Discussing audio amplifier design -- BJT, discrete

[Jon Kirwan]

On Fri, 5 Feb 2010 00:33:53 -0500, "Paul E. Schoen"
<paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:iqomm597nf576piiskbpq3m14uuiheont3@4ax.com...

Hmm. Totally new thoughts. So that's what the term
"regulation" means. It's about the transformer design? And
here I was off and away on the capacitive-filtered ripple
side. Well, that's still useful to have gone back to,
anyway.

I'm not buying the 0.7V diode drop, yet. At peak currents
near 10 times larger than average load currents, I have to
imagine more than 0.7V drop with anything silicon and not
schottky. Do they use schottky's? (Leakage comes to mind.)

Silicon diodes are the norm except for high power, high efficiency, high
frequency, and low voltage. But they do have forward drops of 0.7 to 0.6
volts at normal operating temperatures, and when drawing minimal current,
as is the case at the waveform peak under no load conditions. Even with a
capacitor, the diode current drops to near zero at the voltage peak. A
different result is expected if there is inductance, of course.

I _do_ understand that the diode current drops to zero at
peak. That's clear from the derivative term (cos()
function.) The current into the capacitor is based upon
dV/dt and if that is zero, so is the current.

I think I'm gathering your point better now, though. I was
talking about _under_ high current load, but you were ahead
of me and addressing yourself to the much lower currents that
take place _after_ the peak spike current occurs but _before_
it reaches the near-zero current at cos(pi/2), where the
voltage continues to rise but the rate of change is slow and
so only a little current is required to keep pace.

Thanks for that clue. It makes sense and I missed thinking
about it when I last spoke. I see it, now.

There is a separate regulation spec for the DC output. It is typically much
worse than the regulation of the transformer, as the capacitors quickly
discharge between peaks and can be charged up only as quickly as the
transformer and diodes allow during the conduction cycle. So we use big
capacitors and linear regulators, or resort to a switching supply.

The two rails (prior to any linear reg) rise up and down (the
ripple) in phase with each other, if I'm imagining it right.
Which means that the (+) rail and the (-) rail will move in
concert. Which makes me suspect that it may be okay for the
output stage. What's your take on that? How much ripple on
the rails is okay?

But if you are lucky enough to have three phase power, you can design a DC
supply with no capacitors and get something like 6% regulation (and
ripple). This is SOP for really high power DC, like 10kVA.

That's way out of my league.

Okay. So the 25V was specifying the peak, not the bottom
side. And that is unloaded, basically. Which brings up the
question of what exactly does 15% regulation _actually_ mean.
What is the definition of "full load?" Since the peak diode
currents can be quite a lot more than the average load
current from my calculations, that seems to place quite a
burden on the transformer ratings.

Transformers are rated at RMS current, which is pretty much all that
matters for heating effect, and it is mostly related to the resistance of
the copper and the allowable rise in temperature in the core. Efficiency
aside, what matters is the temperature the insulation can withstand before
deteriorating, and usually that is at least 130C, or 100C above ambient.
The smaller the tranny, the better it sheds heat (surface area/volume), so
regulation and efficiency of smaller ones tend to be poorer.

Still, my question seems to remain about the peak currents.
If the peak diode current spike takes place say 30 degrees
prior to the peak voltage and is 10X the average load
current's needs, one cannot ignore that peak current having
to come from a secondary that has ohmic resistance as well as
the possible volt-second problems at 50Hz and 60Hz. Or?

Full load is just the maximum RMS current at which the transformer is
rated. This may be further complicated by duty cycle ratings, which can be
continuous or intermittent. Generally intermittent duty is 50% duty cycle,
with ON times not greater than 30 minutes, at least for larger transformers
with more thermal mass. At 50% duty cycle the output rating is 1.4 times
the true continuous rating. And then the allowable duty cycle is the
inverse of the square of the overload. For the circuit breaker test sets I
design, we specify output up to 10x the continuous rating, at which the
duty cycle is only 1%. But the ON time is limited to about 100 mSec, which
is more than enough to trip a circuit breaker instantaneously, and then you
should wait 10 seconds before doing it again.

