Chapter 14: Isolated Converters: Flyback, Forward, Push-Pull, and Bridge — Power Electronics

Where this sits

Part 3 · DC–DC Converters
Chapter 14 of 30

You should already know

Volt-second balance, the buck and buck–boost gains, and the direct/indirect distinction from Chapter 12 and Chapter 13.

By the end you can

Choose an isolated topology from power level and input range, size a turns ratio and duty ratio together, and explain what a snubber or a reset winding is actually doing.

Time

≈ 55 min reading · ≈ 50 min problems

i What you'll learn

Why galvanic isolation is a safety requirement rather than an engineering preference, and what a converter must do to provide it.
How a turns ratio gives voltage gain for free, releasing the duty ratio to do the regulating.
The flyback: an isolated buck–boost, \( V_o = \dfrac{N_s}{N_p}\cdot\dfrac{D}{1-D}V_s \), and why it stops at about 150 W.
The forward converter: an isolated buck, \( V_o = \dfrac{N_s}{N_p}D\,V_s \), and why it needs a reset winding and \( D < 0.5 \).
Push–pull, half-bridge and full-bridge — how using both halves of the B–H loop doubles what a core can carry.
Core saturation, flux walking and leakage inductance: the three failure modes that dominate isolated-converter design.
How the feedback crosses the isolation barrier without breaking it.

Section 14-1

Two Reasons for a Transformer

Chapters 12 and 13 solved the arithmetic of DC–DC conversion completely: any voltage ratio can be produced, in either direction, with either polarity. Yet almost every mains-connected power supply in existence contains a transformer. There are two reasons, and they are independent.

1 Safety: galvanic isolation

A mains-fed converter has a DC link at roughly 325 V (from 230 V AC) or 170 V (from 120 V AC), and its negative rail is connected to the incoming neutral through the rectifier.
Without isolation, the output's "ground" is referenced to the mains. Touch it and you are the return path.
Safety standards (IEC 60950, IEC 62368-1, IEC 60601 for medical) therefore mandate a barrier between the mains and anything a user can touch, specified in creepage and clearance distances and in the insulation's dielectric withstand — typically 3 kV AC for one minute, and 4 kV for medical.
A transformer provides that barrier by construction: energy crosses as a magnetic field, and no conductive path exists at all.

2 Engineering: the turns ratio does the heavy lifting

Chapter 12 established that a single boost stage runs out of usefulness above a gain of about 4 or 5, because \( I_o/(1-D) \) and the resistive correction defeat it.
A transformer provides any ratio you can wind, at no cost in duty ratio.
So the duty ratio is freed to do what it is good at — regulating over a modest range — while the turns ratio does the gross conversion.
Multiple secondaries cost almost nothing, which is why one supply can produce +12 V, +5 V and −12 V from a single switch.

◇Intuition · gearbox and throttle

A car does not climb a hill by pressing the accelerator to the floor in top gear. It changes down — and then uses a modest throttle opening to control the speed precisely.

The transformer is the gearbox and the duty ratio is the throttle:

Wrong way: ask a boost converter for a gain of 30 by pushing \( D \) to 0.97. Everything about the circuit fights you — currents, timing, loop gain.
Right way: wind a 1:30 transformer for the gross ratio, then run \( D \) around 0.4 and let it move between 0.3 and 0.5 to regulate.

Why the second is better in every respect: currents are reasonable on both sides, the off-time is long enough for the energy transfer to complete, \( dV_o/dD \) is small so the loop is easy, and there is symmetric headroom in both directions for line and load variation.

And a bonus the gearbox analogy predicts: add another output shaft and you get another speed for almost nothing. Add another secondary winding and you get another rail for the cost of some wire.

There is a cost, and it is worth stating plainly before the topologies: a transformer's core carries flux, and flux cannot go on increasing forever. Every isolated topology in this chapter is, in part, an answer to the question how does the core get reset before the next cycle? Section 14-6 makes that the organising idea.

Section recap. A transformer provides a safety barrier that no non-isolated topology can, and a turns ratio that lets the duty ratio stay in its comfortable range. The price is a core whose flux must be reset every cycle — which is what distinguishes the topologies from one another.

Section 14-2

The Flyback Converter

The flyback is the isolated buck–boost, and the resemblance is exact: take a buck–boost, split its inductor into two windings on the same core, and use the second winding to feed the output.

Because of that origin, its "transformer" is not really a transformer. It is a coupled inductor — an energy-storage device with two windings — and the distinction matters more than it sounds.

◇Intuition · the bucket, now with two handles

Chapter 12's buck–boost was a bucket carried between the well and the tank, filled from one and emptied into the other, never touching both. The flyback is the same bucket, but with a handle on each side and a wall between the well and the tank.

Switch on. The primary winding is across the source and current builds up, storing energy in the core. The secondary diode is reverse-biased by design — the windings are deliberately wound in opposition — so nothing reaches the output. The output capacitor carries the load alone.
Switch off. The primary current stops. The stored energy has to go somewhere, so the flux collapses, the secondary voltage reverses, its diode conducts, and the energy is delivered to the output.

Note that the two windings never conduct at the same time. That is what makes it a coupled inductor rather than a transformer — a transformer conducts on both sides simultaneously and stores as little energy as possible; a flyback stores all the transferred energy and conducts on one side at a time.

Which is why a flyback core has an air gap. A gap seems like a defect — it reduces inductance and increases the magnetising current. But energy is stored where the reluctance is, and a gapped core stores far more energy before saturating than an ungapped one. The flyback needs that storage; a forward converter, which stores nothing deliberately, does not have a gap.

Switch on: the primary is across the source

Working

\[ v_{Lp,on} = V_s \quad\text{for a time } DT \]
Switch off: the secondary is across the output The secondary sees \( V_o \). Referred to the primary through the turns ratio \( n = N_p/N_s \):

Working

\[ v_{Lp,off} = -n\,V_o \quad\text{for a time } (1-D)T \]
Volt-second balance on the primary

Result

\[ V_sDT - nV_o(1-D)T = 0 \;\Longrightarrow\; V_o = \frac{1}{n}\cdot\frac{D}{1-D}V_s = \frac{N_s}{N_p}\cdot\frac{D}{1-D}V_s \]

The buck–boost gain multiplied by the turns ratio — and with the polarity now a matter of how you wind the secondary rather than a constraint.
Find the switch stress With the switch off, the primary sees the reflected output voltage on top of the supply:

Working

\[ V_{sw} = V_s + nV_o + V_{spike} \]

That last term is the leakage-inductance spike of Section 14-6, and it is not small.

