John E. Coleman (ARS WA5BXO) wa5bxo2005 at pctechref.com
Fri May 5 16:05:36 EDT 2006

```Correction:
The RMS value of peak on a sine wave is the square root or 0.5
The square root of 2 (1.414) converts it from RMS to Peak.

The Converted formula at the bottom of the page for power is however
correct.

Thanks for overlooking my error.
John

-----Original Message-----
[mailto:amradio-bounces at mailman.qth.net] On Behalf Of John E. Coleman (ARS
WA5BXO)
Sent: Friday, May 05, 2006 2:49 PM
Subject: RE: [AMRadio]Shunt and RF power measurment

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
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

______________________________________________________________
Help: http://mailman.qth.net/mmfaq.html