Safe Operating Area calculation for audio power amps

[Phil Allison]

"Clifford Heath"

Phil Allison wrote:

** Lose the smartarse attitude !!!

I've been biting my tongue real hard for a while here.
I don't need to say what everyone else is already thinking.

** I see now the smartarse attitude is 100% inbuilt.

** Not relevant to your ETI 480 - as the DC rails are a given.
The supply rails set the output power limit into a load.

Not entirely. If I can't make the thing safe at +-37 volts,
I'll buy a different supply tranny. In fact the monster is
likely to be so annoying that I will anyhow. I just thought
that adding another bridged pair might be useful in future.

** I have had quite enough of your sickening bad manners, snipping out the
context and ignoring important points and constantly moving the damn
goalposts.

I sincerely hope the whole POS blows up in your face.




.............. Phil

 

[Clifford Heath]

Phil Allison wrote:

** I have had quite enough of your sickening bad manners,

It's amazing how quickly the troll reverts to its escapist behaviours
when it's finally unable to avoid the exposure of its mathematical
ignorance.

Now that Phil (with his useful practical knowledge) has left the
field, does anyone want to step up to answer the valid theoretical
questions he spent so long evading?

 

[Phil Allison]

"Clifford Heath"

Phil Allison wrote:

** I have had quite enough of your sickening bad manners,

It's amazing how quickly the troll reverts to its escapist behaviours
when it's finally unable to avoid the exposure of its mathematical
ignorance.

** What a fucking arrogant PIG you are !!!

Now that Phil (with his useful practical knowledge) has left the
field, does anyone want to step up to answer the valid theoretical
questions he spent so long evading?

** Your idiot "question" keeps changing and has no answer - since you do
not even faintly comprehend it yourself.

Go straight to hell.




............. Phil

 

[Alan Rutlidge]

"Phil Allison" <philallison@tpg.com.au> wrote in message
news:35opfdF4o8lo7U1@individual.net...

"Clifford Heath"
Phil Allison wrote:

** I have had quite enough of your sickening bad manners,

It's amazing how quickly the troll reverts to its escapist behaviours
when it's finally unable to avoid the exposure of its mathematical
ignorance.


** What a fucking arrogant PIG you are !!!



Now that Phil (with his useful practical knowledge) has left the
field, does anyone want to step up to answer the valid theoretical
questions he spent so long evading?


** Your idiot "question" keeps changing and has no answer - since you
do
not even faintly comprehend it yourself.

Go straight to hell.




............ Phil

Oh Phil dear, why don't you go cry to mommy. You lost - too bad.

 

[Mr.T]

"Phil Allison" <philallison@tpg.com.au> wrote in message
news:3648ccF4ald9gU1@individual.net...

** Why are larger diameter cones more efficient ????

Sometimes they're not. Depends on the design.

Double sound pressure is 6 dB more.

To drive two speakers takes twice the power - so 3dB more input

Ergo - there is a 3dB increase in SPL - for free.

Not really, you've simply changed the radiation pattern. But measured on
axis you are correct.

MrT.

 

[Pooh Bear]

Clifford Heath wrote:

Ok, so I've seen all Leo Simpson's designs and his articles on
designing audio power amps, and I understand the principles
of reactive loads etc. Problem is, I can't find an algebraic
derivation of the power curves he plots, and though I have
derived an expression for it, I can't see how to integrate it
over one half-cycle.

My goal is to bridge two ETI-480's together from a +-37 V
supply, and I'm trying to work out whether the output devices
need to be upgraded. The accepted bridge configurations for
the ETI480 assume it swings to within 5 V of the rail, so
that gives me 255W into a resistive load.

Do folk do the dissipation/SOA calculation numerically, or
can someone point me to an algebraic formula for it?

Hi Clifford,

saw your post in abse so came here to see what the fuss was about.

You can indeed calculate dissipation numerically in out put devices for
waveforms that are open to mathematical ennumeration using integration
like sinewaves. You can also repeat the exercise to see the effect of
load phase lead / lag. You can even account for the device saturation
voltage too.You only need to integrate over a 1/4 cycle btw since all 4
quadrants return the same result.

