[AMRadio]Shunt and RF power measurment |
John E. Coleman (ARS WA5BXO)
wa5bxo2005 at pctechref.com
Fri May 5 15:49:29 EDT 2006
Rick I will apologize ahead of time here because this math may be very easy for you but for some it may not be. So I have explained it below and simplified it at the end. I have done this many times. I have used the Heath Cantenna and mounted the 6AL5 in the little box on the top. You would need to build it into your dummy load area of your BC rig. The measurement of the DC voltmeter and the accuracy of the 50 OHM dummy load are your only error points on HF. See the schematic at this link http://wa5bxo.shacknet.nu/RF%20Power%20Measurement/ One of the best ways to get a really accurate reading is to install 6AL5 diode very close or in the dummy load compartment and connect the plates to the dummy load input connection point and put a ceramic bypass cap on the cathodes. Use a separate filament XFMR for it, as it will have a few hundred volts on the cathode. You may want to use a lantern battery for the filament source if it is just a temporary installation. The capacitor on the cathode of the 6AL5 will charge to the peak of the RF voltage and can be measured with any DC meter. Here is a scenario: DC measured = 300 Volts 300 * .707 = RMS = 212.1 (Erms * Erms) / R = Power (212.1 * 212.1) / 50 = 899.72 Watts The above calculations have a slight error because of the round off of RMS (.707) The RMS value of something is actually the square root of 2. With that in mind the total equation for a 50 ohm dummy load is reduced to: (Epeak *Epeak)/100 = Power Where Epeak is the measured DC on the cathode of the 6AL5 So if the DC measured was 300V the power is 900 Watts You just square you measurement and divide by 100. If you measure 316 Volts DC then you are pretty close to 1000 Watts John, WA5BXO
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