🔑

Flyback converter

\[ V_o = \frac{N_s}{N_p}\cdot\frac{D}{1-D}V_s, \qquad V_{sw} = V_s + \frac{N_p}{N_s}V_o + V_{spike}, \qquad P_o = \frac{1}{2}L_pI_{pk}^2 f \]

That last expression is the one that limits the topology. All the transferred power passes through the core as stored energy, so raising the power means raising \( I_{pk} \) — and the core volume rises with it.

▣In practice · why every phone charger is a flyback

Open any USB charger and you will find a flyback, almost without exception. The reasons are worth listing because they explain a design decision made hundreds of millions of times a year.

Fewest components of any isolated topology. One switch, one coupled inductor, one diode, two capacitors. There is no output inductor at all — the coupled inductor does both jobs.
Wide input range for free. Universal mains means 85–265 V AC, so 120–375 V DC — a 3:1 range. The \( D/(1-D) \) characteristic spans it with \( D \) between about 0.25 and 0.5.
Multiple outputs are nearly free. Each extra rail is one winding and one diode, and all of them are regulated approximately by the same feedback loop.
Inherent short-circuit behaviour. Energy per cycle is capped at \( \tfrac12 L_pI_{pk}^2 \), so a shorted output cannot draw more than the controller's current limit allows.

And the limit that explains why your laptop charger is different. Above roughly 100–150 W the flyback stops making sense: the peak currents are large, the core must store the full cycle energy, the output capacitor sees severe ripple, and efficiency falls. At those powers, designs move to a forward or resonant converter.

The modern refinement is the quasi-resonant flyback, which waits for the drain voltage to ring down to its minimum before turning on. That recovers much of the switching loss, and it is why a 65 W GaN charger can be the size of a matchbox when a 65 W adapter of ten years ago was the size of a paperback.

Section recap. The flyback is a buck–boost whose inductor has been split into two windings, so it stores all the transferred energy and needs a gapped core. \( V_o = (N_s/N_p)\cdot D/(1-D)\cdot V_s \), the switch blocks \( V_s + nV_o \) plus a leakage spike, and the topology runs out at 100–150 W.

Section 14-3

The Forward Converter

If the flyback is an isolated buck–boost, the forward converter is an isolated buck — and it inherits all the buck's advantages, including the absence of a right-half-plane zero.

The structural difference from the flyback is that the secondary is wound in the same sense as the primary, so the output diode conducts while the switch is on. Energy flows straight through the transformer rather than being stored in it.

The flyback stores energy in the core and releases it during the off-time; the forward passes it straight through during the on-time and uses the off-time to reset the core.

Switch on: the secondary drives the output inductor The secondary voltage is \( V_s N_s/N_p \), and the output inductor sits between it and \( V_o \):

Working

\[ v_{Lo,on} = \frac{N_s}{N_p}V_s - V_o \]
Switch off: the output inductor freewheels

Working

\[ v_{Lo,off} = -V_o \]
Volt-second balance on the output inductor

Result

\[ V_o = \frac{N_s}{N_p}\,D\,V_s \]

Exactly the buck's \( V_o = DV_s \), scaled by the turns ratio. So the forward converter inherits the buck's continuous output current, small output capacitor, and — crucially — no right-half-plane zero.

But there is a problem the flyback did not have, and it is fundamental.

The primary's magnetising current still builds up during the on-time — a transformer's magnetising inductance is finite, so some current flows into the core regardless of the load. In the flyback that current was the mechanism, and the off-time discharged it into the secondary. In the forward converter the secondary diode blocks during the off-time, so there is nowhere for the magnetising current to go.

Left alone, the flux would ratchet upward a little every cycle until the core saturates, at which point the primary becomes a short across the supply and the switch is destroyed.

🔑

Forward converter, and the constraint the reset imposes

\[ V_o = \frac{N_s}{N_p}D\,V_s, \qquad D_{max} = \frac{N_r/N_p}{1 + N_r/N_p} \;\xrightarrow{\;N_r = N_p\;}\; D < 0.5 \]

With the usual 1:1 reset winding, the core needs as long to reset as it took to magnetise — so the duty ratio can never exceed 0.5, and the switch must block \( 2V_s \).

◈Why the reset winding costs exactly half the duty range

The reset winding is a third winding, wound in opposition to the primary, with its own diode returning to the supply. Its job is to give the magnetising current a path during the off-time — and, importantly, to return that energy to the source rather than dissipating it.

Volt-second balance decides the timing. The core's flux must return exactly to where it started every cycle:

\[ \underbrace{V_s \cdot DT}_{\text{magnetising}} = \underbrace{V_s\frac{N_p}{N_r}\cdot(1-D)T}_{\text{resetting}} \]

With \( N_r = N_p \) — the usual choice, because the two windings can then be wound bifilar for tight coupling — this reduces to \( D = 1 - D \), so \( D \) must not exceed 0.5. Exceed it and the reset does not complete before the next cycle begins, the residual flux accumulates, and the core saturates within a few cycles.

The trade you can make. Choosing \( N_r < N_p \) resets the core faster and allows a larger \( D \). But the reset winding then develops a higher voltage, and the switch must block \( V_s(1 + N_p/N_r) \):

\( N_r = N_p \): \( D_{max} = 0.50 \), switch blocks \( 2V_s \). The standard choice.
\( N_r = 0.5N_p \): \( D_{max} = 0.33 \), switch blocks \( 3V_s \).
\( N_r = 2N_p \): \( D_{max} = 0.67 \), switch blocks \( 1.5V_s \).

Read the middle row carefully — reducing the reset turns reduces the maximum duty ratio and increases the stress. The relationship runs the opposite way to most people's first guess, and it is worth checking against the formula rather than intuition.