But here's the question, again. The currents via the bridge
diodes are in no way very similar to sine waves. In fact,
they almost suddenly rise up to follow the dV/dt requirement
and then decay out somewhat later on and then very lightly
(as you reminded me above) have any current requirement up to
and perhaps shortly after pi/2. This is a complex waveform
and not easily turned into rms. For example, the average
current might be 1A and the RMS might actually be closer to
4A. Or with an average of 1A the RMS might instead be closer
to 3A or 5A depending upon the capacitors and the phase and
duration of the spike. That's got to give caution when
thinking about these things. It would be very easy, without
going carefully into the waveform itself, to make a
misjudgment about it. And that still ignores _peak_ which
might very well be 12A or 15A or more, despite the 3A rms or
5A rms figure. What drop will that represent?

Seems a lot of factors to balance.

I designed a "Programmable Overload Device", or POD, which takes into
account the current and the time, as well as the actual temperature using a
thermistor, to enforce reasonable duty cycles. Fuses, circuit breakers, and
Motor Overloads do a similar function, but don't fully take into account
all the factors. The intelligence for this is buried in the PIC code, and
is rather involved and yet imperfect. If I could accurately model the
heating and cooling effects of current in a transformer, it would be ideal.
Now that's where theory can really help.

So could you go further here? In other words, let's say I
know that the average load current will be 1.4A, but that the
peak diode current given the bridge/capacitor design will be
15A. The transformer is a 25.2Vrms CT unit. The DC rails
are at -15 and +15, with 2200uF caps on each side to ground,
and the ripple on them is about 3.8V peak to peak (+/-1.9V
around 15V.)

What's the VA rating here? And "regulation" number are you
looking for in the transformer and how does it relate back to
VA and other terms that might be used?

It's really easier (and perhaps even more appropriate) to use a tool such
as LTSpice for this purpose. You could look at all the variables over time,
quantized to steps small enough to minimize error, and finally arrive at a
steady state solution where you may be able to describe such complex
entities as RMS current with an equation, but all you will have done is
spend a lot of time doing what LTSpice does so well and so quickly. So I
cobbed together a simple power supply simulation, which in this case models
part of a power supply that I have been using on my Ortmasters, with a
Signal 241-6-16 transformer. The ASCII file is at the end of this post.

Thanks.

I'm using a voltage doubler circuit on each leg of the 16VCT transformer,
as I need to get at least 17 VDC for 15VDC linear regulators for the analog
portion of the circuit. I figure no more than 20 mA. So for simulation
puposes I use a 1k resistor as the load. The transformer is 32 VA, or 2A at
16V, and I estimate 15% regulation which is a 2.4 V drop at 16V or open
circuit 18.4 VRMS. I'm using a voltage source with 26 volts peak and 1.1
ohms internal resistance. The capacitors are 220uF, and MURS120 diodes. As
a result, I get 22.35 VDC outputs, and the transformer current is 104 mA
RMS, with peaks of about 360 mA.

Just for fun, I changed the output loads to 10 ohms, and I found that the
current is only 345 mA RMS, and the transformer current is 611 mA RMS, with
peaks of about 1 amp. The capacitively coupled design is inherently
current-limited, which can be a good thing.

Okay. Well, I'll take a look. Are you using one of the two
non-linear inductors used to model core saturation?

If you put a capacitor on the output, it eventually charges to the peak
voltage. This is the high limit that must be considered for design. It
may
not be exact, and probably will be a bit lower, because a power
transformer
is usually designed to operate in partial saturation, so the output will
not increase linearly above its design rating.

Ah. Core saturation is __intended__ as part of the design? I
haven't done that one before. What guidance can you give on
that aspect?

