I'm really surprised at all the folks that missed the entire point.
Look at the numbers again that I stated relating to 50 ohms:
70.710 volts x 1.414 amps = ~100 watts
273.861 volts x 5.477 amps = ~1500 watts
I said the measurement technique was to use the scope to measure peak to
peak voltage. Reason being, most folks do not have RF volt meters and the
HB probes using various diodes are as unreliable as they come without proper
calibration techniques. Thus the scope is the easiest way and most accurate
to make measurements in the HF spectrum.
Yes the scope does present P - P values but I have to ask, do most folks
know how to convert P - P values to RMS and do they know what the scope
measures in terms of voltage units? Also I'd expect that few have accurate
RF amp meters available to them.
----- Original Message -----
From: "Ken Brown" <firstname.lastname@example.org>
To: "Discussion of Ten-Tec Equipment" <email@example.com>
Sent: Thursday, June 09, 2005 7:49 PM
Subject: Re: [TenTec] Voltages on 50 ohm Dummy Load
>>I measure the peak to peak value using my scope. Then power equals E
>>squared divided by R. This is good enough for "government work".
> This is wrong! Using peak voltage instead of RMS voltage of the peaks
> will result in a power miscalculation by a factor of two. Using Peak to
> Peak voltage instead of RMS voltage of the peaks would result in a
> miscalculation by a factor of 8!
> Peak voltage is 1.414 X RMS. Peak to peak voltage is 2 X Peak or 2 X
> 1.414 X RMS. When you square this in the power equation your
> miscalculated power would be 8 X the actual power.
> DE N6KB
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