Re: SSTC transformers

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Jimmy Hynes" <jphynes@xxxxxxxxx>


> Actually, thats not correct either.  If you want to be precise, the no
> margin rail-to-rail voltage for a H-bridge that actually switches would
> be 1000V minus Vce (conducting IGBT) not 1000V+Vce.
>
> Dan


Working in the assumption that everything is 'perfect', and current is
always flowing the 'right' way, it is +Vce.

Say we ground the emitter of the lower IGBT, it is at 0v. Vce of the
lower IGBT when on is say 2v. Vce of the upper IGBT is 1000, because
we're pushing this thing to the limit :-o.

Now, the voltage from the upper collector to the lower emitter
(ground) is 2v+ 1000v, or 1002v. This is exactly 1000V+Vce. The same
can be shown for when the upper IGBT is on.

Say you have some really cruddy IGBTs that can withstand 100v when
off, and drop 50v when on. say you give it 90Vdc rails.

They can clearly survive when they're both off, since 100v+100v (we're
assuming everything is perfect, remember? :-P) is 200v, which is >
90v. They can clearly survive when one IGBT is directly shorted, since
100v >90v.

Are you saying that they can only survive 100v-50v = 50v when the
lower IGBT is just 'on'?

Of course, when current is flowing in the opposite direction, the
voltage drop does subtract from your maximum, although that is the
diode's voltage drop, not the IGBTs. Once we take that into account
when we define our "zero margin voltage", we have to take into account
other sources of 'extra' voltage, like voltage spikes due to stray
inducances, and our maximum voltage becomes 1000v - "something
relatively big and complex to calculate"

That said, the 2v doesn't really matter when you look at a real system...