On Fri, 29 Sep 2006 06:30:53 -0700, Frank Gilliland
wrote in
:

On 29 Sep 2006 05:40:05 -0700, "Telstar Electronics"
wrote in
. com:

U-Know-Who wrote:
516 watts PEP, huh? You are ignorant.

http://www.telstar-electronics.com/SkyWave%202879AB.pdf

Actually it will do that... and more, if you will accept compression
distortion. I suggest you great electronics wizards go to
http://www.rf-amplifiers.com/index.php?topic=peak_power and read the
defibition of PEP. Or maybe you think the people at EMPower don't know
what they're talking about either... LOL


Yep, their 'defibition' is right on the money. Now I suggest -you- do
a little math with power dissipation and efficiency ratings.
snip


Since you don't want to do the math yourself, I'll do it for you.....

You claim 516 watts PEP and 55% efficiency. And I'll be super-nice and
say that's at 14 volts and you aren't going into saturation. So that
means:

516 watts / 14 volts = 36.86 amps
36.86 amps / 55% efficiency = 67 amps
67 amps / 2 transistors = 33.5 amps Ic on each transistor.

Well that's a problem since the transistor is rated for an absolute
maximum of 25 amps at the collector. A paradox to be sure. But wait...

An efficiency of 55% means that 45% is dissipated as heat. So:

33.5 amps * 45% = 15.1 amps
15.1 amps * 14 volts = 211 watts

Wheee, that's friggin' hot!!! Especially when the device is only rated
for 250 watts dissipation max, assuming a perfect heat sink! Now a 200
watt soldering gun will glow red in just a few seconds..... should we
take a brief sojourn into reality and calculate junction temperature?
I'm betting it will greatly exceed the max rated 175 degrees. Care to
take that bet, Brian?