Buck converter

Intent

• Find the buck converter’s dc inductor current, current ripple and peak current at an operating point.
• Estimate the output voltage ripple, and know what the ideal formula leaves out.
• State the voltage the switch and the diode must block, and where the buck leaves CCM.

Theory

Circuit and conversion ratio

Buck converter. $Q_1$ and $D_1$ connect the inductor to $V_g$ or to ground; $L$ and $C$ filter the switched voltage into the output $V$ across the load $R$. [Erickson & Maksimović 2020, Ch. 2]

The buck converter connects a switch between the input and the switch node, a diode from the switch node to ground, and an L-C low-pass filter from the switch node to the load [Erickson & Maksimović 2020, Ch. 2]. The filter passes the average of the switch node, so in CCM the conversion ratio equals the duty ratio (buck conversion ratio).

DC inductor current

The output capacitor carries no dc current (charge balance), so the inductor supplies the whole load current:

$$I_L = \frac{V}{R}$$

$I_L$

dc inductor current

$R$

load resistance

$V$

output voltage

Buck 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

charge balance: the output capacitor carries no dc current, so the inductor supplies the whole load current

Source

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

Derivation

step-by-step sympy derivation

Equation id

buck.IL · equations.yaml entry

Current ripple and peak current

While the switch is on, the inductor sees the difference between input and output voltage, and its current rises linearly; while the diode conducts, it sees the output voltage with the opposite sign and falls back [Erickson & Maksimović 2020, Ch. 2]. The half-ripple, and the peak-to-peak ripple in terms of the switching frequency, are

$$\Delta i_L = \frac{\left(V_g - V\right) D T_s}{2 L}$$

$\Delta i_L$

inductor ripple, half peak-to-peak

$D$

duty ratio

$L$

inductance

$T_s$

switching period

$V$

output voltage

$V_g$

input voltage

Buck 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

Convention

$\Delta i_L$ is HALF the peak-to-peak ripple (Erickson)

Source

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

Derivation

step-by-step sympy derivation

Equation id

buck.ripple.iL · equations.yaml entry

$$\Delta i_{L,\mathrm{pp}} = \frac{\left(V_g - V\right) D}{L f_s}$$

$\Delta i_{L,\mathrm{pp}}$

inductor ripple, peak-to-peak

$D$

duty ratio

$L$

inductance

$V$

output voltage

$V_g$

input voltage

$f_s$

switching frequency

Buck inductor current ripple, 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

Convention

peak-to-peak; twice the Erickson $\Delta i_L$

Source

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

Derivation

step-by-step sympy derivation

Equation id

buck.ripple.iL_pp · equations.yaml entry

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

$$I_\mathrm{pk} = I_L + \Delta i_L$$

$I_\mathrm{pk}$

peak current

$\Delta i_L$

inductor ripple, half peak-to-peak

$I_L$

dc inductor current

Peak inductor current: dc value plus half the ripple✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• continuous conduction mode (CCM)
• definition

Notes

the switch and the diode carry this peak current in turn

Source

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

Equation id

ripple.Ipk · equations.yaml entry

Output voltage ripple

The load takes the dc part of the inductor current and the capacitor takes the triangular ripple. During the half-period in which that ripple is positive, the capacitor gains a charge that raises its voltage from its minimum to its maximum [Erickson & Maksimović 2020, Ch. 2]:

$$\Delta v = \frac{\Delta i_L T_s}{8 C}$$

$\Delta v$

output ripple, half peak-to-peak

$C$

capacitance (output capacitor)

$\Delta i_L$

inductor ripple, half peak-to-peak

$T_s$

switching period

Buck 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
• the whole inductor ripple current flows into the output capacitor; the load draws only dc

Convention

$\Delta v$ is HALF the peak-to-peak ripple, like $\Delta i_L$ (Erickson)

Notes

ideal capacitor; a capacitor's series resistance adds a ripple term of its own

Source

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

Derivation

step-by-step sympy derivation

Equation id

buck.ripple.v · equations.yaml entry

The capacitor is fed by a continuous triangular current rather than by pulses, which keeps the buck’s output ripple small for a given capacitance. The derivation is on the derivations page.

