CCM and DCM

Intent

• Explain why the discontinuous conduction mode (DCM) appears at light load when the second switch is a diode.
• Compute the diode-conduction interval and the DCM conversion ratio of the buck, boost and buck-boost.
• Recognise DCM on an oscilloscope from its waveforms.

Theory

Where DCM comes from

In the basic converters the second switch is a diode, which conducts in one direction only. The inductor current is a dc value plus a triangular ripple. When the load gets lighter, the dc value falls but the ripple does not: in CCM the ripple depends only on the applied voltages, $L$ and $T_s$. Once the lower peak of the ripple reaches zero, the diode stops conducting before the period ends, and the inductor current stays at zero until the switch turns on again [Erickson & Maksimović 2020, Ch. 5]. Each period now has three subintervals: the switch conducts for $D T_s$, the diode for $D_2 T_s$, and neither for the remaining $D_3 T_s$. This is the discontinuous conduction mode (DCM). The normal case, in which the inductor current never reaches zero, is the continuous conduction mode (CCM).

Inductor current in CCM, at the boundary $K = K_\mathrm{crit}$, and in DCM, where it rises for $DT_s$, falls to zero in $D_2T_s$ and stays at zero for $D_3T_s$. [Erickson & Maksimović 2020, Ch. 5]

Which mode a converter runs in is decided by the dimensionless parameter K and its critical value; see the K-parameter page.

Analysing DCM

Volt-second balance and charge balance still hold, but two things change [Erickson & Maksimović 2020, Ch. 5]:

1. The diode interval $D_2$ is a new unknown, so a second equation is needed.
2. The inductor current ripple is no longer small compared with its dc value, so the small-ripple approximation must not be applied to the inductor current. Its average is computed from the actual triangular waveform.

For the buck, volt-second balance over the switch and diode intervals gives the diode interval in terms of the conversion ratio:

$$D_2 = \frac{D \left(1 - M\right)}{M}$$

$D_2$

diode-conduction fraction (DCM)

$D$

duty ratio

$M$

conversion ratio

Buck in DCM: diode-conduction interval✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

the idle interval is $D_3 = 1 - D - D_2$; $D_2 \le 1 - D$ holds exactly when $M \ge D$, so a DCM buck always has $M > D$

Source

Erickson & Maksimović 2020, Ch. 5 (The Discontinuous Conduction Mode)

Derivation

step-by-step sympy derivation

Equation id

buck.dcm.D2 · equations.yaml entry

Charge balance then sets the average inductor current, a triangle that starts and ends at zero, equal to the load current. Solving the two conditions together gives the buck DCM conversion ratio:

$$M = \frac{2}{1 + \sqrt{1 + \frac{4 K}{D^{2}}}}$$

$M$

conversion ratio

$D$

duty ratio

$K$

inductor parameter (CCM vs DCM)

Buck conversion ratio, DCM✓ verified by 8 tests

Assumptions, convention and source

Assumptions

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

Convention

$M = V/V_g$; valid for $K < K_\mathrm{crit} = 1 - D$

Source

Erickson & Maksimović 2020, Ch. 5 (The Discontinuous Conduction Mode)

Derivation

step-by-step sympy derivation

Equation id

buck.dcm.M · equations.yaml entry

The boost and the buck-boost follow the same route, with the average diode current set equal to the load current:

$$M = \frac{1 + \sqrt{1 + \frac{4 D^{2}}{K}}}{2}$$

$M$

conversion ratio

$D$

duty ratio

$K$

inductor parameter (CCM vs DCM)

Boost conversion ratio, DCM✓ verified by 8 tests

Assumptions, convention and source

Assumptions

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

Convention

$M = V/V_g$; valid for $K < K_\mathrm{crit} = D(1 - D)^2$

Source

Erickson & Maksimović 2020, Ch. 5 (The Discontinuous Conduction Mode)

Derivation

step-by-step sympy derivation

Equation id

boost.dcm.M · equations.yaml entry

$$M = - \frac{D}{\sqrt{K}}$$

$M$

conversion ratio

$D$

duty ratio

$K$

inductor parameter (CCM vs DCM)

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

Assumptions, convention and source

Assumptions

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

Convention

$M = V/V_g < 0$ (inverting); valid for $K < K_\mathrm{crit} = (1 - D)^2$

Source

Erickson & Maksimović 2020, Ch. 5 (The Discontinuous Conduction Mode)

Derivation

step-by-step sympy derivation

Equation id

buckboost.dcm.M · equations.yaml entry

The step-by-step derivations are on the derivations page.

