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DEI Transistors

144inBama

Sr. Member
Apr 22, 2020
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Lower Alabama
Here's where ohms law fails me in RF.... If you use Ohm's Law with 14.7VDC and say 65A, it gives you .226 ohms and 955.5 watts.... Where does the 50 ohm load (coax) come into play for the calculation? The entire transmission length has a 50 ohm load yet you cannot figure that into the equation.
 
Last edited:

Shockwave

Sr. Member
Sep 19, 2009
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Here's where ohms law fails me in RF.... If you use Ohm's Law with 14.7VDC and say 65A, it gives you .226 ohms and 955.5 watts.... Where does the 50 ohm load (coax) come into play for the calculation? The entire transmission length has a 50 ohm load yet you cannot figure that into the equation.
The amplifier is converting DC input to RF output. The .226 ohm input resistance is what it takes to draw close to a thousand Watts from the DC line. To produce a similar output power into 50 ohms, means the RF voltage is much higher than the DC voltage. Someplace over 200 volts of RF. With approximately 200 volts of RF applied to a 50-ohm antenna system, we are delivering 800 watts of RF to that antenna.
 

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