Switch stress and the mode boundary

The switch blocks the input voltage while it is off, and the diode blocks it while the switch is on [Erickson & Maksimović 2020, Ch. 6]:

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

$V_\mathrm{DS}$

switch voltage while off

$V_g$

input voltage

Buck 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

Source

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

Derivation

step-by-step sympy derivation

Equation id

buck.Vds · equations.yaml entry

Both carry the peak inductor current [Erickson & Maksimović 2020, Ch. 6]. At light load the inductor current reaches zero and the buck enters DCM once $K$ falls below its critical value; its output then rises above $D V_g$ (DCM ratio).

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 while the diode blocks, then the diode carries the inductor’s current while the switch is off [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 D blocks: L lies between the input and the output. The voltage across L averages 12 V. L's current grows: it stores energy. C discharges, then charges. 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 on through D, which conducts. The voltage across L averages −12 V. L's current shrinks: it releases energy. C charges, then discharges. The input supplies nothing.

The operating modes of one steady period of the example buck-basic, 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

At light load the inductor’s current reaches zero before the period ends, and a third mode follows in which neither the switch nor the diode conducts and the capacitor alone feeds the load [Erickson & Maksimović 2020, Ch. 5]:

carries current  no current

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

S conducts and D blocks: L lies between the input and the output. The voltage across L averages 8.27 V. L's current grows: it stores energy. C discharges, then charges. The input delivers power.

(b) Mode 2 switch off (t₁ to t₂: 5 µs to 7.62 µs, 2.62 µs)

S is off: L's current flows on through D, which conducts. The voltage across L averages −15.8 V. L's current shrinks: it releases energy. C charges, then discharges. The input supplies nothing.

(c) Mode 3 idle (t₂ to t₃: 7.62 µs to 10 µs, 2.38 µs)

L's current is zero (DCM): S and D are both off. L carries no current. C discharges. The input supplies nothing.

The operating modes of one steady period of the example buck-light-load, 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

The buck example of the switching-principle page, with its currents and ripple:

Buck example

Symbol	Value	How it is obtained
$V_g$ input voltage	24 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	10 Ω	given
$V$ output voltage	12 V	Output voltage from the conversion ratio
$I_L$ dc inductor current	1.2 A	Buck dc inductor current, CCM
$\Delta i_L$ inductor ripple, half peak-to-peak	300 mA	Buck inductor current ripple (half peak-to-peak), CCM
$\Delta i_{L,\mathrm{pp}}$ inductor ripple, peak-to-peak	600 mA	Buck inductor current ripple, peak-to-peak, CCM
$I_\mathrm{pk}$ peak current	1.5 A	Peak inductor current: dc value plus half the ripple
$\Delta v$ output ripple, half peak-to-peak	37.5 mV	Buck output voltage ripple (half peak-to-peak), CCM
$K$ inductor parameter (CCM vs DCM)	2	Dimensionless inductor parameter K
$K_\mathrm{crit}$ K at the CCM/DCM boundary	0.5	Critical K of the buck converter

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

The half-ripple of 300 mA is small compared with the dc current of 1.2 A, so the small-ripple approximation holds. The switch and the diode must carry 1.5 A at the peak, and the ideal output ripple is 37.5 mV, a small fraction of the output voltage.

Try it

▶ Open in the equation explorer: Buck output voltage ripple (half peak-to-peak), CCM
Preset from the example buck-basic

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

Sweep $C$ over two decades (logarithmic axis): the ripple falls in inverse proportion to the capacitance. Then halve $T_s$: at the same current ripple the voltage ripple halves. In a real converter the current ripple halves too, so doubling the switching frequency divides the output ripple by four.

