Re: Transistor Veb rating... plus Patent Update


"James Arthur" <dagmargoodboat@xxxxxxxxx> wrote in message
news:1171662768.877420.106980@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Feb 16, 3:46 am, "Genome" <mrspamizg...@xxxxxxxxxxx> wrote:
"James Arthur" <dagmargoodb...@xxxxxxxxx> wrote in message

news:1171614683.225166.128670@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx



On Feb 15, 1:01 am, "Genome" <mrspamizg...@xxxxxxxxxxx> wrote:
This could almost be a first in that I don't often ask questions. Some
of
you might remember my mention of a Patent what I have applied for. If
you
don't then an explanation about it is here.

http://www.genomerics.org/patent/patent.html

<snip>

Genome, color me stupid, but it looks like your model switches off in
about 200nS, but IRF specs the device to switch in roughly 1/3rd that
time:

http://www.irf.com/product-info/datasheets/data/irgp50b60pd.pdf

Also, IRF specs the max turn-off loss at 530uJ, versus your
(modelled) 1.56mJ.

I'm sure I've missed something obvious...kindly bludgeon me with
enlightenment. (Not that this affects your loss-saving gadget)

Best,
James Arthur

Yes, you are right. However I do make a statement on the first page....

"Good old LTSpice let's us have a look at the sort of behaviour that
makes
life dull. If you believe the models that is."

It's one of Wins pet moans about the disagreement between spice models
and
reality. I used the IR model because it does manage to demonstrate the
kind
of behaviour I'm trying to overcome. ST models are/were rubbish in that
respect showing little to no turn off losses.

Cheers

DNA

Your disclaimer is clear and noted.

Kundun, I am but a bug, but if you've got a honking FET there
anyhow, why not parallel it with the IGBT, drive both ON
simultaneously, and turn the FET OFF a recovery-time after turning off
the IGBT?

In this way you'd gain the switching speed of a FET, save some
switching loss at turn-on too--further improving efficiency--and
secure the saturation voltage advantages of the IGBT. The FET need
only handle the full load briefly, and so could be small. Drive would
be simple.

Humbly yours,
James Arthur


The grovelling is unecessary.... I am not Welsh so my first name is not Dai.

Yes, what you suggest was in my journey to the final answer. The second
circuit on the first page shows something that would be similar. The I2
current source reduces the IGBT current to zero in the same way a directly
paralleled mosfet would.

However, as suggested, the problem is stored charge. The recovery time you
mention is the time for charge recombination to occur. I do not know much
about this process other than some words for it. However it takes a
relatively long time if left to its own devices. There are processing
methods and such stuff that reduce this called names like lifetime killing
or quenching but they still don't seem to work too well.

I suppose that if you are prepared to sit around waiting for it to happen
then it does work but that is going to be a long and variable time.

This is why I thought about actively removing the charge by applying a
reverse bias to the device. This is done in transistor switching ciruits and
is easy because the base terminal is available. With the IGBT it is not but
by dangling stuff about the place it is possible to get the mosfet to do the
job.

In the first boost circuit that is done by splitting the boost inductor.
That's a bit loose. In the second circuit, on the second page, it is done by
adding the forward converter part which produces a more tightly defined
result.

When I had done it I hunted around for patents that did the same thing. I
found many that use a simple paralleled mosfet, which does not work very
well as far as I am concerned. I did not find any that applied a reverse
bias.... which does work exceptionally well.

There's all sorts of stuff about the relative timing and length of drives
and selection of devices and possibilities for zero voltage switching that
make things a bit more complicated.

By having a fiddle in the circuit on the second page I can get the
'parallel' mosfet losses down to 3W. The IGBT loses about 29W and the mosfet
that does resonant switching of the diode, and other bits, loses 23W. If I
put a SiC diode in there I might get rid of those 23W so total losses for a
2kW PFC stage would be 32W.

Take those figures with a big pinch of salt though.

DNA


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