On Jun 2, 2:49*pm, "Joel Koltner" wrote:
Something I've wondered for awhile now is this: Since you get the same VSWR
with a load both higher than or lower than your "expected" load (we'll call
this Z0, and the actual load Zl), either Zl/Z0 or Z0/Zl respectively, is one
siutation more likely to kill an amp than the other?

When ZlZ0, you tend to (ignoring phase effects here) end up with a higher
voltage sitting on the finals than when Zl=Z0... but less current than the
matched case. *With ZlZ0, it's typically lower voltage but higher current.
Based on what I know from, e.g., power supply design, there's usually more of
a margain from overvoltage than there is from overcurrent, in that you can
often find parts that inherently withstand a 2:1 or better overvoltage margain
with just a bit of extra cost, whereas overcurrent will tend to overheat the
power devices and getting rid of, e.g., 2-4x as much heat (vs. matched case)
can require significantly greater heatsinking which often isn't cheap.

In the extreme case, I'd expect that far more amps will blow when shorted than
when open-circuited. *But I don't really have much experience in this area...
so... could anyone comment on what you tend to see in real designs? *Perhaps
also compare and contrast tube-based amp with solid-state ones?

Thanks you,
---Joel

Mario is right...it's a pretty tricky question, or at least it's
tricky to answer accurately!

I fully agree with him that it's difficult to know what impedance will
be reflected by any particular load, when viewed at the plates/
collectors/drains of the output devices.

Something else to consider: generally too much current (by only a
little, anyway) will cause heating, but it may take many milliseconds
or seconds for the heat to build up to the point of being dangerous.
On the other hand, when you get enough voltage to cause problems, the
problem is quite often instantaneous: within a microsecond or so.
Poof. Automatic protection circuits have a bit easier time handling
excess power (current; heat) than they do excess voltage.

Cheers,
Tom