There is a short answer to whether an SSTC switches softly, and it is davekni's: an SSTC has no primary resonant capacitor, so the primary switches at the peak of the current, and zero current switching is a DRSSTC thing.
That answer is right about most beginners' coils. It is not the whole picture, and the whole picture is a single condition that tells you where any coil sits.
What soft switching would buy
If the load at resonance is purely resistive, the current crosses zero exactly when the voltage changes sign, and the switching is free. Burnett lists what follows:
- switching losses practically disappear;
- no overshoot and no ringing, because
di/dtis just the slope of a sine rather than a step in tens of nanoseconds; - paralleled devices share honestly: matching of switching times matters less, because there is almost no current near the moment of switching;
- series devices barely avalanche.
The condition, derived
The bridge's current is the resonant current, in phase, plus the magnetising current, ninety degrees behind. So the total lags the voltage by
tan(phi) = I_magnetising / I_resonantand putting Burnett's reflected-tank expressions in gives it in one line:
I_resonant / I_magnetising = k^2 · Q
phi = arctan( 1 / (k^2 · Q) )Which lets you see the whole range instead of two camps:
k= 0.5,Q= 100 with no arc:k²Q= 25, lag of 2°. Near perfect.k= 0.5,Q= 20: 5, lag 11°. Still good.k= 0.5,Q= 8 with the arc loading it: 2, lag 27°. Real losses.k= 0.3,Q= 8: 0.7, lag 55°. Bad.k= 0.15,Q= 8: 0.18, lag 80°. Switching almost at the current peak.
Read it as one coil moving down the table. Early in the bang, no arc, high Q, it is nearly soft. The arc grows, kills the Q, and the angle walks right. And a coil with a loose primary at k = 0.15, which is what a DRSSTC uses, never gets there at all.
So the primary is wound tight, not for energy transfer in the abstract but to get onto the right side of this condition. How that winding is actually built is a page of its own.
Burnett accepts a flashover risk that follows from it, and says why:
Spark breakout loads the secondary winding ruining the Q, so resonant rise is not good after spark breakout, therefore a high coupling coefficient is required to obtain maximum transformer action.
After breakout the coil is working less as a resonator and more as a transformer, and coupling is what carries it. Which is the Q = 8 row of the table, arrived at from the other direction.
Two caveats, and one of them looks like a contradiction
The numbers
- The condition:
Q > 1/k². - The lag it produces:
arctan(1 / k²Q), tabulated above from 2° to 80°. - Secondary Q: around 100 to 200 before breakout, about 8 with the arc loading it, and as low as 4 on other people's measurements.
- What switching costs when the condition fails: an FGA60N65SMD at 60 A and 240 kHz gives 187 W of switching loss against 40 W of conduction, about 227 W per device. Switching is nearly five times dearer than conduction.
- Which is why an SSTC takes MOSFETs. The switching happens at large current, so switching loss decides, and an IGBT's is large. Fast IGBTs are only usable at low duty.
What goes wrong
- Hard switching on a coil that calculated as soft. Check
k²Qwith the arc's Q, not the unloaded one. The condition is satisfied before breakout and can fail after it. - A DRSSTC primary geometry reused on an SSTC. At
k= 0.15 the condition cannot be met at any Q you will see, and the bridge switches at the current peak every cycle. - Paralleled devices that share badly only at high power. Same cause. The sharing was being carried by soft switching and the arc took it away.
- IGBTs that survive on the bench and die on the arc. The bench had no load to kill the Q.
- Soft switching that appears late in each bang and not at the start. Working as described. The secondary current needs tens of cycles to build.
Where next
- An SSTC primary is a different part, the winding this condition demands.
- IGBT or MOSFET, the choice this decides.
- Switching at zero is the failure case, the second question.
The SSTC department diagram draws the primary standing beside the secondary on one axis, because being near it is the only way they are coupled at all, and this is the department where near means near.