Flyback converter

Intent

• Compute the flyback’s conversion ratio in CCM and DCM with the 1:n turns-ratio convention.
• Find the switch and diode stress, and the energy the leakage inductance leaves behind at every turn-off.
• Explain why a fixed duty ratio into a fixed output pins the input voltage of a CCM flyback.

Theory

From buck-boost to flyback

Flyback converter. Coupled windings with turns ratio $1:n$ take the place of the buck-boost inductor: energy stored in the magnetizing inductance while $Q_1$ is on reaches the output through $D_1$ while $Q_1$ is off. The dots at opposite ends show that the windings conduct in turn. [Erickson & Maksimović 2020, Ch. 6]

The flyback converter replaces the buck-boost’s inductor by two coupled windings on one core, with turns ratio 1:n, where $n$ is the secondary turns divided by the primary turns. The magnetizing inductance $L_M$, referred to the primary, stores energy while the switch is on; the secondary winding and the diode deliver it to the output while the switch is off. The windings isolate the output and the turns ratio scales the conversion ratio [Erickson & Maksimović 2020, Ch. 6]:

$$M = \frac{n D}{1 - D}$$

$M$

conversion ratio

$D$

duty ratio

$n$

turns ratio, secondary/primary

Flyback conversion ratio, CCM✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• small-ripple approximation

Convention

$M = V/V_g$; 1:n transformer with $n = N_s/N_p$

Source

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

Derivation

step-by-step sympy derivation

Equation id

flyback.ccm.M · equations.yaml entry

$K$ is formed with $L_M$ and the actual load resistance, and the turns ratio then appears in the critical value (flyback critical K). In DCM the ratio no longer depends on the turns ratio [Erickson & Maksimović 2020, Ch. 5]:

$$M = \frac{D}{\sqrt{K}}$$

$M$

conversion ratio

$D$

duty ratio

$K$

inductor parameter (CCM vs DCM)

Flyback conversion ratio, DCM✓ verified by 8 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• discontinuous conduction mode (DCM)
• periodic steady state
• small-ripple approximation
• resistive load R

Convention

1:n transformer, $n = N_s/N_p$; $K$ uses $L_M$ (primary-referred) and the actual load $R$

Notes

independent of $n$: the turns ratio only enters $K_\mathrm{crit}$ and the stresses

Source

Erickson & Maksimović 2020, Ch. 5–6 (DCM; flyback as an isolated buck-boost)

Derivation

step-by-step sympy derivation

Equation id

flyback.dcm.M · equations.yaml entry

CCM currents and ripple

In CCM the flyback behaves like the buck-boost with the secondary quantities referred to the primary. While the diode conducts, the secondary carries the magnetizing current divided by $n$, so charge balance on the output capacitor gives the average magnetizing current [Erickson & Maksimović 2020, Ch. 6]:

$$I_M = \frac{n V}{\left(1 - D\right) R}$$

$I_M$

dc magnetizing current

$D$

duty ratio

$R$

load resistance

$V$

output voltage

$n$

turns ratio, secondary/primary

Flyback magnetizing current (referred to the primary), CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• small-ripple approximation
• resistive load R

Convention

1:n transformer, $n = N_s/N_p$; the secondary carries $I_M/n$ while the diode conducts

Notes

with $n = 1$ this is the buck-boost's inductor current

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits); derived here by charge balance

Derivation

step-by-step sympy derivation

Equation id

flyback.IM · equations.yaml entry

While the switch is on, $V_g$ across $L_M$ sets the magnetizing-current ripple, and the output capacitor alone feeds the load, which sets the output ripple:

$$\Delta i_M = \frac{V_g D T_s}{2 L_M}$$

$\Delta i_M$

magnetizing ripple, half peak-to-peak

$D$

duty ratio

$L_M$

magnetizing inductance (primary)

$T_s$

switching period

$V_g$

input voltage

Flyback magnetizing current ripple, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state

Convention

$\Delta i_M$ is HALF the peak-to-peak ripple (Erickson)

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits); derived here by volt-seconds on L_M

Derivation

step-by-step sympy derivation

Equation id

flyback.ripple.iM · equations.yaml entry

$$\Delta v = \frac{V D T_s}{2 R C}$$

$\Delta v$

output ripple, half peak-to-peak

$C$

capacitance (output capacitor)

$D$

duty ratio

$R$

load resistance

$T_s$

switching period

$V$

output voltage

Flyback output voltage ripple, CCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• continuous conduction mode (CCM)
• periodic steady state
• small-ripple approximation
• resistive load R

Convention

$\Delta v$ is HALF the peak-to-peak ripple (Erickson)

Notes

while the switch is on the diode is off and the capacitor alone supplies the load, as in the buck-boost

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits); derived here by charge balance

Derivation

step-by-step sympy derivation

Equation id

flyback.ripple.v · equations.yaml entry

The converter designer solves these for $L_M$ and $C$.

