The DCM flyback as a loss-free resistor

Intent

• Size a fixed-duty-ratio DCM flyback so that it presents a source’s resistance.
• Find the source amplitude above which it leaves DCM, and know what it does then.
• Know what voltage its switch blocks, and why more power through the same resistance needs more of it.

Theory

Sizing the loss-free resistor

A flyback in DCM with a fixed duty ratio presents the loss-free resistor at its input [Erickson & Maksimović 2020, Ch. 15 (AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode)]. Solved for the magnetizing inductance, that resistance gives the inductance a design needs:

$$L_M = \frac{R_\mathrm{in} D^{2}}{2 f_s}$$

$L_M$

magnetizing inductance (primary)

$D$

duty ratio

$R_\mathrm{in}$

emulated input resistance

$f_s$

switching frequency

Magnetizing inductance of a DCM flyback that presents a given input resistance✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

the loss-free resistor $R_\mathrm{in} = 2 L_M f_s / D^2$ solved for $L_M$; to match a linear source, $R_\mathrm{in} = R_s$

Source

Erickson & Maksimović 2020, Ch. 15 (AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode); derived here; Singer 1990

Derivation

step-by-step sympy derivation

Equation id

lfr.L_M · equations.yaml entry

To match a linear source, the input resistance is set to the source resistance (extraction by a loss-free resistor). A lower switching frequency, or a larger duty ratio, needs a larger inductance for the same resistance. The output voltage does not enter: in DCM the input resistance depends only on the inductance, the switching frequency and the duty ratio.

Where DCM ends

The flyback stays in DCM while its input voltage is below the critical input voltage, which the output voltage, the diode’s drop, the duty ratio and the turns ratio set. A source strong enough to raise the divider voltage (bus voltage of a loss-free resistor) above it drives the flyback into CCM. With the output fixed, the CCM conversion ratio then allows only one input voltage, the critical one [Erickson & Maksimović 2020, Ch. 6 (Converter Circuits)]: the bus is held there, and the flyback becomes a constant-voltage sink, whose share of the available power depends on the source’s amplitude (constant-voltage extraction).

What the switch blocks, and more power

The switch blocks the input voltage plus the output voltage reflected through the turns ratio (switch voltage), and the leakage inductance’s spike adds on top (snubbers and clamps). Through the same input resistance, a larger power needs a higher input voltage:

$$V_g = \sqrt{P R_\mathrm{in}}$$

$V_g$

input voltage

$P$

power

$R_\mathrm{in}$

emulated input resistance

Input voltage at which a loss-free resistor draws a given power✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

more power through the same input resistance needs a higher input voltage, which the switch must also block

Source

Erickson & Maksimović 2020, Ch. 15 (AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode); derived here; Singer 1990

Derivation

step-by-step sympy derivation

Equation id

lfr.Vg_power · equations.yaml entry

so drawing more power through a matched flyback raises its switch’s voltage. That holds up to the critical voltage. There CCM holds the bus, and while the flyback switches, the switch voltage stays at its CCM value whatever power the source then pushes in; the largest power the flyback draws as a resistor is the power a DCM converter draws at the critical voltage.

Worked example

A flyback sized to present the source’s resistance:

Loss-free-resistor flyback example

Symbol	Value	How it is obtained
$V_\mathrm{oc}$ source open-circuit voltage	150 V	given
$R_s$ source resistance	10 kΩ	given
$D$ duty ratio	0.2	given
$f_s$ switching frequency	10 kHz	given
$V$ output voltage	5 V	given
$V_D$ output-diode forward drop	500 mV	given
$n$ turns ratio, secondary/primary	0.1	given
$L_M$ magnetizing inductance (primary)	20 mH	Magnetizing inductance of a DCM flyback that presents a given input resistance
$R_\mathrm{in}$ emulated input resistance	10 kΩ	Input resistance of a DCM buck-boost or flyback (loss-free resistor)
$V_g$ input voltage	75 V	Bus voltage of a loss-free resistor fed by a linear source
$\eta_\mathrm{ext}$ share of the available power extracted	1	Extraction fraction of a loss-free resistor
$P_\mathrm{max}$ largest power available	563 mW	Maximum power from a linear source
$V_{g,\mathrm{crit}}$ input voltage at the CCM/DCM boundary	220 V	Flyback: input voltage at the CCM/DCM boundary (fixed D, fixed output)
$V_\mathrm{DS}$ switch voltage while off	130 V	Flyback switch voltage while off (without the leakage spike)

Operating mode: DCM: the flyback is a loss-free resistor ($V_g < V_{g,\mathrm{crit}}$: holds)

At 0.2 and 10 kHz, a magnetizing inductance of 20 mH presents 10 kΩ, the source’s resistance. The source’s 150 V then divides to 75 V, below the critical input voltage of 220 V: the flyback takes the whole available 563 mW, and its switch blocks 130 V before the leakage spike.

