Operational Amplifiers (Opamp) are integrated devices and encapsulated in several
materials (such as plastic, ceramic, etc ...), and usually consist of dozens
or hundreds (or even more) of interconnected transistors in a way that
reproduce as accurately as possible the characteristics of your original design.
As a rule, we only have access to the external terminals of the operating amplifier, which usually consists of an inverter input, a non-inverting input, a output and more the power terminals, one of positive polarity and another of negative polarity (relative to the other feed terminal).The most commonly used operating amplifiers have 8 external pins,
being the most common and least expensive example, called LM 741. The two letters indicate
the manufacturer and the number identify their characteristics.

For study purposes we are only interested in the input and output terminals.
In the Figura 41-01 we can see the symbology used for an operational amplifier.
The previously described five pins appear and there are two more
(not shown in the figure) that are used
for setting off-set. This feature will not be studied here
because it goes beyond the scope of this work.

2. The Ideal Operational Amplifier

For our studies we will always consider the operational amplifier as IDEAL. In fact, today technology is so advanced in making operating amplifiers, that it is permissible to assume them as
ideal, already that the theoretical results obtained will little differ from the practical results.

The characteristics that form the basis of an ideal operational amplifier can
be summarized in four fundamental laws described below:

What do these laws mean?

The first one says that there is no current flowing through the input terminals. This means that the
input impedance of an ideal operational amplifier is INFINITE.

The second one says that there is no voltage drop between the input terminals. Therefore, we can
assume that the potential difference between the inverting input and the
non-inverting is equal to ZERO.
Summing up:

i_{1} = i_{2} = 0

V_{i} = V_{2} - V_{1} = 0 ⇒ V_{1} = V_{2}

Stay tuned for these two last equations, as they will be very useful in solving problems involving
operating amplifiers.

3. Electric Model of an Ideal Opamp

An operational amplifier, when associated with passive circuits, such as resistors, capacitors, and inductors, can perform some mathematical operations
such as addition, subtraction, multiplication, division, differentiation and integration.

In the Figura 41-02, we see the schematic representation of the electric model of a
operational amplifier that satisfies the laws seen in the previous item.

From this figure (above) it is possible to write the equation that relates the input voltage
with the output voltage. Thus, we have:

V_{o} = A_{v} V_{i}

It is of interest that the operating amplifier works in a linear region, except in special cases. For this to happen we must satisfy the inequality below:

- V ≤ V_{o} ≤ + V

In most circuits, the voltage V varies between 10 and 15 volts, while the voltage gain
A_{v} is bigger than 10^{5}. Note that if in the previous equation we divide all
elements by A_{v}, and remembering that:

V_{i} = V_{2} - V_{1} = V_{o} / A_{v}

Then, making this substitution, we find:

(- V/ A_{v}) ≤ (V_{2} - V_{1}) ≤ (+ V/ A_{v})

Well, since A_{v} has a very large value, so V /A_{v} → 0 and
we can write that:

V_{1} ≅ V_{2}

In our studies we will assume the operational amplifier behaving as an ideal device
( A_{v} → ∞), then we can write that:

V_{1} = V_{2}

We will leave as an example of application of these characteristics the exercise of page 159
from the book [4] Fundamentos de Circuitos Elétricos - SADIKU, Matthew - 2013, where we show
in the Figura 41-03 the circuit and the statement.

Enunciated: considering the operational amplifier as
ideal, calculate the voltage gain at the closed loop K, the output current
i_{o}, as well as the output voltage V_{o} and the output current
i_{o} when V_{s} = 1 volt.

Note that the current i_{1} = 0 implies that the current i_{r}
which passes through the resistor of 40 kΩ, also passes through the resistor of
5 kΩ, characterizing a series circuit. So we can calculate
V_{1} by a simple voltage divider.

V_{1} = V_{o} ( 5 / (5 + 40)) = V_{o} / 9

Moreover, from the circuit we can deduce that:

V_{s} = V_{1} = V_{2}

By doing the proper substitution, we can calculate the voltage gain the closed loop (K)
of the circuit.

K = V_{o} / V_{s} = 9

Let us now calculate the value of i_{o}. Paying attention to the output of the
operating amplifier, we clearly see that there is a relation of currents with i_{o},
that is:

i_{o} = i_{r} + i_{x}

knowing this relationship, we can relate i_{o} with
V_{o}, that is:

i_{o} = V_{o}/ (5+40) + V_{o}/ 20

Note that the first parcela represents i_{r} and the second parcela represents
i_{x}.This is a direct application of the nodal voltage method.

When V_{s} = 1 volt we have that V_{o} = 9 x 1 = 9 volts, because
K = 9 and therefore, by making the numerical substitutions in the previous equation, we find the output current i_{o}, or:

i_{o} = 0.2 + 0.45 = 0.65 mA

So, using simple equations and proper reasoning, we have been able to solve the problem.
Note that the voltage gain of the closed-loop ( K) circuit is totally independent of the
open loop gain (A_{v}) of the operational amplifier, and depends only on the elements that make up the feedback circuit.

4. Buffer

"Buffer" or also known as voltage follower, or casador (adapter) of impedance is a voltage amplifier of unit gain. Its input impedance can be considered infinite
(without any exaggeration) and its output impedance is practically null.
For this reason the buffer circuit is widely used when we want to couple a circuit that has high impedance output to a circuit that has a low input impedance. So what a buffer does is output a "true copy" of the input voltage. Because it has a high input impedance, it does not overload the previous stage. And because it has a very low output impedance, it can power the subsequent stage even if it has a low input impedance.

See the Figura 41-04 for a buffer circuit.

Note that buffer is built from a non-inverting amplifier, doing
R_{f} = 0 e R_{i} = ∞. In this way, the gain is given by
K = 1 + (R_{f} / R_{i}) = 1.

Due to all these interesting buffer features, some books also call it a Voltage-controlled voltage source. Any voltage that is connected to the input will be faithfully reproduced at the output.