I've done this using Mathcad donkey's years back. I actually forget the
result now but I generally double the resistive dissipation figures when
calculating appropriate device dissipation for the real world. Don't
forget to derate the dissipation figures for heatsink temp rise btw !

None of this helps you with SOA though.

The good news is that there are bipolar devices available where
secondary breakdown doesn't occur below about 70V anyway so you needn't
worry about it in your case if you substitute.

What are the original devices ?


Graham

 

[Clifford Heath]

Graham,

saw your post in abse so came here to see what the fuss was about.

Thanks.

With regard to the SOA and the graphs I was originally asking
about, I've had an "aha" moment and realised that the VI curves
for reactive loads are just showing the half-ellipse formed by
phase delay. So I was completely misunderstanding the plots,
which just show Vs-Vout against I. Here I thought there was some
funky conic being calculated. Stoopid, because I understood what
was going on, just not the graphs. So the SOA isn't an issue for
me any more.

Wrt actual RMS power out for a given sine-wave, I think I've
worked that out also - check this reasoning. The phase shift on
the power curve is exactly half the current phase shift. So if the
current lags 45 degrees, the power lags 22.5, and the peak power
is 85% (which is sin(65.5)^2) of the power into a resistive load.
Assuming no clipping of course. This is the result I was trying to
get by integration over sin(x)*sin(x+p)/|Z|.

Graham wrote:

You only need to integrate over a 1/4 cycle btw since all 4
quadrants return the same result.

I can't see that, because with a reactive load, 1/2 cycle of
current contains two segments of the voltage curve which aren't
similar due to the phase shift.

The good news is that there are bipolar devices available where
secondary breakdown doesn't occur below about 70V anyway so you needn't
worry about it in your case if you substitute.
What are the original devices ?

2N3055/2955 - very pedestrian. If they don't survive, I'll be
forced to substitute, though I'll need to stay in a TO-3 can.
In the past, folk have substituted MJ15003/MJ15004. Otherwise
what would you suggest?

Clifford Heath.

 

[Phil Allison]

"Clifford Heath"

With regard to the SOA and the graphs I was originally asking
about, I've had an "aha" moment and realised that the VI curves
for reactive loads are just showing the half-ellipse formed by
phase delay. So I was completely misunderstanding the plots,
which just show Vs-Vout against I. Here I thought there was some
funky conic being calculated. Stoopid, because I understood what
was going on, just not the graphs. So the SOA isn't an issue for
me any more.

** The penny finally drops for CH with a Massive Clang !!!!!


Some posting history:


PA ** Your original question made no sense

CH On the contrary, go and review it.

PA ** I did, many times - it is based on a false assumption.

CH Do you actually have anything to offer on my original question, or
is the maths beyond you?

PA ** Your original question made no sense - as do your other
assertions.

You have fundamentally misunderstood the problems involved.


CH It's amazing how quickly the troll reverts to its escapist behaviours
when it's finally unable to avoid the exposure of its mathematical
ignorance. Now that Phil (with his useful practical knowledge) has left the
field, does anyone want to step up to answer the valid theoretical questions
he spent so long evading?


-------------------------------------



** I suppose it would be dumb of me to expect an apology ......




............... Phil

 

[Pooh Bear]

Clifford Heath wrote:

Graham,
saw your post in abse so came here to see what the fuss was about.

Thanks.

With regard to the SOA and the graphs I was originally asking
about, I've had an "aha" moment and realised that the VI curves
for reactive loads are just showing the half-ellipse formed by
phase delay. So I was completely misunderstanding the plots,
which just show Vs-Vout against I. Here I thought there was some
funky conic being calculated. Stoopid, because I understood what
was going on, just not the graphs. So the SOA isn't an issue for
me any more.

Wrt actual RMS power out for a given sine-wave, I think I've
worked that out also - check this reasoning. The phase shift on
the power curve is exactly half the current phase shift. So if the
current lags 45 degrees, the power lags 22.5, and the peak power
is 85% (which is sin(65.5)^2) of the power into a resistive load.
Assuming no clipping of course. This is the result I was trying to
get by integration over sin(x)*sin(x+p)/|Z|.

Graham wrote:
You only need to integrate over a 1/4 cycle btw since all 4
quadrants return the same result.

