M6 · Match a source with a DCM flyback

Goal

• Work out the power a linear source can give and the input resistance of a flyback in DCM with a fixed duty ratio.
• Choose the magnetizing inductance that matches the two, and check the bus voltage and the power at the match.
• Find the limits of the match: a mismatched duty ratio, CCM at a strong source, and the switch’s rating.

Before you start

• Read source models, matching and the DCM flyback as a loss-free resistor.
• In DCM with a fixed duty ratio, the flyback’s average input current is proportional to its input voltage: its input is a resistance that passes the power on instead of dissipating it [Singer 1990].

Steps

1. The source

The example source has an open-circuit voltage of 150 V behind 10 kΩ. Work out the most power it can deliver (maximum power from a linear source).

2. The flyback’s input resistance

The flyback runs at $f_s$ = 10 kHz and $D$ = 0.2, with $L_M$ = 20 mH. Work out its input resistance (the loss-free resistor) and compare it with the source’s.

3. Design for the match

Turn the question round: at this duty ratio and switching frequency, which magnetizing inductance gives an input resistance equal to the source’s (magnetizing inductance for a resistance)?

4. The bus at the match

The two resistances divide the source voltage (bus voltage of a loss-free resistor). Work out the bus voltage, then open the source matcher with the example and read the same numbers there:

▶ Open in the source matcher: Example linear source feeding a fixed-duty-ratio DCM flyback
Preset from the example source-lfr

5. A mismatched duty ratio

Double $D$ in the tool. Predict the new input resistance and the share of the available power the flyback takes (extraction by a loss-free resistor), then compare with the tool.

6. Where the match ends

Set $D$ back and raise the source’s open-circuit voltage. Find the amplitude at which the bus reaches the critical voltage (critical input voltage) and CCM holds it there. Read the switch voltage at that point against the switch’s rating of 400 V.

Acceptance criteria

• The most power the example source can deliver.
  Your value (W)Check
  Show the example's value

  562.5 mW (from source-lfr: Example linear source feeding a fixed-duty-ratio DCM flyback)

• The flyback's input resistance at the example's $D$, $f_s$ and $L_M$.
  Your value (Ω)Check
  Show the example's value

  10 kΩ (from source-lfr: Example linear source feeding a fixed-duty-ratio DCM flyback)

• The magnetizing inductance that matches the source at that $D$ and $f_s$.
  Your value (H)Check
  Show the example's value

  20 mH (from lfr-flyback: Loss-free-resistor flyback example)

• The bus voltage at the match.
  Your value (V)Check
  Show the example's value

  75 V (from source-lfr: Example linear source feeding a fixed-duty-ratio DCM flyback)

• With $D$ doubled, your predicted input resistance and share of the power agree with the source matcher.
• The source amplitude at which CCM takes over the bus, and the switch voltage there against its rating.

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Gotchas

• The input resistance goes with the inverse square of the duty ratio: a small error in $D$ moves the match a lot.
• The flyback is a resistor only in DCM. Above the critical voltage CCM holds the bus, and the flyback becomes a constant-voltage sink whose share of the power depends on the source’s amplitude.
• A larger power through the same resistance needs a higher bus voltage, and the switch blocks the bus plus the reflected output voltage. Check the rating at the strongest source.

Go deeper

• Paper Realization of loss-free resistive elements — S. Singer · retrieved 2026-09-24
  The loss-free resistor: a converter that draws power like a resistor and delivers it instead of dissipating it.
• 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.

Quiz

1. A linear source of 20 V behind 100 ohm. What is the most power it can deliver?

   1. 4 W
   2. 1 W
   3. 2 W
   4. 0.2 W
   Answer and explanation

   Correct answer: B Into a load equal to its own resistance: 10 V across 100 ohm, 0.1 A, 1 W. That is $V_\mathrm{oc}^2$ over $4 R_s$.

2. A DCM flyback with a fixed duty ratio has $L_M$ = 1 mH, $f_s$ = 50 kHz and $D$ = 0.5. What is its input resistance, $2 L_M f_s$ over $D^2$?

   1. 100 ohm
   2. 200 ohm
   3. 400 ohm
   4. 800 ohm
   Answer and explanation

   Correct answer: C 2 times 1 mH times 50 kHz is 100 ohm, and over 0.25 that is 400 ohm.

3. The duty ratio of that flyback doubles. Its input resistance:

   1. Halves
   2. Falls to a quarter
   3. Doubles
   4. Stays the same
   Answer and explanation

   Correct answer: B The input resistance goes with the inverse square of the duty ratio.

4. The flyback's input resistance matches the source's. What is the bus voltage?

   1. The open-circuit voltage
   2. Half the open-circuit voltage
   3. Twice the open-circuit voltage
   4. The critical voltage
   Answer and explanation

   Correct answer: B The two equal resistances divide the source voltage in half, and the source then delivers its most power.

5. The source gets stronger until the bus would exceed the flyback's critical voltage. What happens?

   1. The flyback keeps acting as a resistor
   2. The flyback enters CCM, which holds the bus at the critical voltage: it becomes a constant-voltage sink
   3. The switch turns off
   4. The input resistance doubles
   Answer and explanation

   Correct answer: B In CCM the conversion ratio ties the input to the fixed output at this duty ratio, so the bus stays at the critical voltage. The share of the power then depends on the source's amplitude.

6. Why does a loss-free resistor take the same share of the available power at any source amplitude?

   1. Because it is a resistance: the divider ratio does not depend on the amplitude
   2. Because the duty ratio follows the amplitude
   3. Because the switch voltage is constant
   4. It does not: the share grows with the amplitude
   Answer and explanation

   Correct answer: A A resistance and the source resistance divide the voltage in a fixed ratio, so the power and the available power scale together with the square of the amplitude.