Bench exercise

On a low-voltage buck evaluation board, measure the inductor current with a current probe and read its peak-to-peak ripple; compare it with the peak-to-peak formula at the measured input voltage, output voltage and switching frequency. Then measure the output ripple across the capacitor terminals with a short probe ground connection. Expect more ripple than the ideal formula predicts: the capacitor’s series resistance and inductance add their own terms.

Gotchas

• The capacitor’s series resistance adds ripple. It adds a voltage equal to that resistance times the peak-to-peak ripple current, in phase with the current. Compare it with the capacitive term above: depending on the capacitor, either can dominate.
• Size the inductor for the peak current, not the dc current. Above its saturation flux density a magnetic core loses its permeability and the inductance collapses [Erickson & Maksimović 2020, Ch. 10]; the peak inductor current must stay below that limit.
• The ripple formula assumes that the capacitor takes all the ripple. That holds when the capacitor’s impedance at the switching frequency is much lower than the load’s.

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 Buck Power Stages in Switchmode Power Supplies (SLVA057) — E. Rogers, Texas Instruments · retrieved 2026-09-23
  Steady-state and small-signal analysis of the buck power stage in CCM and DCM, with component requirements.
• Application note Basic Calculation of a Buck Converter's Power Stage (SLVA477) — B. Hauke, Texas Instruments · retrieved 2026-09-23
  A compact CCM design-calculation checklist: duty ratio, inductor ripple, capacitors.

Quiz

1. A buck converter delivers $5\,\mathrm{V}$ into a $2.5\,\Omega$ load in CCM. What is its dc inductor current?

   1. $0.5\,\mathrm{A}$
   2. $1\,\mathrm{A}$
   3. $2\,\mathrm{A}$
   4. It depends on the duty ratio
   Answer and explanation

   Correct answer: C The output capacitor carries no dc current, so the inductor supplies the whole load current, the output voltage divided by the load resistance.

2. The dc inductor current is $2\,\mathrm{A}$ and the half-ripple $\Delta i_L$ is $0.4\,\mathrm{A}$. What peak current must the switch carry?

   1. $2\,\mathrm{A}$
   2. $2.2\,\mathrm{A}$
   3. $2.4\,\mathrm{A}$
   4. $2.8\,\mathrm{A}$
   Answer and explanation

   Correct answer: C The peak is the dc value plus half the peak-to-peak ripple, and $\Delta i_L$ already is that half.

3. The switching frequency of a CCM buck is doubled, with the same $L$, $C$, duty ratio and load. What happens to the ideal output voltage ripple?

   1. It halves
   2. It falls to a quarter
   3. It is unchanged
   4. It doubles
   Answer and explanation

   Correct answer: B The current ripple is proportional to $T_s$, and the voltage ripple is proportional to the current ripple times $T_s$, so it scales with $T_s^2$.

4. What voltage must the diode of an ideal buck converter block while the switch is on?

   1. The output voltage
   2. The input voltage
   3. The difference between input and output voltage
   4. Twice the input voltage
   Answer and explanation

   Correct answer: B With the switch on, the switch node sits at $V_g$ while the diode anode is grounded, so the diode blocks $V_g$.

5. Why is the output ripple of a buck converter small compared with that of a boost with the same output capacitance?

   1. The buck switches faster
   2. The buck output capacitor receives a continuous triangular current, while the boost capacitor must supply the load alone during the on-interval
   3. The buck has no diode
   4. The buck output voltage is lower
   Answer and explanation

   Correct answer: B In the buck the inductor feeds the output continuously and only the small triangular ripple reaches the capacitor. In the boost, the diode current pulses, and the capacitor alone supplies the load during $D T_s$.

6. A buck capacitor has a series resistance. Its measured output ripple is larger than the ideal formula predicts. Why?

   1. The series resistance adds a ripple voltage equal to the resistance times the peak-to-peak ripple current
   2. The series resistance lowers the switching frequency
   3. The series resistance increases the inductor ripple
   4. The series resistance makes the converter enter DCM
   Answer and explanation

   Correct answer: A The ripple current flows through the capacitor and its series resistance; the resistive drop adds to the capacitive ripple and can be the larger of the two.