Alternatives used in practice: an RCD clamp (simple, but dissipates the magnetising energy), active clamp (a small MOSFET and capacitor that recycle it and allow \( D > 0.5 \)), and the two-switch forward, which clamps the primary with two diodes so the switches only ever see \( V_s \) — the most robust arrangement, and common at higher powers.

▲Common mistake

"The forward converter's transformer is like the flyback's, so it needs an air gap too."

Wrong Gapping a forward transformer "for safety". The gap lowers the magnetising inductance, which increases the magnetising current and makes reset harder.

Right A forward transformer wants the highest magnetising inductance it can get — an ungapped core with high permeability.

The reasoning, in one line each:

A flyback must store energy in the core, and energy is stored in the gap. No gap means saturation at a low current. The gap is essential.
A forward converter must not store energy — every joule it stores is magnetising current that has to be reset, and reset costs duty ratio and switch voltage. The magnetising current is pure overhead.

The general principle worth carrying: in an indirect converter (flyback, buck–boost), the magnetic component is an energy store and wants a gap. In a direct converter (forward, push–pull, bridge), it is a true transformer and wants none.

This is also why the two components look different in a catalogue. A flyback transformer is specified by its inductance and saturation current, like an inductor. A forward transformer is specified by its turns ratio, its volt-second product and its leakage — like a transformer.

Section recap. The forward converter is an isolated buck: \( V_o = (N_s/N_p)DV_s \), continuous output current, no RHP zero. The magnetising current has nowhere to go during the off-time, so a reset winding is needed — and with the usual 1:1 winding it caps \( D \) at 0.5 and forces the switch to block \( 2V_s \).

Section 14-4

Push–Pull, Half-Bridge and Full-Bridge

Both topologies so far drive the core in one direction only. The flux goes up during the on-time and returns to zero during the off-time — so only half of the core's available flux swing is ever used.

The remaining three topologies drive the core in both directions, alternately. That doubles the usable flux swing, which for a given core and frequency roughly doubles the power it can handle.

◇Intuition · rowing with both arms

A single-ended converter is like rowing with one oar and then lifting it clear to reset. Half the stroke does work; the other half is recovery.

A push–pull converter uses both. Switch A drives the core one way, switch B drives it the other, and both strokes deliver power.

What that buys, concretely:

Twice the flux swing. The core traverses from \( -B_{max} \) to \( +B_{max} \) instead of 0 to \( +B_{max} \), so \( \Delta B \) doubles and, since \( V = N A \, dB/dt \), so does the power for the same core.
No reset winding. Each half-cycle resets the core for the other. The reset is the work.
Twice the effective output frequency. The output sees a pulse from each half, so the output filter operates at \( 2f \) and can be smaller.

And what it costs: more switches, and a new failure mode. If the two half-cycles are not perfectly symmetric — slightly different on-times, slightly different device drops — the core does not return exactly to where it started. A small offset accumulates cycle after cycle until the core saturates. This is flux walking, and Section 14-6 deals with it, because it destroys converters that otherwise look correct on paper.

The five isolated topologies compared
Topology	Switches	Switch blocks	Gain	Core use	Typical power
Flyback	1	\(V_s + nV_o\)	\(\dfrac{N_s}{N_p}\dfrac{D}{1-D}\)	One quadrant, gapped	up to ~150 W
Forward	1	\(2V_s\)	\(\dfrac{N_s}{N_p}D\)	One quadrant	100–500 W
Push–pull	2	\(2V_s\)	\(2\dfrac{N_s}{N_p}D\)	Two quadrants	200 W–1 kW
Half-bridge	2	\(V_s\)	\(\dfrac{N_s}{N_p}D\)	Two quadrants	200 W–2 kW
Full-bridge	4	\(V_s\)	\(2\dfrac{N_s}{N_p}D\)	Two quadrants	500 W upward

Interactive · turns ratio and duty ratio together

A 12 V output from a rectified mains DC link. Choose the turns ratio and see what duty ratio each topology then needs across the whole universal-input range — and whether it stays inside its legal limits.

Turns ratio N_p:N_s 16 : 1

DC link voltage 325 V

Duty ratio required at this operating point, and whether it is legal
Topology	Gain law	Required D	Limit	Verdict
Flyback	\(\frac{1}{n}\frac{D}{1-D}\)	—	\(D < 0.75\)	—
Forward	\(\frac{1}{n}D\)	—	\(D < 0.5\)	—
Half-bridge	\(\frac{1}{n}D\)	—	\(D < 0.5\)	—
Full-bridge	\(\frac{2}{n}D\)	—	\(D < 0.5\)	—

Reflected V_o192 V

Flyback switch517 V

Forward switch650 V

Bridge switch325 V

—

◈Why the half-bridge halves the switch stress and the push–pull does not

Both use two switches and both drive the core in two directions. Yet a push–pull switch must block \( 2V_s \) and a half-bridge switch only \( V_s \). The difference is where the transformer is connected.

Push–pull. The primary is centre-tapped, with the tap at \( +V_s \) and a switch at each end. When switch A is on, its end is pulled to ground — so that half-winding has \( V_s \) across it. By transformer action the other half develops \( V_s \) too, on top of the \( V_s \) already at the tap. Switch B therefore sits at \( 2V_s \).

Half-bridge. Two capacitors split the supply, creating a mid-point at \( V_s/2 \). The primary connects between that mid-point and the junction of two switches. Each switch only ever connects one end of the primary to a rail, so neither can see more than \( V_s \). The primary sees \( \pm V_s/2 \).

The trade, stated plainly. The half-bridge halves the voltage stress and halves the volt-seconds available to the primary — so for the same output it needs twice the primary current, or a turns ratio half as large. Voltage stress traded for current stress.

Which is why the choice follows the input voltage:

Low input (12 V, 48 V): the current is already high and the voltage stress is trivial. Push–pull wins — \( 2V_s = 96 \) V is an easy device.
Mains input (325 V DC): \( 2V_s = 650 \) V, and above 600 V devices become markedly more expensive and slower. Half-bridge or full-bridge wins.

And the full-bridge combines both advantages: each switch blocks only \( V_s \), and the primary sees the full \( \pm V_s \). Four switches and two more gate drives, but at a kilowatt that is a small fraction of the cost — which is why almost every high-power isolated converter is a full-bridge.

