D
Dimitrij Klingbeil
Guest
On 09.03.2016 00:10, Jeroni Paul 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.
Cursitor Doom wrote:
This is a 2W resistor. If it's trying to dissipate more than 2W as
I strongly believe, something's definitely wrong. The problem up
until now has been measuring the dissipation, because we can't use
I^2*R or variations thereof because of the highly noisy/irregular
waveform. So... Here's the clever bit: Measure exactly how long it
takes at present for the resistor to reach say 50'C. I believe it's
around 1 minute, but get the exact time. Then let it cool
completely back to the room ambient temperature. Remove from
circuit. Attach to bench power supply and by means of trial and
error, set the voltage across the resistor to raise it's
temperature to 50'C in 1 minute (will obviously require several
attempts, but no matter). Read off the voltage level which produces
this outcome, then just do V^2/R to find W and see if it exceeds
2W. I'll do it first thing tomorrow!
Good test. It would be more straight to plug the right voltage to
make it dissipate 2W, for 20 ohm that would be 6,3V and check what
temp results in 1 minute. You could use another identical resistor
for the test if you have one lying around to avoid unsoldering.
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.