DL1 20W Dummy Load

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DL1 20W Dummy Load

Bruce Bowman-3
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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RE: DL1 20W Dummy Load

Don Wilhelm-3
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
>
>
--
No virus found in this outgoing message.
Checked by AVG Anti-Virus.
Version: 7.0.308 / Virus Database: 266.11.10 - Release Date: 5/13/2005

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Re: DL1 20W Dummy Load

Robert Friess
In reply to this post by Bruce Bowman-3
The correct formula which accounts for both the center tap detector and the
diode drop appears in the DL1 manual.

73,
Bob, N6CM


----- Original Message -----
From: "Bruce Bowman" <[hidden email]>
To: "Elecraft Mail List" <[hidden email]>
Sent: Saturday, May 14, 2005 1:02 AM
Subject: [Elecraft] DL1 20W Dummy Load


>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
>
> _______________________________________________
> 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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Re: DL1 20W Dummy Load

Robert Friess
In reply to this post by Don Wilhelm-3
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
>>
>>
> --
> No virus found in this outgoing message.
> Checked by AVG Anti-Virus.
> Version: 7.0.308 / Virus Database: 266.11.10 - Release Date: 5/13/2005
>
> _______________________________________________
> 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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