Maximum use of the iron occurs near the maximum flux density. It results in
increased current which actually occurs at 90 degrees to the applied
voltage, so the distortion is not in the form of a flattening of the
voltage waveform but rather like crossover distortion. But it does result
in a somewhat non-linear effect, as it interacts with the resistance of the
windings. See the following for more information:
http://openbookproject.net/electricCircuits/AC/AC_9.html

Will do.

and more about regulation:
http://www.allaboutcircuits.com/vol_2/chpt_9/6.html

It is most pronounced in ferroresonant transformers:
http://www.ustpower.com/Support/Voltage_Regulator_Comparison/Ferroresonant_Transformer_CVT/Constant_Voltage_Transformer_Operation.aspx

Something new, again. Thanks.

Under load, the output will drop, caused by the effects of primary and
secondary coil resistance as well as magnetic effects. These will cause
heating over a period of time, and the coil resistance will increase,
adding to the effect until a point of equilibrium is reached based on the
ambient conditions and removal of heat via conduction, convection, and
radiation.

Now that, I understand and worry about.

That's why most designs are made with a generous safety factor so you do
not need to worry about these effects. They can be predicted approximately
and that is good enough.

And I'm not designing to save every nickel. Which helps. I
can see that the design issues become very peculiar when
every shaved penny is desired while still preserving design
function. Of course, there is more money available for the
required extra design time then, too, I suppose.

Hehe. I want to _learn_ to design to specified criteria,
have a comprehensive view of the theoretical concepts
involved, and that means I need to only pick the first one.
The 'quickly' is unimportant -- one to two years is good
enough. The 'cheaply' is equally unimportant. If it costs
me 10 times as much in terms of parts and time as it would
just buying something commercial, buying a commercial
solution will teach me exactly zero about what I need to
learn to design what my daughter needs. And there is NOTHING
on the market to get there, either. No one else has my
problem. Or few do.

It might be worthwhile to discuss those details here to dig up some ideas.

My daughter enjoys turning the volume to maximum on an
amplifier and then curling up into a ball nearby or else
running away from it and not coming back to change it
downwards. We come in running and turn things down many
times a day. (She turns them up again, sometime later on.)

What I'd like is to limit the maximum volume for her -- but
given any fixed output amplifier that is pretty easy to
achieve, as I can insert dummy resistances in series with the
speakers to get that (proper wattages, of course.) The other
issue is that the volume needs to mute after some adjustable
set time (on order of minutes), but only after the last time
the volume control is touched. And any slight motion of it
should restore the volume as last set. She knows how to work
a knob and will quickly learn the behavior. I'm considering
the use of an optical quadrature knob for this.

Separately, I need to modify a microwave oven control. She
destroys them, fairly routinely. So I have a nice collection
of the transformers! But I also know that the magnetron and
transformer and waveguide parts are fairly standardized, it
seems. The controls are very custom, but that merely means I
can design my own and fit that to the standard interface
available for the magnetron section. Later project that will
_also_ incorporate a temperature sensor system I'm working on
right now that allows me to measure, in situ, in the oven
chamber -- uses phosphor thermometry to get the job done and
works well in this application. But that is more than a year
out, right now.

This is a "give a person a fish and they eat for a day, teach
a person to fish and they eat for the rest of their lives"
thing.

I've heard it said that, "teach a man to fish, and he'll spend all day in a
boat drinking beer!"

hehe. It's also said that for every saying there is an equal
and opposite saying. One of those fundamental forces, you
know?

The digressions are great! I am NOT in a rush to build,
though. I'm wanting to engage the math and learn what can be
achieved by deducing from parsimonous theory. Then test a
few things on the bench, ask questions, learn some more. Etc.
So theory _and_ practical approaches are important. Not one,
or the other, but both!!

Pendulum motion is well understood. One might either have a
practical knowledge about it and some tables and just go with
that. Probably, lots of folks making pendulum clocks stop
there and go no further and are none the worse for that. It
is similarly very easy to develop the infinite series that
describes it (or use the sqrt(L/g) proportionality as a first
order approximation or for small starting angles) from the
simple differentials involved and to take an entirely
theoretical approach, as well.