Section recap. Driving the core in both directions doubles the usable flux swing and removes the need for a reset winding. Push–pull is best at low input voltage; half-bridge halves the switch voltage at the cost of doubled current; full-bridge gives both advantages for four switches, and dominates above about 500 W.

Section 14-5

Feedback Across the Barrier

An isolated converter has a controller on the primary side, at mains potential, and an output on the secondary side that a user may touch. The controller needs to know the output voltage. Connecting a wire between them would defeat the entire purpose.

There are three standard answers, and the choice affects accuracy, cost and safety certification.

Getting the feedback signal across the isolation barrier
Method	How it works	Accuracy	Cost	Used in
Optocoupler + shunt reference	A TL431 on the secondary compares \(V_o\) with a reference and drives an LED; a phototransistor on the primary receives it	±1% or better	Low	The overwhelming majority of supplies
Auxiliary winding (primary-side regulation)	A third winding reflects the output voltage during the off-time; the controller samples it	±5%	Lowest — no optocoupler at all	Low-cost chargers, LED drivers
Signal transformer or digital isolator	A small pulse transformer or capacitive/magnetic isolator carries a modulated signal	Excellent	Highest	Digitally controlled and high-reliability supplies

✦Think about it

Primary-side regulation removes the optocoupler entirely — fewer parts, no ageing LED, lower cost. Why has it not displaced the optocoupler everywhere?

Three reasons, in order of importance

1. It measures the wrong thing. The auxiliary winding reflects the secondary winding's voltage, which is the output voltage plus the diode drop, plus the drop across the winding resistance, plus the drop across the output capacitor's ESR. All three vary with load current and temperature. So regulation degrades from ±1% to ±5%, which is fine for a phone charger and unacceptable for a 3.3 V logic rail with a ±3% window.

2. It can only sample at one instant. The auxiliary winding only reflects the output while the secondary diode is conducting. The controller must sample near the end of that interval, just before the current reaches zero, and only once per cycle. That limits the loop bandwidth and makes the transient response noticeably slower.

3. Leakage inductance corrupts the sample. Immediately after turn-off, the auxiliary winding rings violently from the leakage-inductance spike. The controller must blank that interval — and the blanking time is a compromise: too short and it samples noise, too long and it misses the useful window at light load.

Where each one belongs:

Primary-side regulation: chargers, LED drivers, anything where ±5% is fine and cost dominates.
Optocoupler: anything needing tight regulation or good transient response — which is most things.
Digital isolators: where the control loop is digital anyway, or where the optocoupler's current-transfer-ratio degradation over 10–15 years is a reliability concern.

A worthwhile aside: the optocoupler's CTR falls with age and temperature, typically 20–50% over a product's life. A loop designed with no margin for that will slowly drift out of specification — a classic field failure that never appears on the bench.

Section recap. An optocoupler with a shunt reference is the default and gives ±1%. Primary-side regulation removes it at the cost of ±5%, a slower loop and sampling difficulties. Digital isolators are used where a digital loop or long-term reliability justifies the cost.

Section 14-6

Three Ways an Isolated Converter Fails

Non-isolated converters fail in fairly obvious ways — a device over-rated, a loop unstable, a capacitor cooked. Isolated converters add three failure modes that come from the magnetics, and all three destroy hardware rather than merely degrading performance.

1 Core saturation

What happens: the flux density reaches \( B_{sat} \). Above it, permeability collapses, so inductance collapses, so \( di/dt = v/L \) becomes enormous. The primary is effectively a short across the supply.
How fast: microseconds. Faster than most over-current protection, which is why cycle-by-cycle current limiting is standard.
What causes it: too many volt-seconds — excessive \( D \), excessive \( V_s \), a failed reset, or a core running too hot (\( B_{sat} \) falls with temperature, typically 30% from 25 °C to 100 °C).
Design margin: operate at \( B_{max} \approx 0.6\,B_{sat} \) at the highest temperature and worst-case volt-seconds.

2 Flux walking

What happens: in a symmetric topology, a small imbalance between the two half-cycles leaves a little residual flux. Next cycle it happens again. The operating point creeps toward one end of the B–H loop until it saturates.
What causes it: unequal on-times (gate-drive propagation mismatch), unequal device voltage drops, unequal winding resistances. Differences of a few percent are enough.
The classic remedy: a DC blocking capacitor in series with the primary. Any DC component of the applied voltage charges it, and it self-corrects. Standard on half-bridges, where it also serves as one of the splitter capacitors.
The better remedy: current-mode control. The controller terminates each pulse on peak current rather than on time, so an asymmetry that would increase the flux automatically shortens that pulse. This is the main reason current-mode control dominates isolated converter design.

3 Leakage inductance

What it is: the fraction of the primary's flux that does not link the secondary — typically 1–3% of the magnetising inductance. It is not coupled, so at turn-off its energy has nowhere to go.
What happens: \( \tfrac12 L_{lk}I_{pk}^2 \) is dumped into the switch's output capacitance, producing a voltage spike that can exceed the supply several times over, plus severe ringing.
The remedy: a snubber or clamp. An RCD clamp catches the spike and dissipates the energy in a resistor. An active clamp recovers it. Either way, the clamp voltage must be set above the reflected output voltage or it will conduct continuously and waste real power.
The prevention: winding technique. Interleaving the primary and secondary (half the primary, the secondary, then the other half) can cut leakage by a factor of four — at the cost of higher interwinding capacitance, which worsens common-mode EMI. The two goals oppose each other, and the resolution is one of the genuinely skilled parts of transformer design.

▲Common mistake

"Set the flyback's clamp voltage as low as possible to protect the switch."

Wrong Clamping just above \( V_s \), so the switch never sees a spike — and finding the clamp resistor dissipating tens of watts.

Right The clamp must sit above \( V_s + nV_o \), typically at \( V_s + 1.5\,nV_o \). Below that it conducts during normal operation and burns the delivered power.

Why. During the off-time the primary legitimately sits at \( V_s + nV_o \) — that is the reflected output voltage, and it is present for the whole of the energy transfer. A clamp set below it conducts throughout that interval and diverts power meant for the load into a resistor.

