Boost converter - Discontinuous conduction mode (DCM)

  • August 28, 2026
  • 1 hour to complete
  • Workbench v1.9

Introduction

In the previous experiment, Boost converter - Continuous conduction mode (CCM) was analyzed in continuous-conduction mode (CCM), i.e., the inductor current was always greater than 0. In this experiment, the boost converter operation under discontinuous-conduction mode (DCM) is analyzed. Unlike CCM, under DCM the inductor current briefly goes to 0 during each switching cycle.

Theoretical background

A boost converter is shown below, with the transistor and the diode forming the two-position switch of the power pole.

Boost converter

Similar to CCM operation, under DCM, turning on the transistor increases the inductor current, and when the transistor is turned off, the inductor current decreases. During the off period, as the energy stored in the inductor is transferred to the output capacitor, the inductor current continues to fall until it reaches zero. If the next switching cycle begins before the current goes to zero, the converter is said to operate under CCM. Otherwise, if the current goes to zero and remains zero until the next switching cycle, the converter is said to operate under DCM. For a given transistor switching function waveform q(t), shown below, with a switch duty ratio d in steady state under DCM, the waveform of the voltage vA at the current port, for an ideal boost converter, is as shown below.

Ideal boost converter steady-state operation waveforms showing the gating signal, power-pole voltage, inductor current, and diode current

When the switch is ON, i.e., q(t) = 1, vA = 0. When the switch is OFF but the inductor current iL is not yet zero, vA = Vo. This is because the diode is conducting during this period. When the switch is OFF and the inductor current iL is zero, vA = Vin. This is because the diode is no longer conducting and it blocks the difference between the output and input voltage across it.

Since, under steady-state conditions, the average current through the output capacitor is zero and the average diode current Idiode equals the output current Io

\[I_{\mathrm{diode}} = I_{o} = \frac{V_{o}}{R}\tag{1}\]

At the boundary of CCM and DCM, the average diode current Idiode,crit equals

\[I_{\mathrm{diode},\mathrm{crit}} = \frac{1 - d}{2}I_{L,pk}\tag{2}\]

During the switch ON time period d × Ts, the inductor current rises from 0 to IL,pk. During this period, the voltage across the inductor is Vin = (1 - d) × Vo. Thus, the peak inductor current is

\[I_{L,pk} = (1 - d) \times \frac{V_{o}}{L} \times d \times T_{s}\tag{3}\]

From Eqns. 1 through 3, solving for the load resistance value at which the boost converter is at the border of CCM and DCM

\[R_{\mathrm{crit}} = \frac{2L}{T_{s} \times d \times (1 - d)^{2}}\tag{4}\]

During CCM the relationship between the input and the output voltage is

\[V_{o} = \frac{V_{\mathrm{in}}}{1 - d}\tag{5}\]

This relationship is not valid under DCM. The new relationship can be derived by assuming that the converter is ideal and that there is no power loss within the converter. Therefore,

\[P_{\mathrm{in}} = P_{\mathrm{out}}\]
\[\Rightarrow V_{\mathrm{in}} \times I_{\mathrm{in}}= \frac{V_{o}^{2}}{R}\tag{6}\]

From the input/inductor current waveform above, the average input current is given by

\[I_{\mathrm{in}} = \frac{1}{2} \times (d + d_{\mathrm{off},1}) \times I_{L,pk}\tag{7}\]

From the inductor current waveform, the peak inductor current is given by

\[I_{L,pk} = d \times T_{s} \times \frac{V_{\mathrm{in}}}{L}\tag{8}\]

From the diode current waveform, the average output current is given by

\[I_{o} = \frac{V_{o}}{R} = \frac{1}{2} \times I_{L,pk} \times d_{\mathrm{off},1}\tag{9}\]

Substituting Eqns. 7 through 9 into Eqn. 6

\[V_{\mathrm{in}} \times \frac{1}{2} \times (d + d_{\mathrm{off},1}) \times d \times T_{s} \times \frac{V_{\mathrm{in}}}{L} = \frac{V_{o}^{2}}{R}\]
\[\Rightarrow \frac{V_{\mathrm{in}}^{2} \times d^{2} \times T_{s}}{2L} + \frac{V_{\mathrm{in}} \times V_{o}}{R} = \frac{V_{o}^{2}}{R}\]
\[\Rightarrow V_{o}^{2} - V_{\mathrm{in}} \times V_{o}- M \times V_{\mathrm{in}}^{2} = 0\]

where \(M = \frac{R \times T_{s}}{2L} \times d^{2}\).

