Boost converter

Intent

• Find the boost converter’s inductor current, current ripple and peak current at an operating point.
• Estimate its output voltage ripple, and explain why it is larger than a buck’s.
• State the switch and diode stress, and the two effects that limit a boost at high duty ratio.

Theory

Circuit and conversion ratio

Boost converter. While $Q_1$ is on, $V_g$ drives the current in $L$ up; while it is off, $L$ and $V_g$ together supply the output through $D_1$, so $V > V_g$. [Erickson & Maksimović 2020, Ch. 2]

The boost converter places the inductor in series with the input, a switch from the inductor’s far end to ground, and a diode from that node to the output capacitor and load [Erickson & Maksimović 2020, Ch. 2]. Storing energy in the inductor while the switch is on and releasing it on top of the input voltage while the diode conducts raises the output above the input (boost conversion ratio).

DC inductor current

The inductor carries the input current. Charge balance on the output capacitor, which the diode feeds only during the off-interval, gives

$$I_L = \frac{V}{\left(1 - D\right) R}$$

$I_L$

dc inductor current

$D$

duty ratio

$R$

load resistance

$V$

output voltage

Boost dc inductor current, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

the inductor carries the input current, larger than the load current by the factor $1/(1 - D)$

Source

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

Derivation

step-by-step sympy derivation

Equation id

boost.IL · equations.yaml entry

so the inductor current exceeds the load current by the factor $1/(1 - D)$: at high step-up the inductor, the switch and the diode carry much more current than the load.

Current ripple and peak current

While the switch is on, the inductor sees the input voltage and its current rises linearly [Erickson & Maksimović 2020, Ch. 2]:

$$\Delta i_L = \frac{V_g D T_s}{2 L}$$

$\Delta i_L$

inductor ripple, half peak-to-peak

$D$

duty ratio

$L$

inductance

$T_s$

switching period

$V_g$

input voltage

Boost inductor current ripple (half peak-to-peak), CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Source

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

Derivation

step-by-step sympy derivation

Equation id

boost.ripple.iL · equations.yaml entry

The switch and the diode carry the inductor current in turn, so both see its peak value (peak current).

Output voltage ripple

The diode current reaches the output in pulses. While the switch is on, the capacitor alone supplies the load, and its voltage falls [Erickson & Maksimović 2020, Ch. 2]:

$$\Delta v = \frac{V D T_s}{2 R C}$$

$\Delta v$

output ripple, half peak-to-peak

$C$

capacitance (output capacitor)

$D$

duty ratio

$R$

load resistance

$T_s$

switching period

$V$

output voltage

Boost output voltage ripple (half peak-to-peak), CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

while the switch is on, the capacitor alone supplies the load

Source

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

Derivation

step-by-step sympy derivation

Equation id

boost.ripple.v · equations.yaml entry

Because the capacitor supplies the full load current for part of every period, instead of absorbing only a small triangular ripple, the boost needs more output capacitance than a buck for the same ripple. The derivation is on the derivations page.

Stress and limits

With the switch off, the switch blocks the output voltage; with the switch on, the diode blocks it [Erickson & Maksimović 2020, Ch. 6]:

$$V_\mathrm{DS} = V$$

$V_\mathrm{DS}$

switch voltage while off

$V$

output voltage

Boost switch voltage while off (also the diode's blocking voltage while on)✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits)

Derivation

step-by-step sympy derivation

Equation id

boost.Vds · equations.yaml entry

Two effects limit the boost at high duty ratio. The inductor’s winding resistance makes the ratio peak and then collapse (ratio with winding resistance), and the control-to-output transfer function has a right-half-plane zero that moves down as $D$ rises (RHP zero).

Operating modes

This section looks at one steady period of the example, mode by mode. The waveforms come from the site’s simulator, whose operating-mode view draws the same for any values. In CCM the period has two modes: the switch conducts and the inductor charges from the input while the capacitor alone feeds the load, then the switch is off and the diode carries the inductor’s current to the output [Erickson & Maksimović 2020, Ch. 2].

carries current  no current

(a) Mode 1 switch on (t₀ to t₁: 0 s to 5 µs, 5 µs)

S conducts and connects L's output end to ground. D blocks. The voltage across L averages 12 V. L's current grows: it stores energy. C discharges. The input delivers power.

(b) Mode 2 switch off (t₁ to t₂: 5 µs to 10 µs, 5 µs)

S is off: L's current flows through D into the output. The voltage across L averages −12 V. L's current shrinks: it releases energy. C charges. The input delivers power.