How to set it properly:

Compute the reflected voltage \( nV_o \).
Set the clamp to catch at roughly \( 1.4 \) to \( 1.6 \) times that above \( V_s \) — enough headroom that it only conducts on the leakage spike.
Choose the switch rating from \( V_s(max) + \) clamp level \( + \) a safety margin, typically 20%.

Worked instance. \( V_s = 375 \) V (high line), \( n = 16 \), \( V_o = 12 \) V, so \( nV_o = 192 \) V. Clamp at \( 375 + 1.5(192) = 663 \) V. Add 20% and you need an 800 V MOSFET — which is exactly why flyback designs for universal mains use 650–800 V devices rather than the 500 V the DC link alone would suggest.

The real trade-off in the turns ratio. A higher \( n \) gives a lower duty ratio and better secondary utilisation, but a higher reflected voltage and therefore a more expensive switch. A lower \( n \) does the reverse. The optimum usually lands where \( nV_o \approx 0.5\,V_{s(max)} \) — which is why flyback turns ratios cluster where they do.

Section recap. Core saturation kills in microseconds and is prevented by volt-second margin; flux walking creeps up over many cycles and is cured by a DC blocking capacitor or current-mode control; leakage inductance produces a turn-off spike that must be clamped above the reflected output voltage, not below it.

Section 14-7

Worked Examples

1 A universal-input flyback

Problem. Design the turns ratio and duty range for a 12 V, 3 A flyback from universal mains (85–265 V AC, so 120–375 V DC), switching at 65 kHz. Then find the switch voltage rating required.

Step 1 — choose the reflected voltage. The usual guideline is \( nV_o \approx 0.5\,V_{s(max)} \), balancing duty ratio against switch stress:

Working

\[ nV_o \approx 0.5(375) = 188\ \text{V} \;\Longrightarrow\; n = \frac{188}{12} \approx 16 \]

Step 2 — find the duty range.

Working

\[ D = \frac{nV_o}{nV_o + V_s} = \frac{192}{192 + V_s} \]

\[ V_s = 120\ \text{V}: \; D = 0.615 \qquad V_s = 375\ \text{V}: \; D = 0.339 \]

Both comfortable — well inside the 0.75 practical ceiling, with margin for transients.

Step 3 — the switch rating.

Working

\[ V_{clamp} = V_{s(max)} + 1.5\,nV_o = 375 + 1.5(192) = 663\ \text{V} \]

\[ V_{rating} \ge 1.2(663) = 796\ \text{V} \;\Longrightarrow\; \text{specify an 800 V MOSFET} \]

Sanity check. 36 W output at 65 kHz means 0.55 mJ per cycle. With \( P_o = \tfrac12 L_pI_{pk}^2f \) and a typical \( I_{pk} \) of 1.5 A, \( L_p \approx 490\ \mu\text{H} \) — an entirely ordinary value for an EF25 core, which confirms the design is in sensible territory.

2 A forward converter and its reset constraint

Problem. A forward converter produces 5 V at 20 A from a 48 V telecom bus (42–56 V), at 200 kHz, with a 1:1 reset winding. Find the turns ratio, the duty range, and the switch rating.

The reset winding caps \( D \) at 0.5, so design the maximum duty ratio at the minimum input, leaving margin — say \( D_{max} = 0.45 \) at \( V_s = 42 \) V.

Working

\[ V_o = \frac{N_s}{N_p}DV_s \;\Longrightarrow\; \frac{N_s}{N_p} = \frac{V_o}{D_{max}V_{s(min)}} = \frac{5}{0.45(42)} = 0.265 \]

\[ \text{Choose } N_p{:}N_s = 4{:}1 \;\Longrightarrow\; N_s/N_p = 0.25 \]

\[ D = \frac{V_o}{(N_s/N_p)V_s} = \frac{5}{0.25\,V_s} = \frac{20}{V_s} \]

\[ V_s = 42\ \text{V}: \; D = 0.476 \qquad V_s = 56\ \text{V}: \; D = 0.357 \]

The low-line case is uncomfortably close to the limit. 0.476 against a hard ceiling of 0.5 leaves only 5% margin — and any transient dip below 42 V, or any extra drop in the diode and windings, will push it over. Two fixes:

Use \( N_p{:}N_s = 3.5{:}1 \), giving \( D = 0.42 \) at low line. More margin, at the cost of higher secondary voltage and so a higher-rated output diode.
Use an active-clamp forward, which recycles the magnetising energy and permits \( D > 0.5 \). More parts, better efficiency, and the standard answer in modern telecom supplies.

Switch rating. With a 1:1 reset, \( V_{sw} = 2V_{s(max)} = 112 \) V. Add 30% for the leakage spike and ringing and specify a 150 V MOSFET.

3 Why the same supply changes topology at 500 W

Problem. A 12 V supply from a 325 V DC link is required at 100 W, then at 500 W, then at 2 kW. Recommend a topology for each and justify the changes.

100 W — flyback.

Fewest parts, no output inductor, cheapest by a clear margin.
\( P = \tfrac12 L_pI_{pk}^2f \) with \( I_{pk} \approx 2 \) A is manageable, and the core is small.
The output capacitor sees roughly \( I_o\sqrt{D/(1-D)} \approx 6 \) A RMS at 8.3 A output — high but survivable with parallel ceramics.

500 W — two-switch forward, or half-bridge.

A flyback here would need \( I_{pk} \approx 10 \) A and store 6 mJ per cycle — a large gapped core, and an output capacitor bank sized by ripple current rather than capacitance.
A forward converter passes energy straight through, so the transformer is a genuine transformer: smaller core, no gap, much lower peak currents.
The output inductor makes the output current continuous, cutting the capacitor's ripple burden by roughly a factor of five.
Two-switch rather than single-switch, because each device then blocks only \( V_s = 325 \) V instead of 650 V — cheaper, faster devices and no reset winding.

2 kW — full-bridge, phase-shifted.

