more power out measurement

[RobertG]

I have a dummy load that I made up out of 10x 510 ohm resistors in parallel
[51 ohms, 30w]. I checked this with my MFJ 259B that says it's flat on all
the HF bands. I then hook this load directly to the output jack and measure
the voltage across it using the K2 RF probe and a DMM. I calculate the
expected RMS voltage as a function of power. I then increase K2/100 power
until I see the expected voltage [power], and then set R26 so that the LCD
shows the "correct" power.

This all sounds okay in theory. Does any one have a comment on the accuracy
of the setting using this method and/or "hidden variables" that I may have
missed? Thanks much.

...robert #5957

Robert G. Strickland PhD ABPH - KE2WY
[hidden email]
Syracuse, New York  USA

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[Don Wilhelm-3]

Robert,

All that sounds fine - your theory is impeccable!
There are 2 limitations - first is that the voltage breakdown of the
diode forces you to do this calibration at a power under 20 watts or so,
and second, the wattmeter diodes may not be exactly linear which will
produce errors at the higher end of the power range.

The wattmeter scaling resistors in the KPA100 are 1% tolerance, so the
ratio is quite good.  I suggest that you use a power of about 10 watts
when doing that calibration.  Remove the power from the KPA100 and power
only the base K2 to make things easier.  Calibrate at 10 watts and then
the scaling resistors should make the calibration at 100 watts almost as
accurate.

If you try to make the calibration at 20 watts and make even small
measurement errors, those errors multiply at 100 watts, but relying on
the accuracy of the scaling resistors and calibrating at 10 watts will
yield a far greater accuracy than attempting to calibrate at say 20
watts (or anywhere on th elow end of the 'high power' range.

Try it, I believe you will find the results quite acceptable.

I frequently use that method when calibrating a wattmeter, but instead
of trying to adjust the power to a certain value, I read the RF voltage
on my 'scope, do a quick calculation and set the pot to the result of
the calculation.  Reading the Peak to Peak RF voltage, I can quickly
calculate the power as V^2/8R (the denominator is 400 for a 50 ohm
load), or if one is using a zero-to-peak indicating diode detector (like
the one used in the DL1) connected directly across the load, the power
is V^2/2R - if the power is greater than 5 watts, the diode drop becomes
negligible in the equations.

73,
Don W3FPR

Robert G. Strickland wrote:

> I have a dummy load that I made up out of 10x 510 ohm resistors in
> parallel [51 ohms, 30w]. I checked this with my MFJ 259B that says it's
> flat on all the HF bands. I then hook this load directly to the output
> jack and measure the voltage across it using the K2 RF probe and a DMM.
> I calculate the expected RMS voltage as a function of power. I then
> increase K2/100 power until I see the expected voltage [power], and then
> set R26 so that the LCD shows the "correct" power.
>
> This all sounds okay in theory. Does any one have a comment on the
> accuracy of the setting using this method and/or "hidden variables" that
> I may have missed? Thanks much.
>
> ...robert #5957
>
> Robert G. Strickland PhD ABPH - KE2WY
> [hidden email]
> Syracuse, New York  USA
>
> _______________________________________________
> Elecraft mailing list
> Post to: [hidden email]
> You must be a subscriber to post to the list.
> Subscriber Info (Addr. Change, sub, unsub etc.):
> http://mailman.qth.net/mailman/listinfo/elecraft   
> Help: http://mailman.qth.net/subscribers.htm
> Elecraft web page: http://www.elecraft.com
>

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