What changes in DCM

• The conversion ratio depends on the load. In CCM, $M$ is a function of $D$ alone; in DCM it also depends on $K$, and so on $R$. As the load is removed, $K$ goes to zero: the buck output rises towards $V_g$, and the boost and buck-boost outputs rise without bound. An open-loop converter must therefore never run without a load.
• The peak inductor current no longer depends on the load. Each period starts from zero current, so the peak is set by the voltage applied during $D T_s$, by $L$ and by $T_s$.
• The switch node rings during the idle interval. A real switch node has capacitance, from the switch and the diode. When the inductor current reaches zero, that capacitance and the inductor form a lightly damped L-C circuit, which rings at

$$f_\mathrm{ring} = \frac{1}{2 \pi \sqrt{L_M C_\mathrm{node}}}$$

$f_\mathrm{ring}$

ringing frequency

$C_\mathrm{node}$

switch-node capacitance

$L_M$

magnetizing inductance (primary)

Ringing frequency of the switch node in the DCM idle interval✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• undamped L-C resonance

Notes

after the diode current reaches zero, $L_M$ rings with the switch-node capacitance

Source

Erickson & Maksimović 2020, Ch. 8 (Converter Transfer Functions: resonant frequency of an L-C pair)

Derivation

step-by-step sympy derivation

Equation id

dcm.ring.f · equations.yaml entry

Worked example

The buck example from the switching-principle page, now with a ten-times lighter load:

Buck example at light load

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
$R$ load resistance	100 Ω	given
$C$ capacitance (output capacitor)	10 µF	given
$T_s$ switching period	10 µs	Switching period
$K$ inductor parameter (CCM vs DCM)	0.2	Dimensionless inductor parameter K
$K_\mathrm{crit}$ K at the CCM/DCM boundary	0.5	Critical K of the buck converter
$L_\mathrm{crit}$ critical inductance (CCM/DCM boundary)	250 µH	Critical inductance at the CCM/DCM boundary
${M}_{\mathrm{ccm}}$ conversion ratio	0.5	Buck conversion ratio, CCM
${M}_{\mathrm{dcm}}$ conversion ratio	0.656	Buck conversion ratio, DCM
$D_2$ diode-conduction fraction (DCM)	0.262	Buck in DCM: diode-conduction interval
${V}_{\mathrm{dcm}}$ output voltage	15.7 V	Output voltage from the conversion ratio

Operating mode: DCM ($K > K_\mathrm{crit}$: does not hold)

The CCM formula would predict an output ratio of 0.5. In DCM the ratio is 0.656, so the output is 15.7 V rather than half the input. The diode conducts for a fraction 0.262 of the period; together with the switch’s 0.5, that leaves an idle interval.

Try it

▶ Open in the equation explorer: Buck conversion ratio, DCM
Preset from the example buck-light-load

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

Sweep $K$ over several decades (logarithmic axis). At small $K$ the ratio approaches 1. As $K$ grows the ratio falls, and it equals the CCM value $D$ exactly where $K$ reaches $K_\mathrm{crit}$. Beyond that point the converter is in CCM and this formula no longer applies, although the explorer keeps plotting it.

Bench exercise

Take a buck converter with a diode rectifier, such as a low-voltage evaluation board with an asynchronous controller in fixed-frequency mode, and a resistive load you can vary. Watch the inductor current with a current probe, and the switch node with a voltage probe, while you lighten the load. Note the load at which the current first touches zero, the appearance of the idle interval, and the ringing on the switch node during it. Check that, with the duty ratio held fixed, the output voltage rises as the load gets lighter.