Switch and diode stress

While the switch is off, the secondary is clamped to the output plus the diode drop, and that voltage appears on the primary divided by $n$. The switch therefore blocks the input voltage plus this reflected voltage [Erickson & Maksimović 2020, Ch. 6]:

$$V_\mathrm{DS} = V_g + \frac{V + V_D}{n}$$

$V_\mathrm{DS}$

switch voltage while off

$V$

output voltage

$V_D$

output-diode forward drop

$V_g$

input voltage

$n$

turns ratio, secondary/primary

Flyback switch voltage while off (without the leakage spike)✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• periodic steady state
• output diode modelled as a constant forward drop V_D

Notes

the leakage-inductance spike adds on top; see $E_\mathrm{lk}$

Source

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

Derivation

step-by-step sympy derivation

Equation id

flyback.Vds_off · equations.yaml entry

While the switch is on, the secondary sees $n V_g$, and the diode blocks it on top of the output voltage:

$$V_R = V + n V_g$$

$V_R$

diode reverse voltage

$V$

output voltage

$V_g$

input voltage

$n$

turns ratio, secondary/primary

Flyback output-diode reverse voltage✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• periodic steady state

Source

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

Derivation

step-by-step sympy derivation

Equation id

flyback.diode.VR · equations.yaml entry

A smaller $n$ lowers the diode stress but raises the reflected voltage on the switch.

DCM peak current and the loss-free resistor

In DCM the magnetizing current starts from zero every period, so its peak depends only on the input voltage, the on-time and $L_M$ [Erickson & Maksimović 2020, Ch. 5]:

$$I_\mathrm{pk} = \frac{V_g D}{L_M f_s}$$

$I_\mathrm{pk}$

peak current

$D$

duty ratio

$L_M$

magnetizing inductance (primary)

$V_g$

input voltage

$f_s$

switching frequency

Flyback peak magnetizing current, DCM✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

in DCM the magnetizing current starts every cycle from zero

Source

Erickson & Maksimović 2020, Ch. 5 (The Discontinuous Conduction Mode)

Derivation

step-by-step sympy derivation

Equation id

flyback.Ipk.dcm · equations.yaml entry

The energy stored per period sets the input power, and the input behaves as a loss-free resistor (DCM input power, input resistance).

Leakage inductance

Not all of the primary flux links the secondary. The leakage inductance carries the primary current at turn-off, and its energy cannot reach the output [Erickson & Maksimović 2020, Ch. 4]:

$$E_\mathrm{lk} = \frac{L_\mathrm{lk} I_\mathrm{pk}^{2}}{2}$$

$E_\mathrm{lk}$

leakage energy per turn-off

$I_\mathrm{pk}$

peak current

$L_\mathrm{lk}$

leakage inductance (primary)

Energy stored in the leakage inductance at turn-off✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• linear (unsaturated) magnetic material

Notes

this energy cannot reach the secondary; a clamp or snubber must absorb it

Source

Erickson & Maksimović 2020, Ch. 4 (Switch Realization: losses from energy stored in leakage and stray inductances)

Derivation

step-by-step sympy derivation

Equation id

flyback.leak.E · equations.yaml entry

A clamp or snubber must take this energy in every period. A dissipative clamp turns it into heat, so the resulting power is a lower bound on the clamp’s loss; a clamp that returns the energy to the input or the output processes this power without dissipating it:

$$P_\mathrm{lk} = E_\mathrm{lk} f_s$$

$P_\mathrm{lk}$

leakage energy times f_s

$E_\mathrm{lk}$

leakage energy per turn-off

$f_s$

switching frequency

Leakage-energy power✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• periodic steady state
• all of the stored energy is dissipated every cycle

Notes

a lower bound for the loss of a dissipative clamp, which also absorbs energy from the reflected voltage and so dissipates more; a clamp that returns the energy to the input or the output processes this power instead of dissipating it

Source

Erickson & Maksimović 2020, Ch. 4 (Switch Realization)

Derivation

step-by-step sympy derivation

Equation id

flyback.leak.P · equations.yaml entry

The clamp check sizes a TVS or RCD clamp for this energy: the clamp voltage, the switch voltage and the clamp’s dissipation.

