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" <[EMAIL PROTECTED]>
To: "Bruce Bowman" <[EMAIL PROTECTED]>; "Elecraft Mail List"
<[email protected]>
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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