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Re: [TenTec] Model 1225 SWR and Wattmeter Kit

To: "'Discussion of Ten-Tec Equipment'" <tentec@contesting.com>
Subject: Re: [TenTec] Model 1225 SWR and Wattmeter Kit
From: "Rick - DJ0IP / NJ0IP" <Rick@DJ0IP.de>
Reply-to: Discussion of Ten-Tec Equipment <tentec@contesting.com>
Date: Tue, 22 Jul 2014 20:56:09 +0200
List-post: <tentec@contesting.com">mailto:tentec@contesting.com>
In James' defense, I would like to say that he is sincerely trying to learn
this stuff, and without a solid technical background, it is not as easy as
some of you might think.  Just reading the ARRL Handbook is not always
enough to learn and understand the subject matter.  I can assure you that
James has read the entire ARRL Handbook.  But that does not equate to
understanding everything.

There are a lot of people here who don't even try to learn.  At least James
is trying and we should applaud him for that.  
Carl, thank you for explaining it in simple terms.  Your explanation was
excellent.
In fact the entire thread was quite informative.

I do not understand all of this stuff either, so you guys can lecture me too
if you wish.
My degree was in Business and Administration, not engineering.  
I too have an Extra Class, but that doesn't mean I understood everything.

The information presented in this thread was excellent; thanks.  But please
don't be too critical of those of us who are trying to understand, but not
coming along as fast as some of the others.

73 - Rick, DJ0IP
(Nr. Frankfurt am Main)


-----Original Message-----
From: TenTec [mailto:tentec-bounces@contesting.com] On Behalf Of Jim Brown
Sent: Tuesday, July 22, 2014 6:36 PM
To: tentec@contesting.com
Subject: Re: [TenTec] Model 1225 SWR and Wattmeter Kit

On 7/22/2014 5:03 AM, Carl Moreschi wrote:
> The LP-100A has two connections to the sensor.  One is for current and 
> one is for voltage.  Power is simply the product of voltage times 
> current.  So for any load impedance, power will always be calculated 
> correctly.
>
> Most less expensive meters only measure voltage.  They calculate power 
> as Voltage squared divided by 50.  They always assume the load is a 
> pure 50 ohms.  But with a 1.5 swr, you could have a 75 ohm load or a
> 33 ohm load or a complex load with both real and imaginary components.  
> If you have a 33 ohm load, your power will read high.  If you have a 
> 75 ohm load, your power will read low. The point is unless the swr is 
> exactly 50 ohms, your power measurement will be wrong by a lot.

There's a third component of the power computation -- the PHASE angle
between the voltage and current, and the LP100A measures that too.
> On 7/22/2014 4:37 AM, K8JHR wrote:
>> Gee... Sorta condescending and demeaning. And not particularly helpful.
>> I often forget ham radio is only for experts with careers in 
>> electrical engineering - the rest of us hobbyists are merely 
>> tolerated, perhaps only because we buy stuff and keep prices down.

You've got an Extra Class ticket. This is NOT advanced EE stuff. The
equation for the computation of power is covered in the earliest chapter on
the fundamentals of electrical circuits in the ARRL Handbook.  If you
bothered to study that (as opposed to cramming for the exam questions) you
would know that.

>>
>> OK, I read the N8LP web page and, I am sorry because I do not see 
>> where it addresses Carl's point about some meters being less accurate 
>> if the load varies from 50 ohms, or how the LP-100 is immune from that
problem.
>> Perhaps someone can explain this on the E-brand reflector... As far 
>> as I can see, both units might be affected similarly as they will be 
>> used exactly in the same way. I suppose I missed it, but I tried.

Like I said in my earlier post, N8LP's home page has a rough description of
the LP100A. To see detail and the specs, you must click on the LP100A link.

> And while I am at it, this whole "contest" thing seems a bit overblown.
> How many contests have been lost because the the operator could not 
> determine amplifier power with sufficient accuracy?

With an SWR of 1.5:1, the error in the power calculation would be +/-
1.8 dB.  Several years ago, N6ZFO did some statistical analysis showing that
1dB translates into a 2.6% increase in our score in the ARRL Sweepstakes. So
that 1.8 dB error becomes a 4.5% better (or worse) score.  In the recent
WRTC competition, the difference between 1st and 2nd place was 3.7%, between
2nd and 3rd was 4.3%, and between 3rd and 4th was 1.3%.

When you're loud at the other end, a couple of dB doesn't matter, but when
you're close to the other guy's noise level, it's the difference between a
QSO and no QSO, or between getting him on the first call, or several calls.
That translates into score.

73, Jim K9YC

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