Forward converter

Intent

• Compute the forward converter’s conversion ratio and output inductor ripple.
• Explain why the transformer must be reset in every period, and how the reset winding limits the duty ratio.
• Find the switch stress set by the reset winding, and the trade-off it implies.

Theory

A transformer-isolated buck

Forward converter with a reset winding. Primary and secondary conduct together (dots at the same end, turns ratio $1:n$). While $Q_1$ is off, the reset winding ($n_r$ turns per primary turn) returns the magnetizing energy to $V_g$ through $D_r$. $D_1$, $D_2$, $L$ and $C$ form a buck-like output stage. [Erickson & Maksimović 2020, Ch. 6]

In the forward converter the transformer passes energy directly while the switch is on: the secondary applies $n V_g$ to a buck-type output stage (a diode, a freewheeling diode and an L-C filter). Unlike the flyback’s coupled inductor, the transformer is not meant to store energy [Erickson & Maksimović 2020, Ch. 6]. The ratio is the buck’s, scaled by the turns ratio:

$$M = n D$$

$M$

conversion ratio

$D$

duty ratio

$n$

turns ratio, secondary/primary

Forward 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

$M = V/V_g$; $n = N_s/N_p$; valid only while $D \le D_\mathrm{max}$ (reset limit)

Source

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

Derivation

step-by-step sympy derivation

Equation id

forward.ccm.M · equations.yaml entry

The output inductor carries the load current, as in the buck (dc inductor current). While the switch is on it sees $n V_g - V$:

$$\Delta i_L = \frac{\left(n V_g - V\right) D T_s}{2 L}$$

$\Delta i_L$

inductor ripple, half peak-to-peak

$D$

duty ratio

$L$

inductance

$T_s$

switching period

$V$

output voltage

$V_g$

input voltage

$n$

turns ratio, secondary/primary

Forward converter output inductor ripple (half peak-to-peak), 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

Notes

the output stage is a buck converter fed with $n V_g$ while the switch is on

Source

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

Derivation

step-by-step sympy derivation

Equation id

forward.ripple.iL · equations.yaml entry

Transformer reset

A real transformer has a finite magnetizing inductance, so the magnetizing current rises while the switch is on. It must return to zero before the next period, or it grows from period to period until the core saturates [Erickson & Maksimović 2020, Ch. 6]. A reset winding, with $n_r$ its turns divided by the primary turns, returns the magnetizing energy to the input while the switch is off. Volt-second balance on the magnetizing inductance then limits the duty ratio:

$$D_\mathrm{max} = \frac{1}{1 + n_r}$$

$D_\mathrm{max}$

largest duty ratio

$n_r$

reset-winding turns ratio

Forward converter: maximum duty ratio set by transformer reset✓ verified by 8 tests

Assumptions, convention and source

Assumptions

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

Convention

reset winding with $N_r$ turns returns the magnetizing energy to $V_g$; $n_r = 1$ gives $D_\mathrm{max} = 0.5$

Source

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

Derivation

step-by-step sympy derivation

Equation id

forward.reset.Dmax · equations.yaml entry

During the reset the reset winding clamps the primary at the input voltage scaled by $1/n_r$, with reversed polarity, and the switch blocks

$$V_\mathrm{DS} = V_g \left(1 + \frac{1}{n_r}\right)$$

$V_\mathrm{DS}$

switch voltage while off

$V_g$

input voltage

$n_r$

reset-winding turns ratio

Forward converter switch voltage during transformer reset✓ verified by 7 tests

Assumptions, convention and source

Assumptions

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

Notes

the reset winding clamps the primary at $-V_g/n_r$; with $n_r = 1$ the switch blocks twice the input voltage

Source

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

Derivation

step-by-step sympy derivation

Equation id

forward.Vds · equations.yaml entry

More reset turns (larger $n_r$) lower the switch stress but also lower the maximum duty ratio, and fewer turns do the opposite. With $n_r = 1$ the duty ratio is limited to one half and the switch blocks twice the input voltage.

Operating modes

This section looks at one steady period of the example, mode by mode. The waveforms come from the site’s simulator, whose operating-mode view draws the same for any values. While the switch conducts, the secondary drives the output inductor’s current through the rectifier diode and the magnetizing current rises. When the switch turns off, the output inductor freewheels through the other diode while the magnetizing current flows through the reset winding back into the input; once it reaches zero, only the freewheeling is left until the period ends [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, and the secondary drives the current through D₁ and L into the output. D₂ and the reset diode D₃ block. The voltage across L averages 14.4 V. L's current grows: it stores energy. The voltage across L_M averages 48 V. L_M's current grows: it stores energy. C discharges, then charges. The input delivers power.

(b) Mode 2 reset and freewheeling (t₁ to t₂: 4 µs to 8 µs, 4 µs)

S is off: the magnetizing current flows through the reset winding and D₃ back into the input, and S blocks the input voltage plus the reset winding's voltage reflected to the primary. L freewheels through D₂. The voltage across L averages −9.6 V. L's current shrinks: it releases energy. The voltage across L_M averages −48 V. L_M's current shrinks: it releases energy. L_M's current reaches zero: the core has reset. C charges, then discharges. The input takes energy back.

