In article ,
"Martin H. Eastburn" wrote:
I've measured the current with metering - just over 100 amps with a tip.
Tricky measuring either voltage or current due to the magnitude of values
and the way the tip changes resistance with heat.
How did you measure the current? This is a very low resistance circuit,
so no standard inline ammeter is going to work without significantly
increasing the loop resistance, and thus decreasing the current.
One way is to measure both the open circuit voltage and the line voltage
- their ratio is the the same as the transformer turns ratio. As
discussed below, for the Wen 250 this ratio is 435:1.
Then measure the current in the 120-volt supply circuit, and multiply by
the turns ratio.
I did this with my Wen 250. It draws 2.2 amps (with a cold tip), so the
tip current is (2.2)(435)= 957 amps.
This is a little low for the 0.175 volts, as this would imply
(0.175)(957)= 167 watts, yet (120)(2.2)= 264 watts. One would expect at
least 90% efficiency in such a transformer, which did not get hot, so
the 167 watts cannot be correct.
Some better measurements are in order. I believe the 264 watts because
that measurement is direct, and exceeds the manufacturer's claim of 250
watts.
The best approach should be to measure the voltage across the tip at the
same time as the 120-volt current is measured.
Joe Gwinn
Martin
Martin Eastburn
@ home at Lions' Lair with our computer lionslair at consolidated dot net
NRA LOH & Endowment Member
NRA Second Amendment Task Force Charter Founder
Joseph Gwinn wrote:
In article ,
Don Foreman wrote:
On Sun, 1 Jan 2006 22:21:57 -0600, "Ron Moore"
wrote:
I was thinking of the welder mech that is used to link batteries into
packs.
Don't know why I was thinking that. I'm sorry, I not sure what you were
referring to that is not DC output, unless it is the resistance welder and
not the charger.
Right, it's the resistance soldering device, soldering guns and
otherwise. The Wassco I mentioned is also AC. Capacitive discharge
DC works great for spotwelding batteries.
Gunner has mentioned soldering heavy wires by pressing a 250-watt gun
(sans tip) up against the splice and pulling the trigger, using the
current to heat the wires. I've never tried that, but it sounds
neat. The tip on a 250 watt gun is a few inches of about 1/8"
square copper and it get real hot right quick.
I recently measured the voltage across the tip of my 250-watt Wen
soldering gun - it's 0.175 volts rms or so (starts lower, then rises as
the copper tip heats and increases in resistance). To achieve the
stated 250 watts, the current will be 250/0.175= 1,429 amps. This could
almost be used for spot welding (where 4,000 amps is more common), and
the resulting magnetic field can pick up small pieces of iron (~50
grams).
For the record, the Wen Soldering Gun (Model 250, bought in 1963)
consists of a big AC power transformer with a one-turn secondary made of
3/8 inch brass rod. The open-circuit voltage (with tip removed) is
0.275 volts rms, for 118.9 volts rms in, so the transformer winding
ratio is 435:1. The gun was manufactured under US patents 2,701,835 and
2,680,187. Go to http://www.pat2pdf.org/ to get copies.
The Weller soldering gun is almost identical, although the tip of the
Wen is instead made of 0.164" diameter (~AWG #6) plated round copper
wire.
Joe Gwinn
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