But I'm interested in more than that. Theory by itself lacks
reality. Reality by itself lacks meaning sans theory. The
two go together like hand in glove, though. Building even
the most simple ones using a peg-in-hole method leads to the
discovery of still more interesting effects, if you know some
theory. For example, the rocking of the pin itself in the
larger hole has a measurable impact of perhaps as much as 2
or 3 percent. It's useful to know that and understand it.
Once that mechanism is itself understood, one can then dig
even deeper to find more subtle (and possibly useful) effects
to continue improvements. A practitioner lacking even the
basic theory might accidentally happen upon some idea, of
course. And a theoretician lacking practical reality to
interfere might accidentally imagine some realistic effect to
pursue, too. But it really takes a marriage of both to make
quick work of progress forward, I think.

Since theory is primary, I like to pursue that part of it
earlier and move to experience once I have the mental tools
required to make sense of the data that results. Without
theory, data is pure noise. Without the theory of a sphere,
even the gentle curvature at the horizon "seen" my a mountain
climber is just so much useless noise to them. But _with_
that theory, the data _means_ much.

I think I had problems in the EE program at Johns Hopkins because it was
too theoretical for my mindset, and I had fundamental problems with
advanced calculus.

Which I find little other than a relaxing joy.

I aced the lab courses and helped others because I had
already designed and built many circuits. But, looking back, I see where
having a stronger grasp of theory would have helped. I still design
circuits with a highly empirical approach, using rule of thumb and
experience to choose components. Now that SPICE is freely available I find
it fascinating to try different values and placements and configurations
"just to see what happens". And I learn by looking at the time domain
simulation plots and determining what may have caused certain glitches or
oscillations that I did not foresee.

What I'm going to say is NOT a judgment in any way. In fact,
many of those I've enjoyed working with have had problems
mastering math and theory and yet are extremely creative and
they find the right crutches to help them where they are
weaker and get the right job done, rightly. One I'm thinking
about right now would use Excel at every turn, almost. But
he had an instinct about _what_ to put into Excel and what
questions to ask and what data to collect. I trusted his
judgment and only very rarely grilled him to see if I could
find holes. And his work habits were like mine -- work all
hours, night and day, when people were depending upon us. I
could call him up with some problem I was wrestling with some
Sunday afternoon and he'd jump over to the lab, dropping
everything else, and simply go, go, and go until we'd
clobbered the problem together. We had each others' backs
and I enjoyed that a lot.

An interesting place where theory has served me well,
personally, regarding one field of instrumentation. I was
simply playing with some different ideas and mathematical
derivations where I had no idea where they'd take me and
landed upon a method that has so far been unexplored in the
literature. It's self-calibrating for offset and gain
because of the _shapes_ that reside in the manifold I was
playing in and that means this can be made _cheaply_, as in a
factor of 10 less than beforehand. I worked with one of the
two physicists writing the seminal papers in the field (back
in the 1950's) and I know that he didn't know about it.

It's only possible to "see" it from a mathematical vision and
it is incredible to see it perform in practice, auto-
calibrating offset and gain. That can be applied as often as
required to control for temperature and time drift, as well.
It was a stunning insight which I am sure I could NOT have
had with LTspice games. Not possible to hack into this
insight. The odds against it are astronomically high.

My talents are more in the realm of imagination and thinking outside the
box. And sometimes it has gotten me into trouble. But I have also sometimes
been able to make a lot of progress in a short period of time. I think some
aspects of design are more of an art than a science, and I look for a sort
of elegance in the finished design of a circuit, even in the placement of
components on the schematic, and also in their placement on a PCB.

I have some small imaginations from time to time (like the
above one discovered while playing), but I enjoy drilling
down into details. My wife is the other way round and has a
stunningly brilliant way of combining things in new ways
every single day and yet no willingness to painstakingly dig
into them. Her insights aren't merely interesting, either.
They are on-target. She is the one who saw this economic
disaster heading our way and forced us to sell our properties
"right away" in early 2005. We actually _made_ a fair amount
of money in the last 5 years because she forced us to buy and
sell oil at the right times and Euro-denominated securites,
and so on. She is like a spider in a web who feels _real_
vibrations that mean something important and knows which to
ignore, too. (I am still mystified by it. And I've known
her from before I have any memory -- she is two years older
and actually was asked to babysit me when I was little. So
that long, at least.)

