Thanks to all that responded... I was pretty sure that I could not melt
the solder joints BUT I had to ask just make sure.. and to clarify the
600-800 watts is from a AMP.. but there where two solder joints on the
banana jacks (not the plugs) that came loose so I'm taking the blame on
that..apparently "poor workmanship" on me.
So I did a better job this time and we will see
Thanks again
73
John
KO4XJ
Jim Miller KG0KP wrote:
> I had to read this twice to figure out what you were talking about.
> First I thought that was WAY TOO BIG of a gun to use for soldering LOL.
>
> Anyway, we are talking about resistance of the solder connection as
> that is the only thing that is going to create heat to attempt to melt
> the solder and since solder is used on many antenna or connectors in
> the radio world and in many many connections in the critical somputers
> that the "resistance" of the solder is NOT a factor or even close.
> First of all, the solder is filling in around what should be metal to
> metal contact so the distance between the metal to metal contact, IF
> ANY, is minimal and, for the most part, that is the only solder
> involved in the electrical current path.
>
> OK, the resistivity of lead at 20 degrees centigrade is 10 microhms
> per centimeter. Lead turns to a liquid at 333 degrees Centigrade. At
> that point is has a resistivity of 95 microhms per centimeter.
>
> The resistivity of tin at 20 degrees centigrade is 11.5 microhms per
> centimeter. Tin turns to a liquid at 235 degrees Centigrade. At that
> point is has a resistivity of 47.6 microhms per centimeter.
>
> Now we mix Tin(60)/Lead(40) and get 10.6 microhms per centimeter. Now
> the distance we are talking about is NOT 1 cm. Not even one mm but
> lets use that anyway ( more like 0 to 1/10 mm). So, we have 1.06
> microhms. Now it will take approximately 1,000,000 volts to get an
> amp at that point. Lets say you can manage 10,000 volts by some super
> driver, OK, that gives you 10 milliohms. Power disipated = .010
> squared times 10,000 or .001 10 watts = 3.12 BTU per watt gives 31 BTU
> dispiation. You can't reach any of those example measurements and
> even if you did, you can't raise the solder to 275 degrees (60/40
> melting temperature without figuring the flux) with 31 BTU/hr.
>
> I worked in computers for 38 years and solder resistance was never a
> consideration (with the exception of "cold" solder joints and they
> don't happen when you know how to solder properly). We had timing
> adjustments required to 2/10 of a nanosecond (i.e. 8 inches of wire =
> 1 nanosecond in wire) we adjsuted by adding about 3/4 inch of wire for
> a 1/10 nanosecond adjustment. Solder resistance = not.
>
> Don't worry about it.
> 73, de Jim KG0KP
>
>
> ----- Original Message ----- From: "John Hudson" <jd_hudson@comcast.net>
> To: <towertalk@contesting.com>
> Sent: Tuesday, May 12, 2009 9:41 AM
> Subject: [TowerTalk] 60/40 solder
>
>
>> Will 600-800 watts (AL-811H) melt 60/40 solder?
>>
>> I made a jumper for my ladder line, LARGE banana plugs on the end
>> plugged into a disconnect homebrew box with large jacks. The lugs on the
>> jacks came un-soldered.. don't know if it was from the power, or from
>> moving the ladder line and they broke loose or a bad solder job !
>>
>> Anyway 60/40 is all I had that day
>>
>> I'm going to resolder and see what happens
>>
>> Thanks
>>
>> John
>> KO4XJ
>>
>>
>> _______________________________________________
>>
>>
>>
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>
>
>
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