Fixed duty ratio into a fixed output

If the output is held by a battery or a regulated bus, and the duty ratio is fixed, CCM volt-second balance leaves only one input voltage that fits:

$$V_{g,\mathrm{crit}} = \frac{\left(V + V_D\right) \left(1 - D\right)}{n D}$$

$V_{g,\mathrm{crit}}$

input voltage at the CCM/DCM boundary

$D$

duty ratio

$V$

output voltage

$V_D$

output-diode forward drop

$n$

turns ratio, secondary/primary

Flyback: input voltage at the CCM/DCM boundary (fixed D, fixed output)✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• at the CCM/DCM boundary
• periodic steady state
• output held at a fixed voltage V (battery or regulated bus)
• output diode modelled as a constant forward drop V_D

Notes

below $V_{g,\mathrm{crit}}$ the converter is in DCM; a source-driven bus that rises to $V_{g,\mathrm{crit}}$ is pinned there by CCM volt-second balance

Source

Erickson & Maksimović 2020, Ch. 6 (Converter Circuits); derived here by volt-second balance

Derivation

step-by-step sympy derivation

Equation id

flyback.V_crit · equations.yaml entry

Below it the flyback runs in DCM and draws power like a resistor. When a current-limited source drives the input, its voltage rises until it reaches this value, where CCM begins and holds the input there: the converter then acts as a constant-voltage sink. At that point the switch blocks

$$V_\mathrm{DS} = \frac{V + V_D}{n D}$$

$V_\mathrm{DS}$

switch voltage while off

$D$

duty ratio

$V$

output voltage

$V_D$

output-diode forward drop

$n$

turns ratio, secondary/primary

Flyback switch voltage with the input pinned at the boundary✓ verified by 7 tests

Assumptions, convention and source

Assumptions

• ideal (lossless) switches and passive components
• at the CCM/DCM boundary
• periodic steady state
• output held at a fixed voltage V (battery or regulated bus)
• output diode modelled as a constant forward drop V_D

Convention

evaluated at $V_g = V_{g,\mathrm{crit}}$

Source

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

Derivation

step-by-step sympy derivation

Equation id

flyback.Vds_clamped · equations.yaml entry

Operating modes

This section looks at one steady period of each example, mode by mode. The waveforms come from the site’s simulator, whose operating-mode view draws the same for any values. In CCM the period has two modes: the switch conducts and the magnetizing inductance charges from the input while the output diode blocks, then the switch is off and the magnetizing current flows out of the secondary through the diode [Erickson & Maksimović 2020, Ch. 6].

carries current  no current

(a) Mode 1 switch on (t₀ to t₁: 0 s to 4 µs, 4 µs)

S conducts: the primary winding is connected across the input. The secondary's dot is at its lower end, so D is reverse biased and no current flows in the secondary. The voltage across L_M averages 48 V. L_M's current grows: it stores energy. C discharges. The input delivers power.

(b) Mode 2 switch off (t₁ to t₂: 4 µs to 10 µs, 6 µs)

S is off: the magnetizing current moves to the secondary, where D carries it, divided by n, into the output. The voltage across L_M averages −32 V. L_M's current shrinks: it releases energy. C charges, then discharges. The input supplies nothing.

The operating modes of one steady period of the example flyback-ccm, from the site's simulator: the key waveforms with the modes' boundaries $t_0, t_1, \ldots$, then the circuit in each mode, its current-carrying branches in colour. ▶ Change the values in the simulator

In DCM the magnetizing current reaches zero before the period ends, and a third mode follows in which neither the switch nor the diode conducts [Erickson & Maksimović 2020, Ch. 5]:

carries current  no current

(a) Mode 1 switch on (t₀ to t₁: 0 s to 3 µs, 3 µs)

S conducts: the primary winding is connected across the input. The secondary's dot is at its lower end, so D is reverse biased and no current flows in the secondary. The voltage across L_M averages 48 V. L_M's current grows: it stores energy. C discharges. The input delivers power.

