Volt-second and charge balance

Intent

• State inductor volt-second balance and capacitor charge balance, and why they hold in steady state.
• Use them to find the dc conversion ratio of a converter in continuous conduction mode (CCM).
• Recognise when the balance conditions do not apply.

Theory

Volt-second balance

In periodic steady state every waveform repeats from one switching period to the next, so the inductor current ends each period where it began. The inductor voltage integrated over one period equals $L$ times that net change in current, which is zero: the average inductor voltage over a period is zero, $\langle v_L \rangle = 0$ [Erickson & Maksimović 2020, Ch. 2]. This is inductor volt-second balance.

Buck in CCM. The inductor voltage $v_L$ is $V_g - V$ while the switch is on and $-V$ while it is off. In steady state the two shaded volt-second areas are equal, so $i_L$ ends each period where it began. $\Delta i_L$ is half the peak-to-peak ripple. [Erickson & Maksimović 2020, Ch. 2]

With the small-ripple approximation, the inductor voltage takes one constant value during the on-interval $D T_s$ and another during the off-interval $(1 - D) T_s$. Balance fixes the second in terms of the first:

$$v_{L,\mathrm{off}} = - \frac{D v_{L,\mathrm{on}}}{1 - D}$$

$v_{L,\mathrm{off}}$

inductor voltage, switch off

$D$

duty ratio

$v_{L,\mathrm{on}}$

inductor voltage, switch on

Volt-second balance: inductor voltage during the off-interval✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• periodic steady state
• small-ripple approximation

Notes

two-interval form of $\langle v_L \rangle = 0$

Source

Erickson & Maksimović 2020, Ch. 2 (Principles of Steady-State Converter Analysis)

Derivation

step-by-step sympy derivation

Equation id

vsb.v_off · equations.yaml entry

The positive volt-seconds applied while the switch is on must be cancelled by negative volt-seconds while it is off.

When the average is not zero

Over one period, the inductor current changes by the average inductor voltage times the period, over the inductance:

$$\Delta i = \frac{\langle v_L \rangle T_s}{L}$$

$\Delta i$

change of inductor current

$L$

inductance

$T_s$

switching period

$\langle v_L \rangle$

average inductor voltage over a period

Change of the inductor current over one period✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• linear (unsaturated) magnetic material

Notes

zero in a periodic steady state (volt-second balance); while the average is not zero, the current changes from one period to the next, as in a start-up or after a load step

Source

Erickson & Maksimović 2020, Ch. 2 (Principles of Steady-State Converter Analysis); derived here

Derivation

step-by-step sympy derivation

Equation id

vsb.drift · equations.yaml entry

During a transient (a start-up, a load or line step) this change carries the converter towards its new steady state, period by period. A circuit in which the average cannot reach zero has no steady state at all: a converter feeding a fixed output voltage in CCM, at a duty ratio above the one that balances it, gains the same inductor current every period, without bound [Erickson & Maksimović 2020, Ch. 2]. Below that duty ratio the current falls to zero within the period instead, and the converter settles in DCM, where the idle interval makes the average zero [Erickson & Maksimović 2020, Ch. 5].

Charge balance

The same argument applies to a capacitor: its voltage ends each period where it began, so the net charge delivered to it is zero and the average capacitor current is zero, $\langle i_C \rangle = 0$. This is capacitor charge balance [Erickson & Maksimović 2020, Ch. 2]. In two-interval form:

$$i_{C,\mathrm{off}} = - \frac{D i_{C,\mathrm{on}}}{1 - D}$$

$i_{C,\mathrm{off}}$

capacitor current, switch off

$D$

duty ratio

$i_{C,\mathrm{on}}$

capacitor current, switch on

Charge balance: capacitor current during the off-interval✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• periodic steady state
• small-ripple approximation

Notes

two-interval form of $\langle i_C \rangle = 0$

Source

Erickson & Maksimović 2020, Ch. 2 (Principles of Steady-State Converter Analysis)

Derivation

step-by-step sympy derivation

Equation id

csb.i_off · equations.yaml entry

Volt-second balance gives the voltage relations of a converter; charge balance gives its dc currents.

