A design case

Intent

Follow one harvesting design from the source to the switch.
Estimate the average power a loss-free resistor takes from a source whose amplitude follows a slow envelope.
Check the design at the envelope’s peak: DCM, and the switch’s voltage.

Theory

The source

The case’s source is linear: an open-circuit voltage behind a large source resistance, whose amplitude rises and falls along a slow envelope, the same half sine over and over. At each instant it can give at most the maximum power of a linear source, a quantity that follows the square of the envelope.

The converter

A fixed-duty-ratio flyback in DCM, sized as a loss-free resistor equal to the source resistance (magnetizing inductance of the loss-free resistor), holds its input at half the source’s voltage and takes the whole available power at every instant (extraction by a loss-free resistor), as long as two things hold. It must stay in DCM at the envelope’s peak, where the bus is highest (critical input voltage). And the envelope must be slow against the bus’s own time constant, the bus capacitance times the source resistance in parallel with the input resistance, so that the bus follows the envelope [Alexander & Sadiku 2017, Ch. 7 (First-Order Circuits)].

The average power

Averaged over the envelope, the power is half its value at the peak, since the square of a sine averages to half its peak’s [Alexander & Sadiku 2017, Ch. 11]:

\bar{P} = \frac{\eta_\mathrm{ext} \hat{V}_\mathrm{oc}^{2}}{8 R_s}

$\bar{P}$: power averaged over the envelope
$R_s$: source resistance
$\hat{V}_\mathrm{oc}$: peak open-circuit voltage
$\eta_\mathrm{ext}$: share of the available power extracted

Average power a loss-free resistor takes from a source whose open-circuit voltage follows a sinusoidal envelope✓ verified by 8 tests

Assumptions, convention and source

Assumptions

linear (Thevenin) source V_oc behind R_s
envelope slow against the bus time constant, and the converter a loss-free resistor (DCM) throughout

Notes

the power

\eta_\mathrm{ext} V_\mathrm{oc}^2/(4 R_s)

follows the square of the envelope, and the square of a sine (or of a rectified sine) averages to half its peak's

Source

Alexander & Sadiku 2017, Sec. 4.8 (Maximum Power Transfer) and Ch. 11 (AC Power Analysis); derived here

Derivation

step-by-step sympy derivation

Equation id

lfr.P_env · equations.yaml entry

The source matcher simulates such an envelope cycle by cycle, and shows the bus held at the critical voltage where the peak would drive the flyback into CCM.

The switch

While the flyback switches, the envelope’s peak sets the highest voltage its switch blocks: the bus plus the reflected output (switch voltage), and the leakage spike on top (snubbers and clamps). A switch rated for the average, or for the envelope’s start, fails at its peak. While it does not switch, at start-up, in an undervoltage lockout or after a fault, nothing loads the bus, which charges to the source’s open-circuit peak, and switching then starts from there: the bus capacitor must take that peak, and the switch that peak plus the reflected output, unless a clamp holds the bus lower.

Worked example

The case, at the envelope’s peak and on average:

Harvesting source example

Symbol	Value	How it is obtained
$\hat{V}_\mathrm{oc}$ peak open-circuit voltage	1 kV	given
$R_s$ source resistance	1 MΩ	given
$T_\mathrm{env}$ envelope half-wave	10 s	given
$f_s$ switching frequency	1 kHz	given
$D$ duty ratio	0.05	given
$V$ output voltage	5 V	given
$V_D$ output-diode forward drop	500 mV	given
$n$ turns ratio, secondary/primary	0.1	given
$P_\mathrm{max}$ largest power available	250 mW	Maximum power from a linear source
$L_M$ magnetizing inductance (primary)	1.25 H	Magnetizing inductance of a DCM flyback that presents a given input resistance
$R_\mathrm{in}$ emulated input resistance	1 MΩ	Input resistance of a DCM buck-boost or flyback (loss-free resistor)
$V_g$ input voltage	500 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
$\bar{P}$ power averaged over the envelope	125 mW	Average power a loss-free resistor takes from a source whose open-circuit voltage follows a sinusoidal envelope
$V_{g,\mathrm{crit}}$ input voltage at the CCM/DCM boundary	1.04 kV	Flyback: input voltage at the CCM/DCM boundary (fixed D, fixed output)
$V_\mathrm{DS}$ switch voltage while off	555 V	Flyback switch voltage while off (without the leakage spike)

Operating mode: DCM at the envelope's peak: a loss-free resistor throughout ( $V_g < V_{g,\mathrm{crit}}$ : holds)

The source rises to 1 kV behind 1 MΩ, along an envelope of half-sines that each last 10 s; at the peak it can give 250 mW. At 1 kHz and 0.05, a magnetizing inductance of 1.25 H presents 1 MΩ, so the flyback takes the whole available power at every instant: 125 mW on average. At the peak the bus reaches 500 V, below the critical 1.045 kV, and the switch blocks 555 V before the leakage spike.

