

  • How would you solve this?
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  • \$\begingroup\$ Perhaps you are overthinking this; at DC the gain is trivial to find (assuming you know the gain equation). Once you have that, everything else falls into place. \$\endgroup\$ – Peter Smith May 13 '18 at 15:43
  • \$\begingroup\$ @PeterSmith, haha that is definitely right. I don't know how many papers I've wasted writing down the same equations over and over again. Then, in the non-inverting op amp the gain would simply be \$\frac{Rf}{R1}+1\$. But I just wanted to solve it as a non-ideal op amp as requested in the book then all got messy haha. \$\endgroup\$ – André Yuhai May 13 '18 at 15:51
  • \$\begingroup\$ The reference direction for \$V_d\$ is not the same in your 2nd model as it was in the first model. And your equations match the 2nd model, so that you have made a circuit with positive feedback instead of negative feedback. \$\endgroup\$ – The Photon May 13 '18 at 16:40
  • \$\begingroup\$ @ThePhoton thanks for pointing it out, I corrected it. \$\endgroup\$ – André Yuhai May 13 '18 at 17:19
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    \$\begingroup\$ I1 shouldn't be where it is. The negative terminal of the VCVS should be connected to ground, not to an input of the op-amp. \$\endgroup\$ – The Photon May 13 '18 at 17:47
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Assuming this is a problem from a book and we can assume ideal parameters to the opamp where no specifications are otherwise given, then:

schematic

simulate this circuit – Schematic created using CircuitLab

Assuming the equivalent circuit at the top for the opamp, the nodal equations are:

$$\begin{align*} \frac{V_X}{R_1}+\frac{V_X}{R_2}+\frac{V_X}{R_{IN}}&=\frac{V_O}{R_1}+\frac{V_S}{R_{IN}}\\\\ \frac{V_O}{R_1}+\frac{V_O}{R_3}+\frac{V_O}{R_{OUT}}&=\frac{V_X}{R_1}+\frac{\left(V_S-V_X\right)\cdot A_\text{OL}}{R_{OUT}}\\\\\therefore\\\\V_X&\approx 0.999955\cdot V_S\\\\V_O&\approx 8.999593\cdot V_S \end{align*}$$

The script I used in sympy (worth getting) is:

var('r1 r2 r3 ri ro aol vo vx vs')
e1=Eq(vx/r1+vx/r2+vx/ri,vo/r1+vs/ri)
e2=Eq(vo/r1+vo/r3+vo/ro,vx/r1+(vs-vx)*aol/ro)
ea=solve([e1,e2],[vx,vo])
    { vo: r1*r3*vs*(aol*(r1*r2 + r1*ri + r2*ri) - r2*(aol*r1 - ro))/(r2*r3*ri*(aol*r1 - ro) + (r1*r2 + r1*ri + r2*ri)*(r1*r3 + r1*ro + r3*ro)),
      vx: r1*r2*vs*(aol*r3*ri + r1*r3 + r1*ro + r3*ro)/(r2*r3*ri*(aol*r1 - ro) + (r1*r2 + r1*ri + r2*ri)*(r1*r3 + r1*ro + r3*ro))
    }
ea[vx].subs({aol:2e5,ri:2e6,ro:50,r1:40e3,r2:5e3,r3:20e3})
0.999954839544092*vs
ea[vo].subs({aol:2e5,ri:2e6,ro:50,r1:40e3,r2:5e3,r3:20e3})
8.99959265268771*vs

Just nodal is enough here. The main thing is to figure out the opamp model from the specs you were given.

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  • \$\begingroup\$ I like how you simplify these circuits but I wonder how we could say that \${V_O}/{R_1}\$. Wouldn't it be \${V_O-V_X}/{R_1}\$ for example and also for the other ones? \$\endgroup\$ – André Yuhai May 14 '18 at 8:31
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    \$\begingroup\$ @AndréYuhai That's the way it is usually taught. I consider the bookkeeping of it harder and therefore more prone to mistakes. So I use a different concept that is used in a Spice program I learned from, reading the code. Currents on the left "spill outward from the node." Currents on the right "spill inward into the node." I can only speak for myself, but I make zero errors in sign now. \$\endgroup\$ – jonk May 14 '18 at 8:37
  • \$\begingroup\$ Your method looks simple and, as you mentioned, less likely to make a mistake. But I better stick to the way they teach it for now until I've practiced enough then I may try your method. :) Op amps are already a hassle for me, haha. Thank you! \$\endgroup\$ – André Yuhai May 14 '18 at 12:39
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In this type of circuit, one easy way is to apply superposition and determine the control variable, \$\epsilon\$. To express it, we will apply superposition to this circuit featuring 1 controlled source: first we consider \$V_{in}=0\$ and second, we will consider the op-amp output \$\epsilon A_{OL}\$ equal to 0. In the first case, the circuit is below:

enter image description here

If you do the maths ok, you can determine \$\epsilon_1\$ whose value is in the Mathcad sheet at the end. Now, put \$V_{in}\$ back in place and consider \$\epsilon A_{OL}\$ equals 0:

enter image description here

Again, if you do the maths ok, you can determine \$\epsilon_2\$ whose value is in the Mathcad sheet at the end. You have the final \$\epsilon\$ value by writing \$\epsilon=\epsilon_1+\epsilon_2\$ and solve for \$\epsilon\$. That is what Mathcad did for us.

Now that you are there, you are almost done. Looking at the circuit, you see that \$V_{out}=\epsilon A_{OL} - i_1R_o\$. \$i_1\$ is the sum of two currents that you easily determine. You end-up with an equation featuring \$V_{out}\$, \$V_{in}\$ and \$\epsilon\$ that you now have on hand. If Mathcad does the job ok, for a 1-V input voltage, the output is exactly 8.99183 V for an open-loop gain of 10k:

enter image description here

The Mathcad file is here but I won't simplify equations, it's already quite late over here : )

enter image description here

These techniques using superposition are part of the fast analytical circuits techniques or FACTs. A solving technique that I encourage students and EEs to acquire.

Edit: I did not realize that I did not use the exact values given in the post. If I plug these in the sheet, I have a gain of 8.99959 as Monsieur jonk found.

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The gain is quite simply Rf/Ri + 1, so the gain is 9.

Vo = 9V for 1V in.

Given that, then the output current is the feedback path (450 microamp) + output current (650 microamp) for a total of 1.1mA. The output resistance of the amplifier is negligible in this situation and may be ignored.

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The 741 opamp design is 50 years old and it is not powered in your circuit. Some have lower gain and some have lower input resistance. Some 40k resistors are 41k and others are 39k. Some 5k resistors are ..... Your circuit does not null the input offset voltage. Why be extremely accurate? The closed loop gain is simply 1 + (40k/5k)= 9 times.

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  • \$\begingroup\$ I didn't design this. It is just a problem in a book called "Fundamentals of Electric Circuits". So I was just trying to solve it the hard way and see the result but I failed. :) \$\endgroup\$ – André Yuhai May 13 '18 at 15:54

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