Matching

Intent

Work out the share of a linear source’s available power that a constant voltage and a loss-free resistor take.
See why a converter between the source and a battery lets the source work at its best voltage.
Know that a converter in DCM with a fixed duty ratio is a loss-free resistor, and what it does as the source’s amplitude changes.

Theory

A constant voltage

A battery, or a bus held at a fixed voltage $V_c$ , connected straight to a linear source draws the current that the difference between the two voltages drives through $R_s$ . The power it takes is that current times its own voltage [Alexander & Sadiku 2017, Sec. 4.8 (Maximum Power Transfer)]:

P = \frac{V_c \left(V_\mathrm{oc} - V_c\right)}{R_s}

$P$: power
$R_s$: source resistance
$V_c$: sink voltage
$V_\mathrm{oc}$: source open-circuit voltage

Power from a linear source into a constant-voltage sink✓ verified by 8 tests

Assumptions, convention and source

Assumptions

linear (Thevenin) source V_oc behind R_s

Notes

the source current

(V_\mathrm{oc} - V_c)/R_s

times the sink voltage; negative when

V_c > V_\mathrm{oc}

, when the sink returns energy to the source

Source

Alexander & Sadiku 2017, Sec. 4.8 (Maximum Power Transfer); derived here

Derivation

step-by-step sympy derivation

Equation id

src.cv_power · equations.yaml entry

As a share of the available power (maximum power from a linear source):

\eta_\mathrm{ext} = \frac{4 V_c \left(V_\mathrm{oc} - V_c\right)}{V_\mathrm{oc}^{2}}

$\eta_\mathrm{ext}$: share of the available power extracted
$V_c$: sink voltage
$V_\mathrm{oc}$: source open-circuit voltage

Extraction fraction of a constant-voltage sink✓ verified by 8 tests

Assumptions, convention and source

Assumptions

linear (Thevenin) source V_oc behind R_s

Notes

power into the sink as a fraction of

P_\mathrm{max}

; equals 1 only at

V_c = V_\mathrm{oc}/2

Source

Alexander & Sadiku 2017, Sec. 4.8 (Maximum Power Transfer); derived here

Derivation

step-by-step sympy derivation

Equation id

src.cv_extraction · equations.yaml entry

It takes the whole of it only at half the open-circuit voltage. A battery’s voltage is set by its charge, not by the source, and a harvester’s amplitude varies: connected straight, the two rarely meet at that point.

A converter in between

A converter between the source and the battery separates their voltages: its input can sit where the source gives the most while its output stays at the battery’s voltage. Ottman et al. ran a step-down converter in DCM between a piezoelectric element and a battery at the duty ratio that gives the most power, and report that the converter increases the harvested power “by approximately 325%” [Ottman et al. 2003].

The loss-free resistor

In DCM with a fixed duty ratio, a buck-boost or flyback draws an average input current proportional to its input voltage: its input is a resistance, the loss-free resistor, which passes the power on instead of dissipating it [Singer 1990] [Erickson & Maksimović 2020, Ch. 15 (AC and DC Equivalent Circuit Modeling of the Discontinuous Conduction Mode)]. With a linear source, that resistance and the source resistance divide the source voltage:

V_g = \frac{V_\mathrm{oc} R_\mathrm{in}}{R_s + R_\mathrm{in}}

$V_g$: input voltage
$R_\mathrm{in}$: emulated input resistance
$R_s$: source resistance
$V_\mathrm{oc}$: source open-circuit voltage

Bus voltage of a loss-free resistor fed by a linear source✓ verified by 8 tests

Assumptions, convention and source

Assumptions

linear (Thevenin) source V_oc behind R_s
discontinuous conduction mode (DCM)
periodic steady state

Notes

a converter whose input is a resistance divides the source voltage with the source resistance

Source

Alexander & Sadiku 2017, Sec. 4.8 (Maximum Power Transfer); derived here

Derivation

step-by-step sympy derivation

Equation id

lfr.Vg · equations.yaml entry

and the converter takes the same share of the available power at any source amplitude [Alexander & Sadiku 2017, Sec. 4.8 (Maximum Power Transfer)]:

\eta_\mathrm{ext} = \frac{4 R_s R_\mathrm{in}}{\left(R_s + R_\mathrm{in}\right)^{2}}

$\eta_\mathrm{ext}$: share of the available power extracted
$R_\mathrm{in}$: emulated input resistance
$R_s$: source resistance

Extraction fraction of a loss-free resistor✓ verified by 8 tests

Assumptions, convention and source

Assumptions

linear (Thevenin) source V_oc behind R_s
discontinuous conduction mode (DCM)
periodic steady state

Notes

power into the loss-free resistor as a fraction of

P_\mathrm{max}

; independent of the source amplitude, and 1 when

R_\mathrm{in} = R_s

Source

Alexander & Sadiku 2017, Sec. 4.8 (Maximum Power Transfer); derived here

Derivation

step-by-step sympy derivation

Equation id

lfr.eta · equations.yaml entry

The share is the whole of it when the input resistance equals the source resistance, and falls slowly on either side. A fixed duty ratio therefore keeps a linear source at its maximum power point as its amplitude changes, without measuring anything, as long as the converter stays in DCM (the DCM flyback).

