Note 10Power Rev 1 Watch the video · 2:09

Zero-voltage switching

What is zero-voltage switching, and what does it save?

Turning a switch on only after a current has already pulled the voltage across it to zero. A pair of 80 V FETs switching 10 A at 48 V then loses 0.45 µJ per turn-on, instead of 2.6 µJ.

The figure turns one FET on twice. Hard, it takes on the current while 48 V is still across it, and for a few nanoseconds it heats at up to 860 W.

At zero volts, a current drains that voltage away before the gate comes on. By the time the FET conducts, there is almost nothing across it to turn into heat.

One FET turning on

Hard, then at zero volts

A TPN19008QM, its gate driven from 10 V through 10 Ω. Hard, 10 A comes into it. At zero volts, a 0.4 µH inductor first pulls 10 A out of it, then turns around. 0 ns is the switch above it letting go.

Where the FET sits

The cell inside most converters

The FET never works alone. Most power converters pair it with a second switch and pass an inductor’s current back and forth between the two.

Only one switch is on at a time. Between turns comes a short dead time, here 50 ns, when both are off and the inductor’s current has to find its own way.

Each switch also has capacitance across it, drawn as C. At 48 V the pair holds Q=33.5 nC, and turning a switch on means moving that charge.

The way the inductor’s current flows as the upper switch lets go decides how the lower one turns on. Flowing into the node, the current holds it at 48 V, so the lower switch turns on hard.

Flowing out of the node, the current drains the lower switch’s C and fills the upper one’s. The voltage falls to zero by itself, before the gate comes on.

The switching cell

Both switches are TPN19008QM, 80 V FETs, with their gates left out. The arrows are the two ways the inductor’s current can flow as the upper switch lets go.

A capacitance that changes with voltage

Both switches, from the fitted model

The C in Fig. 10 isn’t one number. A FET’s capacitance is largest near zero volts and falls as the voltage across it rises, in this one from 783 pF at 1 V to 227 pF at 48 V.

That’s why this page counts charge instead. Q adds the capacitance up over every volt the switch swings through, for both switches:

Q=2∫48 V0CdV

=2×16.7 nC=33.5 nC

The capacitance at the node

Lower switch
Upper switch
Both

The upper switch has 48 V less whatever the lower one has, so its curve is the lower one’s, flipped. Below, the energy one switch stores is the shaded area, 0.32 µJ. A fixed capacitor holding the same charge would follow the dashed line and store the whole triangle, 0.40 µJ.

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Output capacitance

Half of QV, whatever the curve

Field marks. App notes model the curve as C=C_oss(V_oss/V)ⁿ, with n between ⅓ and ½. This FET’s fitted model follows n = 0.37.

Look-alike. The energy a capacitance stores, E=∫48 V0VCdV, is the shaded area in Fig. 10b. It’s less than ½QV because most of the capacitance sits at low voltage, so both switches together store 0.65 µJ at 48 V.

Range. Together the two curves are symmetric about 24 V, 1,012 pF at 1 V and at 47 V. On average the charge moves against half the supply, so moving Q costs ½QV, 0.80 µJ, whatever the curve’s shape.

Habitat. A hard turn-on burns that ½QV in its channel. A zero-voltage swing borrows it from the current instead.

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Where a hard turn-on loses energy

10 A into the node, and out of it

Start with the hard turn-on, the one most converters make every cycle. Its 2.6 µJ comes from four places, and turning on at zero volts removes three of them.

Where it goes

Overlap
Capacitance
Diode recovery
Dead time

Both switches’ heat. Diode recovery is the difference from the same turn-on run with a body diode that stores no charge.

242

Hard switching

Four costs, three avoidable

Field marks. For a few nanoseconds the switch carries current with voltage still across it: 1.2 µJ. The textbook sketch, E=½VI(t_i+t_v), gives 2.1 µJ, but the loop’s inductance lets the voltage start falling early.

Look-alike. Turning on also empties the switch’s capacitance and refills the other’s through its channel: ½QV=0.80 µJ. With no current at all, the turn-on still loses 0.84 µJ.