Solving the above quadratic equation yields

\[V_{o} = \frac{V_{\mathrm{in}}}{2} \times \left(1 + \sqrt{1 + 4M}\right) = M_{\mathrm{calc}} \times V_{\mathrm{in}}\tag{10}\]

where

\[M_{\mathrm{calc}} = \frac{1 + \sqrt{1 + 4M}}{2}\tag{11}\]

The above waveform was for an ideal boost converter. One stark difference between the ideal and actual response is the observed oscillation in vA and the inductor current iL during the doff,2Ts time period, as shown below.

Practical boost converter steady-state operation waveforms showing the gating signal, power-pole voltage, and inductor current

This is due to the fact that both the diode’s and the switch’s parasitic capacitances, along with the boost converter’s inductor, form an LC tank circuit as discussed in Buck converter - Discontinuous conduction mode (DCM).

Preparing the Workbench model

This model is exactly the same as the one used earlier in Boost converter - Continuous conduction mode (CCM).

  1. Copy the folder containing the pre-built example project for this experiment, usually located at C:\Program Files (x86)\Sciamble\WorkBench v1\Examples\CUSPLab\BasicPowerElectronics\Experiment6, and paste it in a location where the user has permission to edit and save files, such as the Desktop folder.

  2. Launch the Workbench application.

  3. Open the pre-built example project that was pasted in Step 1.

  4. Open the Workbench model logo Boost model within the project.

  5. Ensure that the PWM switching frequency on the Device configuration page in the project properties is set to 20 kHz.

  6. Ensure that the PWM channel is set to the Si power pole, i.e., Channel is set to 2.

Preparing the setup

This setup is exactly the same as the one used earlier in Boost converter - Continuous conduction mode (CCM).

    warningWarning

    Before proceeding, ensure that the isolated power supply is powered down and that the USB cable is disconnected.

  1. Magnetics card connection: Replace the existing magnetics card with the Buck/Boost/Buck-boost magnetics card, making sure that all six pins are properly aligned and in contact with the power-pole board, adjusting the angle if necessary.

    warningWarning

    Never leave the magnetics card unscrewed. If contact is lost while the converter is running, interrupting the inductor current can produce a very high voltage, potentially damaging the converter or creating a safety hazard.

  2. Rheostat setting: Set the slider such that the resistance across the two closest rheostat terminals is 40 Ω.

  3. Power connections:

    The boost-converter connections are the same as those of the buck converter, with the input and output terminals swapped.

    1. Connect the I+/I- terminals to the rheostat load:

      1. Rheostat rail: ● I+ (Red)

      2. Rheostat terminal closest to the rail: ● I− (Black)

    2. Connect the O+/O- terminals to the isolated power supply:

      1. DC +ve: ● O+ (Red)

      2. DC −ve: ● O− (Black)

    3. Connect the power supply ground to the ground terminal on the board:

      Ground: ● GND (Green)

  4. DSO connections:

    1. Connect DSO channel 1 probe to Ii. Set the following options, if they are supported by the DSO:

      1. Set the probe to 1x.

      2. Set the measurement type as Current.

      3. Set the scaling factor to 4x.

      4. Set the offset at 1.5 V or 6 A.

      5. Set termination to 1 MHz.

      6. Set the channel as inverted.

    2. Connect DSO channel 2 probe to Io. Use the same settings as Channel 1.

    3. Connect DSO channel 3 probe to Vi using the 20:1 attenuator. Set the following options, if they are supported by the DSO:

      1. Set the probe to 1x.

      2. Set the measurement type as Voltage.

      3. Set the scaling factor to 20x.

      4. Set the offset at 0 V.

      5. Set termination to 1 MHz.

      6. Set the channel as default/non-inverted.

    4. Connect DSO channel 4 probe to Sa using the 20:1 attenuator. Use the same settings as Channel 3.

    please noteNote

    Since the input and output power terminals are swapped, Channel 2, which is connected to Io, actually measures the input/inductor current. Channel 1, which is connected to Ii, actually measures the diode current, and Channel 3, which is connected to Vi+, actually measures the output voltage.

