CB gear for sale
 

"N8WWM" wrote in message
news:1097882528.56/Y//wsOcl7ntLQduxcVQ@teranews...
32 pill mobile amp. Swings 7000 watts. $2000 firm.

Be LOUD and PROUD!





How much drive does it require?

Bada Bing wrote:

"N8WWM" wrote in message
news:1097882528.56/Y//wsOcl7ntLQduxcVQ@teranews...

32 pill mobile amp. Swings 7000 watts. $2000 firm.

Be LOUD and PROUD!






How much drive does it require?


And, HOW BIG of a 440 Volt 3 Phase Sub-Station is required?
The corona off the gutters ought to remove the need for a yard
light, tho ! :)

--

A typical efficiency is about 50%, possibly less at these current levels, so
figure14000W input, or 1000A at 14V? I'm going to bet that there's some
significant I^2R loss in that system.

Of course on the new car systems with the higher voltages, things will get a
lot easier.

--
KC6ETE Dave's Engineering Page, www.dvanhorn.org
Microcontroller Consultant, specializing in Atmel AVR

--

"NN7KexNOSPAMk7zfg" wrote:
Bada Bing wrote:

"N8WWM" wrote in message
news:1097882528.56/Y//wsOcl7ntLQduxcVQ@teranews...

32 pill mobile amp. Swings 7000 watts. $2000 firm.

Be LOUD and PROUD!






How much drive does it require?


And, HOW BIG of a 440 Volt 3 Phase Sub-Station is required?
The corona off the gutters ought to remove the need for a yard
light, tho ! :)

I'm pretty sure that's a mobile amp Doug is selling. Must take quite
a bank of alternators.

Steveo wrote:
"NN7KexNOSPAMk7zfg" wrote:

Bada Bing wrote:


"N8WWM" wrote in message
news:1097882528.56/Y//wsOcl7ntLQduxcVQ@teranews...


32 pill mobile amp. Swings 7000 watts. $2000 firm.

Be LOUD and PROUD!






How much drive does it require?



And, HOW BIG of a 440 Volt 3 Phase Sub-Station is required?
The corona off the gutters ought to remove the need for a yard
light, tho ! :)


I'm pretty sure that's a mobile amp Doug is selling. Must take quite
a bank of alternators.
More like a WELDER, GENERATOR or, one hell of an extension cord!
and, the antenna would need to be about 5 inches in diameter
to keep from melting!

7kW at 50 ohms isn't THAT much current.

What's the feedline connector?

--
KC6ETE Dave's Engineering Page, www.dvanhorn.org
Microcontroller Consultant, specializing in Atmel AVR

"Dave VanHorn" wrote in message
...
7kW at 50 ohms isn't THAT much current.

What's the feedline connector?


BNC should work nicely.

I forget the math?
P=IxE and E=IxR so P=IxIxR? 7KW=IxIx50 7000/50=IxI 140=IxI
140=me twice
Damn now I've got a headache!


Dave VanHorn wrote:

7kW at 50 ohms isn't THAT much current.

What's the feedline connector?

I forget the math?
P=IxE and E=IxR so P=IxIxR? 7KW=IxIx50 7000/50=IxI 140=IxI
140=me twice
Damn now I've got a headache!

P=I^2R so assuming a properly matched load, P/50=I^2 and I is about 12A.

--
KC6ETE Dave's Engineering Page, www.dvanhorn.org
Microcontroller Consultant, specializing in Atmel AVR

On Mon, 18 Oct 2004 10:39:02 -0500, "Dave VanHorn"
wrote in :

I forget the math?
P=IxE and E=IxR so P=IxIxR? 7KW=IxIx50 7000/50=IxI 140=IxI
140=me twice
Damn now I've got a headache!

P=I^2R so assuming a properly matched load, P/50=I^2 and I is about 12A.


That's 12 amps @ xxMHz -- don't forget about skin effect. At 30 MHz
with 12 AWG copper you will end up radiating about 22 watts per foot
in the form of heat. And to put that into perspective, most small
soldering irons are 15-30 watts.






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