(b) Mode 2 switch off (t₁ to t₂: 3 µs to 4.77 µs, 1.77 µs)

S is off: the magnetizing current moves to the secondary, where D carries it, divided by n, into the output. The voltage across L_M averages −81.5 V. L_M's current shrinks: it releases energy. C charges, then discharges. The input supplies nothing.

(c) Mode 3 idle (t₂ to t₃: 4.77 µs to 10 µs, 5.23 µs)

The magnetizing current is zero (DCM): S and D are both off. L_M carries no current. C discharges. The input supplies nothing.

The operating modes of one steady period of the example flyback-dcm, from the site's simulator: the key waveforms with the modes' boundaries $t_0, t_1, \ldots$, then the circuit in each mode, its current-carrying branches in colour. ▶ Change the values in the simulator

Worked example

Three flyback examples: one in CCM with a resistive load, one in DCM, and one charging a fixed output at a fixed duty ratio.

Flyback example in CCM, ideal diode

Symbol	Value	How it is obtained
$V_g$ input voltage	48 V	given
$n$ turns ratio, secondary/primary	0.25	given
$D$ duty ratio	0.4	given
$f_s$ switching frequency	100 kHz	given
$L_M$ magnetizing inductance (primary)	200 µH	given
$R$ load resistance	5 Ω	given
$V_D$ output-diode forward drop	0 V	given
$M$ conversion ratio	0.167	Flyback conversion ratio, CCM
$V$ output voltage	8 V	Output voltage from the conversion ratio
$K$ inductor parameter (CCM vs DCM)	8	Dimensionless inductor parameter K
$K_\mathrm{crit}$ K at the CCM/DCM boundary	5.76	Critical K of the flyback converter
$V_\mathrm{DS}$ switch voltage while off	80 V	Flyback switch voltage while off (without the leakage spike)
$V_R$ diode reverse voltage	20 V	Flyback output-diode reverse voltage

Operating mode: CCM ($K > K_\mathrm{crit}$: holds)

Flyback example in DCM, ideal diode

Symbol	Value	How it is obtained
$V_g$ input voltage	48 V	given
$n$ turns ratio, secondary/primary	0.25	given
$D$ duty ratio	0.3	given
$f_s$ switching frequency	100 kHz	given
$L_M$ magnetizing inductance (primary)	50 µH	given
$R$ load resistance	20 Ω	given
$V_D$ output-diode forward drop	0 V	given
$L_\mathrm{lk}$ leakage inductance (primary)	1 µH	given
$K$ inductor parameter (CCM vs DCM)	0.5	Dimensionless inductor parameter K
$K_\mathrm{crit}$ K at the CCM/DCM boundary	7.84	Critical K of the flyback converter
$M$ conversion ratio	0.424	Flyback conversion ratio, DCM
$V$ output voltage	20.4 V	Output voltage from the conversion ratio
$I_\mathrm{pk}$ peak current	2.88 A	Flyback peak magnetizing current, DCM
$V_\mathrm{DS}$ switch voltage while off	129 V	Flyback switch voltage while off (without the leakage spike)
$E_\mathrm{lk}$ leakage energy per turn-off	4.15 µJ	Energy stored in the leakage inductance at turn-off
$P_\mathrm{lk}$ leakage energy times fs	415 mW	Leakage-energy power

Operating mode: DCM ($K > K_\mathrm{crit}$: does not hold)

Example flyback charging a fixed output at a fixed duty ratio

Symbol	Value	How it is obtained
$V$ output voltage	12 V	given
$V_D$ output-diode forward drop	500 mV	given
$D$ duty ratio	0.3	given
$n$ turns ratio, secondary/primary	0.25	given
$V_{g,\mathrm{crit}}$ input voltage at the CCM/DCM boundary	117 V	Flyback: input voltage at the CCM/DCM boundary (fixed D, fixed output)
$V_\mathrm{DS}$ switch voltage while off	167 V	Flyback switch voltage with the input pinned at the boundary

In the CCM example the reflected voltage adds to the input, so the switch blocks 80 V although the input is 48 V. In the DCM example the leakage inductance of 1 µH alone means that a dissipative clamp loses at least 415 mW.

Try it

▶ Open in the equation explorer: Flyback switch voltage while off (without the leakage spike)
Preset from the example flyback-ccm

▶ Open in the simulator: Flyback
Preset from the example flyback-dcm

Sweep the turns ratio $n$: a smaller $n$ raises the reflected voltage and the switch stress. Then open the diode stress equation and sweep $n$ again to see the opposite trend.