The core is driven in both directions, so it carries twice the power of a single-ended design of the same size.
Each switch blocks only \( V_s \), and the primary sees the full \( \pm V_s \) — the best combination available.
167 A on the secondary demands synchronous rectification; Schottky diodes would dissipate over 80 W.
Phase-shifted control allows zero-voltage switching by using the leakage inductance to resonate with the device capacitance, so switching loss largely disappears — the difference between 92% and 96% efficiency, which at 2 kW is 80 W of heat.

The pattern. Each change is forced by a quantity that grew faster than the topology could accommodate: stored energy first, then device voltage stress, then switching loss. Recognising which one binds is how topology selection is actually done.

4 Diagnosing a saturating transformer

Problem. A push–pull converter works at low power but fails within seconds at full load. The primary current waveform shows a normal ramp that suddenly turns almost vertical near the end of each pulse. What is happening, and what are the candidate causes?

Diagnosis: the core is saturating. The signature is unmistakable — \( di/dt = v/L \), so when \( L \) collapses at \( B_{sat} \), the current rises almost without limit. The "hockey stick" current waveform means saturation and essentially nothing else.

Why it appears only at full load narrows the cause considerably:

Flux walking — the most likely cause in a push–pull. A small asymmetry between the half-cycles biases the operating point, and at high current there is no longer room for the bias. Check by comparing the two switches' on-times on an oscilloscope, and their \( V_{DS(on)} \) drops.
Insufficient volt-second margin. At full load the duty ratio is largest and \( B_{max} \) with it. Verify \( B_{max} = V_sD T/(2N_pA_e) \) against the datasheet \( B_{sat} \) at operating temperature, not at 25 °C.
Thermal. \( B_{sat} \) for ferrite falls roughly 30% from 25 °C to 100 °C. A design with 20% margin on a cold core has none on a hot one — and full load is when it is hot.
Turns count wrong. Worth eliminating early: measure the primary inductance and compare with \( A_L N^2 \).

The fixes, in the order to try them:

Add a DC blocking capacitor in series with the primary. Cheapest, and cures flux walking directly.
Switch to current-mode control, which terminates each pulse on peak current and so self-corrects asymmetry cycle by cycle.
Add turns or a larger core to increase the volt-second capability.

What not to do: raise the current limit so the protection stops tripping. It is reporting a real fault.

Section 14-8

Summary & Formula Sheet

Chapter 14 in five sentences:

A transformer provides a safety barrier no non-isolated topology can, and a turns ratio that frees the duty ratio to regulate rather than to convert.
The flyback is an isolated buck–boost — it stores all the energy, needs a gapped core, and runs out at 100–150 W.
The forward is an isolated buck — energy passes straight through, so no RHP zero and small magnetics, but the magnetising current must be reset, which caps \( D \) at 0.5 with a 1:1 winding.
Driving the core in both directions doubles what it can carry: push–pull for low input voltage, half-bridge and full-bridge for mains.
Three magnetics failure modes dominate: saturation (microseconds), flux walking (many cycles), and the leakage spike (every turn-off).

Formula sheet · Chapter 14

Flyback gainisolated buck–boost

\( V_o = \dfrac{N_s}{N_p}\cdot\dfrac{D}{1-D}V_s \)

Flyback duty ratio\(n = N_p/N_s\)

\( D = \dfrac{nV_o}{nV_o + V_s} \)

Flyback switch stressclamp above \(V_s + nV_o\)

\( V_{sw} = V_s + nV_o + V_{spike} \)

Flyback powerwhy it caps out near 150 W

\( P_o = \tfrac12 L_p I_{pk}^2 f \)

Forward gainisolated buck; no RHP zero

\( V_o = \dfrac{N_s}{N_p}D\,V_s \)

Forward reset limit\(N_r = N_p \Rightarrow D < 0.5\)

\( D_{max} = \dfrac{N_r/N_p}{1 + N_r/N_p} \)

Forward switch stress1:1 reset winding

\( V_{sw} = V_s\left(1 + \dfrac{N_p}{N_r}\right) = 2V_s \)

Push–pull gain\(D\) per switch, \(<0.5\)

\( V_o = 2\dfrac{N_s}{N_p}D\,V_s \)

Half-bridge gainprimary sees \(\pm V_s/2\)

\( V_o = \dfrac{N_s}{N_p}D\,V_s \)

Full-bridge gainprimary sees \(\pm V_s\)

\( V_o = 2\dfrac{N_s}{N_p}D\,V_s \)

Peak flux densitykeep \(\le 0.6B_{sat}\) when hot

\( B_{max} = \dfrac{V_s\,DT}{N_p A_e} \)

Leakage spike energywhat the clamp must absorb

\( W_{lk} = \tfrac12 L_{lk}I_{pk}^2 \)

Optimum flyback turns ratiobalances \(D\) against switch cost

\( nV_o \approx 0.5\,V_{s(max)} \)

Key terms

Galvanic isolation: No conductive path between input and output. Required by safety standards for mains-connected equipment.
Creepage and clearance: The minimum distances across a surface and through air between isolated conductors. Set by standards from the working voltage and pollution degree.
Flyback converter: Isolated buck–boost. Stores energy in a gapped coupled inductor; windings never conduct simultaneously.
Forward converter: Isolated buck. Energy passes straight through an ungapped transformer during the on-time; needs a reset mechanism.
Reset winding: A third winding that returns the magnetising energy to the source during the off-time. With \(N_r = N_p\) it caps \(D\) at 0.5.
Magnetising current: The current that establishes flux in the core, independent of load. Useful in a flyback; pure overhead in a forward converter.
Flux walking: Progressive drift of the core's operating point caused by asymmetry between half-cycles in a push–pull or bridge converter.
Leakage inductance: Primary flux that does not link the secondary. Its stored energy produces the turn-off voltage spike.
RCD clamp: Resistor, capacitor and diode network that catches the leakage spike and dissipates its energy. Must clamp above \(V_s + nV_o\).
Active clamp: A clamp that recycles the leakage and magnetising energy instead of dissipating it, and allows \(D > 0.5\) in a forward converter.
Primary-side regulation: Sensing the output through an auxiliary winding, removing the optocoupler at the cost of accuracy and loop speed.
Phase-shifted full bridge: A full bridge whose two legs are shifted in phase, using leakage inductance to achieve zero-voltage switching.

