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

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? ...

A fairly tricky question

You are right that voltage margin is usually less expensive to obtain
than current margin, but in RF devices this is not always true; e.g.,
the classic MRF316 can withstand 9 amps and 35 volts, for a nominal
output (VHF) power of 80 W, and is guaranteed to survive a 30:1 mismatch
(with any phase angle), but other transistors are less tolerant - e.g.
the widely known MRF455 is rated for 18V (only) and 15A for a 60W power
(HF).

But the main question is that the output matching network transforms
load impedances in a way not easy for the final user to predict, so a
short on output may appear to the device as something very different,
whilst a reactive load could well appear like a short or open circuit or
something like that. Adding to this the transformation of load impedance
caused by the transmission line, we can hardly know the difference
between the effects of a load impedance and those of another - and this
is why the specifications are usually given in terms of generic mismatch.

As for the difference in behaviour between solid state and tube amps,
the latter are notoriously more tolerant because failures in tubes are
usually rather "slow" (causes are mainly overheating, but also
evaporation of electrodes due to excessive current or disruptive
discharge due to overvoltage, and each of these causes requires a
certain amount of time to reach "final damage"), while the fusion of a
part of a semiconductor device (due to excessive current or power
dissipation, or formation of hot spots) or the disruptive discharge in a
junction (usually due to avalanche effect) are very fast phenomena,
which kill the device in very little time, or even in almost no time at
all... but all this is true at more or less any working frequency, not
only in RF.

Hoping to have shed some little light...
--
73 es 51 de i3hev, op. mario

Non è Radioamatore, se non gli fuma il saldatore!
- Campagna 2006 "Il Radioamatore non è uno che ascolta la radio"

it.hobby.radioamatori.moderato
http://digilander.libero.it/hamweb
http://digilander.libero.it/esperantovenezia

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

Thanks Mario, that's quite insightful!

On Jun 2, 5:49*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?

Many simplest solid-state amps "SWR protection" circuitry is in fact
"overvoltage protection" circuitry (e.g. ZlZ0).

The output transformer and low-pass network can also be seriously
overstressed when ZlZ0.

For a while, thermal runaway was a big problem with bipolar continuous
duty (especially FM) amps when ZlZ0. Today most solid-state amps
(even if not bipolar) have overtemp and overcurrent protection as a
belt-and-suspenders on top of temperature-dependent bias. Overtemp by
itself won't stop the worst transformed loads but with overcurrent
protection the past couple generations of HF finals aren't so easy to
kill.

BTW... my Ten Tec Triton IV (1976 vintage) still has the original
final transistors.

For a lot of medium-power homebrew gear the cost of any final
protection is more than the cost of the finals. Witness the many
IRF510 amps. For a while on my bench I was burning through IRF510's
like they were goin out of style :-)

Tim N3QE