Isolated Converters: Flyback, Forward and Bridge

Choose the turns ratio. The flyback's ratio is the buck–boost's, scaled by the turns:

Why cap \(D\) near 0.45? Because the reflected voltage adds to the input at the switch, and a larger duty ratio means a larger reflection. Problem 3 shows the same trade in a forward converter, where it is a hard limit rather than a choice.

The actual duty range with 8:1:

Device stresses. The switch must block the input plus the reflected output:

The spike of Problem 6 typically adds 50 to 100 V, so a 650 V MOSFET is the usual choice — 600 V is uncomfortably tight. Note that \(V_R\) is fixed by the turns ratio alone, independent of the operating point.

The output diode sees the reverse of the same picture:

The magnetising inductance, sized at low line where the currents are worst:

That average is over the whole period; during the on-time the current is higher by \(1/D\):

Now the flyback's weak point — the output capacitor. The secondary delivers current only during the off-time, so the capacitor supplies the entire load for \(DT_s\):

Four and a half amps of RMS ripple for a 5 A output — nearly one-for-one. This is the flyback's structural cost: no output inductor means the capacitor does all the filtering, and it must be several low-ESR parts in parallel. Problem 2 shows what an output inductor buys.

The forward is a buck behind a transformer, so its ratio is linear in \(D\):

A duty range of 0.144 to 0.45, similar to the flyback's. But note the ratio is 4.5:1 rather than 8:1 — the forward needs fewer primary turns for the same output, because it transfers during the on-time rather than storing.

The switch stress is the forward's penalty. During reset, the reset winding returns the magnetising energy to the input, and with \(N_r = N_p\) that puts the full input voltage back across the switch:

Against 471 V for the flyback. A 750 V requirement means a 900 V or 1000 V device, whose \(R_{DS(on)}\) is far worse — and that is why the single-switch forward is uncommon at universal input, and why the two-switch variant of Problem 5 exists.

The output inductor, sized at high line where the ripple is worst — the buck rule of Set 18:

And the payoff. With an inductor feeding the output, the capacitor sees only a triangular ripple:

An order of magnitude. That single row is why forward converters dominate wherever output ripple or capacitor lifetime matters — and why the flyback survives only where its simplicity outweighs it.

The two compared:

Read the last two rows together. A flyback with three outputs needs three diodes and three capacitors; a forward needs three diodes, three inductors and three capacitors — and the flyback's cross-regulation is better, because all outputs are fed from the same collapsing flux.

Where each belongs:

The crossover is around 75 W and it is set by the capacitor. Below that, 4.5 A of ripple current is manageable with two or three ceramics; above it, the capacitor bank costs more than the forward converter's extra inductor and reset winding.

Why reset is mandatory. During the on-time the primary carries a magnetising current that builds flux in the core:

Unlike a flyback, that flux is not delivered to the secondary — the forward's secondary conducts only while the primary does, and the magnetising component is left behind. If it is not removed each cycle it accumulates, and after a few cycles the core saturates and the switch sees a short circuit.

The reset winding drives the flux back. Volt-second balance on the core requires:

Reset must finish before the next cycle begins:

And the switch pays for it. During reset the reflected reset voltage adds to the input:

A clean inverse trade: whatever is gained in duty ratio is paid in switch voltage, and the product \(D_{max}\times V_S/V_{in}\) is exactly 1 in every row. There is no free choice here, only a preference.

Why 1:1 is chosen. Two practical reasons beyond the compromise:

The reset winding must be closely coupled to the primary or its own leakage produces a spike far worse than the reset voltage itself. Winding it bifilar with the primary is easy when the turns are equal and awkward when they are not. And (ii) a 1:1 ratio means the reset winding can use the same wire gauge and the same layer, costing almost nothing.

The alternatives:

The last two rows are what modern designs use. A two-switch forward clamps the primary between the rail and ground with two diodes, so neither switch ever exceeds \(V_{in}\) — halving the voltage requirement at the cost of a second switch and a high-side driver. An active clamp goes further, recycling the magnetising energy into a capacitor that then provides zero-voltage switching.

Applying it to the Problem 2 design:

Two 500 V MOSFETs cost less than one 900 V part and conduct far better, since \(R_{DS(on)}\) scales roughly as \(V_{BR}^{2.5}\). That is why the two-switch forward, not the single-switch version, is the standard 100 to 500 W topology.

How a bridge differs. Four switches apply \(+V_{in}\), then zero, then \(-V_{in}\), then zero — so the transformer is excited symmetrically and resets itself. No reset winding, and the output sees two pulses per switching period:

The turns ratio, set at minimum input:

A narrow duty range, because the input comes from a regulated PFC bus rather than a raw rectified mains. That is typical — bridges usually sit behind the boost PFC stage of Set 19, Problem 4.