Check yourself

Test Yourself

Chapter 14 · six questions answers hidden until you ask

Why does a flyback transformer need an air gap when a forward transformer must not have one?

Because they are storing entirely different amounts of energy, on purpose.

Where magnetic energy is actually stored. Energy density is \( \tfrac12 BH \), and \( H = B/\mu \). Ferrite has \( \mu_r \) of a few thousand, so for the same \( B \), \( H \) in the air gap is thousands of times larger than in the core. Essentially all the energy is in the gap, even though the gap is a fraction of a millimetre.

The flyback must store energy. Every joule delivered to the output is first held in the core during the on-time — \( P_o = \tfrac12 L_pI_{pk}^2f \). Without a gap the core saturates at a small current and can hold almost nothing. The gap is what makes the topology possible.

The forward converter must not. Its transformer conducts on both sides at once, so power passes straight through. Any energy it stores is magnetising current — current that produces flux but delivers nothing to the load, and that must then be reset. Adding a gap lowers \( L_m \), which raises the magnetising current, which makes reset harder and wastes conduction loss for no return.

The two components are therefore specified differently:

Flyback: specify primary inductance and saturation current, like an inductor. Gapped, with \( L_p \) chosen from the required stored energy.
Forward: specify turns ratio, volt-second product and leakage. Ungapped, with \( L_m \) as high as the core allows.

The general rule: indirect converters store, so they gap. Direct converters transfer, so they do not.

A forward converter's duty ratio is limited to 0.5. Why can a flyback happily run at 0.6, when both are single-switch isolated converters?

Because the two use the off-time for different purposes, and only one of them has a separate reset to fit in.

The forward converter's off-time has two jobs: the output inductor freewheels, and the transformer's magnetising current must be reset through the reset winding. With \( N_r = N_p \), resetting takes exactly as long as magnetising did, so \( t_{reset} = t_{on} \), and \( t_{on} + t_{reset} \le T \) gives \( D \le 0.5 \) immediately.

The flyback's off-time has one job, and it is the reset. Delivering the stored energy to the secondary is resetting the core — the flux falls from its peak back to its starting value while the output diode conducts. The two are the same event, so there is no separate reset interval to accommodate.

What limits the flyback instead:

The energy must be fully delivered within \( (1-D)T \), or the converter enters continuous conduction mode — legal and common, but it changes the dynamics and introduces an RHP zero.
The switch stress \( V_s + nV_o \) rises with the turns ratio needed for a high \( D \).
Practical designs stop around \( D = 0.7 \) to \( 0.75 \), and it is a soft economic limit rather than a hard physical one.

The way to remember it: ask what the core is doing during the off-time. In a flyback it is delivering power. In a forward converter it is doing nothing useful — just resetting — which is precisely why that topology feels wasteful of duty ratio, and why active-clamp variants exist to recover it.

A half-bridge and a push–pull both use two switches on the same core. Why is the half-bridge preferred for mains input and the push–pull for a 48 V bus?

Because they trade voltage stress against current stress in opposite directions, and which trade is favourable depends entirely on the input voltage.

Push–pull: centre-tapped primary, tap at \( +V_s \). Each switch blocks \( 2V_s \). The primary half-winding sees the full \( V_s \).

Half-bridge: split-capacitor mid-point. Each switch blocks only \( V_s \). The primary sees \( \pm V_s/2 \) — half the volt-seconds, so twice the primary current for the same power.

At 48 V input:

Push–pull switches block 96 V. A 150 V MOSFET is cheap, fast and has very low \( R_{DS(on)} \).
Half-bridge switches block 48 V but carry twice the current, and \( I^2R \) doubles the conduction loss.
Push–pull wins: the voltage stress is trivial to accommodate and the lower current is worth having.

At 325 V input:

Push–pull switches block 650 V, plus a leakage spike — so 800 V or 900 V devices, which are markedly more expensive and have \( R_{DS(on)} \) several times higher for the same die.
Half-bridge switches block 325 V, so 500 V or 600 V devices — a mature, cheap, high-performance category.
Half-bridge wins, and comfortably.

The underlying reason: \( R_{DS(on)} \) scales roughly as \( V_{BR}^{2.5} \). Doubling the required blocking voltage costs about 5.7 times the on-resistance for the same die area. That is a steep penalty, and it is why the crossover between the two topologies sits somewhere around 100 V input.

The bonus for the half-bridge: the splitter capacitors block DC, so flux walking is cured by construction rather than by an added component.

Why does a flyback stop being sensible above about 150 W, when the equations contain no such limit?

Because \( P_o = \tfrac12 L_pI_{pk}^2 f \) forces every quantity in the converter upward together, and each of them hits a practical wall.

1. Core size. Stored energy sets core volume. Doubling the power doubles the energy per cycle, and the core grows with it. At a few hundred watts a flyback transformer becomes larger and more expensive than a forward transformer of the same rating — often by a factor of two or three.

2. Peak currents. Because energy is stored and released rather than passed through, the peak current is roughly twice the average, and the RMS is about \( 1.15 \times \) the CCM equivalent. Conduction losses rise as the square. At 500 W the primary peak might be 10 A where a forward converter would need 4 A.

3. Output capacitor ripple. This is often the binding constraint and the least anticipated. The flyback has no output inductor, so the capacitor absorbs the entire pulsed secondary current. Its RMS ripple is approximately \( I_o\sqrt{D/(1-D)} \) — at 12 V, 40 A and \( D = 0.4 \), that is 33 A RMS. No reasonable capacitor bank handles that.

4. Leakage energy. \( \tfrac12 L_{lk}I_{pk}^2 \) scales with the square of the peak current. At high power the clamp dissipates tens of watts, and efficiency falls several points.

What a forward converter changes. Energy passes straight through, so the core is smaller; the output inductor makes the output current continuous, so the capacitor's ripple burden falls by roughly a factor of five; the peak currents are lower; and the leakage energy is smaller in proportion.

The boundary is economic, not physical. Kilowatt flybacks exist — some traction and welding supplies use them — but they are chosen for specific reasons (multiple outputs, wide input range, tolerance of short circuits) that outweigh the penalties. For a general-purpose supply, 150 W is where the arithmetic turns.