Buck

While the switch is on, the inductor sits between $V_g$ and the output, so its voltage is $V_g - V$. While the diode conducts, the inductor’s input end is grounded and its voltage is $-V$. Balancing the two gives the buck conversion ratio [Erickson & Maksimović 2020, Ch. 2]:

$$M = D$$

$M$

conversion ratio

$D$

duty ratio

Buck conversion ratio, CCM✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• small-ripple approximation

Convention

$M = V/V_g$

Source

Erickson & Maksimović 2020, Ch. 2 (Principles of Steady-State Converter Analysis)

Derivation

step-by-step sympy derivation

Equation id

buck.ccm.M · equations.yaml entry

Boost

The boost places the inductor at the input. With the switch on, the inductor sees $V_g$; with the diode conducting, it sees $V_g - V$. For these to balance, $V$ must exceed $V_g$:

$$M = \frac{1}{1 - D}$$

$M$

conversion ratio

$D$

duty ratio

Boost conversion ratio, CCM✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• small-ripple approximation

Convention

$M = V/V_g$

Source

Erickson & Maksimović 2020, Ch. 2 (Principles of Steady-State Converter Analysis)

Derivation

step-by-step sympy derivation

Equation id

boost.ccm.M · equations.yaml entry

Buck-boost

The buck-boost connects the inductor to $V_g$ while the switch is on, and to the output while the diode conducts. Its output is inverted, so the ratio is negative; its magnitude is below 1 for $D < 0.5$ and above 1 for $D > 0.5$:

$$M = - \frac{D}{1 - D}$$

$M$

conversion ratio

$D$

duty ratio

Buck-boost conversion ratio, CCM✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• small-ripple approximation

Convention

$M = V/V_g$; the output is inverted, so $M < 0$

Source

Erickson & Maksimović 2020, Ch. 2 (Principles of Steady-State Converter Analysis)

Derivation

step-by-step sympy derivation

Equation id

buckboost.ccm.M · equations.yaml entry

The step-by-step sympy derivations of all three ratios are on the derivations page.

Worked example

Two interval values, balanced: an inductor that sees a given voltage, and a capacitor that receives a given current, during the on-interval.

Balance example

Symbol	Value	How it is obtained
$D$ duty ratio	0.25	given
$v_{L,\mathrm{on}}$ inductor voltage, switch on	12 V	given
$i_{C,\mathrm{on}}$ capacitor current, switch on	1 A	given
$v_{L,\mathrm{off}}$ inductor voltage, switch off	−4 V	Volt-second balance: inductor voltage during the off-interval
$i_{C,\mathrm{off}}$ capacitor current, switch off	−333 mA	Charge balance: capacitor current during the off-interval

With $D$ = 0.25, the off-interval is three times as long as the on-interval, so the off-interval voltage has a third of the on-interval magnitude and the opposite sign.

The same balance applied to the boost and buck-boost examples, both at the same duty ratio:

Boost example

Symbol	Value	How it is obtained
$V_g$ input voltage	12 V	given
$D$ duty ratio	0.5	given
${M}_{\mathrm{ideal}}$ conversion ratio	2	Boost conversion ratio, CCM

Buck-boost example

Symbol	Value	How it is obtained
$V_g$ input voltage	12 V	given
$D$ duty ratio	0.5	given
$M$ conversion ratio	−1	Buck-boost conversion ratio, CCM

Try it

▶ Open in the equation explorer: Boost conversion ratio, CCM
Preset from the example boost-basic

▶ Open in the simulator: Boost
Preset from the example boost-basic

Sweep $D$ towards 1 and watch the ideal boost ratio grow without bound. The averaged-models page shows why a real boost cannot follow this curve.

A buck fed into a fixed output at a duty ratio that cannot balance has no steady state: the simulator shows its current running away, the average inductor voltage behind it, and the duty ratio that would balance.

▶ Open in the simulator: Buck
Preset from the example sim-buck-fixed

Bench exercise

On a low-voltage converter evaluation board, measure the inductor voltage with a differential probe (or two probes and the oscilloscope’s math subtraction). In steady state, read the two plateau voltages and the on and off times, and check that the two volt-second areas are equal and opposite. Then step the load and look again during the transient: while the inductor current is changing, the areas are no longer equal.

