OLTC update - primary IGBT loss

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Terry Fritz" <twftesla-at-qwest-dot-net>

Hi Bert, Paul, and All,

I checked it all out and the 3.8nH inductance and 0.4mOhm resistance of the
caps and it is just is not a factor at all.  The leads are a concern but
there are ten in parallel and all that.  However, I think I have "found"
some losses ;-))

I dropped two high voltage high bandwidth scope probes directly between the
CE leads of a center IGBT (on the leads, right up next to the plastic case)
and got this:

	http://hot-streamer-dot-com/temp/OLTC09-02-03.gif

The red line is the CE voltage while the IGBT is on.

Seemed a little high so I tried it with both probes going to the same point
to check if the 2000 amps was getting on these "cheapo" probes"

	http://hot-streamer-dot-com/temp/OLTC09-02-04.gif

Looks fine!  So the IGBT CE waveform is probably very real!

I pulled all the waveforms out to a file (99% of you folks don't need to
click on this musty old set of 40,000 little numbers ;-))
http://hot-streamer-dot-com/temp/OLTC09-02-05.dta

But it all cleans up to the nice CE wave form here:

	http://hot-streamer-dot-com/temp/OLTC09-02-06.gif

We can clearly see where the IGBT kicks it on the high side at about 6
volts and the HEXFRED cuts in at about 5 volts.  Note how the HEXFRED
easily kills the inductive spikes that are tiny in this case.  However, the
wave form is not flat-topped as (some IR data sheets would lead us to
belive.  However, they do give some pessimistic instant current graphs ;-))
 Looks like we have a solid diode drop (rather high) and a DC resistance.
We'll just call the diode drop 6 volts and the resistance above that as
6/2000 = 0.003 ohms.  The current in this case is:

	http://hot-streamer-dot-com/temp/OLTC09-02-07.gif

So at full rated current, this extra resistance is dissipating about 70%
more heat in the IGBT as we would normally think from the data sheets.
That could "mess up" some designers out there using them :o))))  But we
seem to know more and can find where we stand.  Funny how Tesla coil's find
the "real specs" for this stuff ;-))  Happy to know that the heat is going
into the IGBTs where it can be easily handled,  If it were going into the
caps, that would be a 'bad' thing.

So with 6 volts drop and 3mOhms of resistance...  At 4500 amps and 560
volts firing...  The loss is 4500*6 + 4500^2 * 0.003 = 87750 peak watts
(note that 56% of that loss is in this extra IGBT package resistance).  The
peak system power is 4500 * 560 = 2,520,000 watts.  Dividing we get a "Q"
of 2,520,000/87750 = 28.7  Paul predicted 30 yesterday (That is why we
"listen" to good o'l Paul ;-))))  We can now plug our really good Q number
into the equation Rpri =  2 x pi x F x L / Q  to find a "nice equivalent
resistance" of 3.228 ohms of equivalent primary resistance.  Almost all of
that is in the IGBTs.  Cornell-Dubilier has "played" with use coilers
before so they probably double their 0.4mOhm resistance number just for
folks like us :-))))

So we need to compare this to a standard primary circuit using high voltage
and a spark gap.  28nF, 21000 volts, 85kHz, Req=3 ohms,...  Peak current is
314 amps.  Peak power loss is 296,000 watts.  Peak system power is 314 *
21000 = 6.6Mwatts.  Q = 22.3!  Oh goody,  My primary has a lot higher
current and higher Q than a normal Tesla coil!!  Q = 28 at 4500 amps!!  So
it would appear that the basic OLTC machine is sound.  It's that darn
secondary...  Paul predicted a Q of 45 when "we" thought it should be 200...

BTW - This peak power loss divided by peak system power seems very useful
as does the equation  Rpri =  2 x pi x F x L / Q.  We can probably carve
those into our desk tops ;-)))

I see a very large toroid in my future... :-))))  Primary seems fine as
long as I can keep whipping off turns...

Many thank to Paul for guiding my brain through all this ;-))

Cheers,

	Terry


At 10:10 PM 9/2/2002 -0500, you wrote:
>Hi Terry,
>
>Did you ever make an estimates of capacitor ESR?? Gotta' believe those
>flimsy leads also add a fair amount of series resistance in the primary
>circuit...
>
>-- Bert -- 
>-- 
>Bert Hickman
>Stoneridge Engineering
>"Electromagically" (TM) Shrunken Coins!
>http://www.teslamania-dot-com
>
>Tesla list wrote:
>> 
>> Original poster: "Terry Fritz" <twftesla-at-qwest-dot-net>
>> 
>> Hi Paul,
>> 
>> I was not able to find any time to work on this yeasterday.  But today...
>> 
>> The 20,40,60...  Volts I gave is a "metered" voltage.  The actual firing
>> voltage is twice that number.
>> 
>> If  Rpri =  2 x pi x F x L / Q
>> 
>> Vfire   Qpri    Rpri
>> 
>> 40      4.7     0.0197
>> 80      7.3     0.0127
>> 120     10.0    0.00927
>> 160     12.1    0.00766
>> 200     14.0    0.00662
>> 
>> So the graph looks like:
>> 
>>         http://hot-streamer-dot-com/temp/OLTC09-02-01.gif
>> 
>> If one uses your equations below which seem to fit very well:
>> 
>>         http://hot-streamer-dot-com/temp/OLTC09-02-02.gif
>> 
>> Looks like I can expect a Q of 21.4 at 4.3mOhms for Rpri.
>> 
>> Vds Seems high here at 8 volts.  Probably not too surprising given that so
>> many terms are at work in the "real" system.  However, if we pump 2000 amps
>> peak into the system, the loss for Vds is:
>> 
>> 8*2081 = 16648 watts peak
>> 
>> While the loss for Rpri is:
>> 
>> 2081^2 x 0.00662 = 28668 watts
>> 
>> The system peak power is about 200*2081 = 416kW
>> 
>> Of course, 416000 /(16648 + 28668) = Q = 9.18 here in my guess work.  Off
>> by the square root of two...
>> 
>> I will have to think about if Rpri or Vds has any reason to be so high.
>> Have to think about cap and Lpri second order resonances and other horrific
>> things...   Probably still have enough drive there in any case, but I just
>> like to know exactly "why"...
>> 
>> Cheers,
>> 
>>         Terry
><SNIP>
>