Linear Circuit Analysis


Ideal Operational Amplifiers

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 $\gt 1 \: {\class{mjunit}M\Omega}$), low output resistance ($\lt 100 \: {\class{mjunit}\Omega}$), and very high gain ($\gt 10^6$), and can often be approximated as ideal OpAmps.

A1 A2
Fig. 1. Simplified representations of an OpAmp.
How to Solve Problems with Ideal OpAmps

Since an OpAmp has four terminals, its electrical behavior is described using three equations. For comparison, a resistor (which has two terminals) is characterized by a single equation (Ohm’s law), while a transistor (which has three-terminals) requires two equations (the Ebers–Moll equations). Note that in this count we do not include the Kirchhoff’s Current Law (KCL) equation, which states that the algebraic sum of the currents leaving the terminals of any device must be zero. If we had included KCL, the number of independent equations that describe a device is equal to the number of terminals. Using the notations shown in Fig. 2, we have:

  • Infinite input resistance which implies the following two equations
$$\begin{equation}I_{+} = I_{-} = 0\end{equation}$$
  • Infinite gain which implies that
$$\begin{equation}V_{+} = V_{-}\end{equation}$$

A1 I+ I- Iout V+ V-
Fig. 2. Currents and potentials in ideal OpAmp.
Using nodal analysis

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:

A. Write the voltage constrained equations for each voltage source (one equation per source). Also, add a voltage constrained equation for each OpAmp, by writing that the potential of the positive input terminal equals the potential of the negative terminal, $v_{+}=v_{-}$ (which results from the infinite gain condition).

B. Write one equation for each control variable.

C. Write KCL equations for regular nodes and use that the currents going through the input terminals of the OpAmps are equal to zero (the infinite input resistance condition). Notice that the node containing the output terminal of the OpAmp is not a regular node and we skip writing KCL for this node.

D. Write KCL equations for supernodes.

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.

Computing gain, $G$

A common question that arises when analyzing circuits with OpAmps is the computation of the gain, 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 (for example, a speaker in an audio amplifier). Since OpAmps are linear devices, the output voltage is proportional to the input voltage and the gain does not depend on $V_{in}$.

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

Inverter

A standard configuration of OpAmps is the inverting amplifier shown in Fig. 3. The nodal equations for nodes $v_1$, $v_2$, and $v_3$ are $$\begin{equation}V_{in}=v_2\end{equation}$$ $$\begin{equation}v_1=0\end{equation}$$ $$\begin{equation}\frac{v_1-v_2}{R_2} + \frac{v_1-v_3}{R_1}=0\end{equation}$$ 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 $V_{out}$ is inverted with respect to the input signal.

R2 R1 A + Vout RL Vin v1 v3 v2
Fig. 3. Inverting amplifier.
Follower

Another standard configuration of OpAmps is the non-inverting amplifier (also called a voltage follower) 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 values of resistors $R_1$ and $R_2$, and $V_{out}$ is directly proportional to the input signal.

A R1 R2 + Vout RL Vin v3 v2 v1
Fig. 4. Non-inverting amplifier.
Unit-gain amplifier

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 their very large input impedance (infinite for an ideal OpAmp), because they draw no current from the input source.

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

A + Vout RL Vin
Fig. 5. Unit-gain amplifier.
Examples of Solved Problems
See also

Non-ideal OpAmps
Nodal analysis

Read more

Operational amplifier