On 09.03.2016 18:48, Cursitor Doom wrote:
On Wed, 09 Mar 2016 00:20:35 +0100, Dimitrij Klingbeil wrote:

There's no need to unsolder, neither to actively avoid unsoldering.
That resistor is connected through a diode on the board. Just apply
the proper polarity signal, and the diode will take care of the
isolation.

When I removed it from circuit I was happy to give it 2W worth of
power and see what happened. It just got hotter and hotter until I
had to pull the plug at 120'C because my probe won't go any higher.
In contrast, a known good 20ohm 2W resistor fed with the same 6.3V
rose to 70'C and stopped at that. I have now replaced the old brown
'thing' with a 6W ceramic power resistor which after half an hour of
running hooked up to the scope and under load only gets up to 40'C
which I'm entirely happy with. The transformer is making a nice
gentle, barely audible hiss; not that rasping sound it used to before
your diode was replaced; all the DC Vouts are fine and there's no
acrid scorching smell coming from the board any more; just a pleasant
warm 'smell' of electronics behaving themselves.

Hi

Looks like you've got it going back to normal operation, finally!

The whole thing might just have been victim of a resistor with a "hot
channel" (old carbon resistors will sometimes do this without reason).

Or an improperly replaced one - likely the same result. No matter if it
broke down with a low resistance path or if it got replaced improperly.

You can check if that was the case (if you still kept the resistor) by
hooking it up to a power supply with an ammeter. If you see it heat up
and then suddenly the amperage rising and the resistor starting to heat
even faster (as if that was an NTC in there) then it very likely has
developed a hot channel (a small area not quite unlike an arc path that
conducts lots of current as it overheats).

If that's the case, it may explain several things. For once, this
resistor was intended to provide a field reset function for the
inductor. The circuit would likely require the reset to be quite
"complete" (the field decaying to zero) between switching cycles.

If the resistance is lower than normal, the field will decay slower, and
a resistor with a hot channel would likely slow the winding reset long
enough to be "incomplete" - that is, by the time that the next switching
cycle arrives, there would be still significant remaining current
flowing through the inductor and the diode.

This can be a condition that the power circuit might not like. There
is only a moderately fast diode in there - the original BY208 is not
really fast. As long as the winding reset is complete, that might not
matter because the diode would already be "out of charge carriers" when
the next pulse arrives, so there would not be much reverse recovery.

If the reset happens to be incomplete because of the resistor breaking
down, then the diode would still be conducting when the pulse comes and
that would bring its (quite long) reverse recovery time into equation.
During that reverse recovery time the diode would pass current and the
resistor (and subsequent circuitry too, up to the resonance LC) would be
getting "hammered" with large 800 V voltage spikes from the switcher,
thus it would be subjected to high peak currents because of the spikes.

Of course, hammering the resistor with enormous voltage spikes won't be
very good for its reliability, and since it was already breaking down,
would also accelerate the process a lot (and stress other parts too).

Dimitrij