Worked example

A linear source, into a battery and into a loss-free resistor:

Source matching example

Symbol	Value	How it is obtained
$V_\mathrm{oc}$ source open-circuit voltage	10 V	given
$R_s$ source resistance	25 Ω	given
$V_c$ sink voltage	3 V	given
$R_\mathrm{in}$ emulated input resistance	50 Ω	given
$P_\mathrm{max}$ largest power available	1 W	Maximum power from a linear source
${P}_{\mathrm{cv}}$ power	840 mW	Power from a linear source into a constant-voltage sink
${\eta_\mathrm{ext}}_{\mathrm{cv}}$ share of the available power extracted	0.84	Extraction fraction of a constant-voltage sink
$V_g$ input voltage	6.67 V	Bus voltage of a loss-free resistor fed by a linear source
${\eta_\mathrm{ext}}_{\mathrm{lfr}}$ share of the available power extracted	0.889	Extraction fraction of a loss-free resistor

The source gives at most 1 W. A battery at 3 V connected straight to it takes 840 mW, a share of 0.84. A loss-free resistor of 50 Ω holds the input at 6.67 V and takes a share of 0.889, and it keeps that share at any source amplitude.

Try it

▶ Open in the equation explorer: Extraction fraction of a constant-voltage sink
Preset from the example match-basic

Sweep the voltage a constant-voltage load holds: the share peaks at half the open-circuit voltage and falls to zero at both ends.

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

Sweep the input resistance: a factor of two either side of the source resistance still takes eight ninths of the available power.

Bench exercise

Build a linear source from a bench supply and a series power resistor. Connect an electronic load in constant-voltage mode, a battery’s stand-in, step its voltage from zero to the supply’s, and record the current at each step. Plot the power against the voltage and find where it peaks; compare with half the supply’s voltage. Then replace the load with a converter whose duty ratio is fixed in DCM, and measure its input voltage and current at two supply voltages: their ratio, the input resistance, stays the same.

Gotchas

A battery connected straight to a source works at the battery’s voltage, not at the source’s best (constant-voltage extraction).
The loss-free resistor holds only in DCM (loss-free resistor); a strong source can drive the converter into CCM (the DCM flyback).
Half the open-circuit voltage is the best point of a linear source only. A source that is not linear has its own (MPPT).
A fixed resistance keeps its share as the amplitude changes; a fixed voltage does not (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.
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

A linear source of 10 V behind 25 ohm charges a battery held at 5 V, connected straight to it. What share of the available power does the battery take?
1. Half
2. All of it
3. A quarter
4. Three quarters
Answer and explanation
Correct answer: B The battery sits at half the open-circuit voltage, where a constant voltage takes the whole available power: 0.2 A times 5 V is 1 W, and the available power is 100 over 100, also 1 W.
The same source charges a battery held at 2 V instead. What share does it take?
1. 0.2
2. 0.8
3. 0.64
4. 0.36
Answer and explanation
Correct answer: C Four times 2 times 8 over 10 squared: 64 over 100. The battery draws 8 V over 25 ohm, 0.32 A, at 2 V: 0.64 W of the available 1 W.
A converter that is a loss-free resistor of 100 ohm draws from the same 25 ohm source. What share does it take?
1. 0.64
2. 0.8
3. 0.25
4. 1
Answer and explanation
Correct answer: A Four times 25 times 100 over 125 squared: 10000 over 15625, which is 0.64.
The source's open-circuit voltage doubles. What happens to the loss-free resistor's share?
1. It doubles
2. It halves
3. It falls to a quarter
4. It stays the same
Answer and explanation
Correct answer: D The two resistances divide any source voltage in the same ratio: the share depends only on the resistances.
The 10 V source doubles to 20 V while the battery stays at 5 V. What share does the battery take now?
1. All of it
2. 0.75
3. 0.5
4. 0.25
Answer and explanation
Correct answer: B Four times 5 times 15 over 20 squared: 300 over 400. The share depends on the battery's voltage against the source's, so with the battery fixed it changes whenever the source's amplitude does.
Which converter input is a loss-free resistor?
1. A buck in CCM
2. Any converter with a fixed duty ratio
3. A buck-boost or a flyback in DCM with a fixed duty ratio
4. A boost in CCM with a fixed duty ratio
Answer and explanation
Correct answer: C In DCM with a fixed duty ratio, a buck-boost or flyback draws an average input current proportional to its input voltage, and passes the power on.

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