Problem 42-1    Source:  Example 5-1 - page 157 - ALEXANDER, Charles K., SADIKU, Matthew N. O. - Book: Fundamentos de Circuitos Elétricos - 5ª Edição - Ed. McGraw Hill - 2013.

An opamp LM741 has an open mesh voltage gain of 2 x 10⁵ input resistance of 2 MΩ and output resistance of 50 ohms . The configuration circuit is shown in Figura 42-01.1, where R_f = 20 kΩ and R_i = 10 kΩ. Use the real model of an opamp.

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

b) Determine the current i₂ when V_s = 2 volts.

Solution of the Problem 42-1

Attention:  The solution to this problem has been adapted from existing on pages 157 and 158 of the book mentioned above.

Item a

See in the figure below the circuit that was used as reference to solve the problem using the real model. From the problem data, it is known that the internal components of the Opamp LM 741 are:   R_i = 2 MΩ, R₀ = 50 Ω and A_v = 2 x 10⁵.

The most practical method that has been studied to solve this type of problem is nodal analysis. Thus, for the node V₁, you can write the relation:

Multiplying the two members by 2000 x 10³, we obtain the relation:

Note that doing the approach 301 V₁ ≅ 300 V₁ and dividing the two members by 100, we obtain the following expression:

Finally, we can write the relation:

Analyzing the node V_o, we can find a second equation that solve the problem. Like this:

In the figure above, note that V_i has the positive polarity on earth, while V₁ = R_i i_i. We concluded that V₁ = - V_i. Knowing that A_v = 200 000, we can substitute in the above equation and after some simplifications obtain:

However, as the problem calls for the relationship between V_o and V_s, just use the two previous equations,  eq. 42-01.1  and   eq. 42-01.2, and we get:

Finally, we determined the gain in closed loop of the circuit, that is:

Using the equation for an ideal amplifier, the gain of the amplifier would be K = -2. Note that, the difference between the values is insignificant. Therefore, it is proved that there is no need to resort to a real circuit to solve the problem. Just use the equations of an ideal circuit.

Item b

We know that:

With these values, you can calculate the value of V₁:

Then, by substituting their numerical values, we find V₁, or:

And finally we calculated i₂ by the relationship observed in the circuit:

Pay attention to the fact that the output voltage V_o is negative, indicating that there is an inversion of 180 ° in the output signal in relation to the input signal V_s. This is what happens when the input signal is injected into the inverter input of Opamp, as is the case with this circuit.