Inductor and Capacitor Behavior in Switching Circuits

Hello! Welcome to the first lesson in our course on Zero Voltage Switching (ZVS). Your goal is to understand ZVS to better review and optimize PCBs, and this course is designed to get you there, focusing on the practical concepts behind key ZVS topologies.

This first lesson covers the most fundamental building block for any switching power converter: the voltage-current relationships for inductors and capacitors. While you're already familiar with these components, we'll refresh our understanding specifically in the context of the fast-switching environments found in power supplies. Mastering how these components behave under these conditions is the first essential step toward analyzing converter losses and, ultimately, understanding how ZVS helps to eliminate them.

Let's begin.

The Capacitor: Resisting Voltage Change

Capacitors are fundamental to power converters, most often used to filter the output and provide a stable DC voltage. Their behavior is governed by one key equation.

To refresh this core concept, let's start with a short video that derives the capacitor's voltage-current relationship from first principles.

Capacitors and Inductors (Circuits for Beginners #19)

This video from the 'Circuits for Beginners' series by Aaron Danner offers a clear introduction to capacitors, explaining how they store charge and what that means for the relationship between the voltage across them and the current flowing through them.

Please watch the first part of the video, from the beginning until 3:47. Focus on the final equation presented and the logic behind it.

As the video explains, the defining relationship for a capacitor is:

I_{C} (t) = C \frac{d V _{C} ( t )}{d t}

Where:

$I_{C} (t)$ is the instantaneous current flowing through the capacitor.
$C$ is the capacitance in Farads.
$\frac{d V _{C} ( t )}{d t}$ is the rate of change of voltage across the capacitor.

In a switching converter, this equation tells us two important things:

A rapid change in voltage ( $\frac{d V _{C}}{d t}$ is large) requires a very large current. This is why we can't instantaneously change the voltage across a capacitor—it would require infinite current.
To change the capacitor's voltage, we must supply a net current over time. This is precisely its function as a filter: it absorbs and supplies current as needed to smooth out voltage fluctuations, resisting fast changes.

The Principle of Capacitor Amp-Second Balance

For a converter operating in a stable, repeating cycle (steady state), there's a powerful analysis principle that stems from this relationship.

DC-DC Power Converters

The textbook 'DC-DC Power Converters' by Prof. Robert W. Erickson provides a concise definition of this principle, known as 'capacitor amp-second balance' or 'charge balance'.

Please read the short paragraph in the 'Analysis of converter waveforms' section that begins 'The principle of inductor volt-second balance...'. Focus on the definition for the capacitor, which is the second principle mentioned.

The principle of capacitor charge balance states that the average current flowing through an ideal capacitor over one complete switching period must be zero. If the average current were not zero, the capacitor would continuously accumulate or lose charge, causing its voltage to drift up or down indefinitely. In a steady-state converter, this doesn't happen. The current that flows in during one part of the cycle must equal the current that flows out during another part.

This is visualized perfectly by the capacitor's role in smoothing the output of a buck converter.

Capacitor Current and Voltage in a Buck Converter — The top graph shows the AC ripple current flowing into the capacitor, `i_c(t)`. In steady state, the charge flowing in (positive current area) equals the charge flowing out (negative current area) over one cycle. The bottom graph shows the resulting small voltage ripple, `Δv`, this charge transfer creates on the DC output voltage.

The Inductor: Resisting Current Change

Inductors are the other key energy-handling component in a switching converter. They store energy in a magnetic field and are typically used to transfer energy from the input to the output.

Let's continue with the video from earlier to review the inductor's fundamental equation.

Capacitors and Inductors (Circuits for Beginners #19)

The second half of the same video explains the inductor's behavior, its relationship to magnetic flux, and its governing voltage-current equation.

Watch from 3:47 to the end (6:08). Pay close attention to the final equation derived: V = L * dI/dt.

The inductor's behavior is described by this equation:

V_{L} (t) = L \frac{d I _{L} ( t )}{d t}

Where:

$V_{L} (t)$ is the instantaneous voltage across the inductor.
$L$ is the inductance in Henries.
$\frac{d I _{L} ( t )}{d t}$ is the rate of change of current through the inductor.

Rearranging this gives us $\frac{d I _{L}}{d t} = \frac{V _{L}}{L}$ , which is the most useful form for understanding switching converters. It tells us that applying a constant voltage across an inductor results in a linearly changing current. This is the characteristic "current ramp" you see in the inductor waveforms of all basic converters.

The following short video provides a concrete example of this effect.

Inductor Voltage

This example from the channel Voltverse demonstrates exactly how a changing current (a triangular waveform) relates to the voltage across the inductor (a square waveform). This is precisely the behavior we see in simple DC-DC converters.

Watch the example segment from 2:12 to 3:54. Notice how a positive, constant rate of change in current corresponds to a constant positive voltage, and a negative rate of change corresponds to a constant negative voltage.

The Principle of Inductor Volt-Second Balance

Similar to the capacitor, the inductor has a core principle for steady-state analysis. In the same paragraph you read earlier from the 'DC-DC Power Converters' text, this principle was defined.

