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Re: [TenTec] Voltages on 50 ohm Dummy Load

To: Discussion of Ten-Tec Equipment <tentec@contesting.com>
Subject: Re: [TenTec] Voltages on 50 ohm Dummy Load
From: "Martin, AA6E" <martin.ewing@gmail.com>
Reply-to: "Martin, AA6E" <martin.ewing@gmail.com>,Discussion of Ten-Tec Equipment <tentec@contesting.com>
Date: Wed, 8 Jun 2005 21:04:46 -0400
List-post: <mailto:tentec@contesting.com>

If you've got 274 volts sinewave peak-to-peak, that's .707 * (274 / 2)
= 96.8 volts RMS.  Across 50 ohms, my number is P = 96.8**2/50 = 187
Watts.  RMS volts are what you want, and it's the voltage swing from
0, not peak to peak. (RMS volts means "take the square root of the
_average_ of the square of the voltage" over one or more cycles.)

You're right that 1500 watts into 50 ohms is apx 274 volts (at any
moment), but power averaged over a cycle is what we and the FCC
usually talk about.  To get the average right, we need to use RMS
volts = .707 x peak volts (from zero).  For Vrms = 274 volts, Vpeak is
387, and Vpeak-to-peak is 774.

If you want to use Vpeak-to-peak (it can be easier to measure), the
formula would be

Power (watts rms) = 0.125 * (Vpp)**2 / R 
for 50 ohms,   P(watts rms) = .0025 * (Vpp)**2

(This is only right if you don't have much harmonic distortion.)

Your RF ammeter measures RMS amps if it's the thermocouple (heating) type.

Hope this helps.  This is an interesting thread, but I forget how it's
related to TenTec.

73 Martin AA6E

On 6/8/05, Robert & Linda McGraw K4TAX <RMcGraw@blomand.net> wrote:
> 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".
> Looking at it another way:
> 100 watts across 50.0 ohms produces 70.710 volts and a current of 1.4142
> amps.
> 1500 watts across 50.0 ohm produces 273.8613 volts and a current of 5.477
> amps.
> Another approach, I have a known good RF amp meter that I use with my dummy
> loads.  Comes in handy.
> Again, This is good enough for "government work".
> 73
> Bob, K4TAX
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