Everyone has their strengths and we all need each other.
There is no one right way to be and we each have to find our
own paths. But it is really nice when you aer lucky enough
to surround yourself with people with different penchants and
skills, each having each others' backs and filling in where
it counts without disingenuousness. We really do need each
other in healthy complementary ways. None of us are whole,
by ourselves.

And thanks so very much by the way,
Jon

 

[Jon Kirwan]

On Fri, 5 Feb 2010 14:47:16 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Thu, 04 Feb 2010 17:16:59 -0800, I wrote:

snip
Since theory is primary, I like to pursue that part of it
earlier and move to experience once I have the mental tools
required to make sense of the data that results.
snip

Okay. On second thought... enough theory. I think it's time
for practice. I already have triple output power supplies,
but using them wouldn't be true to the actual amplifier
situation. And any testing of distortions needs to cope with
that reality.

So I'm moving forward on the power supply rails. I need to
scarf around and see what I have available. I'll post what I
find, the resulting design and thinking, photos perhaps, and
the results of testing with static loads. Once that is done,
I'd like some advice about the next step, though. But until
then, I'll just focus on getting that part put to bed. That
much I can do right now.

I've decided that your kick in the butt, Paul, was what I
needed. I have enough in mind to move out of the thinking
stage and into trying some different alternatives. I'll get
going.

Thanks,
Jon

There's this saying "Practice without theory is blind and theory
without practice is lame".

I hope I made clear my recognition that _both_ are vital. If
science processes hadn't caught wind of this centuries ago,
we'd still be in the dark ages I suspect.

You've made it clear that you want to
thoroughly understand the hows and whys of amplifier design from
mathematical models. I have no quarrel with that approach and I
also use it myself within the limits of my own capability - *up
to a point*. But there comes a point at which striving for
absolute precision solely from theory results in diminishing
returns.

Yes. As a neophyte, it's hard to know where that boundary is
at, though. It's like the old software law. The first
software attempt does far too little and there are many
complaints and suggestions and flaws in it. The second
attempt does way too much and is overburdened and complex
almost to the point of uselessness. It's only by the time
you've done the third one that you get it close to right.

I'm trying to get past the first attempt and do too much for
my second. Kind of a shortcut to getting to that sweet third
one, I suppose.

Take the case of the pendulum you brought up earlier. The basic
theory is well established, but to predict the behaviour of a
practical pendulum with 100% precision will require taking into
account the effects of so many factors that it may well be
impossible. E.g., the aerodynamics of the pendulum's shape
including minute irreguarities on its surface, the exact strength
and orientation of the earth's magnetic field at the location and
its effect on traces of magnetic materials in the alloy, friction
with suspended particles in the air in addition to the air
itself, friction at the point of suspension and elasticity of the
suspension, etc., etc. Even if all these influencing factors are
included in the equation, the physical values to be entered can
never be measured with 100% accuracy.

Theory let's you "see." Start with the basic pendulum theory
with pendulums where angle=sin(angle), nearly. It allows you
to observe such pendulum behavior with some understanding.
The swings are not just random rockings back and forth or
something that appears vaguely "regular," but now instead
they are something to expect and recognize and "see." For a
time, it's all you need. Then someone develops a better
clock. Suddenly, you uncover the fact that various pendulums
are NOT nearly as well predicted as they should be. They
vary by 3% or so, which keeping the swing angles in good
control is far beyond the tolerances of the theory.

But without the prior theory, and better clocks, you wouldn't
even notice. And now "seeing" you might start a new
exploration to discover some new principles to add. And in
doing so, you may now have a chance to discover yet another
anomaly that resides still deeper yet.

