An operational amplifier (or OpAmp) is a voltage amplifier with a differential input, a single-ended output, and very high gain. Operational amplifiers are widely used in electronics for signal amplification, filter design (for instance to avoid using bulky inductors), analog calculators, analog-to-digital converters, differentiators, integrators, etc.

An ideal OpAmp is an OpAmp with infinite input resistance $R_{in}=\infty$, zero output resistance $R_{out}=0$, and infinite gain $A=\infty$. Many commercial OpAmps have relatively high input resistance (usually $\gt 1 \: {\textcolor{gray}M\Omega}$), low output resitance ($\lt 100 \: {\textcolor{gray}\Omega}$) and very high gain ($\gt 10^6$) and can be often approximated as ideal OpAmps.

OpAmps are 4-terminal devices, unually represented in simplified form as shown in Fig. 1. Since the ideal OpAmp is a 4-terminal device the algebraic sum of the currents flowing out through the 4-terminals is equal to 0 (Kirchhoff's current law). Therefore, even when representing OpAmps with only 3 electrodes (like the right diagram in Fig. 1), we are still assuming that the OpAmp contains a 4th electrode (the ground electrode) in order to satisfy Kirchhoff's current law.

A few standard configurations with ideal OpAmps are presented below.

Since OpAmps have 4 terminals (excluding the lines that power the dependent voltage source) they can be described electrically using three equations (notice that a resistor, which is a 2-terminal device, is described using one equation-Ohm's law; a transistor, which is a 3- terminal deivice, is described using 2 equations, etc.). Using the notations shown in Fig. 2, we have:

• Infite input resistance which implies the folowing two equations

• Infite gain which implies that

It is usually easier to write the nodal equations for a circuit containing OpAmps than the mesh equations. When using nodal analysis, we follow the same algorithm presented in the DC and AC circuit analysis sections. In particular, we:

Then, we solve the above system of nodal analysis equations and compute the values of the nodal potentials and control variables. After this, we write the equations for the sought variables as a function of the nodal potentials.

A few examples with ideal OpAmps are presented below.

A question that often comes out when solving problems with OpAmps is to compute the gain of a circuit, which is defined as $$\begin{equation}G=\frac{V_{out}}{V_{in}}\end{equation}$$ where $V_{out}$ and $V_{in}$ are the output and input voltages of the circuit. The output voltage is usually the voltage across a load impedance that can be the speaker of an audio amplifier or the input of another amplifier or processing unit. The input voltage could be a microphone or the output of the previous stage of the amplifier. Since OpAmps are linear devices the output voltage is proportial with the input voltage and the gain does depend on $V_{in}$.

Next, we present a few common configurations using OpAmps. Problems with OpAmps can often be solved by identifying the configuration of the OpAmp in the circuit (i.e. inverter or follower) and applygin the final equations for the gain, derived below.

A standard configuration for OpAmps is the inverting amplifier shown in Fig. 3. The nodal equations for nodes $v_1$, $v_2$, and $v_3$ are $$\begin{equation}V_{in}=v_1\end{equation}$$ $$\begin{equation}v_1=0\end{equation}$$ $$\begin{equation}\frac{v_1-v_2}{R_2} + \frac{v_1-v_3}{R_1}=0\end{equation}$$ The above system of equations can be solved for $v_1$, $v_2$, and $v_3$. Then, we can compute the sought variable $V_{out}=v_3$ and obtain $$\begin{equation}\frac{V_{out}}{V_{in}}=-\frac{R_2}{R_1} \end{equation}$$ We notice that the total gain of the inverter configuration depends only on the value of resistors $R_1$ and $R_2$ and $V_{out}$ in inverted with respect to the input signal.

Another standard configuration for OpAmps is the non-inverting amplifier (follower) shown in Fig. 4. The nodal equations for nodes $v_1$, $v_2$, and $v_3$ are $$\begin{equation}V_{in}=v_1\end{equation}$$ $$\begin{equation}v_1=v_3\end{equation}$$ $$\begin{equation}\frac{v_3-v_2}{R_2} + \frac{v_3}{R_1}=0\end{equation}$$ The above system of equations can be solved for $v_1$, $v_2$, and $v_3$. Then, we can compute the sought variable $V_{out}=v_3$ and obtain $$\begin{equation}\frac{V_{out}}{V_{in}}=1+\frac{R_2}{R_1}\end{equation}$$ We notice that the total gain of the follower configuration depends only on the value of resistors $R_1$ adn $R_2$ and $V_{out}$ is directly proportioal with the input signal.

A particular case of the follower configuration is the unit-gain amplifier shown in Fig. 5. In this case $R_2=0$, $R_1=\infty$ and $$\begin{equation}V_{out}=V_{in}\end{equation}$$ An important benefit of unit gain amplifiers is that the input impedance is very large (infinity in the case of ideal OpAmps), because they do not draw any current from the input source.

Notice that we could also build a "negative" unit gain amplifier using the inverter configuration.