The operating modes of one steady period of the example boost-ideal, from the site's simulator: the key waveforms with the modes' boundaries $t_0, t_1, \ldots$, then the circuit in each mode, its current-carrying branches in colour. ▶ Change the values in the simulator

Worked example

An ideal boost example:

Ideal boost example

Symbol	Value	How it is obtained
$V_g$ input voltage	12 V	given
$D$ duty ratio	0.5	given
$f_s$ switching frequency	100 kHz	given
$L$ inductance	100 µH	given
$C$ capacitance (output capacitor)	10 µF	given
$R$ load resistance	20 Ω	given
$V$ output voltage	24 V	Output voltage from the conversion ratio
$I_L$ dc inductor current	2.4 A	Boost dc inductor current, CCM
$\Delta i_L$ inductor ripple, half peak-to-peak	300 mA	Boost inductor current ripple (half peak-to-peak), CCM
$I_\mathrm{pk}$ peak current	2.7 A	Peak inductor current: dc value plus half the ripple
$\Delta v$ output ripple, half peak-to-peak	300 mV	Boost output voltage ripple (half peak-to-peak), CCM
$K$ inductor parameter (CCM vs DCM)	1	Dimensionless inductor parameter K
$K_\mathrm{crit}$ K at the CCM/DCM boundary	0.125	Critical K of the boost converter

Operating mode: CCM ($K > K_\mathrm{crit}$: holds)

At this duty ratio the inductor current, 2.4 A, is twice the load current, and the switch and diode must carry 2.7 A at the peak. With the same capacitance as the buck example, the output ripple of 300 mV is several times larger, because the capacitor alone feeds the load for half of every period.

Try it

▶ Open in the equation explorer: Boost dc inductor current, CCM
Preset from the example boost-ideal

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

Sweep $D$ from 0 to 1: at a fixed output voltage and load, the inductor current grows without bound as $D$ approaches 1.

Bench exercise

On a low-power boost evaluation board supplied from a current-limited bench supply, measure the input current with a current probe: it is the inductor current, continuous and triangular. Then measure the diode current, or the output capacitor current, and see the pulses. Compare the measured output ripple with the prediction at the measured duty ratio and load, then halve the load resistance and check that the ripple doubles.

Gotchas

• The switch cannot disconnect the load. Even with the switch held off, the inductor and the diode connect the input to the output, so the output follows the input and the capacitor charges through them at power-up.
• Low input voltage means high current. At a fixed output voltage and power, halving the input voltage doubles both the step-up ratio and the input current; the inductor, the switch and the diode must be rated for that current plus the half-ripple.
• Past the peak of the lossy ratio, more duty ratio gives less output. 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.
• 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.
• Application note Basic Calculation of a Boost Converter's Power Stage (SLVA372) — B. Hauke, Texas Instruments · retrieved 2026-09-23
  The boost version of the same checklist.

Quiz

1. An ideal CCM boost delivers $24\,\mathrm{V}$ into $12\,\Omega$ at $D = 0.5$. What is its dc inductor current?

   1. $1\,\mathrm{A}$
   2. $2\,\mathrm{A}$
   3. $4\,\mathrm{A}$
   4. $8\,\mathrm{A}$
   Answer and explanation

   Correct answer: C The load current is $2\,\mathrm{A}$, and the inductor carries the input current, larger by the factor $1/(1 - D)$, which is 2 here.

2. Which interval sets the output voltage ripple of a boost converter?

   1. The off-interval, when the diode charges the capacitor
   2. The on-interval, when the capacitor alone supplies the load
   3. The dead time between the two intervals
   4. Neither: the ripple is set by the inductor current ripple alone
   Answer and explanation

   Correct answer: B While the switch is on, the diode is off and the capacitor supplies the full load current by itself, so its voltage falls by $2\Delta v$.

3. What voltage must the switch of an ideal boost block while it is off?

   1. The input voltage
   2. The output voltage
   3. The sum of input and output voltage
   4. The difference between output and input voltage
   Answer and explanation

   Correct answer: B With the diode conducting, the switch node is connected to the output, so the switch blocks $V$.

4. The boost switch is held permanently off. What is the output voltage (ideal diode)?

   1. Zero
   2. The input voltage
   3. Twice the input voltage
   4. Undefined: the output floats
   Answer and explanation

   Correct answer: B The inductor and the diode still connect the input to the output, so in steady state the output follows the input; the switch cannot disconnect the load.

5. At a fixed output voltage and output power, the input voltage of a boost is halved. What happens to the dc inductor current?

   1. It halves
   2. It stays the same
   3. It doubles
   4. It quadruples
   Answer and explanation

   Correct answer: C The inductor carries the input current, and the input power, equal to the output power in the ideal converter, is the input voltage times the input current.

6. Why does a boost need more output capacitance than a buck for the same ripple?

   1. Its switching frequency is lower
   2. Its capacitor must supply the full load current during every on-interval, instead of absorbing only a small triangular ripple
   3. Its inductor is larger
   4. Its output voltage is higher
   Answer and explanation

   Correct answer: B In the buck the inductor feeds the output continuously; in the boost the diode current arrives in pulses, and between pulses the capacitor alone carries the load.