Range. Through the dead time the other switch’s body diode carried the current, 0.36 µJ. Its stored charge then comes out at full voltage, 0.24 µJ.

Habitat. At zero volts nearly all that’s left is the dead time, 0.43 µJ. It goes to the other switch letting go and to this one’s own diode.

243

Why a faster gate can’t fix it

The hard turn-on at six gate resistors

The usual fix for switching loss is a faster gate. Driving it through less resistance shortens the overlap, so the loss falls from 6.5 µJ at 47 Ω to 2.0 µJ at 1 Ω.

Then it stops falling. A faster gate can’t shrink the capacitance’s ½QV or the dead time, and the diode’s recovery grows. The faster edge also makes the other switch ring to 81 V, past its 80 V rating.

Getting under that floor takes a different idea: get rid of the voltage before the switch turns on.

Gate resistor against loss

The hard turn-on at 48 V and 10 A. The gate always turns off through 2 Ω.

How a current empties the switch

1 A out of the node, 50 ns of dead time

Run the inductor’s current out of the node and it does that for the switch. Here is the turn-on with just 1 A, slow enough to watch.

Let go. The other switch turns off.
Swing. The current pulls the charge out of the switch and into the other one, at a steady rate.
Diode. At zero the switch’s own body diode catches the current.
On. The gate comes on with nothing across the switch.

The swing moves Q at the rate the current flows, so the time it takes is

t=QI

=33.5 nC1 A=33.5 ns

Simulated, the swing takes 32.4 ns. Its energy goes back into the current instead of into heat.

Q already adds the capacitance up over every volt, so the rule holds whatever the curve’s shape.

The opening’s current was 10 A, so its swing took only 4.0 ns.

One zero-voltage turn-on

Across the switch
Through it

Negative current flows backward through the switch, out of the node.

How much current it takes

Every current against three dead times

The swing has to finish inside the dead time, or the switch turns on with voltage still across it. That sets the least current that will do.

Too little leaves voltage for the turn-on to burn. Too much only keeps the diode conducting longer, which is why 10 A loses more than 1 A.

Current at the node

0.5 A out

Dead time

Enough current

0.5 A out: 11 V left as it turns on

I≥Qt_dead=33.5 nC50 ns=0.67 A

Across the switch
Its channel starts to conduct

Lost: 0.04 µJ
Across it as it turns on: 11 V
The swing took: Not done
Q ÷ I: 67.0 ns

Out means the current leaves the node, the zero-voltage way. 10 A in is the hard turn-on. The least lost here is 0.01 µJ, at 0.5 A with 100 ns.

When an inductor makes the current

1 µH, the switch held off

So far the current held steady through the swing, as it does in a large inductor. A small inductor’s current falls as the voltage swings against it.

The swing then runs on the energy the inductor stores, and the rule becomes one of energy instead of time. The energy to cover is ½QV, 0.80 µJ, more than the 0.65 µJ the capacitances store.

Three starting currents

1.14 A
1.27 A
1.60 A

The inductor runs from the node to a steady 48 V, carrying its current out of the node until the other switch lets go at 0 ns. 1.14 A is what the stored-energy rule asks for, 1.27 A what ½QV asks for.

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Resonant swing

The inductor’s energy pays for the swing

Field marks. The swing reaches zero only if ½LI²≥½QV. For 1 µH that means I≥QV/L=1.27 A. Simulated, 1.27 A bottoms out 0.3 V short of zero.

Look-alike. Many app notes size the inductor by what the capacitances store instead, ½LI²≥2E, which asks for only 1.14 A. Simulated, that stops 4.0 V short and swings back up.

Range. At just enough current the swing takes a quarter of a resonant cycle, t=π/2LC. With C = Q/V that’s 41.5 ns. Simulated, 41.2 ns.

Habitat. The opening used 0.4 µH carrying 10 A, which stores 20 µJ, 25 times what the swing needs. Phase-shifted full bridges size their resonant inductor by energy too. TI’s SLUA107A counts what the capacitances store, the smaller figure.

