You may be confusing PEP with RMS and instantaneous voltage. If you
have 1500 watts of power at any instant (fraction of a cycle) in a 50
ohm load, you have 274 volts, from P = V**2/R. So far so good. When
we normally talk about power in an AC (RF) context, we average over a
whole RF cycle. It turns out that that gives you 274 volts RMS (root
mean square) for a power level of 1500 watts. (The peak voltage
during the cycle is sqrt(2) higher = 387 volts, assuming no
harmonics.)
Now PEP refers to peak envelope power -- peaks due to voice
modulation. Voice modulation is much slower than the RF cycle. At a
voice peak and 1500 watts PEP, you would have 1500 watts of RMS power
(averaged over an RF cycle) and 274 volts RMS or 387 volts peak. (It's
the peak voltage times the possible VSWR that tells you what the
voltage rating of your coax, connectors and other RF components needs
to be.)
Another way to think about RMS voltage and power is that they refer to
the "heating value". In other words, 274 volts RMS into 50 ohms
delivers 1500 watts RMS, the same as 274 volts of DC into 50 ohms
delivers a steady 1500 watts. The load resistor gets just as hot
either way.
73 Martin AA6E
On 6/8/05, Williams, Barry <Bnwilliams@varco.com> wrote:
> Hello All,
>
> I need to calibrate the watt meters on my Titan, Omni VI+ and 253 Tuner.
> I have a Drake 2 kw PEP dummy load and a nice 200 mHz storage scope. In
> order to measure output, should I measure RMS, or Peak voltage, (if I
> use a sine wave input), and what are the voltages for 100 watts and the
> legal limit?
>
> My guess for a sine wave would be that for 1500 watts PEP out, I should
> see a peak voltage of about 274 volts, and 70 volts for 100 watts, if
> the dummy load is truly 50 ohms. Is this correct?
>
> Thanks,
> Barry
>
>
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>
--
martin.ewing@gmail.com
http://blog.aa6e.net
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