Bench exercise

On a low-power flyback evaluation board, watch the switch voltage while the switch is off. You should see a short spike at turn-off (the leakage energy being clamped), then a plateau at the input voltage plus the reflected output voltage, and, in DCM, ringing once the secondary current has reached zero. Measure the plateau and compare it with the switch-stress equation at the measured input and output voltages.

Gotchas

• The leakage energy has to go somewhere. Without a clamp or snubber the turn-off spike rises until something breaks down. A dissipative clamp loses at least the leakage power above, and a clamp that recovers the energy must still be rated to process it [Erickson & Maksimović 2020, Ch. 4].
• An open-loop flyback must not lose its load. In DCM the output rises as the load is removed (DCM ratio grows as $K$ falls), and with it the reflected voltage, until the clamp limits the switch voltage.
• A fixed-duty-ratio flyback from a current-limited source pins its input. Once CCM begins, the input sits at the critical voltage above instead of rising further.

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.
• Course Converter Circuits — University of Colorado Boulder (Coursera) · retrieved 2026-09-23
  DCM analysis and the isolated converter circuits, following the same textbook.
• 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.
• Video channel Sam Ben-Yaakov — power electronics videos — S. Ben-Yaakov · retrieved 2026-09-23
  Intuitive, equation-light explanations of converter behaviour.

Quiz

1. An ideal CCM flyback with turns ratio 1:n, $n = 0.5$, runs at $D = 0.5$ from $24\,\mathrm{V}$. What is its output voltage?

   1. $6\,\mathrm{V}$
   2. $12\,\mathrm{V}$
   3. $24\,\mathrm{V}$
   4. $48\,\mathrm{V}$
   Answer and explanation

   Correct answer: B The ratio is $n D/(1 - D)$, the buck-boost ratio scaled by the turns ratio: $0.5$ here, so the output is half the input.

2. A flyback runs from $48\,\mathrm{V}$ with $n = 0.25$ and a $12\,\mathrm{V}$ output (ideal diode). What does the switch block while off, without the leakage spike?

   1. $60\,\mathrm{V}$
   2. $96\,\mathrm{V}$
   3. $120\,\mathrm{V}$
   4. $144\,\mathrm{V}$
   Answer and explanation

   Correct answer: B The output voltage appears on the primary divided by $n$, which is $48\,\mathrm{V}$ here, and adds to the input voltage.

3. In the same flyback, what reverse voltage does the output diode block while the switch is on?

   1. $12\,\mathrm{V}$
   2. $24\,\mathrm{V}$
   3. $48\,\mathrm{V}$
   4. $60\,\mathrm{V}$
   Answer and explanation

   Correct answer: B The secondary sees $n V_g$, which is $12\,\mathrm{V}$ here, and the diode blocks it on top of the output voltage.

4. How does the DCM conversion ratio of a flyback depend on the turns ratio, with $K$ formed from $L_M$ and the actual load?

   1. It is proportional to $n$
   2. It is proportional to $1/n$
   3. It does not depend on $n$
   4. It is proportional to $n^2$
   Answer and explanation

   Correct answer: C In DCM the energy stored per period in $L_M$ is delivered to the load whatever the turns ratio, and the ratio is $D/\sqrt{K}$.

5. A flyback has $2\,\mu\mathrm{H}$ of leakage inductance and turns off $3\,\mathrm{A}$ at $100\,\mathrm{kHz}$. At least how much power must a dissipative clamp dissipate?

   1. $0.09\,\mathrm{W}$
   2. $0.45\,\mathrm{W}$
   3. $0.9\,\mathrm{W}$
   4. $1.8\,\mathrm{W}$
   Answer and explanation

   Correct answer: C The leakage stores half of $L_\mathrm{lk}$ times the square of the current, $9\,\mu\mathrm{J}$ per turn-off, and that happens once per period. A dissipative clamp turns all of it into heat.

6. A fixed-duty-ratio flyback charges a battery from a current-limited source. The source voltage stops rising at one value. Why?

   1. The source reaches its current limit
   2. At that input voltage the flyback enters CCM, and CCM volt-second balance with a fixed output and duty ratio allows only that input voltage
   3. The clamp starts to conduct
   4. The transformer saturates
   Answer and explanation

   Correct answer: B Below the critical voltage the flyback is in DCM and behaves as a resistor. At the critical voltage CCM begins, and volt-second balance pins the input: the converter becomes a constant-voltage sink.