Re: Can a computer virus kill the CPU?

Ideal power supplies meet a V-I curve that is a rectangle. The
upper right corner is where power supply outputs maximum power. In
reality, power supplies do not operate well at that corner. This is a
point of excessive current. Therefore proper foldback current limiting
is designed so that power supply never gets into this 'corner'. All
this is responsibility of the power supply designer; user need not
worry about this condition. But manufacturer should state in specs
that overcurrent protection exists.

Proper overcurrent protection is not found in some supplies as some
third party testing demonstrates. TomsHardware.com (?) could not test
some power supplies becasue those supplies failed before full current
was output. IOW power supply was defective by design. Current
liimiting was set too high assuming current limiting even existed.

Yes, some power supplies are defective - will burn on too much load.
But then such supplies typically also don't provide numerical specs
that claim otherwise.

The lurker may not know what any of those spec numbers mean. But if
a vendor will not provide a long list of spec numbers, then that is the
red flag. That red flag says those who do know what numbers mean
cannot warn others.

Overcurrent and other hardware design defects are possible if
component manufacturer does not provide numerical specs - therefore
need not meet those industry requirements.

Same applies to hardware damaged by a virus. If a virus damages
hardware, then hardware is defective by design. What did manufacturer
numerical specs say? Well if specs do not claim to meet industry
standards, then why should hardware survive? Missing specs may identify
the exception. Virus should not damage properly designed hardware.

Robert Mabee wrote:
Current is limited at somewhat above the rated value; if the load
has a low resistance then it pulls the voltage down from the nominal;
when the voltage is significantly below nominal the current limit is
reduced, further reducing voltage -- this is the foldback part of the
curve. Current can be quite low into a short circuit but must be
sufficient at all intermediate voltages to power the expected load
(which may act considerably worse than a resistor) and charge up
capacitors at the required ramp-up rate. Thus, the current into a
weak short can be 90% of the limit at 90% of nominal voltage, or 81%
of rated power all dissipated in the immediate vicinity of the screw
that went missing. (I'm assuming the actual load isn't much, maybe
because the CPU is in a power-saving mode, or the customer bought way
too big a power supply because "more is better".)
...

That's not even implementable for shorts applied with power on, only
for shorts present before power so the power supply will stay at low
voltage and current. There is enough energy stored in capacitors to
vaporize a printed-circuit trace regardless of anything the power
supply can do.

Perhaps we are talking about different things. I gladly concede that
the power supply won't be damaged, or can and should be designed such
that it won't. I suppose this only costs series and bypass diodes,
and higher voltage rating on main output rectifier.
...

Misses the worst case for the power supply transistor, though. Heat
dissipated in a linear regulator (typically used for all but the highest
current outputs) peaks at a voltage somewhat below nominal, into an
overload that isn't a full short circuit.

Again, the load may be damaged by conditions that the supply withstands.
Printed circuits aren't designed to survive much more current than they
consume.
...

We should stick closer to that topic. ...

.