Gotchas

• Synchronous rectification removes DCM. DCM arises because the diode cannot carry reverse current. A transistor driven as the complement of the main switch conducts in both directions, so the inductor current simply goes negative at light load and the converter stays in CCM [Erickson & Maksimović 2020, Ch. 5].
• “Light load” is relative. Whether a given load is light depends on $L$, $T_s$ and $D$ as well as on $R$; compare $K$ with $K_\mathrm{crit}$.
• Do not use CCM formulas in DCM. Ripple, peak current and conversion ratio all follow different expressions once the current reaches zero.

Go deeper

• Book chapter The Discontinuous Conduction Mode (Ch. 5) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  Origin of DCM, the K and K_crit boundary and the DCM conversion ratios derived with charge balance.
• Course Converter Circuits — University of Colorado Boulder (Coursera) · retrieved 2026-09-23
  DCM analysis and the isolated converter circuits, following the same textbook.
• 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.

Quiz

1. Why does a buck converter with a diode rectifier enter DCM at light load?

   1. The switching frequency drops at light load
   2. The dc inductor current falls while the ripple stays the same, so the current reaches zero and the diode cannot carry it negative
   3. The duty ratio becomes larger than 0.5
   4. The output capacitor discharges completely
   Answer and explanation

   Correct answer: B In CCM the ripple depends only on the applied voltages, $L$ and $T_s$, not on the load. When the dc current falls below the half-ripple, the lower peak would be negative; the diode blocks reverse current, so the current stays at zero for the rest of the period.

2. In DCM, which assumption used for CCM analysis must be dropped for the inductor current?

   1. Volt-second balance
   2. Charge balance
   3. The small-ripple approximation
   4. Periodic steady state
   Answer and explanation

   Correct answer: C In DCM the ripple is as large as the current itself, so the inductor current cannot be replaced by its dc value. Balance conditions still hold because the circuit is still in periodic steady state.

3. An open-loop buck converter in DCM has its load resistance increased. What happens to the output voltage at a fixed duty ratio?

   1. It rises, towards the input voltage
   2. It stays at $D V_g$
   3. It falls towards zero
   4. It oscillates at the switching frequency
   Answer and explanation

   Correct answer: A Increasing $R$ lowers $K$, and the DCM ratio rises towards 1 as $K$ goes to zero. Only in CCM is the ratio fixed by $D$ alone.

4. A buck converter runs at $D = 0.5$ and its output measures $0.7$ times the input. What can you conclude?

   1. The converter is in CCM with a lossy inductor
   2. The duty ratio measurement is wrong
   3. The converter is in DCM
   4. The converter is in CCM and at no load
   Answer and explanation

   Correct answer: C In CCM an ideal buck gives exactly $D$, and losses only lower the ratio. A ratio above $D$ means the inductor current reaches zero: the converter is in DCM, where $M > D$.

5. What causes the ringing on the switch node of a DCM converter during the idle interval?

   1. Reverse recovery of the diode at turn-on of the switch
   2. The inductor resonating with the capacitance at the switch node once its current has reached zero
   3. Resonance of the output capacitor with its ESR
   4. Noise from the control loop
   Answer and explanation

   Correct answer: B With both switch and diode off, the inductor and the switch-node capacitance form a lightly damped L-C circuit, which rings at its resonant frequency $1/(2\pi\sqrt{L C})$.

6. The diode of a buck converter is replaced by a transistor that is driven as the complement of the main switch. What happens at light load?

   1. The converter enters DCM earlier
   2. The converter stays in CCM: the inductor current simply goes negative for part of the period
   3. The output voltage rises towards the input voltage
   4. The switching frequency halves
   Answer and explanation

   Correct answer: B DCM exists because the diode cannot conduct in reverse. A synchronous rectifier conducts in both directions, so the current reverses instead of stopping, and the CCM relationships continue to hold.