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2:09 · With sound

Zero-voltage switching

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

Read Note 10

Transcript

Zero-voltage switching has been a game changer for half bridges.

But what is it, exactly? Let's go over it.

Every time a FET turns on, its voltage has to fall and its current has to rise.

With hard switching, the current rises first, while the full voltage is still across it.

Voltage times current is power, and that overlap turns into heat.

Zero-voltage switching brings the voltage to zero before the gate turns on.

Then the current rises with nothing across it, so almost no heat.

Many topologies do this, from LLC converters to active-clamp flybacks.

Let's explore one of them, in a motor driver.

This half bridge drives one phase of a motor from 48 V, at 10 A.

Hard switching: the low side lets go, and the motor current moves to its diode.

Then the high side turns on with 48 V across it.

Until it pulls the node up, it carries the full current and the full voltage at once.

That's 10.8 µJ of heat in one turn-on.

Turn the gate current up, and it drops to 1.6 µJ.

But the low-side FET rings right up to its 80 V rating.

One fix: a small inductor and a two-way switch, fed from the bus midpoint.

It builds up about 6 A more than the motor draws.

When the low side lets go, that extra current swings the node up to 48 V by itself.

The high side then turns on with nothing across it.

About 1 µJ, counting the assist's own losses, and the low side peaks at 52 V.

What about turning on the low side?

When the high side lets go, the motor's own 10 A swings the node down to zero.

The low side then turns on with nothing across it, no assist needed.

When the current flows into the node, the low side's turn-on is the hard one.

The assist then runs in reverse, building up about 6 A more than the motor pushes in.

That surplus swings the node down, and the low side turns on at zero volts.

About 1 µJ again, so both FETs turn on at zero volts either way.