Linear Circuit Analysis
1. Introduction
2. Basic Concepts
- Currents and voltages
- Linear circuits
- Linear components
- Loops and nodes
- Series and parallel
- R, L & C combinations
- V & I combinations
- Power and energy
3. Simple Circuits
- Ohm's law
- Kirchhoff's current law
- Kirchhoff's voltage law
- Single loop circuits
- Single node-pair circuits
- Voltage division
- Current division
4. Nodal and Mesh Analysis
5. Additional Analysis Techniques
- Superposition
- Source transformation
- The $V_{test}/I_{test}$ method
- Norton equivalent
- Thévenin equivalent
- Max power transfer
6. AC Analysis
7. Operational Amplifiers
8. Laplace Transforms
9. Time-Dependent Circuits
- Introduction
- First-order transients
- Nodal analysis
- Mesh analysis
- Laplace transforms
- Additional techniques
10. Two-port networks
Ideal Operational Amplifiers
An ideal OmAmp is an OpAmp with input resistance $R_{in}=\infty$, output resistance $R_{out}=0$, and gain $A=\infty$. In practice, real OpAmps have relatively high input resistance (usually over $1 M\Omega$), low output resitance (below $100 \Omega$) and very high gain ($over $10^6$) and can be often approximated with an ideal OpAmp.
OpAmps are unually represented in simplified form, as shown in
. Notice that, just like in the case of non-ideal OpAmp, the OpAmp is a 4-terminal device and, according to KCL the algebraic sum of the currents flowing out through the 4-terminals is equal to 0. Notice again that if we do not show the ground electrode (see the right diagram in ), we are still assuming that OpAmmp contains a 4th (ground electrode); for this reason the algebraic sum of the currents flowing out through the 3-terminals of the OpAmp shown in the right diagram in is not necessarily 0. This does not mean that KCL is not satisfied but it only says that the OpAmp contains a 4th electrode (the electrode connected to the ground) which is not shown on the diagram.A few standard configurations with ideal OpAmps are presented below.
How to solve problems with ideal OpAmps
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
, we have:- Infite input resistance which implies the folowing two equations
- Infite gain which implies that
How to use nodal analysis
When using nodal analysis to solve problems with ideal OpAmps, 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 one voltage constrined equation for each OpAmp, which we impose the potential of the positive input terminal equals the potential of the negative terminal.
B. Write one equation for each control variable.
C. Write KCL equations for regular nodes. Here, we do not write the equation for the node where the output terminal of the OpAmp is connected, as we do not know the value of the current through the output terminal.
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.
How to use mesh analysis
To do.
Inverter
A standard configuration for OpAmps is the inverting amplifier shown in
. The nodal equtions for nodes $v_1$ and $v_2$ are $$\begin{equation}V_{in}=v_1\end{equation}$$ $$\begin{equation}\frac{v_1-v_2}{R_2} + \frac{v_1-v_3}{R_1}=0\end{equation}$$ Since $v_1=0$ and $v_3=V_{out}$ we obtain that $$\begin{equation}\frac{V_{out}}{V_{in}}=-\frac{R_2}{R_1} \end{equation}$$ We notice that in the case of the inverter configuration the total gain depends only on the value of resistors $R_1$ adn $R_2$ and $V_{out}$ in inverted with respect to the input signal.Follower
Another standard configuration for OpAmps is the non-inverting amplifier (follower) shown in
. The nodal equtions for nodes $v_1$ and $v_3$ are $$\begin{equation}V_{in}=v_1\end{equation}$$ $$\begin{equation}\frac{v_3-v_2}{R_2} + \frac{v_3}{R_1}=0\end{equation}$$ Since $v_1=v_3$ and $v_2=V_{out}$ we obtain that $$\begin{equation}\frac{V_{out}}{V_{in}}=1+\frac{R_2}{R_1}\end{equation}$$ We notice that in the case of the follower configuration the total gain depends only on the value of resistors $R_1$ adn $R_2$ and $V_{out}$ is directly proportioal with the input signal.Unit-gain amplifier
A particular case of the fallower configuration is the unit-gain amplifier shown in
. In this case $$\begin{equation}V_{out}=V_{in}\end{equation}$$ 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.