What is the closed-loop gain in this op amp circuit?

Question

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 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.

^{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?

Answer 1

Assuming this is a problem from a book and we can assume ideal parameters to the opamp where no specifications are otherwise given, then:

^{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.

Answer 2

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:

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:

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:

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

Answer 3

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.

Answer 4

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.