poobah said:
😕 😕 😕 😕
Nah..
What about the thermostatic device? If it's one of those magnetic styles that use the curie temperature to disconnect the heat path to the tip, that is indeed an issue. It would try to maintain the proper tip temperature, but dwell below temp would (of course) be longer..ie, the duty cycle would be extended by the lower power capability.
How much energy is involved in the magnetic domain conversion?? Or do we consider it as a lossless system, thereby introducing an unknown variance..
Stand orientation is another issue. What angle, what is the convection cooling rate...that needs room temp number..and the density, so the altitude of the workshop is an issue, as well as the barometric pressure. And, the barrel diameter. Is the tip conical or does it have flat surfaces...
Cheers, John.
PS..heard new riders of the purple sage last night..they are good...
Almost duct taped the sound board operator to a door of a bathroom stall, man did he butcher the vocals..(as in..what vocals)
Now consider if the iron had been used for any period of time, on symmetric AC, and had established true bi-directionality in conduction properties of the filament. How long would it require for the expected adirectionality of the filament to develop when using the diode?
😕
😕
But from what I've read, once burned in, a length of wire has superior conductivity properties for ANY signal, symmetrical or assymetric (otherwise, far from improving its sound, the resulting non-linearity of a cable would result in increased distortion).
The real question is whether the iron is rated at 50W when brand new or after x hours of running in.
The real question is whether the iron is rated at 50W when brand new or after x hours of running in.
can't believe that this thread is now 3pages long, so mr threadstater, would you mind telling us who passed your quiz?😀
You all failed! Have to take into account the sag in the power line voltage when plugged in. 24.5 Watts unless you bought one of those $500 silver power cords, then 24.6 Watts.
No cheating allowed by putting the iron on the floor so the electrons role down the hill easier!
Don
😀
No cheating allowed by putting the iron on the floor so the electrons role down the hill easier!
Don
😀
Of course, some irons have an SCR temp control, so one will either get 0% power with a diode or 100%, depending on polarity. And some better irons have an xfmr, in which case you will get smoke too! (and 1000% power)
Don
Don
tubelab.com said:
Would there not also be just a touch of the 'inductive collapse' coming down the pike as a thermal buildup issue? Just musing a bit, gotta morning coffee in me (so far) , being adventerous.
poobah said:You're dismissing the pioneering works of B. Alderdash (a Brit BTW), and others...
N. Onsense , isn't he from Pakistan or was that India?
What color is this soldering iron, anyway?
Now the REAL question is,Which sounds better?
Joints soldered with an iron with a diode
or
Joints soldered with an iron without a diode
😕
Joints soldered with an iron with a diode
or
Joints soldered with an iron without a diode
😕
"if he (threadstarter) did apply any voltage at all "
Was this the same guy who couldn't get the solder to stick?
Don
Was this the same guy who couldn't get the solder to stick?
Don
Smoking,
I think you're on to something here... that thread left off with pool acid for contact cleaner.
I think you're on to something here... that thread left off with pool acid for contact cleaner.
vax9000 said:
jneutron said:
jneutron said:
A shame that you got the answer wrong then, isn't it? 😉 🙂
Assumptions:
1.) A "perfect diode" has no forward voltage drop at any forward current, and has perfect blocking characteristics.
2.) The soldering iron is not an AC-only device and is therefore not broken by the uni-directional current that will result from the addition of the diode.
3.) The soldering iron presents a purely resistive load.
4.) That "serie with the iron solder" was an accidental typo, and the O/P meant for the diode to be in series with the solder iron. Connecting the diode to the solder you are trying to melt should have no affect on the power drawn by the solder iron.
The AC voltage across the element will be halved. Power is proportional to voltage squared (for a load that is not voltage-dependant), and therefore the AC power is quartered.
For those attempting to solve the problem "intuitively", by saying "half the voltage, therefore half the power", think about it like this: what about the current? Isn't that halved too?
Half the current and half the voltage = quarter the power.
The answer to the question, given the four assumptions above, is 12.5 W.
I'm still having trouble soldering the diode in series with the AC line -- the soldering iron stops working when I cut the wire to splice-in the diode. What am I doing wrong? 
Neither the voltage or current halves. The area under the VI curve is 1/2 the non-rectified case, the average power delivered halves.
Sorry, I thought we were joking around...
4fun: Shaded area = energy. Shaded areas per second = power.
4fun: Shaded area = energy. Shaded areas per second = power.
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