Try it

▶ Open in the equation explorer: Average power a loss-free resistor takes from a source whose open-circuit voltage follows a sinusoidal envelope
Preset from the example harvesting-source

Sweep the envelope’s peak: the average power follows its square.

▶ Open in the equation explorer: Extraction fraction of a loss-free resistor
Preset from the example harvesting-source

Sweep the flyback’s input resistance around the source resistance: the share stays close to the whole within a factor of two.

Bench exercise

Do not build the case at its full voltage on an open bench. Scale it down: divide the source voltage by a factor and keep the resistances, which divides the bus voltage by that factor and the power by its square; the output and its reflection through the transformer stay as they are. Emulate the source with a programmable supply stepping through the envelope behind a resistor equal to the source resistance, and the load with the flyback at its fixed duty ratio. Log the input voltage and current at each step, compute the power, and compare its average over the envelope with the formula, scaled.

Gotchas

An unloaded bus charges to the open-circuit voltage. Behind a large resistance the source’s current is small, but a capacitor across it still reaches the full voltage, and holds it: rate the bus and the switch for it, or clamp the bus.
The average needs a slow envelope. An envelope comparable to the bus’s time constant leaves the bus behind it, and the loss-free resistor’s share no longer holds at every instant.
Check DCM at the peak, not at the average (critical input voltage).
Rate the switch for the peak, and for an unloaded bus (switch voltage).

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.

Quiz

A source's open-circuit voltage rises to 500 V behind 1 Mohm. What is the most power it can give at the peak?
1. 62.5 mW
2. 125 mW
3. 250 mW
4. 31.3 mW
Answer and explanation
Correct answer: A 500 V squared over four times 1e6 ohm: 250000 over 4e6, which is 62.5 mW.
A matched loss-free resistor stays in DCM over a slow, sinusoidal envelope that peaks there. What average power does it take?
1. 62.5 mW
2. 31.3 mW
3. 15.6 mW
4. 44.2 mW
Answer and explanation
Correct answer: B The power follows the square of the envelope, and the square of a sine averages to half its peak's: half of 62.5 mW.
A flyback at 2 kHz and a duty ratio of 0.05 must present 1 Mohm. What magnetizing inductance does it need?
1. 1.25 H
2. 2.5 H
3. 0.625 H
4. 6.25 H
Answer and explanation
Correct answer: C The resistance times the duty ratio squared over twice the switching frequency: 1e6 times 0.0025 over 4000.
Where should DCM be checked for a source whose amplitude follows an envelope?
1. At the envelope's average
2. At the envelope's start
3. Anywhere: DCM does not depend on the source
4. At the envelope's peak, where the bus is highest
Answer and explanation
Correct answer: D The critical input voltage is fixed; the bus is highest at the peak, where CCM would first take over.
The bus capacitance is 1 uF, and both the source resistance and the flyback's input resistance are 1 Mohm. What is the bus's time constant?
1. 0.5 s
2. 1 s
3. 2 s
4. 0.25 s
Answer and explanation
Correct answer: A The capacitance times the two resistances in parallel: 1e-6 times 0.5e6 ohm.
The flyback stops switching and nothing else loads the bus. Where does the bus settle, behind a 1 kV open-circuit voltage and 1 Mohm?
1. At zero: 1 Mohm lets almost no current through
2. At half the open-circuit voltage
3. At the full open-circuit voltage, 1 kV
4. At the critical input voltage
Answer and explanation
Correct answer: C With no load, no current flows and no voltage falls across the source resistance: the capacitor reaches the full open-circuit voltage, and holds it.

The numbers in the examples, presets and exercises on this site are synthetic: round teaching values, not data from any real design.