The principle of inductor volt-second balance states that the average voltage across an ideal inductor over one complete switching period must be zero. If the average voltage were non-zero, the term $\frac{d I _{L}}{d t}$ would have a net positive or negative average, causing the inductor current to ramp up or down to infinity—clearly not a steady state.

This means the "volt-seconds" applied during the "on" time must be cancelled out by the "volt-seconds" applied during the "off" time.

Tying it Together: The Buck Converter Example

Since you're already familiar with the buck converter topology, we can use it to see these principles in action. The diagrams below, from a Microchip application note, show the key waveforms.

Switch Mode Power Supply (SMPS) Topologies

This application note on SMPS topologies provides clear illustrations of the waveforms in a buck converter. These diagrams perfectly visualize the concepts we've just discussed.

Examine Figure 2 in the 'Buck Converter' section. Pay close attention to waveforms (C), the voltage across the inductor, and (E), the current through the inductor. Notice the direct relationship between them.

Let's break down what's happening in those waveforms using our principles:

During $T_{O N}$ : The switch Q1 is closed. The voltage across the inductor $L$ is $V_{L} = V_{I N} - V_{O U T}$ . Since this voltage is positive and constant (assuming small ripple on $V_{O U T}$ ), the inductor current $I_{L}$ ramps up linearly according to $Δ I_{L} = \frac{V _{I N} - V _{O U T}}{L} T_{O N}$ .
During $T_{O F F}$ : The switch Q1 opens, and the diode D1 turns on. The voltage across the inductor is now $V_{L} = - V_{O U T}$ . Since this voltage is negative and constant, the inductor current ramps down linearly: $Δ I_{L} = \frac{- V _{O U T}}{L} T_{O F F}$ .

The principle of volt-second balance demands that the average voltage over the full period $T_{S} = T_{O N} + T_{O F F}$ is zero.

\frac{1}{T _{S}} ((V_{I N} - V_{O U T}) \cdot T_{O N} + (- V_{O U T}) \cdot T_{O F F}) = 0

(V_{I N} - V_{O U T}) \cdot T_{O N} = V_{O U T} \cdot T_{O F F}

Solving this for $V_{O U T}$ and using the definition of duty cycle $D = \frac{T _{O N}}{T _{S}}$ (and thus $\frac{T _{O F F}}{T _{S}} = 1 - D$ ), we get:

V_{I N} \cdot D \cdot T_{S} = V_{O U T} \cdot (D \cdot T_{S} + (1 - D) \cdot T_{S}) = V_{O U T} \cdot T_{S}

V_{O U T} = D \cdot V_{I N}

This shows how the fundamental V-I relationship for the inductor, combined with the volt-second balance principle, directly yields the ideal voltage conversion ratio for the buck converter.

Test your understanding!

Imagine a boost converter (where the inductor is between the input source and the switch). During the switch's ON-time, the voltage across the inductor is $V_{I N}$ . During the OFF-time, it's $V_{I N} - V_{O U T}$ . Using the principle of volt-second balance, can you derive the ideal voltage conversion ratio for a boost converter, $V_{O U T} / V_{I N}$ ?

Show answer

Applying volt-second balance, the average voltage across the inductor must be zero:

\frac{1}{T _{S}} ((V_{I N}) \cdot T_{O N} + (V_{I N} - V_{O U T}) \cdot T_{O F F}) = 0

V_{I N} \cdot T_{O N} = - (V_{I N} - V_{O U T}) \cdot T_{O F F} = (V_{O U T} - V_{I N}) \cdot T_{O F F}

Dividing by $T_{S}$ and substituting $D = T_{O N} / T_{S}$ and $1 - D = T_{O F F} / T_{S}$ :

V_{I N} \cdot D = (V_{O U T} - V_{I N}) \cdot (1 - D)

V_{I N} \cdot D = V_{O U T} (1 - D) - V_{I N} (1 - D)

V_{I N} \cdot (D + 1 - D) = V_{O U T} (1 - D)

V_{I N} = V_{O U T} (1 - D) ⟹ \frac{V _{O U T}}{V _{I N}} = \frac{1}{1 - D}

Conclusion

In this lesson, we refreshed our understanding of the two most important passive components in power electronics and established the key principles for their analysis in switching circuits.

Key Takeaways:

Capacitor: The current through a capacitor is proportional to the rate of change of its voltage ( $I_{C} = C \frac{d V _{C}}{d t}$ ). It acts to smooth voltage. In steady state, the average current through it over a switching cycle is zero (charge balance).
Inductor: The voltage across an inductor is proportional to the rate of change of its current ( $V_{L} = L \frac{d I _{L}}{d t}$ ). It acts to smooth current. In steady state, the average voltage across it over a switching cycle is zero (volt-second balance).
These two balance principles are powerful tools that allow us to quickly determine the ideal relationships in DC-DC converters.

In our next lesson, we will build on this by looking at how to calculate the energy stored in these components, which is critical for understanding the flow of power in a converter.

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