Averaged models

Intent

Build the averaged dc model of a converter and add the inductor’s winding resistance to it.
Predict how that resistance limits the boost converter’s conversion ratio and efficiency.
Describe the DCM switch network as a loss-free resistor and compute its input resistance.

Theory

From switched circuit to dc model

Averaging every circuit equation over one switching period, with the small-ripple approximation, removes the switching action: the averaged inductor-voltage and capacitor-current equations are those of a dc circuit. In that circuit the switch network acts as an ideal dc transformer whose turns ratio is the conversion ratio. It passes power from input to output without loss, and every loss mechanism appears as an ordinary resistor or voltage drop around it [Erickson & Maksimović 2020, Ch. 3]. One model then gives the output voltage, the currents and the efficiency.

Boost with inductor winding resistance

Add the inductor’s winding resistance $R_L$ to the boost. Volt-second balance gains the drop across $R_L$ in both intervals, while charge balance is unchanged; solving the two together gives [Erickson & Maksimović 2020, Ch. 3]:

M = \frac{1}{\left(1 - D\right) \left(1 + \frac{R_L}{\left(1 - D\right)^{2} R}\right)}

$M$: conversion ratio
$D$: duty ratio
$R$: load resistance
$R_L$: inductor winding resistance

Boost conversion ratio with inductor winding resistance, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

continuous conduction mode (CCM)
periodic steady state
small-ripple approximation
inductor winding resistance R_L; all other components ideal
resistive load R

Notes

the second factor collapses the ratio at high

D

M

has a maximum instead of growing without bound

Source

Erickson & Maksimović 2020, Ch. 3 (Steady-State Equivalent Circuit Modeling, Losses, and Efficiency)

Derivation

step-by-step sympy derivation

Equation id

boost.ccm.M_RL · equations.yaml entry

The ideal boost ratio is multiplied by a factor that falls towards zero as $D$ approaches 1: the inductor carries the input current, which grows much faster than the output current as $D$ rises, so the loss in $R_L$ eventually takes all the extra input power. Instead of growing without bound, the conversion ratio reaches a maximum and then falls. The efficiency follows from the same model:

\eta = \frac{1}{1 + \frac{R_L}{\left(1 - D\right)^{2} R}}

$\eta$: efficiency
$D$: duty ratio
$R$: load resistance
$R_L$: inductor winding resistance

Boost efficiency with inductor winding resistance, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

continuous conduction mode (CCM)
periodic steady state
small-ripple approximation
inductor winding resistance R_L; all other components ideal
resistive load R

Source

Erickson & Maksimović 2020, Ch. 3 (Steady-State Equivalent Circuit Modeling, Losses, and Efficiency)

Derivation

step-by-step sympy derivation

Equation id

boost.ccm.eta_RL · equations.yaml entry

$R_L$ is compared with $(1 - D)^2 R$ , the load resistance reflected through the dc transformer to the inductor’s side. At high $D$ this reflected resistance is small, so even a small winding resistance costs efficiency. The derivation is on the derivations page.

DCM: the loss-free resistor

In DCM the inductor current starts from zero in every period. The energy stored during the on-interval, and so the power drawn from the input, then depends only on the input voltage, the duty ratio, the inductance and the switching frequency. For the buck-boost and the flyback, with the magnetizing inductance $L_M$ :

P = \frac{V_g^{2} D^{2}}{2 L_M f_s}

$P$: power
$D$: duty ratio
$L_M$: magnetizing inductance (primary)
$V_g$: input voltage
$f_s$: switching frequency

Power drawn by a DCM buck-boost or flyback✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

the energy

\tfrac12 L_M I_\mathrm{pk}^2

is stored once per cycle, independent of the load

Source

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

Derivation

step-by-step sympy derivation

Equation id

dcm.P_in · equations.yaml entry

The average input current is therefore proportional to the input voltage, and the input port behaves like a resistor [Singer 1990] [Erickson & Maksimović 2020, Ch. 15]:

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

$R_\mathrm{in}$: emulated input resistance
$D$: duty ratio
$L_M$: magnetizing inductance (primary)
$f_s$: switching frequency

Input resistance of a DCM buck-boost or flyback (loss-free resistor)✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

the average input current is proportional to

V_g

; the power is passed on, not dissipated

Source

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

Derivation

step-by-step sympy derivation

Equation id

lfr.R_in · equations.yaml entry

This resistor does not dissipate the power it absorbs: it passes that power to the output. The averaged DCM switch network is therefore modelled as a loss-free resistor, a property that energy-harvesting interfaces use to present a resistive load to a source. The source matcher matches a flyback’s loss-free resistor to a linear source.

Worked example

The boost example, with a winding resistance:

Boost example

Symbol	Value	How it is obtained
$V_g$ input voltage	12 V	given
$D$ duty ratio	0.5	given
$R$ load resistance	20 Ω	given
$R_L$ inductor winding resistance	200 mΩ	given
${M}_{\mathrm{ideal}}$ conversion ratio	2	Boost conversion ratio, CCM
$M$ conversion ratio	1.92	Boost conversion ratio with inductor winding resistance, CCM
$\eta$ efficiency	0.962	Boost efficiency with inductor winding resistance, CCM
$V$ output voltage	23.1 V	Output voltage from the conversion ratio

A winding resistance of 200 mΩ, only a hundredth of the load resistance, lowers the ratio from 2 to 1.92 and the efficiency to 0.962, because at this duty ratio it is compared with a quarter of the load resistance.

