[BC] trouble with an old BC-1G need advice

[dynotherm at earthlink.net]

The assumption that an unmeasured line connected to a BC-1G
is presenting 50 ohms to the box is the April Fools joke. The 
1G won't care much if its load is 40 or 60 ohms. It will work 
reasonably well. Theoretical speculations in this case are a 
waste of b/w. This is a PRACTICAL problem.

Bill, there has been good advice in this thread, but you do
need to measure the tower base Z and ATU input Z. Then, when
you are sure all is OK, you need to "level" the line. IOW,
get two RFA's that will measure 6 or 8 amps, test them in
series, put one at each end of the line and adjust the SHUNT 
branch of the ATU to give you essentially the same current at 
both ends of the line. Unless you happen to have a line that 
is about 180 degrees of electrical length (unusual to have 
that), when you level the line you will be close enough.

To others reading this thread - there are better ways, but
this will work for Bill to get him working correctly and
it is not subject to much error if another RFA is tested in 
series with the BC-1G line current meter into the built-in
dummy load, then used at the ATU input to get the line level.

Bill, if you have a significant change in tower base Z from your 
license you MUST file a new resistance with the FCC. This is 
not optional. See 73.54(c) and 73.51. 73.51 is for indirect 
power measurement using plate voltage and current in case you
have to do it that way temporarily.

Phil Alexander, CSRE, AMD




-----Original Message-----
>From: RichardBJohnson at comcast.net
>Sent: Apr 2, 2008 1:44 PM
>To: broadcast at radiolists.net
>Cc: BOYDSIRBOYD at aol.com
>Subject: Re: [BC] trouble with an old BC-1G need advice
>
>I'm assuming this is not an April Fools joke.
>
>Ammeter reading =  square root of [power you want divided by the resistance]
>
>I = sqrt(P/R)
>
>If you want 1250 watts and the line impedance is 50 ohms it's:
>I = sqrt(1250/50)
>I = sqrt(25)
>i = 5 amperes
>
>Maybe you want 1000 watts exactly. It's:
>I = sqrt(1000/50)
>I =  sqrt(20)
>I = 4.47 amperes
>
>Use this math as a simple prototype to substitute your numbers.
>It works for the tower as well. Let's pretend that the tower impedance
>was 31 + j16. Ignore the "j stuff, that's reactance.
>
>Let's say you want the maximum power allowed there, and it is
>1000 watts. Then:
>
>I = sqrt(1000 / 31)
>I = sqrt(32.258)
>I = 5.68 amperes
>
>--
>Cheers,
>Richard B. Johnson
>Read about my book
>http://www.LymanSchool.org
>
>
> -------------- Original message ----------------------
>From: BOYDSIRBOYD at aol.com
>> as per the owner it has always been that way,, there are two other  engineers 
>>  
>> who have worked on this thing in the past and they all say  the same thing. 
>> i'm  
>> now at a loss as to what needs to be done.
>>  
>> and CHUCK  is the best in the world and another engineer he had used  at one 
>> time had worked for GATES and was the best on these transmitters. and the  
>> readings all had to be the same as the normal readings to be right. or it didn't  
>> work like it was suppose to. that's why i need advice and help.
>>  
>> bill
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