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What it saves

One turn-on every cycle

A converter pays its turn-on loss every cycle, so the loss grows with switching frequency: P=Ef.

At low frequency the two turn-ons cost little either way. Above a few hundred kilohertz the hard one costs more than carrying the current.

Loss against frequency

Hard
Zero-voltage
Carrying 10 A

The two turn-ons from the top of the page, 2.6 µJ and 0.45 µJ.

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Zero-voltage switching

Nearly six times the frequency for the same loss

Field marks. At 100 kHz a hard turn-on costs 0.26 W and a zero-voltage one 0.05 W. At 1 MHz, 2.6 W and 0.45 W.

Look-alike. Carrying 10 A through one switch or the other costs 1.5 W either way. Hard switching matches it at 568 kHz, zero-voltage at 3.2 MHz.

Range. Most of what’s left is diode conduction. Just enough current and a dead time to match get it down to 0.01 µJ.

Habitat. The current has to come from somewhere, and making it costs. An assist cell simulated on moteus-c1’s leg cut its turn-on from 1.6 µJ to 1.0 µJ, not to zero.

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Zero-voltage circuits

Where the swinging current comes from

Every turn-on on this page used a current that was already there. A real converter has to arrange one, and the key sorts the usual ways.

Key to zero-voltage circuits

Where does the swinging current come from?

1aThe inductor’s current already flows out of the switch when the other lets goFree, like a buck’s low side
1bIt flows into the switchGo to 2
2aA small inductor can let its current reverse every cycleTriangular current mode
2bIt can’tGo to 3
3aA transformer is in the loopGo to 4
3bThere isn’t oneAn assist cell
4aIts magnetizing current is big enoughLLC, active-clamp flyback
4bIts leakage inductance has to supply itPhase-shifted full bridge

Each still has to meet one of the two rules on this page: enough current to move Q inside the dead time, or, from a small inductor, enough energy.

Conditions

Switches: Two Toshiba TPN19008QM, 80 V, with a model fitted by hand to the datasheet, as in Note 4. At 48 V each holds 16.7 nC.
Driver: An ideal 10 V source, on through 10 Ω, or 1 to 47 Ω in Fig. 10d, and off through 2 Ω, with 3 nH in each gate loop. 50 ns of dead time, 25 and 100 ns in the explorer.
Copper: 1.3 nH from the 1 µF capacitor around the two switches
Current: 10 A, held steady over the edge, into the node or out of it. In the opening’s zero-voltage pass, 0.4 µH to a steady 48 V instead, and 1 µH in Fig. 10f.
Heat: The opening shows the lower FET’s own. Everywhere else it is both switches’.
Lost: What both switches dissipate beyond carrying the current, from the other switch’s gate command until this one is on and settled 250 ns later. What their capacitance stores and gives back isn’t counted, and nor is the gate driver’s energy.
Solver: ngspice, at 27 °C

Simulated only. Nothing on this page has been measured on a board yet.

Limits

81 V is the FET model’s breakdown. At gate resistors of 4.7 Ω and below the other switch’s ring reaches it, so its real peak would be higher.
The model’s capacitance and body diode are fitted to the datasheet’s curves, not measured.
A large inductor’s current changes a little over an edge. Here it is held steady.
The overlap is what’s left of the hard turn-on’s loss after the other three parts.

Files

zsim/general.pyThe switching cell, every run on this page and how the loss is counted
gatesim/fets.pyThe fitted FET model, from Note 4

Sources

TI SLUA107A, phase-shifted full-bridge zero-voltage design
Infineon, primary-side MOSFET selection for LLC
TI SLUA871, the active-clamp flyback
Haider and others, zero-voltage triangular current mode, 2020
De Doncker and Lyons, US 5,047,913, the assist cell
Brendan Graham, ZVS critical primary current and effective drain-source capacitance
Kasper, Burkart, Deboy and Kolar, ZVS of power MOSFETs revisited, 2016
ngspice, the circuit simulator

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Rev 1 · 3 Nov 2026

Rev 13 Nov 2026First published

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