The switch stress is the headline result:

Not \(2V_{in}\), as in the single-switch forward. Each switch is clamped between the rail and ground by the bridge structure itself, so no reflected voltage is added. That halving is the whole reason bridges scale.

The switch currents:

Each switch conducts for only half the effective duty, since the two diagonal pairs alternate. So four devices each carry 2.08 A rather than one carrying 3.47 — and the \(I^2R\) loss splits accordingly, the same arithmetic as the interleaving of Set 18.

The output filter benefits from the doubled ripple frequency:

Only 16 µH for a 20 A output, because the filter sees 200 kHz rather than 100. A forward converter at the same frequency would need twice the inductance for the same ripple.

Why bridges dominate above 500 W:

The second row is easy to miss and is worth as much as the first. A forward converter's core operates from zero to \(B_{max}\) and back; a bridge swings from \(-B_{max}\) to \(+B_{max}\), so the same core handles twice the volt-seconds. Set 24 works that through.

What bridges cost. Four switches means four gate drives, two of them high-side:

Set 7's dead-time and bootstrap analysis applies to each leg. And the shoot-through risk is doubled — two legs, each of which can destroy the converter if both devices conduct together. At 60 W that complexity is absurd; at 1 kW it is unavoidable.

Define the figure of merit. Component count is a poor guide; what determines cost and loss is how much voltage and current each switch must handle:

A switch rated for high voltage and high current is expensive and lossy in both dimensions, so their product is a reasonable proxy for silicon cost — and it exposes differences that counting transistors hides.

Flyback, from Problem 1:

Single-switch forward, from Problem 2. The primary current is the reflected output current plus a small magnetising term:

Two-switch forward, where the clamp diodes hold each switch to the rail:

The single-switch forward is worse than the flyback on this measure despite being the more sophisticated topology, entirely because of the \(2V_{in}\) reset stress. Add a second switch and it becomes the best of the three by a factor of two.

The complete selection guide:

Every entry above 75 W has a switch stress of exactly \(V_{in}\), and that is not a coincidence — any topology that reflects a voltage back onto its switch is uncompetitive above the point where device cost matters.

The half bridge, and why it sits where it does. Two switches apply \(\pm V_{in}/2\) to the transformer:

Same voltage stress as a full bridge but twice the current, because only half the input voltage reaches the transformer. So a half bridge suits mid power, where current is manageable, and a full bridge takes over when the current becomes the constraint — typically around 500 W.

Two considerations the VA figure misses:

The last row connects back to Set 19. A flyback is a buck–boost, so it inherits the right-half-plane zero and its bandwidth limit; a forward is a buck, and has neither. For a supply that must respond quickly to load steps, that alone can decide the choice.

Where the problem comes from. Not all the primary flux links the secondary. When the switch turns off, the magnetising current transfers to the secondary but the leakage current has nowhere to go:

Two per cent of a 60 W output, dumped into the switch's drain node every cycle. With nowhere to go, it charges the drain capacitance to whatever voltage is needed — hundreds of volts above the 471 V of Problem 1, which is why the device would fail without a clamp.

The RCD clamp catches the drain at a chosen voltage and burns the energy in a resistor. Choose the clamp at 1.5 times the reflected voltage:

The clamp dissipates more than the leakage energy alone, because while the drain is held above \(V_R\) the transformer keeps pushing current into the clamp:

Three times the leakage energy — 6% of the output power, wasted. And the multiplier is \(V_{clamp}/(V_{clamp}-V_R)\), which grows without bound as the clamp voltage approaches \(V_R\). Clamping tightly to protect the switch costs efficiency very rapidly.

Size the components:

For 10% ripple on the clamp capacitor. Note that \(R\) must be rated for nearly 4 W — it is a real power component, not a signal resistor, and it is often several resistors in series to share the voltage.

The trade in clamp voltage:

Efficiency wants a high clamp; the device wants a low one. The 1.5× choice is the usual compromise, and reducing the leakage inductance is a far better answer than optimising the clamp — which is a magnetics problem, addressed in Set 24.

Now compare DCM. The boundary magnetising inductance at low line and full load:

Choosing \(L_m = 200\ \mu\text{H}\) puts the converter firmly in DCM, where it becomes the energy pump of Set 20:

What DCM costs and buys:

The fourth row is the one that decides it at this power. Leakage energy goes as \(I_{pk}^2\), so doubling the peak current quadruples the snubber loss — which would take the clamp from 3.7 W to over 15 W. DCM is excellent below about 20 W where the absolute numbers are small, and unusable at 60.