Comparing the basic converters

Intent

• Choose between the basic converters by conversion range, isolation and control difficulty.
• Compare how hard each converter works its transistor, with one number: the transistor utilization.
• Know which converters have a right-half-plane zero and which draw pulsed input or output current.

Theory

Conversion range and isolation

The buck only steps down and the boost only steps up (buck, boost). The buck-boost does both, with an inverted output (buck-boost). The flyback and the forward converter add a transformer: the output is isolated and the turns ratio scales the range (flyback, forward) [Erickson & Maksimović 2020, Ch. 6].

Transistor stress and utilization

A converter that needs a transistor rated for a high voltage and a high rms current to deliver a given power uses that transistor poorly. One figure of merit is the transistor utilization: the load power divided by the product of the transistor’s peak voltage and its rms current [Erickson & Maksimović 2020, Ch. 6]. With small ripple the transistor current is a flat pulse lasting $D T_s$, and the utilization depends on the duty ratio alone:

$$U = \sqrt{D}$$

$U$

transistor utilization

$D$

duty ratio

Transistor utilization of the buck converter, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• rms currents computed with small ripple (flat-topped current pulses)

Notes

load power over the transistor's peak voltage $V_g$ times its rms current

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits)

Derivation

step-by-step sympy derivation

Equation id

util.buck · equations.yaml entry

$$U = \frac{1 - D}{\sqrt{D}}$$

$U$

transistor utilization

$D$

duty ratio

Transistor utilization of the boost converter, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• rms currents computed with small ripple (flat-topped current pulses)

Notes

the transistor blocks $V$ and carries the inductor current during $D T_s$

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits)

Derivation

step-by-step sympy derivation

Equation id

util.boost · equations.yaml entry

$$U = \left(1 - D\right) \sqrt{D}$$

$U$

transistor utilization

$D$

duty ratio

Transistor utilization of the buck-boost and flyback converters, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• rms currents computed with small ripple (flat-topped current pulses)

Notes

the transistor blocks $V_g + V$ (flyback: $V_g + V/n$); the turns ratio cancels, and the maximum is at $D = 1/3$

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits)

Derivation

step-by-step sympy derivation

Equation id

util.buckboost · equations.yaml entry

$$U = \frac{\sqrt{D}}{1 + \frac{1}{n_r}}$$

$U$

transistor utilization

$D$

duty ratio

$n_r$

reset-winding turns ratio

Transistor utilization of the forward converter with a reset winding, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• rms currents computed with small ripple (flat-topped current pulses)

Notes

magnetizing current neglected; the duty ratio must stay below the reset limit

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits)

Derivation

step-by-step sympy derivation

Equation id

util.forward · equations.yaml entry

The buck approaches full utilization at large duty ratios. The boost is best at small duty ratios, where it steps up only a little. The buck-boost and the flyback are poorest, because the transistor blocks the sum of input and reflected output voltage; their utilization peaks at $D = 1/3$. A forward converter with $n_r = 1$ is limited to $D \le 0.5$ and blocks twice the input voltage, which caps its utilization too. The derivations are on the derivations page.

Control and current waveforms

The buck and the forward converter feed the output through an L-C filter and have no right-half-plane zero. The boost, the buck-boost and the flyback deliver energy to the output only during the off-interval, which creates the right-half-plane zero that limits their loop bandwidth. The buck draws a pulsed input current and delivers a continuous output current; the boost does the opposite; the buck-boost and the flyback pulse at both ports [Erickson & Maksimović 2020, Ch. 6].

Worked example

The same duty ratio in every topology:

Comparison example at one duty ratio

Symbol	Value	How it is obtained
$D$ duty ratio	0.4	given
$n_r$ reset-winding turns ratio	1	given
${M}_{\mathrm{buck}}$ conversion ratio	0.4	Buck conversion ratio, CCM
${M}_{\mathrm{boost}}$ conversion ratio	1.67	Boost conversion ratio, CCM
${M}_{\mathrm{bb}}$ conversion ratio	−0.667	Buck-boost conversion ratio, CCM
${U}_{\mathrm{buck}}$ transistor utilization	0.632	Transistor utilization of the buck converter, CCM
${U}_{\mathrm{boost}}$ transistor utilization	0.949	Transistor utilization of the boost converter, CCM
${U}_{\mathrm{bb}}$ transistor utilization	0.379	Transistor utilization of the buck-boost and flyback converters, CCM
${U}_{\mathrm{fwd}}$ transistor utilization	0.316	Transistor utilization of the forward converter with a reset winding, CCM

At this duty ratio the buck’s utilization, 0.632, is well above that of the buck-boost, 0.379: for the same load power, the buck-boost needs a transistor with a larger product of voltage rating and current rating.