The DCM buck-boost example, seen from its input:

DCM buck-boost example

Symbol	Value	How it is obtained
$V_g$ input voltage	10 V	given
$D$ duty ratio	0.5	given
$L_M$ magnetizing inductance (primary)	100 µH	given
$f_s$ switching frequency	100 kHz	given
$P$ power	1.25 W	Power drawn by a DCM buck-boost or flyback
$R_\mathrm{in}$ emulated input resistance	80 Ω	Input resistance of a DCM buck-boost or flyback (loss-free resistor)

Try it

▶ Open in the equation explorer: Boost conversion ratio with inductor winding resistance, CCM
Preset from the example boost-basic

▶ Open in the simulator: Boost
Preset from the example boost-basic

Sweep $D$ from 0 to 1: the ratio peaks and then collapses. Increase $R_L$ and watch the peak drop and move to a lower duty ratio.

Bench exercise

On a low-power boost converter supplied from a current-limited bench supply, measure the inductor’s dc resistance with a four-wire method first. Then, at a fixed resistive load, step the duty ratio up in open loop while you measure the input and output power with multimeters. Plot the conversion ratio and the efficiency against $D$ and compare them with the two equations above, using your measured $R_L$ . The switch and diode add further losses, so the measured efficiency will sit below the prediction.

Gotchas

Past the maximum, more duty gives less output. Beyond the peak of the ratio, raising $D$ lowers $V$ . A feedback loop that assumes the opposite sees its gain change sign there.
$R_L$ is the dc winding resistance only. Core loss, ac winding loss and switching loss are not in this model; each adds its own loss element [Erickson & Maksimović 2020, Ch. 3].
The loss-free resistor holds only in DCM. Once the current no longer returns to zero in each period, the input current is no longer proportional to the input voltage.

Go deeper

Book chapter Steady-State Equivalent Circuit Modeling, Losses, and Efficiency (Ch. 3) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
DC transformer models and how winding and semiconductor resistances limit the conversion ratio and efficiency.
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.
Course Introduction to Power Electronics — University of Colorado Boulder (Coursera) · retrieved 2026-09-23
First course of the CU Boulder specialization: steady-state analysis, losses and efficiency.

Quiz

In the averaged dc model of a converter in CCM, what does the switch network become?
1. A resistor equal to the load resistance
2. An ideal dc transformer whose turns ratio is the conversion ratio
3. A short circuit
4. A current source equal to the inductor current
Answer and explanation
Correct answer: B Averaged over a period, the switches transfer power from input to output without loss, scaling voltage by $M(D)$ and current by its inverse, like an ideal transformer that also works at dc.
In a boost converter with inductor winding resistance, what happens to the conversion ratio as $D$ approaches 1?
1. It grows without bound, as in the ideal case
2. It settles at a constant value
3. It reaches a maximum and then falls towards zero
4. It oscillates
Answer and explanation
Correct answer: C The ideal ratio is multiplied by a factor that falls to zero as $D$ approaches 1: the inductor current grows so fast that the loss in the winding resistance takes all the additional input power.
A boost converter runs at $D = 0.5$ with $R_L = 0.1\,\Omega$ and $R = 10\,\Omega$ . What is its efficiency, if only $R_L$ causes loss?
1. About 99 %
2. About 96 %
3. About 90 %
4. About 80 %
Answer and explanation
Correct answer: B $R_L$ is compared with $(1 - D)^2 R$ , a quarter of $10\,\Omega$ . The efficiency is $1/(1 + 0.1/2.5)$ , about 0.96.
Why does the same winding resistance cost more efficiency in a boost at high duty ratio?
1. The winding heats up and its resistance rises
2. The switching frequency increases
3. It is compared with $(1 - D)^2 R$ , the load resistance reflected to the inductor side, which shrinks as $D$ grows
4. The output voltage falls
Answer and explanation
Correct answer: C Through the dc transformer the load appears as $(1 - D)^2 R$ on the inductor side, in series with $R_L$ . The efficiency is the share of power that reaches this reflected resistance.
In DCM, how does the input port of a buck-boost converter behave, averaged over a period?
1. Like a resistor whose absorbed power is passed to the output instead of dissipated
2. Like a constant-current sink
3. Like a constant-voltage source
4. Like an open circuit
Answer and explanation
Correct answer: A Each period starts from zero current, so the average input current is proportional to the input voltage: a resistance. The energy is stored in the inductor and delivered to the output, not dissipated, hence a loss-free resistor.
A DCM buck-boost has $L_M = 50\,\mu\mathrm{H}$ , $f_s = 100\,\mathrm{kHz}$ and $D = 0.5$ . What is its input resistance?
1. $20\,\Omega$
2. $40\,\Omega$
3. $80\,\Omega$
4. $160\,\Omega$
Answer and explanation
Correct answer: B The input resistance is $2 L_M f_s/D^2$ : twice $50\,\mu\mathrm{H}$ times $100\,\mathrm{kHz}$ is $10\,\Omega$ , divided by 0.25.

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