RE: OLTC update - primary IGBT loss

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

Original poster: "John H. Couture by way of Terry Fritz <twftesla-at-qwest-dot-net>" <couturejh-at-mgte-dot-com>


Terry, All -

The gif figure  shown below of a real world dampened curve can be used to
find the Q Factor using the 10% method indicated in my Tesla Coil Notebook.
It is only an approximation so I doubt you will find this in any other
publication. The curve has the typical exponential form of the dampened
wave. All you have to do is measure the the curve on your computer screen
and find the number of cycles it takes to reach the 10% amplitude. It is
only an approximation but your calculated Q can then be compared with the
real world test curve.

   Q = 3.1416/log dec and

   Q = 3.1416/ln(10) times the number of cycles to 10% amplitude.

The curve appears to be roughly about 14 cycles to 10%. This would give a Q
Factor of about

   Q = 3.1416/ln(10) * 14 = 19.10

This Q is pretty close to the 22 to 28 you are finding. Could the
difference be due to the resistance you selected and it should be slightly
changed? It looks like it is close to Paul's
         79/4 factor = 19.75 Q Factor

Also, the approximate percent reduction per cycle can be found as follows:

   Log dec = 3.1416/Q = 3.1416/19.01 = .1653

   a1/a3 = e^.1653 = 1.1797

   Percent reduction = 100/1.1797 = 84.77%

John Couture

--------------------------


-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Monday, September 02, 2002 10:31 PM
To: tesla-at-pupman-dot-com
Subject: OLTC update - primary IGBT loss


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

Hi Bert, Paul, and All,

----------------------------  snip


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