Posted by
Robert Friess on
May 14, 2005; 4:21pm
URL: http://elecraft.85.s1.nabble.com/DL1-20W-Dummy-Load-tp378360p378362.html
Not quite, V^2/25 only accounts for the power in the bottom half of the
dummy load. The top half dissipates an equal amount. The formula in the
manual accounts for all of this.
73,
Bob, N6CM
----- Original Message -----
From: "W3FPR - Don Wilhelm" <
[hidden email]>
To: "Bruce Bowman" <
[hidden email]>; "Elecraft Mail List"
<
[hidden email]>
Sent: Saturday, May 14, 2005 7:29 AM
Subject: RE: [Elecraft] DL1 20W Dummy Load
> Bruce,
>
> Look in the DL1 instruction manual - there is a graph of power vs. output
> voltage properly compensated for the diode curve.
>
> The first page of the DL1 manual provides the formula of P=((Vx1.414)
> +0.15)^2/50 -- for those with sharp eyes, there is an extra right paren
> in
> th eequation in the manual. Simplifying the formula to V^2/25 will work
> at
> higher power levels where the diode drop becomes insignificant (about 2
> watts and above).
>
> 73,
> Don W3FPR
>
>> -----Original Message-----
>
>>
>> I was doing a calibration of a KAT2 I just built and had a significant
>> error to adjust for Forward power (displayed power was ~30% of
>> measured). When I looked closely at the DL1 I noticed the trace to the
>> detector comes from the center of the 50 ohm load. That means the
>> detected voltage is across 1/2 the load (25 ohms) and changes the power
>> calc to V^2/25 = P vs. of V^2/50 = P. Can someone confirm this?
>>
>> Thanks...
>>
>> Bruce Bowman NM5B
>> Sasnta Fe, NM
>>
>>
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