I can't see that, because with a reactive load, 1/2 cycle of
current contains two segments of the voltage curve which aren't
similar due to the phase shift.

Maybe you're right about that. I'd have to fire up the model to check.

It's kinda academic when using mathcad or whatever since it does all the hard
work for you anyway.

The good news is that there are bipolar devices available where
secondary breakdown doesn't occur below about 70V anyway so you needn't
worry about it in your case if you substitute.
What are the original devices ?

2N3055/2955 - very pedestrian. If they don't survive, I'll be
forced to substitute, though I'll need to stay in a TO-3 can.
In the past, folk have substituted MJ15003/MJ15004. Otherwise
what would you suggest?

If you want TO-3 ( as it seems ) there's nothing better than MJ15003/4 ( at
this voltage ) as substitutes. Highly reliable devices. I've designed them
into products in the past and they're just fine.


Graham

 

[Phil Allison]

"Pooh Bear"

If you want TO-3 ( as it seems ) there's nothing better than MJ15003/4 (
at
this voltage ) as substitutes. Highly reliable devices. I've designed them
into products in the past and they're just fine.

** Says a pommie smartasre who has never bought those types from Dick
Smith Electronics ( who sold various fakes in the past and have now dropped
them) or Jaycar who sell "MOSPEC" brand or WES who still sell fakes
......




.............. Phil

 

[Clifford Heath]

Phil Allison wrote:

PA ** Your original question made no sense

You keep saying this, and it's true it was based on a false
assumption (that the graphs had anything to do with average
dissipation), but it wasn't a nonsense question.

If you'd told me to read up on secondary breakdown, I think I
would have got it sooner. For thirty years I've thought that
"thermal runaway" was a process that took minutes, not ms.
Hence my confusion about needing to calculate average power.

When I suggested integrating the instanteous power formula over
a half-cycle, it wasn't nonsense - but in the sinewave case with
a single reactance it turns out to be unnecessary. I don't think
it deserved to be described as "drivel". Graham's answered my
question though by saying that real engineers use Mathcad or the
like for a numerical solution to such problems.

Sorry for being so thick. I just wanted to know how to calculate
whether the amp would be safe. You didn't seem to want to help
with that part, but I shouldn't have abused you for it. Your other
comments were useful and mostly to-the-point. So yes, take that
as an apology.

BTW, I returned the Jaycar speaker (the box was unopened). They
were pretty reluctant (past their 7-day policy), being quite
convinced that the particular speaker would be fine, possibly
more robust that many instrument speakers (they sell the same unit
in 4 ohms for car subs which take a bit of punishment). But they
did it - after I mentioned "statutory warrantee regarding
fitness-for-purpose". I have yet to choose another, though WES
have a 12" in the same range as your suggested 15" (which doesn't
suit the cab).

What do you think of Neil Kennedy's assertion that a bass amp &
speaker shouldn't clip or distort disproportionately?

Clifford Heath.

 

[Clifford Heath]

Phil Allison wrote:

** Says a pommie smartasre who has never bought those types from Dick
Smith Electronics ( who sold various fakes in the past and have now dropped
them) or Jaycar who sell "MOSPEC" brand or WES who still sell fakes

Where in Aus can you buy real ones?

Why did you call Graham a smartasre?

 

[Phil Allison]

"Clifford Heath"

Phil Allison wrote:

PA ** Your original question made no sense

You keep saying this, and it's true it was based on a false
assumption (that the graphs had anything to do with average
dissipation), but it wasn't a nonsense question.

** Fraid it was - in relation to your concerns.

If you'd told me to read up on secondary breakdown, I think I
would have got it sooner. For thirty years I've thought that
"thermal runaway" was a process that took minutes, not ms.

** You were right the first time.

Thermal runaway and second breakdown are two quite different things.

When I suggested integrating the instanteous power formula over
a half-cycle, it wasn't nonsense - but in the sinewave case with
a single reactance it turns out to be unnecessary.

** As I told you - several times.

I don't think
it deserved to be described as "drivel". Graham's answered my
question though by saying that real engineers use Mathcad or the
like for a numerical solution to such problems.