  5. Jumper settings:

    1. Insert jumpers ❶ and ❺ (●) to bypass the external current measurement.

    2. Insert the GaN/Si jumper ❷ (●) to Sa/Sw. This connects the buck-converter inductor to the midpoint of the Si power pole.

    3. Jumper ❸ (●) for the GaN FET’s external diode may remain inserted or removed, since this experiment does not use the GaN power pole.

    4. Insert jumper ❹ (●) to connect the drains of the top switches to the positive DC bus (Vin+).

    5. Remove jumper ❻ (●) to disconnect the capacitors across the switches.

  6. Connect the USB cable to the power-pole board and the computer.

  7. DC power supply settings:

    1. Make sure that the DC power supply is fully turned down to 0 V prior to turning on the supply.

    2. Turn on the power supply and gradually ramp up the voltage from 0 V to 15 V.

    3. If the option is available, set the power supply current limit at 4.5 A.

The final wiring should look similar to this:

Boost converter wiring diagram

Real-time open-loop control of boost converter - varying load resistance
Running the setup:

  1. Click on the Numerical simulation to Real-time mode transition button icon in the top dock of Workbench to transition from the simulation mode to the real-time mode.

  2. Click Numerical simulation and real-time prototyping Run button to run the control algorithm in real time.

  3. Gradually increment the duty cycle from 0 to 0.2.

  4. Gradually reduce the load resistance until the peak inductor current exceeds 3 A or an overcurrent fault occurs, using at least nine increments to make measurements: four in CCM, one at the boundary of CCM and DCM, and four in DCM.

    At the boundary of CCM and DCM, the inductor current (DSO channel 2) goes to zero for just an instant.

  5. Make the following measurements:

    1. Make a note of the voltage and current values displayed on Workbench for each incremental change in the load resistance.

    2. Observe the DSO waveforms and make a copy of the voltage across the switch (channel 4), the output voltage (channel 3), the inductor current (channel 2), and the diode current (channel 1) for each incremental change in the load resistance. Adjust the time base to show between 4 and 10 switching cycles.

    3. In DCM, make a note of the frequency of the ringing component in the measured voltage across the switch (channel 4).

  6. Click on Numerical simulation and real-time prototyping Stop button to stop the model.

If required, repeat the same experiment using the GaN power pole instead of the Si power pole, as demonstrated in Switching characteristic of Si MOSFET/GaN FET and diode.

Lab report and reading assignment

  1. Attach the DSO waveforms showing the voltage across the switch, the output voltage, the inductor current, and the diode current for varying load resistance.

  2. For a duty cycle of 0.2, enter the measured values from the Workbench screen capture and calculate the following values:

    Vin (V) Vo (V) Io (A) R (Ω) Mcalc Mact

    where Mcalc is the calculated voltage-conversion ratio. It is \(M_{\mathrm{calc}} = \frac{1}{1-d}\) in CCM and is given by Eqn. 11 in DCM. The measured voltage-conversion ratio is \(M_{\mathrm{act}} = \frac{V_{o}}{V_{\mathrm{in}}}\) under both CCM and DCM. Load resistance \(R = \frac{V_{o}}{I_{o}}\).

    Plot Mcalc and Mact as a function of the output current Io.

  3. Compute the Rcrit using Eqn. 4. Use the inductance value estimated in Boost converter - Continuous conduction mode (CCM) experiment. Compare this result with the measured value in the table above.

Reference

  1. "Power Electronics, A First Course," Ned Mohan and Siddharth Raju, Wiley Publication.



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