You cannot open the top and bottom devices of a leg together: that is a short across the bus, it is called shoot-through, and it kills switches instantly. The cure is a pause, and the pause creates the next problem.
What it is and why
The pause between closing one device and opening the other is dead time. With both devices off, the current in the inductance has nowhere to go, and interrupting current in an inductance is the thing you must never do: you get back EMF, the same effect as a relay or a motor.
So each device gets a diode across it, a freewheel diode, which gives the current a path and clamps the overshoot to the bus.
Where the current goes during that pause depends on your tuning, and this is why detuning is dangerous rather than merely inefficient. Burnett's three cases:
- Bridge above resonance. The current changes sign after the pause, and the freewheel diodes carry it. Relatively gentle, and this is where zero voltage switching is possible.
- Bridge exactly at resonance. The current is near zero and the diodes barely work. The best case.
- Bridge below resonance. The current changes sign before the pause, so it goes through the diodes, and their reverse recovery hits the switches.
Burnett: most of the stresses that a solid-state driver experiences in this application are a result of large currents being commutated between switches and freewheel diodes.
What you decide
How long the pause is
Criterion: the minimum that reliably prevents shoot-through, and not a nanosecond more. What actually sets it is the gate loop, and it has to be normalised as a fraction of the period rather than in nanoseconds, since the same setting is a fifth of the story at 150 kHz and a quarter of the period at
- That is its own piece.
Which diode does the freewheeling
Every MOSFET has a parasitic diode between drain and source, so it looks as though the freewheel diode comes free. It is a trap.
Burnett: the MOSFET body diode is a side effect of the fabrication process and is not a particularly good diode. Its reverse recovery time is very long.
The classical fix, where it matters: a Schottky in series with the drain, so the body diode can never conduct at all, with a fast external diode across the pair. A Schottky stores no charge and turns off immediately, and the bonus is that the reverse current no longer passes through the transistor die.
What to do if the external diodes still cannot keep up
Slow the switches down, with more gate resistance. The diode gets more time to recover, the reverse current peak falls, and less interference is radiated. It is the opposite of the instinct and it is Burnett's advice.
What will get you
And keep the tuning accurate, because then the current in the pause is near zero anyway and none of this is being stressed.
The numbers
- Voltage margin on the devices: about a third. Running from 320 V means 600 V parts.
- The safe side of resonance: above it. Below it the diodes commutate into the switches.
- Dead time: the minimum that prevents shoot-through, expressed as a fraction of the period.
What goes wrong
- Devices die at one tuning and survive at another. Running below resonance, where reverse recovery lands on the switches.
- Shoot-through despite a generous dead time. The body diode, not the timing.
- Large overshoot on the bridge output. Missing or slow freewheel path; the current in the inductance had nowhere to go.
- Adding dead time made it worse. It would. See the loop above.
- Radiated interference that arrived with a faster gate drive. Reverse recovery. Slowing the switches is the fix here, not speeding them up.
Where next
- IGBT or MOSFET, which device type to reach for, and its body diode is part of that answer.
- The gate loop is an RLC, which sets the dead time in the first place.
- An SSTC primary is a different part, where the worked primary current for choosing devices lives.
Attribution: the three cases and the body diode argument are Burnett's.