Power Measurement - Where am I wrong?

[Francis Belliveau]

I do not have a power meter so I figured that I could compute power from a
peak to peak measurement using my scope.  This should be accurate enough for
ballpark knowledge.
However, the measured voltage increases as frequency increases with the 10
meter voltage being nearly double the 80 meter voltage.  This would mean
power increases with frequency unless the formula I was going to use is
missing a frequency dependent component.

I have a KAT2 installed and I used a good 50 ohm resistive load.

My math follows:

Vrms = Vp-p/(2*sqrt(2))
Vrms = Irms*R assuming purely resistive load
or Irms = Vrms / R

if Prms = Irms*Vrms
substituting yields
Prms = (Vp-p * Vp-p) / (8 * R)

I know that Pave = Prms / 2
but that does not help much because I would still get more than 10 Watts out
on 10 meters when set for 5 watts out.

Where did I go wrong?

Fran


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[Ron D'Eau Claire-2]

Yes, you have the right formula, Fran.

Power = (Vpk-pk)^2/8R
Where R is the load resistance.

It is *not* frequency sensitive, but hardware often is.

Either your scope probe is not properly frequency-compensated or your load
is not a good, solid 50 ohm non-reactive dummy load. Note that it's very
hard to try to make these measurements into a real antenna, even if the SWR
indicator says 1:1. That's because most SWR meters are no where near
accurate enough.

First step, I'd bypass the KAT2 in case it's inserting reactance, trying to
"adjust" the tuning into a 50 ohm dummy load.

Ron AC7AC

-----Original Message-----
I do not have a power meter so I figured that I could compute power from a
peak to peak measurement using my scope.  This should be accurate enough for
ballpark knowledge. However, the measured voltage increases as frequency
increases with the 10 meter voltage being nearly double the 80 meter
voltage.  This would mean power increases with frequency unless the formula
I was going to use is missing a frequency dependent component.

I have a KAT2 installed and I used a good 50 ohm resistive load.

My math follows:

Vrms = Vp-p/(2*sqrt(2))
Vrms = Irms*R assuming purely resistive load
or Irms = Vrms / R

if Prms = Irms*Vrms
substituting yields
Prms = (Vp-p * Vp-p) / (8 * R)

I know that Pave = Prms / 2
but that does not help much because I would still get more than 10 Watts out
on 10 meters when set for 5 watts out.

Where did I go wrong?

Fran


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[Don Wilhelm-3]

Fran,

You have the calculations correct - I use Vp-p^2/400 all the time for a 50
ohm load.  I use the 'scope probe connected directly across the dummy load
(the 'scope probe has a short grounding lead).

It is quite unlikely that the power output really climbs with power, so
there must be something strange going on with your 'scope and its probes.
My Tek 465B with 150 MHz 10x probes (with the probe compensators corrected
to the 'scope's internal square wave signal) can easily be trusted to 30 MHz
or maybe even 50 MHz, but that is just what I have here - I don't know what
you are using.

Are you certain you are using a good 50 ohm resistive dummy load?  If you
have any doubts, sweep it with an antenna analyzer to be certain - don't
have an analyzer?, then take your dummy load to someone who has one and do a
quick sweep on it.

A quick sanity check on your 'scope is to use an RF Probe and a DMM to
measure the RF Voltage at the dummy load - compare that voltage reading to
what you are seeing on the 'scope.  Due to it's simplicity a properly
constructed RF Probe is usually quite accurate up to 30 MHz or more if the
tip to crystal path length and the ground leads are short.  Actually, most
RF Probes are more accurate than many wattmeters provided you have a good 50
ohm resistive dummy load.

73,
Don W3FPR

> -----Original Message-----
>
> I do not have a power meter so I figured that I could compute power from a
> peak to peak measurement using my scope.  This should be accurate
> enough for
> ballpark knowledge.
> However, the measured voltage increases as frequency increases with the 10
> meter voltage being nearly double the 80 meter voltage.  This would mean
> power increases with frequency unless the formula I was going to use is
> missing a frequency dependent component.
>
> I have a KAT2 installed and I used a good 50 ohm resistive load.
>
> My math follows:
>
> Vrms = Vp-p/(2*sqrt(2))
> Vrms = Irms*R assuming purely resistive load
> or Irms = Vrms / R
>
> if Prms = Irms*Vrms
> substituting yields
> Prms = (Vp-p * Vp-p) / (8 * R)
>
> I know that Pave = Prms / 2
> but that does not help much because I would still get more than
> 10 Watts out
> on 10 meters when set for 5 watts out.
>
> Where did I go wrong?
>
> Fran
>
>

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[Francis Belliveau]

All,

Thanks to all who answered me either direct or on the list.

Based upon the answers I would guess that I need to pay better attention to
lead length.  I did not make any attempts to keep things short.

I will also think about Ron's suggestion to bypass the KAT2.

I'll post the final results in a couple weeks (my time is sparse) when I
complete the tests.

73,
Fran


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