[Elecraft] Power measurement with a 'scope

Sat Feb 16 07:36:00 2002

Thanks to all with the Waveformsize / 2 / root 2 =3D Vrms. However, =
that's the same measurement as the RF Probe/DMM measurement, and I think =
my meter is more accurate than I can read the 'scope. So, I have a 100 =
MHz 'scope, but I do power measurements with the DMM. Sound right?

As to the math abuse. Any error measurement can go both ways unless =
specified otherwise. 5% error means +5% or less or -5% or more. It could =
be expressed by reducing the target by 5% (we'll use the 15 volts from =
your example) 14.25 v, +10%, -0%. There is no correct way to call this a =
10% error BOTH ways, which would mean it's really 20% error which would =
mean 40% error which would mean my K2 puts out a KW.

Your example states 5W and 50ohm. So, a measured 14.25 is 10% low =
[(15.8-14.25)/15.8]. Going to 10% high, we have 17.40v. These yield =
power readings of 4.06 (19% low) and 6.06 (21% high).

So, your idea was correct but the example had some problems.

As to the Cantenna resistance at HF, how do I measure that without an =
accurate power meter? I figured it wouldn't have any reactive components =
at all and the DC ohmmeter would do.

Dan / WG4S +- 3 letters / K2 #2456 +- 0=20

-----Original Message-----=20
From: [email protected]
[mailto:[email protected]]On Behalf Of Ron D' Eau Claire
Sent: Saturday, February 16, 2002 12:40 AM
To: Elecraft
Subject: RE: [Elecraft] Power measurement with a 'scope

> Now, it's true, if you don't know whether your 5% voltage error
> is high or
> low, then the resulting power measurement could be 10% low or 10% =
high,
> giving, in some sense, a possible 20% error spread.  But, of course, =
then
> the voltage measurement would, in the same sense, have been a 10% =
error,
> not a 5% error, and, again, the power error would be twice that of the
> voltage.
>
> dave belsley, w1euy

Thanks, Dave. I wasn't being rigorous in my example and I apologize.

The point I was making is that because the voltage is squared in the =
process
of calculating the power, the accuracy of the voltage measurement is =
much
more important to the final result than one might realize at first.

For example, if you are making exactly 5 watts out and measure 14.24 =
volts
instead of 15 volts rms (reading 5% low) or 15.75 volts instead of 15 =
volts
rms(reading 5% high) then the power calculation at 50 ohms load would be
4.06 watts at the lower side and 4.96 watts on the high side. That's a =
total
range of error of just about 20%. That's pretty bad when one starts out
thinking that 5% accuracy is surely adequate for a good, precise power
measurement.

Ron AC7AC
K2 # 1289

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