On 29 Sep 2006 06:39:35 -0700, "Telstar Electronics"
wrote in
. com:

Frank Gilliland wrote:
BTW, how
did you get your distortion figures?

The 1dB compression point was calculated using an HP spectrum analyzer
I have access to at work. It was also used to get the harmonic
readings.


So how did you calculate the "1dB compression point"?

Frank Gilliland wrote:
So how did you calculate the "1dB compression point"?

RF power in vs RF power out. When the gain reduces by 1dB... you're
there.

www.telstar-electronics.com

On 29 Sep 2006 06:49:47 -0700, "Telstar Electronics"
wrote in
. com:

Frank Gilliland wrote:
So how did you calculate the "1dB compression point"?

RF power in vs RF power out. When the gain reduces by 1dB... you're
there.


But how do you measure power with a spectrum analyzer?

And have you done the math yet for your power dissipation problem?

On 29 Sep 2006 06:47:15 -0700, "Telstar Electronics"
wrote in
. com:

Frank Gilliland wrote:
Where's any 2SC2879 data sheet that specifies its "1% compression
point"?

Frank, I'm surprised at you.
It doesn't... that parameter doesn't apply to the transistor itself...
only a final amplifier. After all, how would Toshiba know what
frequency or class of operation were being used?


Well, they designed the transistor for a specific application, and the
test circuit is rated at 28MHz, so I'm pretty sure they knew exactly
how it was going to be used.

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?

Frank Gilliland wrote:
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.

Amplifier efficiency isn't based on peak power (PEP) you putz... see
http://www.rf-amplifiers.com/index.php?topic=dc_input

Are you really this stupid?... or are you just putting us on?

www.telstar-electronics.com

On 29 Sep 2006 08:01:56 -0700, "Telstar Electronics"
wrote in
.com:

Frank Gilliland wrote:
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.

Amplifier efficiency isn't based on peak power (PEP) you putz... see
http://www.rf-amplifiers.com/index.php?topic=dc_input


Once again you didn't understand what I wrote. I wasn't calculating
amplifier efficiency; I calculated peak collector current -based on-
your very own amplifier efficiency claim. Would you rather I use the
transistor's collector efficiency of 35%?


Are you really this stupid?... or are you just putting us on?


I have thought of asking the same question of you, but I haven't
because you would ignore it just like all the other tough questions.
That, and I already know the answer.

Frank Gilliland wrote:
I have thought of asking the same question of you, but I haven't
because you would ignore it just like all the other tough questions.
That, and I already know the answer.

Great response... you really are a piece of work... LOL

www.telstar-electronics.com

Hello TE:

What is the harmonic content with your amp?

Do you have a filter in the output circuit?

Just wondering, here waiting for the Fedex guy.

Jay in the Mojave


Telstar Electronics wrote:

Frank Gilliland wrote:

BTW, how
did you get your distortion figures?


The 1dB compression point was calculated using an HP spectrum analyzer
I have access to at work. It was also used to get the harmonic
readings.

www.telstar-electronics.com

On 29 Sep 2006 08:18:43 -0700, "Telstar Electronics"
wrote in
. com:

Frank Gilliland wrote:
I have thought of asking the same question of you, but I haven't
because you would ignore it just like all the other tough questions.
That, and I already know the answer.

Great response... you really are a piece of work... LOL


Ok, we'll use the collector efficiency.....

516 watts / 14 volts = 36.86 amps
36.86 amps / 35% efficiency = 105.3 amps
105.3 amps / 2 transistors = 52.65 amps Ic on each transistor.

And since 65% is dissipated as heat,

52.65 amps * 65% = 34.22 amps (Scary!)
34.22 amps * 14 volts = 479 watts.... HOLY ****!!! You better have a
BIG MF HEAT SINK!!! Are these transistors water-cooled?