Try it

▶ Open in the equation explorer: Transistor utilization of the buck-boost and flyback converters, CCM
Preset from the example comparison-basic

Sweep $D$ from 0 to 1 and find the maximum of the buck-boost and flyback utilization. Then open the buck and boost utilization equations and compare the curves.

Bench exercise

With a current probe and an oscilloscope that computes rms values, measure the transistor current and the drain-source voltage of two evaluation boards of different topologies delivering similar power. Compute each transistor’s utilization from the peak voltage, the rms current and the measured output power, and compare the results with the equations above at the measured duty ratios.

Gotchas

• Utilization compares transistors only. It leaves out diode stress, magnetics size, switching loss and the capacitors’ ripple currents; use it as one criterion among several.
• The small-ripple assumption flatters converters with large ripple. A large current ripple raises the rms current above the flat-pulse value and lowers the real utilization.
• Isolation is a requirement, not a figure of merit. If the application needs galvanic isolation, only the transformer-based topologies qualify, whatever their utilization.

Go deeper

• Book chapter Converter Circuits (Ch. 6) — R. W. Erickson, D. Maksimović · retrieved 2026-09-23
  How the basic topologies relate, and transformer-isolated versions including the flyback and forward converters.
• Lecture notes 6.334 Lecture Notes — D. Perreault, MIT OpenCourseWare · retrieved 2026-09-23
  Chapter-style notes on dc/dc converters, magnetics and isolated converters.
• Course Converter Circuits — University of Colorado Boulder (Coursera) · retrieved 2026-09-23
  DCM analysis and the isolated converter circuits, following the same textbook.

Quiz

1. An application needs $5\,\mathrm{V}$ from an input that varies between $3\,\mathrm{V}$ and $9\,\mathrm{V}$, without isolation. Which single basic converter can do it?

   1. The buck
   2. The boost
   3. The buck-boost (accepting an inverted output)
   4. None of the basic converters
   Answer and explanation

   Correct answer: C The output must be above the input at low line and below it at high line; only the buck-boost steps both up and down, at the price of an inverted output.

2. What is the transistor utilization of an ideal CCM buck at $D = 0.25$?

   1. 0.25
   2. 0.5
   3. 0.75
   4. 1
   Answer and explanation

   Correct answer: B For the buck the utilization is $\sqrt{D}$: the transistor blocks $V_g$, carries the load current as a pulse of duty ratio $D$, and the load power is $D V_g I$.

3. At which duty ratio is the utilization of the buck-boost and the flyback largest?

   1. $D = 1/4$
   2. $D = 1/3$
   3. $D = 1/2$
   4. $D = 2/3$
   Answer and explanation

   Correct answer: B The utilization is $(1 - D)\sqrt{D}$; its derivative vanishes where $1 - 3D$ is zero.

4. Which of these converters has a right-half-plane zero in its control-to-output transfer function in CCM?

   1. The buck
   2. The forward converter
   3. The flyback
   4. None of them
   Answer and explanation

   Correct answer: C The flyback, like the boost and the buck-boost, delivers energy to the output only during the off-interval; the buck and the forward converter feed the output through an L-C filter while the switch is on.

5. Why does the transistor utilization of the flyback not depend on its turns ratio?

   1. Because the transformer stores no energy
   2. Because at a given duty ratio the transistor sees the same peak voltage and the same rms current whatever $n$ is; the turns ratio only scales the output voltage and current
   3. Because the flyback always runs at $D = 1/3$
   4. Because the leakage inductance cancels the turns ratio
   Answer and explanation

   Correct answer: B In CCM the transistor blocks $V_g + V/n$, which equals $V_g/(1 - D)$, and for a given load power its current does not depend on $n$ either; the turns ratio changes only the output side.

6. Which converter delivers a continuous current to its output capacitor and draws a pulsed current from its input?

   1. The buck
   2. The boost
   3. The buck-boost
   4. The flyback
   Answer and explanation

   Correct answer: A The buck's inductor sits between the switch node and the output, so the output sees the continuous inductor current, while the input sees it only while the switch is on.