Try it

▶ Open in the equation explorer: Magnetizing inductance of a DCM flyback that presents a given input resistance
Preset from the example lfr-flyback

Sweep the duty ratio: the inductance that presents the same resistance grows as its square.

▶ Open in the simulator: Flyback
Preset from the example flyback-source

A stronger source into the same kind of flyback: the bus rises until CCM begins and settles at the critical input voltage.

Bench exercise

Build the flyback with its duty ratio fixed and a bus capacitor across its input, large enough that with the series resistor it averages the pulsed input current over many switching periods, and feed it from a bench supply through a series resistor equal to its computed input resistance. Measure the input voltage: it should sit at half the supply’s voltage. Raise the supply slowly until the input voltage stops following it; that is the critical input voltage, to compare with the formula. The supply must reach twice that voltage, several hundred volts with the example’s numbers, and the bus capacitor holds what it reaches: keep the circuit enclosed, and discharge the capacitor through a resistor before touching it. A larger turns ratio n (N_s/N_p) lowers the critical voltage (critical input voltage) and keeps the exercise at a safer voltage. At each step, capture the switch’s drain voltage with a probe rated for it (probes), and read the plateau before the leakage spike against the formula.

Gotchas

• Above the critical input voltage the flyback no longer matches. CCM holds the bus there (critical input voltage).
• The switch voltage includes the reflected output, and the leakage spike adds to it (switch voltage).
• While the flyback does not switch, the bus charges to the open-circuit voltage: at start-up, in an undervoltage lockout or after a fault, nothing draws on the source. The switch then blocks that voltage plus the reflected output: rate it for that, or clamp the bus (the design case).
• More power through the same resistance needs a higher voltage (input voltage for a given power).
• The match is forgiving. The share is flat at its top: a small error in the inductance moves the input resistance by the same fraction, and the share falls by only about a quarter of that fraction squared (extraction by a loss-free resistor).

Go deeper

• Book chapter AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode (Ch. 15) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  The DCM averaged switch model, in which the input behaves as a loss-free resistor.
• 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.
• Paper Optimized piezoelectric energy harvesting circuit using step-down converter in discontinuous conduction mode — G. K. Ottman, H. F. Hofmann, G. A. Lesieutre · retrieved 2026-09-24
  A step-down converter in DCM between a piezoelectric element and a battery, run at the duty ratio that harvests the most.

Quiz

1. A flyback at 5 kHz and a duty ratio of 0.1 must present 20 kohm, the source resistance, in DCM. What magnetizing inductance does it need?

   1. 20 mH
   2. 40 mH
   3. 2 mH
   4. 200 mH
   Answer and explanation

   Correct answer: A The input resistance times the duty ratio squared over twice the switching frequency: 2e4 times 0.01 over 1e4, which is 0.02 H.

2. The duty ratio is doubled to 0.2 with the same switching frequency and resistance. What inductance is needed now?

   1. 40 mH
   2. 10 mH
   3. 80 mH
   4. 20 mH
   Answer and explanation

   Correct answer: C The inductance grows as the square of the duty ratio: four times 20 mH.

3. Where does a DCM flyback matched to a linear source hold its input voltage?

   1. At the source's open-circuit voltage
   2. At the critical input voltage
   3. At the output voltage times the turns ratio
   4. At half the source's open-circuit voltage
   Answer and explanation

   Correct answer: D Its input resistance equals the source resistance, so the two divide the source voltage in half.

4. The source grows until the divider voltage would exceed the critical input voltage. What does the flyback do?

   1. It enters CCM, which holds its input at the critical voltage
   2. It stays in DCM and its input keeps rising
   3. It stops switching
   4. Its output voltage rises with the source
   Answer and explanation

   Correct answer: A With the output fixed, the CCM conversion ratio allows only the critical input voltage: the bus is held there, and the flyback becomes a constant-voltage sink.

5. A flyback with a 5 V output, a 0.5 V diode, a duty ratio of 0.2 and a turns ratio of 0.1 sits at 75 V. What is its critical input voltage, and is it in DCM?

   1. 55 V, CCM
   2. 220 V, DCM
   3. 27.5 V, CCM
   4. 440 V, DCM
   Answer and explanation

   Correct answer: B 5.5 V times 0.8 over (0.1 times 0.2), which is 220 V; the input is below it, so the flyback is in DCM.

6. What voltage does its switch block, before the leakage spike?

   1. 75 V
   2. 80.5 V
   3. 130 V
   4. 550 V
   Answer and explanation

   Correct answer: C The input voltage plus the output and diode voltage reflected through the turns ratio: 75 V plus 5.5 V over 0.1.

7. The magnetizing inductance comes out 10 % low. About how much of the available power does the flyback lose?

   1. 10 %
   2. 5 %
   3. 1 %
   4. A quarter of a percent
   Answer and explanation

   Correct answer: D The input resistance is 10 % low too, and at the top of the share a fraction off costs about a quarter of its square: 4 times 0.9 over 1.9 squared is 0.997.