LC natural response - intuition

We develop an intuition for the natural response of the inductor-capacitor $(\text{LC})$ circuit.

Circuits with two energy storage elements (capacitors or inductors) are called second-order systems. The voltage and current rock back-and-forth, or oscillate. Second-order systems are where sine waves come from in analog circuits.

After you have a good mental image of what’s going on, the next article is a formal derivation of the LC natural response.

Written by Willy McAllister.

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Contents

• First-order systems
• Second-order systems
• Predict the natural response
• Mechanical analogy: pendulum

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First-order systems

Up to now we’ve looked at first-order circuits like the RC natural response and the RL natural response. These have a single energy-storage element, one $\text{C}$ or $\text{L}$. The natural response of first-order circuits has an exponential shape that “slumps” to its final value. The original stored energy is eventually dissipated by the resistor.

The $\text{RC}$ natural response is described by a first-order differential equation.

Second-order systems

Now we look at a circuit with two ideal energy-storage elements and no resistor. Circuits with two storage elements are second-order systems because they produce equations with second derivatives.

Second-order systems are the simplest systems that rock back and forth in time, or oscillate. The classic mechanical second-order system is a pendulum. You provide some initial energy by pulling the pendulum to the side. When you let it go it oscillates back and forth. In electronics, the classic oscillating system is the $\text{LC}$ circuit.

Predict the natural response

Circuit conditions just after the switch closes.

Let’s say the capacitor has an initial voltage, which means it is storing some charge, $q = \text C\,\text V_0$. Think of $q$ as positive charge sitting on the top plate of the capacitor. We are going to reason through this by tracking what happens to $q$.

There is no initial current in the inductor, and therefore no current in the capacitor, either. The charge is just sitting there on the capacitor. What happens when the switch closes and we let the circuit “do whatever it wants”? (Its natural response.)

The inductor starts with $0$ current. All of a sudden the initial voltage, $v = \text V_0$ shows up when the switch closes. This voltage generates a current in the inductor, and it starts storing energy in its surrounding magnetic field. Where does that current (flowing charge) come from? It comes from the capacitor, of course.

Over at the capacitor, charge (current) flows out of the top plate, shown in blue below. It flows through the inductor and around to the bottom capacitor plate. As charge leaves the top plate, $q=\text C\,v$ is going down. That tells us $v$ has to be going down, as shown in orange.

A little after closing the switch, current is rising and voltage is falling.

Eventually, we reach a state where the charge on the top plate is the same as the bottom plate. The voltage across the capacitor therefore falls to $0$.

Over at the inductor there is a current flowing, even though the voltage is $0$, because the energy stored in the inductor’s magnetic field keeps it flowing. (Current does not abruptly drop to $0$ when the voltage reaches $0$.)

The voltage eventually falls to $0$ as the amount of charge on the top and bottom capacitor plates becomes the same. At the same time, the current in the inductor reaches its highest value. That current continues to pump the remaining charge onto the bottom plate of the capacitor.

The inductor current continues to move charge from the top plate to the bottom. Now there is more positive charge on the bottom plate than the top, so the voltage actually reverses sign and becomes negative.

As charge continues to build up on the bottom plate, it repels against the arrival of new charge from the inductor current. The inductor current bends over and starts to drop back toward $0$.

The inductor continues heaping positive charge on the bottom plate of the capacitor, so the voltage goes negative.

After a little while, the voltage reaches a peak negative value when all the charge has flowed to the bottom plate. The voltage will be the negative of whatever the capacitor started at. Charge stops moving for a brief moment, so the current crosses $0$.

After all the available charge has flowed to the bottom plate, the voltage reaches its negative peak, and the current falls to $0$.

The previous image is almost identical to where we started. The current is back to zero, and the voltage is at a peak value. The peak happens to be the negative of where we started. We can go back to the start of the story and tell it again just the same, except with charge moving from the bottom plate of the capacitor back to the top. Here’s the end result after one full cycle,

The second half of the cycle is the same as the first, but with charge moving from the bottom plate of the capacitor back to the top plate.

The exchange of charge between top and bottom plates continues forever. The rate of oscillation (the frequency), is determined by the value of $\text L$ and $\text C$. We will discover how that works when we do the formal derivation of the $\text{LC}$ natural response in the next article.

Mechanical analogy: pendulum

The $\text{LC}$ circuit is analogous to a mechanical oscillator, the frictionless swinging pendulum.

• Voltage $v(t)$ is the analog of position. We measure position of the pendulum as it swings away from the center point. The distance is $0$ $(v=0)$ when the pendulum hangs straight down. It swings to $v=+\text V_0$ or $-\text V_0$ at either extreme position.

• Current $i(t)$ is the analog of velocity. The pendulum moves fastest at the mid-point $(i=\text I_{max})$. It is motionless $(i=0)$ at end of its swing, just for an instant.

• The initial voltage $+\text V_0$ corresponds to how far we pull the pendulum to the right before letting go.

Letting go of the pendulum corresponds to closing the switch. What happens next is the natural response. If the pivot point is frictionless and there is no air resistance, the pendulum swings forever.

The $\text{LC}$ circuit (and the pendulum) trade voltage and current back and forth in a sine wave pattern. We see a timing difference of $1/4$ of a cycle between them. When either is $0$, the other is $\pm$peak value.

(Full disclosure: The pendulum swings in a sine wave if the distance from the rest position is small. If the initial displacement is large, the answer is not exactly a sine wave.)