Problem 41-1    Source:  Problem adapted from the text on page 169 -    THOMAS, Roland E. , ROSA, Albert J. , THOUSSAINT, Gregory J. -   Book: The Analysis & Design of Linear Circuits - 6ª Edição - Ed. John Willey & Sons, Inc. - 2009.

Assume that the operational amplifier in the circuit of the Figura 41-01.1 has an infinite input impedance and a voltage gain very large, but finite.

a) Determine the closed loop gain, V_o / V_s.

b) What is the effect of the feedback loop on the output impedance R_o of the operational amplifier.

Solution of the Problem 41-1

Attention:  The solution to this problem has been adapted from the existing text on page 167/168 of the book mentioned above.

Item a

For an ideal operational amplifier its voltage gain is considered as infinity. However, it is known that every real operating amplifier has a finite voltage gain. In this approach it was adopted that the amplifier operating voltage has a very large voltage gain but finite. As these devices currently have an input impedance of 10⁶ or even 10¹² ohms, it eliminated the input impedance in the circuit analysis.

Looking closely at the circuit you will notice that the voltage at the input non-inverting is determined by the independent source V_s. For a very high input circuit impedance, this means that the current at the input is zero. Then, to determine the voltage V₂, in the invertor input, we used a voltage divider with R_i and R_f, or:

On the other hand, we can determineV_o making a resistive divider with R_o, R_i e R_f. And remembering that V₁ = V_s:

Substituting the value of V₂ found in the first equation in the latter and after an algebraic arrangement in the expression, we get:

Let's analyze this equation when we have a finite value for A_v. We must consider that the value of R_o is much smaller when compared with the value of R_i + R_f. Then the same can be despised. Thus, with this approximation we can rewrite the previous equation, arriving at:

Where K is the gain the closed loop, already seen and given by:

Note that in the equation found for V_o, making A_v → ∞, the fraction is reduced to a value equal to K . Exactly the voltage gain of an ideal amplifier the closed mesh. So, the approximation made is very good when compared to the ideal template. A good choice is to limit the gain value to the mesh closed K, at least 1% of the value of A_v.

Item b

The output impedance of an amplifier circuit is affected by the feedback. To analyze this situation, the Thevenin equivalent is used for circuit output. The open circuit voltage and the short-circuit current must be calculated. The open circuit voltage, that is, Thévenin's voltage has already been calculated and will be repeated here.

To calculate the short-circuit current, the output must be shorted and calculate the current through it. With the output short, obviously V₂ = 0 amd V₁ = V_s. See how the circuit was in the Figura 41-01.2.

Then, from the circuit, the current of short-circuit is:

With this data we can calculate the Thevenin resistance, or:

Finally, making the necessary substitutions we find:

Note that for the case of K to be much smaller than A_v the equation can be simplified in such a way that the output impedance of the circuit is practically equal to zero. This can be easily understood by analyzing the simplified equation shown below.

For an operational amplifier like the LM 741, it is known that R_o = 50 Ω and A_v = 2 x 10⁵. Using this opamp in a circuit where the closed loop gain is of the order of K = 100, the output impedance of the LM 741 will be in the order of 0.025 Ω. Practically zero.