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Static_killer77
New member
Username: Static_killer77
Post Number: 8
Registered: 10-2005
| | Posted on Saturday, June 07, 2008 - 5:43 pm: | 
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I've been seeing this electrical law being used by many amplifier builders for quite some time now. Seems someone somewhere got someone believing that DC current multiplied by DC Voltage somehow equals RF Power output. Tisk Tisk. Example; if you look on Texas Star's web page for instance, it claims the the 500v makes 500 watts of power at 45 amps. Now, if use the standard 13.8 voltage and plug it in, you get 621 watts out. That would be great, but... if turn the equation around; 500 watts divided by the 45 amps, then it only requires a little over 11 volts to achieve the same 500 watts. Another examlpe is my radio. At a full dwaw of 3.00 amps, at 13.75 volts in my truck, I get a solid 10 watt dead key. But if you do the math, my radio should dead key 41 watts. That ain't even close to 10. So if any of the gurus out there wanna try and shine some light(besides telling me it'a selling technique ) I'd love to here from ya! Thanks all, Static |
Bruce
Senior Member
Username: Bruce
Post Number: 4734
Registered: 9-2003
| | Posted on Sunday, June 08, 2008 - 7:44 am: |
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E * I = INPUT POWER
SO if it has 621 watts INPUT and 500 output it is 80% efficient. Can't BE DONE WITH A TRUE LINEAR.
That amp must be running class "C" not AB or B and class "C" is not a linear amp .......
BTW my home built CLASS "C" 29.600 MHZ FM amp for 10 meters runs 325 watts out on a bird meter at about 40 amps.
On 6 since 66
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Static_killer77
New member
Username: Static_killer77
Post Number: 9
Registered: 10-2005
| | Posted on Saturday, July 12, 2008 - 1:36 am: |
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The Texas Star 500v is advertised as class AB-1, so it can't be class "B", or "C".
BTW here's their link. www.texasstar.com/export/specifications.html
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Drzuo
Junior Member
Username: Drzuo
Post Number: 20
Registered: 1-2006
| | Posted on Monday, July 14, 2008 - 10:35 pm: |
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remember you are also comparing RMS DC current with peak AC watts the whole DC to AC thing. Pat, you explain I'm tired. |
Patzerozero
Senior Member
Username: Patzerozero
Post Number: 4343
Registered: 7-2004
| | Posted on Saturday, July 19, 2008 - 8:31 pm: |
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http://www.qsl.net/km5kg/loads.htm
not exactly the answer to the question asked, but an explanation of why TS says what it says, yet those numbers can be either attained, exceeded, or not, based on swr and whether load impedance is high or low. that's why some see 600+ from a dx500 & others see just over 400. the 400 reading is a more realistic number for the transistors when the load is matched across the board & theoretically MOST of the watts out of the box gets out of the antenna. a 600 watt number is showing that out of the box, but maybe 200 or more is coming back rather then going out via the antenna. load impedance also affects current draw. so 13.8 volts can produce 500 watts, but with varying draw based on how well the antenna system is matched to everything else. |
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