M2 · Find the critical K in the simulator

Goal

• Find, by experiment, the load resistance at which a buck and a boost enter DCM.
• Turn that resistance into the dimensionless $K$ and compare it with the critical value $K_\mathrm{crit}(D)$ of each converter.
• See what the output voltage does once the converter is in DCM.

Before you start

• Read CCM and DCM and the K parameter.
• Know the definition of $K$ (the K parameter) and the critical values of the buck (buck) and the boost (boost).

Steps

1. The buck’s boundary

Open the simulator with the buck example:

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

It starts in CCM with $R$ = 10 Ω. Raise $R$ step by step and watch the detected mode above the charts. It comes from the simulated inductor current, which reaches zero within the period in DCM. It is not computed from $K$. Close in on the resistance where the mode changes, from both sides.

2. K at the boundary

With $L$ = 100 µH and $f_s$ = 100 kHz, work out $K$ at the resistance you found and enter it below. The simulator also prints $K$ and $K_\mathrm{crit}$ next to the mode, so you can compare the equation’s boundary with the one the current shows.

3. The boost’s boundary

Do the same with the boost example, which starts at $R$ = 20 Ω:

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

At the same duty ratio its critical value is much smaller than the buck’s, so it stays in CCM up to a much lighter load.

4. The output in DCM

Open the buck at a light load, $R$ = 100 Ω:

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

Read the output voltage in the load table below the charts and enter it below. It lies above $D V_g$, because in DCM the conversion ratio depends on $K$ too (the buck's DCM ratio).

5. The boost’s worst duty ratio

The boost’s critical value is not monotonic in $D$. Sweep $D$ in the equation explorer and find where it peaks:

▶ Open in the equation explorer: Critical K of the boost converter
Preset from the example boost-ideal

Acceptance criteria

• $K$ where the buck example's detected mode changes from CCM to DCM, worked out from the resistance there.
  Your valueCheck
  Show the example's value

  0.5 (from buck-basic: Buck example)

• The same for the boost example.
  Your valueCheck
  Show the example's value

  0.125 (from boost-ideal: Ideal boost example)

• At both boundaries, the mode the simulated current shows agrees with the critical value the equations give.
• The output voltage of the buck at the light load, read in the simulator's load table.
  Your value (V)Check
  Show the example's value

  15.74 V (from buck-light-load: Buck example at light load)

• The duty ratio at which the boost's $K_\mathrm{crit}$ is largest, found in the explorer.

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Gotchas

• $K$ rises with the switching frequency and the inductance and falls with the resistance, so changing $f_s$ or $L$ moves the boundary as much as changing the load does.
• Near the boundary the current only just touches zero, and the smallest change can flip the mode. Take the resistance halfway between the last CCM and the first DCM setting.
• In DCM the output depends on the load. Holding it at a set point then takes a different duty ratio, which a regulator finds by itself.

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.

Quiz

1. A buck has $L$ = 50 µH, $f_s$ = 200 kHz and a load of 20 ohm. What is $K$?

   1. 0.5
   2. 1
   3. 2
   4. 0.25
   Answer and explanation

   Correct answer: B $K$ is $2L$ over $R T_s$, that is $2 L f_s$ over $R$: 2 times 50 µH times 200 kHz is 20 ohm, and over 20 ohm that gives 1.

2. That buck runs at $D$ = 0.75. Is it in CCM?

   1. Yes: $K$ = 1 is above $K_\mathrm{crit}$ = 0.25
   2. No: $K$ = 1 is below $K_\mathrm{crit}$ = 1.33
   3. Only if the output capacitor is large
   4. It is exactly at the boundary
   Answer and explanation

   Correct answer: A The buck's critical value is $1 - D$ = 0.25, and CCM needs $K > K_\mathrm{crit}$.

3. The load of a buck in CCM gets lighter and lighter. What happens first?

   1. The inductor saturates
   2. The output voltage falls
   3. The inductor current reaches zero before the period ends: the converter enters DCM
   4. The ripple of the inductor current grows
   Answer and explanation

   Correct answer: C A larger $R$ lowers $K$. The ripple does not depend on the load, so as the average current falls, the valley of the current reaches zero, and below $K_\mathrm{crit}$ it stays there for part of the period.

4. At which duty ratio is the boost's critical value largest?

   1. $D$ = 1/2
   2. $D$ = 1/3
   3. $D$ = 2/3
   4. It rises all the way to $D$ = 1
   Answer and explanation

   Correct answer: B The boost's critical value, $D(1-D)^2$, has zero slope at $D$ = 1/3, where it is 4/27. A boost with $K$ above 4/27 stays in CCM at every duty ratio.

5. A buck at $D$ = 0.5 enters DCM as its load gets lighter. Compared with $D V_g$, its output voltage is:

   1. Higher
   2. Lower
   3. The same: the ratio is $D$ in both modes
   4. Zero
   Answer and explanation

   Correct answer: A In DCM the ratio depends on $K$ as well, and rises from $D$ towards 1 as the load gets lighter.

6. You double the switching frequency of a buck that sits just inside DCM. What happens?

   1. Nothing: the boundary depends only on $D$
   2. $K$ doubles, and the converter can move back into CCM
   3. $K$ halves, and it moves deeper into DCM
   4. The duty ratio doubles
   Answer and explanation

   Correct answer: B $K$, which is $2 L f_s$ over $R$, grows with $f_s$: the ripple shrinks, so the valley of the current lifts off zero.