> Thanks for the reply. The rated input voltage I was supplying was a bit
> higher than what it was designed for (states on the Powerbook 7.5V 2A max)
> and I was giving it just under 10v (using a battery pack that supplied a 3A
> max supply). Normally semiconductors are a bit more tolerant to a bit more
> voltage but you might be correct. As far as the amperage deal. Simple
> electronics specifies that a unit will only consume as much current as it
> needs. So supplying 10A of power would net the same results as supplying 2A.
> There are a few exceptions to the rule, such as a AMM meter which is set
> inline of a lead. But for the most part components won't be damaged by
> excessive current.
There are many unfortunate statements above ("simple electronics specifies..." is so
wrong that it isn't even wrong!). Short story: There are many, many components in
electronics which behave nonlinearly. Even a battery, if
charged with seemingly small overvoltages, can exhibit catastrophic behaviors. A diode
is a device where, if you apply a mere extra 60mV (that's 0.06 volts) or so directly
across it, the current can go up by a factor of 10!
It does not "know" what current it "needs", and will happily, destructively
demonstrate that truth for you.
Another way to look at it: You applied nearly a 50% overvoltage. If it were a
resistive load, the current would have gone up by ~1.5x, too, for a power increase to
~2x the nominal value -- that's a lot, even for a resistor.
But you weren't feeding a resistor -- it's a much more complicated thing!
--
Prof. Thomas H. Lee
Center for Integrated Systems, CIS-205
420 Via Palou Mall
Stanford University
Stanford, CA 94305-4070
http://www-smirc.stanford.edu
650-725-3709 ph, -3383 fax
--
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