** But Graham is not a real engineer - he is an ex DJ posing as one.

In fact, the task is easily done with a pocket calculator or even a slide
rule. You just need to realise that amp dissipation is always simply the DC
supply watts MINUS load watts taken over one cycle.

If you calculate the peak load current then that is *exactly* the same as
the peak DC supply current - the waveform is half sine in each DC rail and
the *average* is 0.32 of the peak. Just multiply the average rail current
by the rail voltage to get DC power in watts. The load power is easy -
just the rms current squared times the resistance.

For your case with a 37 volt rail and 100 watts at 4 ohms - peak current is
7.07 amps so the DC power is 84 watts per rail or 168 watts total. The
output is 100 watts so the dissipation is 68 watts, divided by four makes
it 17 watts per device. You should repeat this calculation for a variety of
power levels and plot the results to find the power level that creates the
most amp dissipation.

With a resistive load, maximum amplifier dissipation happens at about 42 %
of max output power. However, if reactance is added in series with the load
creating a 45 degree phase angle, the maximum dissipation shifts up to
full power and is just a tad less than before.

Sorry for being so thick. I just wanted to know how to calculate
whether the amp would be safe. You didn't seem to want to help
with that part, but I shouldn't have abused you for it.

** I did try to help - but you did not listen.

When you ask a technical question, on a non simple matter, you MUST not
specify the kind of answer you want - that often precludes supplying the
right answer.

Whether a pair of ETI 480s operating * IN BRIDGE MODE * is safe from
failure depends totally on the particular speaker load it has to drive, the
way the amp is to be very operated and what the chance of ever suffering a
low impedance or short circuit is. The latter may be sudden death of the
fuses are not fast acting enough.

BTW In the ETI 480 - the BD139/140 driver transistors are MORE likely to
fail due to over current or SOA limits being exceeded than the outputs are.
The SC 480 is the better amp.

Your other comments were useful and mostly to-the-point. So yes, take that
as an apology.

** Hooray !

BTW, I returned the Jaycar speaker (the box was unopened). They
were pretty reluctant (past their 7-day policy), being quite
convinced that the particular speaker would be fine, possibly
more robust that many instrument speakers (they sell the same unit
in 4 ohms for car subs which take a bit of punishment).

** Those dudes would not have the slightest idea.

Do any of *them* design or even repair guitar amps for a living ??

But they
did it - after I mentioned "statutory warrantee regarding
fitness-for-purpose". I have yet to choose another, though WES
have a 12" in the same range as your suggested 15" (which doesn't
suit the cab).

** For bass guitar, the more cone area the better the result. Thrashing
one 12 inch driver to within an inch of its life is not the best way to go -
very few commercial amp use that idea. Using two or even four 12 inch
drivers is far better, and actually increases the efficiency ( in dBs per
watt) so far more SPL is achieved than with one higher powered driver.

As you have seen, decent instrument speakers have around 100 dB/watt
sensitivity, while hi-fi woofers are some 8 to 10 dB less efficient. That
is the same as going from a 10 watt amp to a 100 watt one !!

Using four 100 dB /watt speakers will give 106 dB /watt efficiency (
anywhere on axis) - equal to four times amp power on top - so equal to
a massive forty times amp power.

There are several other reasons to prefer instrument speakers, plus they are
designed for the job.




............... Phil

 

[Phil Allison]

"Clifford Heath"

Phil Allison wrote:
** Says a pommie smartasre who has never bought those types from Dick
Smith Electronics ( who sold various fakes in the past and have now
dropped them) or Jaycar who sell "MOSPEC" brand or WES who still
sell fakes

Where in Aus can you buy real ones?

** Farnell - or the actual importers.

Why did you call Graham a smartasre?

** Are you complaining about a typo ???





.............. Phil

 

[David]

On Sun, 30 Jan 2005 18:45:40 +1100, Phil Allison wrote:

Using four 100 dB /watt speakers will give 106 dB /watt efficiency (
anywhere on axis) - equal to four times amp power on top - so equal to
a massive forty times amp power.

How does using four speakers increase the efficiency? Surely using four
speakers will require four times the power to produce 106dB, therefore the
efficiency would still be 100db/watt ?

