Carrying this two small steps further, for quick calculation:
P = (V(p-p)/20)^2
so that measuring transmitter power through a 30 dB attenuator (a popular
choice for nmr labs, make sure it's a 20 watt attenuator or larger) then
6.3 vpp corresponds to 100 watts into the 30 dB
a popular power level for high resolution spectrometers
Finally, each factor of 2 away from 6.3 vpp represents a factor of 4 away
from 100 watts power.
All this assumes measurements at 50 ohm impedance level. Make sure the
'scope is also set for 50 ohms input. It's still amazing how many
pitfalls and conditions can come into a basic measurement like this.
- Alan Olson
At 6:03 PM -0500 6/10/98, Charles G. Fry wrote:
>Woody has a typo; the last equation is easy to remember:
>
> P = V(RMS)^2 / R
>
> V(RMS) = V(p-p)/[2*sqrt(2)] ...peak-to-peak is always bigger than RMS!
>
> then if R=50 ohms
>
> P = V(p-p)^2 / 400
>
>Cheers,
>
>
>
------------------------------------------------
From Alan Olson, Instrument Engineer
National Institutes of Health, NINDS
10 CENTER DR, Room B1D 125, MSC 1060
Bethesda, Maryland 20892-1060
Phone: 301 496 8139, FAX: 301 402 0119
e-mail: awo@helix.nih.gov, olsonc@cvn.net(Alan)
voice mail: 1 800 952 7951 + when it answers: 68139
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