(c) Mode 3 freewheeling (t₂ to t₃: 8 µs to 10 µs, 2 µs)

The core has reset: the magnetizing current is zero and D₃ blocks. L freewheels through D₂. The voltage across L averages −9.6 V. L's current shrinks: it releases energy. L_M carries no current. C discharges. The input supplies nothing.

The operating modes of one steady period of the example forward-basic, 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

A forward converter example with equal primary and reset turns:

Forward converter example

Symbol	Value	How it is obtained
$V_g$ input voltage	48 V	given
$n$ turns ratio, secondary/primary	0.5	given
$n_r$ reset-winding turns ratio	1	given
$D$ duty ratio	0.4	given
$f_s$ switching frequency	100 kHz	given
$L$ inductance	100 µH	given
$R$ load resistance	5 Ω	given
$D_\mathrm{max}$ largest duty ratio	0.5	Forward converter: maximum duty ratio set by transformer reset
$M$ conversion ratio	0.2	Forward conversion ratio, CCM
$V$ output voltage	9.6 V	Output voltage from the conversion ratio
$V_\mathrm{DS}$ switch voltage while off	96 V	Forward converter switch voltage during transformer reset
$\Delta i_L$ inductor ripple, half peak-to-peak	288 mA	Forward converter output inductor ripple (half peak-to-peak), CCM
$I_L$ dc inductor current	1.92 A	Buck dc inductor current, CCM

Operating mode: the transformer resets ($D < D_\mathrm{max}$: holds)

At the chosen duty ratio of 0.4, below the limit of 0.5, the transformer resets in every period, and the switch must block 96 V.

Try it

▶ Open in the equation explorer: Forward converter switch voltage during transformer reset
Preset from the example forward-basic

▶ Open in the simulator: Forward
Preset from the example forward-basic

Sweep $n_r$: the switch stress falls towards the input voltage as the reset winding gets more turns. Then open the reset-limit equation and sweep $n_r$ again to see the maximum duty ratio fall.

Bench exercise

On a single-switch forward evaluation board, watch the switch voltage. While the switch is off, you should see a plateau at the reset voltage for as long as the magnetizing current takes to return to zero, followed by ringing that settles at the input voltage. Measure the plateau duration at two duty ratios and check that the reset time grows in proportion to the on-time.

Gotchas

• Exceeding the reset limit saturates the core. If the duty ratio is pushed past the limit, even during a transient, the magnetizing current does not return to zero and grows from period to period [Erickson & Maksimović 2020, Ch. 6]; a duty-ratio clamp in the controller prevents this.
• The reset winding sets the switch rating. Choosing $n_r$ fixes both the maximum duty ratio and the switch stress.
• Output stage equals buck. Output ripple, peak current and the CCM/DCM boundary of the output inductor follow the buck page, with $n V_g$ in place of $V_g$.

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.

Quiz

1. An ideal forward converter with $n = 0.25$ runs at $D = 0.4$ from $48\,\mathrm{V}$. What is its output voltage?

   1. $4.8\,\mathrm{V}$
   2. $12\,\mathrm{V}$
   3. $19.2\,\mathrm{V}$
   4. $8\,\mathrm{V}$
   Answer and explanation

   Correct answer: A The forward converter is a buck fed with $n V_g$: the output is $n D V_g$.

2. Why must the forward converter's transformer be reset in every switching period?

   1. To discharge the output capacitor
   2. Because otherwise the magnetizing current grows from period to period until the core saturates
   3. To reverse the output polarity
   4. To limit the leakage inductance
   Answer and explanation

   Correct answer: B The magnetizing current rises while the switch is on. Volt-second balance on the magnetizing inductance requires a reset interval that brings it back to zero before the next period.

3. A forward converter has equal primary and reset turns ($n_r = 1$). What is its maximum duty ratio?

   1. 0.25
   2. 0.5
   3. 0.67
   4. 1
   Answer and explanation

   Correct answer: B The reset winding applies the input voltage in reverse, so the reset takes as long as the on-time, and the two must fit in one period.

4. In that converter, fed from $48\,\mathrm{V}$, what does the switch block during the reset?

   1. $48\,\mathrm{V}$
   2. $72\,\mathrm{V}$
   3. $96\,\mathrm{V}$
   4. $144\,\mathrm{V}$
   Answer and explanation

   Correct answer: C The reset winding clamps the primary at $-V_g/n_r$, and the switch blocks $V_g$ minus that: twice the input voltage for $n_r = 1$.

5. The reset winding is given more turns (larger $n_r$). What happens?

   1. Both the switch stress and the maximum duty ratio rise
   2. The switch stress falls and the maximum duty ratio falls
   3. The switch stress rises and the maximum duty ratio falls
   4. Nothing changes
   Answer and explanation

   Correct answer: B A larger $n_r$ lowers the reset voltage on the primary, so the switch blocks less, but the reset then takes longer, which lowers the duty-ratio limit.

6. How does the transformer's role differ between the forward and the flyback converter?

   1. There is no difference
   2. The forward transformer passes energy directly while the switch is on; the flyback's coupled inductor stores it and releases it while the switch is off
   3. The forward transformer stores energy; the flyback transformer does not
   4. Only the flyback transformer provides isolation
   Answer and explanation

   Correct answer: B In the forward converter the secondary conducts during the on-interval, so energy flows straight through; in the flyback the diode conducts during the off-interval, delivering the energy stored in $L_M$.