A designer sets a flyback's RCD clamp to catch at 400 V, with a 325 V link and a 192 V reflected voltage. What happens?

The clamp conducts during normal operation and dissipates a large fraction of the converter's output power.

Where the numbers go wrong. During the off-time the primary legitimately sits at

\[ V_s + nV_o = 325 + 192 = 517\ \text{V} \]

for the whole of the energy-transfer interval. That is not a spike; it is the reflected output voltage, and it is exactly how the energy gets to the secondary. A clamp set at 400 V conducts throughout that interval.

How much power it wastes. The clamp diverts current whenever the primary tries to exceed 400 V, which is essentially all of the off-time. Rather than catching a brief spike, it becomes a parallel load taking a substantial share of the transferred energy — comfortably tens of watts in a 36 W supply, which is to say the converter cannot regulate at all.

The symptoms on the bench are unmistakable once you know them: the clamp resistor is far too hot to touch, the output cannot reach its set voltage at full load, efficiency is dismal, and the drain waveform shows a flat plateau at the clamp level for the whole off-time rather than a brief spike decaying to a plateau at 517 V.

The correct setting:

\[ V_{clamp} = V_s + 1.5\,nV_o = 325 + 288 = 613\ \text{V} \]

and the switch is then rated at \( 1.2 \times 613 \approx 740 \) V, so an 800 V part.

The principle to carry: a clamp must sit above every voltage the circuit produces legitimately, and catch only what it produces parasitically. Setting it from the supply voltage alone, without adding the reflected output, is one of the most common flyback design errors — and the one that most reliably produces a converter that appears to work at no load and collapses under it.

Why does current-mode control cure flux walking, when voltage-mode control does not?

Because current-mode control terminates each pulse on a measurement of the flux, while voltage-mode control terminates it on a clock.

What flux walking is. In a push–pull or bridge converter, half-cycle A magnetises the core one way and half-cycle B the other. If A applies slightly more volt-seconds than B — a few nanoseconds of extra on-time, or a slightly lower device drop — the core does not return to where it started. The offset accumulates every cycle until the core saturates.

Voltage-mode control cannot see it. The pulse ends when the sawtooth crosses the error voltage — a timing decision that knows nothing about the core. If half-cycle A is systematically longer, it stays longer, and the drift continues unopposed.

Current-mode control does see it, because in a transformer the primary current is proportional to the flux (plus the reflected load current). The pulse ends when the sensed current reaches a threshold. So:

If half-cycle A pushed the flux too far positive, the next A half-cycle starts from a higher flux, so its current reaches the threshold sooner, and the pulse is shorter.
The correction is automatic, cycle by cycle, with no additional components.

This is negative feedback on the flux itself, and it is the main reason current-mode control dominates isolated converter design — not the loop-shaping advantages usually cited first.

Two caveats worth knowing:

Current-mode control is unstable above \( D = 0.5 \) without slope compensation — a ramp added to the sensed current. This is why every current-mode controller IC has a slope-compensation pin or an internal ramp.
It corrects volt-second asymmetry but not asymmetry in the sensing path itself. A current-sense transformer with different behaviour on the two half-cycles will introduce its own walk.

The belt-and-braces approach, used in most production designs, is current-mode control and a DC blocking capacitor in series with the primary. Neither is expensive, and the failure they prevent is catastrophic.

Practice

Problems

Three habits for isolated converter problems:

Choose the turns ratio first, from the worst-case duty ratio you are willing to run at. Everything else follows.
Check the switch stress at the highest input voltage, and remember to add the reflected output and the leakage spike.
Ask what resets the core. If you cannot answer, the design is not finished.

Problems 1–5 are direct application; 6–9 need judgement; 10–12 are design questions worth discussing in a tutorial.

A flyback produces 15 V at 2 A from a 300 V DC link with \( N_p{:}N_s = 20{:}1 \), at 100 kHz. Find \( D \), the switch blocking voltage before the spike, and the primary inductance if the peak primary current is 1.2 A.
A forward converter with \( N_p{:}N_s = 6{:}1 \) and a 1:1 reset winding runs from 400 V. Find the output voltage at \( D = 0.4 \), the switch rating, and the maximum output voltage the topology permits.
A push–pull converter produces 24 V at 10 A from 48 V with \( N_p{:}N_s = 1{:}1 \). Find the duty ratio per switch and the blocking voltage each device must withstand.
A half-bridge produces 12 V from 325 V at \( D = 0.4 \). Find the required turns ratio, and compare it with the ratio a full bridge would need for the same output and duty ratio.
A flyback transformer has \( L_p = 600\ \mu\text{H} \) and \( L_{lk} = 12\ \mu\text{H} \), with \( I_{pk} = 2 \) A at 65 kHz. Find the leakage energy per cycle and the power the clamp must dissipate.
A universal-input flyback (120–375 V DC) delivers 20 V at 3 A. Choose a turns ratio, justify it, find the duty range, and specify the switch voltage rating with reasons.
Explain why a forward converter has no right-half-plane zero while a flyback in continuous conduction does, and state what that means for the two converters' achievable loop bandwidths.
A 500 W supply is to be built from a 325 V link. Compare a two-switch forward and a half-bridge on device count, device ratings, transformer utilisation, control complexity and cost. Recommend one.
A push–pull converter saturates at full load but not at half load. List four candidate causes, state the measurement that would distinguish each, and give the remedy for each.
Design a 60 W flyback for 12 V output from universal mains at 65 kHz. Specify the turns ratio, duty range, primary inductance, peak current, switch voltage rating and clamp level. State the output capacitor's RMS ripple current and comment on what type of capacitor is required.
A telecom supply must produce 3.3 V at 30 A from a 36–72 V bus with better than ±2% regulation and fast transient response. Compare an active-clamp forward converter with a phase-shifted full bridge, and recommend one with reasons covering efficiency, part count, transformer utilisation and control.
An engineer proposes replacing a flyback's optocoupler feedback with primary-side regulation to save cost in a 5 V, 2 A charger. Analyse the consequences for regulation accuracy, transient response, no-load behaviour and cable-drop compensation, and state under what output specification the change would be acceptable.