Gotchas

• Balance holds only in steady state. During start-up, load steps and line steps the average inductor voltage is not zero; that non-zero average is exactly what changes the inductor current [Erickson & Maksimović 2020, Ch. 2].
• Do not apply the small-ripple approximation to the ripple you are computing, or to the waveform that causes it. The buck’s output-voltage ripple is caused by the inductor current ripple, so the inductor current cannot be treated as constant when that ripple is computed [Erickson & Maksimović 2020, Ch. 2].
• Losses change the balance equations. Winding resistance and semiconductor drops add terms to the interval voltages and lower the ratio; see the averaged-models page.

Go deeper

• Book chapter Principles of Steady-State Converter Analysis (Ch. 2) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  The standard derivation of volt-second and charge balance, the small-ripple approximation and inductor/capacitor ripple.
• Course Introduction to Power Electronics — University of Colorado Boulder (Coursera) · retrieved 2026-09-23
  First course of the CU Boulder specialization: steady-state analysis, losses and efficiency.
• Application note Understanding Boost Power Stages in Switchmode Power Supplies (SLVA061) — E. Rogers, Texas Instruments · retrieved 2026-09-23
  The boost counterpart: steady-state and small-signal analysis in CCM and DCM.

Quiz

1. Why is the average inductor voltage over one period zero in periodic steady state?

   1. Because the inductor has no resistance
   2. Because the inductor current ends each period at the value it started with, so the integral of $v_L$ over the period is zero
   3. Because the switch is ideal
   4. Because the output capacitor is large
   Answer and explanation

   Correct answer: B The integral of the inductor voltage over a period equals $L$ times the change in inductor current over that period. In periodic steady state the current returns to its starting value, so the integral, and the average, are zero. No ideality assumption is needed.

2. An inductor sees $+10\,\mathrm{V}$ during the on-interval, with $D = 0.2$. What voltage must it see during the off-interval in steady state?

   1. $-10\,\mathrm{V}$
   2. $-2.5\,\mathrm{V}$
   3. $+2.5\,\mathrm{V}$
   4. $-40\,\mathrm{V}$
   Answer and explanation

   Correct answer: B The positive volt-seconds, $10\,\mathrm{V}$ for $0.2\,T_s$, must be cancelled during the off-interval of $0.8\,T_s$, which takes a quarter of the magnitude with the opposite sign.

3. Which condition gives the dc currents in a converter, for example the dc inductor current of a boost?

   1. Inductor volt-second balance
   2. The small-ripple approximation alone
   3. Capacitor charge balance
   4. Kirchhoff's voltage law around the switch
   Answer and explanation

   Correct answer: C Volt-second balance relates voltages; charge balance on the output capacitor relates the currents flowing into it during each interval, and so fixes the dc inductor current in terms of the load current.

4. An ideal boost converter in CCM runs at $D = 0.75$ from a $5\,\mathrm{V}$ input. What is its output voltage?

   1. $6.7\,\mathrm{V}$
   2. $15\,\mathrm{V}$
   3. $20\,\mathrm{V}$
   4. $3.75\,\mathrm{V}$
   Answer and explanation

   Correct answer: C The boost ratio is $1/(1 - D)$, which is 4 at this duty ratio, so the output is four times the input.

5. For which duty ratios is the output magnitude of an ideal CCM buck-boost larger than its input voltage?

   1. $D > 0.5$
   2. $D < 0.5$
   3. For every $D$
   4. For no $D$: it can only step down
   Answer and explanation

   Correct answer: A The magnitude of the ratio is $D/(1 - D)$, which exceeds 1 exactly when $D$ exceeds $1 - D$, that is for $D > 0.5$.

6. During a load step, the average inductor voltage over a switching period is not zero. What does that tell you?

   1. The converter has entered DCM
   2. The small-ripple approximation is violated
   3. The inductor is saturating
   4. The inductor current is changing from one period to the next
   Answer and explanation

   Correct answer: D A non-zero average inductor voltage means a net change of inductor current over the period. Volt-second balance holds only in steady state; during a transient the imbalance is what moves the current to its new value.