David

 

[Phil Allison]

"David"
Phil Allison wrote:

Using four 100 dB /watt speakers will give 106 dB /watt efficiency (
anywhere on axis) - equal to four times amp power on top - so equal
to
a massive forty times amp power.


How does using four speakers increase the efficiency?

** How come a 15 inch speaker is more efficient than a 12 or a 10 or a 6 or
a 5 or a 4 inch one??

Go check maker's data.

Surely using four speakers will require four times the power to produce
106dB,

** Nope.

Ever heard of the "superposition principle" ???????

Based, in this case on the fact that sound pressures add linearly.





............... Phil

 

[David]

On Sun, 30 Jan 2005 22:59:50 +1100, Phil Allison wrote:

"David"
Phil Allison wrote:


Using four 100 dB /watt speakers will give 106 dB /watt efficiency (
anywhere on axis) - equal to four times amp power on top - so equal
to
a massive forty times amp power.


How does using four speakers increase the efficiency?


** How come a 15 inch speaker is more efficient than a 12 or a 10 or a 6 or
a 5 or a 4 inch one??

Go check maker's data.

Whats this go to do with this? You were not talking about different size
speakers.

Surely using four speakers will require four times the power to produce
106dB,


** Nope.

Ever heard of the "superposition principle" ???????

Based, in this case on the fact that sound pressures add linearly.

Exactly, and as there are four speakers, you will need four times the
power, as this adds linearly as well, so the efficiency wont have changed
at all.

David

 

[Phil Allison]

"David"
Phil Allison wrote:

** How come a 15 inch speaker is more efficient than a 12 or a 10 or a 6
or
a 5 or a 4 inch one??

Go check maker's data.


Whats this go to do with this? You were not talking about different size
speakers.

** Why are larger diameter cones more efficient ????

Magic ??

Surely using four speakers will require four times the power to produce
106dB,


** Nope.

Ever heard of the "superposition principle" ???????

Based, in this case on the fact that sound pressures add linearly.


Exactly, and as there are four speakers, you will need four times the
power, as this adds linearly as well, so the efficiency wont have changed
at all.

** Errr - two speakers producing the same sound pressure at a point will
add to double that of one.

This is the superposition principle.

Double sound pressure is 6 dB more.

To drive two speakers takes twice the power - so 3dB more input

Ergo - there is a 3dB increase in SPL - for free.

With four speakers, the pressure increase will be four times - or 12 dB.

But it takes four times the power to do it.

So in that case, 6 db is for free.

So 6 dB more efficient than for one speaker driven by the same electrical
power.

QED.



No go learn some basic physics - you stupid shithead.




.............. Phil

 

[David]

On Mon, 31 Jan 2005 01:13:32 +1100, Phil Allison wrote:

"David"
Phil Allison wrote:

** How come a 15 inch speaker is more efficient than a 12 or a 10 or a 6
or
a 5 or a 4 inch one??

Go check maker's data.


Whats this go to do with this? You were not talking about different size
speakers.


** Why are larger diameter cones more efficient ????

Magic ??



Surely using four speakers will require four times the power to produce
106dB,


** Nope.

Ever heard of the "superposition principle" ???????

Based, in this case on the fact that sound pressures add linearly.


Exactly, and as there are four speakers, you will need four times the
power, as this adds linearly as well, so the efficiency wont have changed
at all.


** Errr - two speakers producing the same sound pressure at a point will
add to double that of one.

This is the superposition principle.

Double sound pressure is 6 dB more.

To drive two speakers takes twice the power - so 3dB more input

Ergo - there is a 3dB increase in SPL - for free.

With four speakers, the pressure increase will be four times - or 12 dB.

But it takes four times the power to do it.

So in that case, 6 db is for free.

So 6 dB more efficient than for one speaker driven by the same electrical
power.

QED.



No go learn some basic physics - you stupid shithead.

"In this house, we obey the laws of thermodynamics!" -Lisa Simpson

 

[Clifford Heath]

Phil Allison wrote:

** Farnell - or the actual importers.

Thanks.

Why did you call Graham a smartasre?
** Are you complaining about a typo ???

No, I just didn't see anything wrong with what he said.
Must be history I guess.