A dead time that is perfectly sensible on a 150 kHz coil eats more than a quarter of the period at 365 kHz. It is the same number of nanoseconds and it is a completely different setting, and the mistake is to carry it across unchanged.
The arithmetic
Two dead zones per cycle, so the fraction of the period they take is 2 · t_dead · f:
- 377 ns, which is one real published figure: 11.3 per cent at 150 kHz, 18.9 at 250, 27.5 at 365, 32.4 at 430.
- 200 ns: 6.0, 10.0, 14.6, 17.2.
- 100 ns, the usual recommendation for a SiC half bridge at high frequency: 3.0, 5.0, 7.3, 8.6.
And on a ramped coil the fraction moves while you watch
Everything above compares one coil against another. A QCW does it to itself inside a single bang, because the working frequency slides down the ramp as the arc loads the secondary — one published trajectory runs 519 kHz down to 463.
The dead time is a fixed number of nanoseconds, so 2 · t_dead · f is largest at the start of the ramp and smallest at the end. A setting chosen by looking at the waveform at full arc is being asked to work at the top of the sweep too, where it costs a bigger share of the period than anything you measured. Set it at the frequency where the fraction is worst, which is the beginning, and it is safe everywhere after that.
Two consequences worth carrying. Dead time that is too short is shoot-through, and shoot-through is a dead-time problem rather than a phase-lead problem, whatever it looks like on the scope. And the floor is not a matter of taste: the bridge needs enough residual current to recharge its own node during the dead time, which is what a phase lead buys.
Measuring your tank's resistance instead of calculating it
There is a method for this that deserves to be better known, and it needs nothing but a scope and a simulator.
Capture the real gate drive waveform. Feed it into SPICE. Run the whole pulse train, and adjust the tank resistance in the model until the decay envelope matches the one on the scope.
One builder's result: about 90 milliohms, which he attributed to skin effect, noting that his 8 AWG conductor was behaving more like 14 AWG.
And the skin effect arithmetic agrees
d = 65.2 / sqrt(f) [mm, in copper]At 150 kHz that is 0.168 mm. For a 3.26 mm conductor, which is 8 AWG, the effective area is pi · D · d = 1.72 mm² against a DC area of 8.37 mm². That is about 15 AWG.
He measured the equivalent of 14. The formula and the bench agree to within a gauge, which is worth knowing both because the method works and because the result is uncomfortable: most of a round conductor is not carrying anything. Copper tube and wide strip, for the same reason.
The three ways to find the secondary's resonance
Since none of the above works if the frequency is wrong:
- A signal generator through a resistor. Start at 1k, go to 10k when you want the dip sharp.
- Ring it down: a nanofarad into the scope, watch the decay.
- An antenna near the topload, looking for the peak.
And then remember that the number you have just measured is the resonance with the primary taken away, and that in the assembled system there is no such mode.
These are other people's measurements and our arithmetic on them. The rig on the DRSSTC department page is drawn to the ratios in these figures rather than to their scale.