I think that was where I was headed, earlier. Countering by
bringing up concepts of perfection (100%) really misses my
point -- that theory is primary to developing meaning and
understanding results. The fact that we do not have 100%
knowledge or the ability to make infinitely precise
measurements does NOT undercut it, at all.

Take the case of the forward drop of the diode in the power
supply that you've been discussing with Paul. This what I did
before personal computers and simulation progs became widely
available: I drew a curve of the diode's V-I characteristics on
graph paper up to the expected peak current. Then I drew a
straight line, approximately following the dynamic curve, from
the peak point down to the voltage axis. I took that voltage as a
constant forward drop and the slope of the line as a constant
series resistor. I then added that resistance to other source
resistances like the transformer winding resistance and either
use it to calculate the rectified and filtered voltage or, more
often, to determine it from a graph such as that in RDH. It also
comes in useful for finding the peak and rms currents. I don't
know if anyone else uses that method or how well it agrees with
theory, but it agrees pretty well with practical measurements.

That is immediately obvious and makes good sense. The
Shockley equation can be readily used in derivative form to
predict the instantaneous resistance slope at any current you
want. And the even simpler V=Vd+R*I equation often used for
diodes (for current values near some nominal I value) works
pretty well within those boundaries. The R can either be
measured as you suggest or else, if the model parameters are
available, easily derived and predicted with some expectation
of closeness should it then _be_ measured.

Your approach makes excellent sense to me.

I don't do this every time I design a power supply. I just make a
mental estimate based partly on theory and partly on past
experience. In short, there's a point at which it makes more
sense to make informed assumptions and approximations even before
doing physical construction.

Which requires experience and practice. Before that is well
acquired, it's good theory that best guides (if available.)

I think I am ready regarding the power supply, though. And
that is probably a good step. Except that I'm still
questioning whether or not the rails can have ripple while
the output stage is tied to them or if it really is necessary
to bother with a crafted linear regulator, as well. (If I
use a regulator, again it will have to be discrete parts and
BJTs, but I think I can do that much, too, with four or five
BJTs per side.)

I guess another part of why I'm thinking through these things
this closely, besides the fact that I don't have a lot of
experience to rely upon and guide me, is simply that I'd like
to "measure twice, cut once."

Jon

 

[Bob Masta]

On Thu, 04 Feb 2010 12:07:16 -0800, Jon Kirwan
<jonk@infinitefactors.org> wrote:

There's another question that comes to mind regarding the
output stage. A lot of talk seems to revolve around
"crossover distortion." Seems almost very first thing folks
talk about when discussing class of operation if not also at
other times.

Seems to me that in a three-rail power supply situation
without an output capacitor involved, the crossover takes
place near the midpoint (ground) voltage between the rails,
at a time when current into the speaker load is also near
zero. (I'm neglecting any thoughts about inductance in the
speaker and physical coupling into the air, for now.) In
other words, where power at the speaker is near zero. Is it
really that important to consider?

I was looking at that terrible large scale gain plot for the
quasicomplementary output stage on the web site recently
mentioned in the thread (the lower curve in Figure 4 on this
link):

http://www.embedded.com/design/206801065?printable=true

(It's not that terrible of a plot, as the variation is from
.96 to .98 with the "normal" middle at .97.)

What's experience say here? Is it really so terrible as to
worry too much about something that takes place near zero
voltage, anyway? I'm just questioning the concern, for now.
I have no understanding about it, at all. Just wondering.

Crossover distortion is a more-or-less fixed
amplitude, so at low signal levels it becomes a
large percentage of THD. Our ears are sensitive
to the relative amplitudes of components, so a
hypothetical fixed amplitude distortion component
might be totally inaudible when it is a low
percentage of a large fundamental, yet be
obnoxious as a larger percentage of a small
fundamental. (This is the same essential problem
as quantization distortion in digital circuits...
you don't hear it on the peaks, only on the quiet
parts.)

Unless you have some application that doesn't
involve soft passages in the signal (like a siren,
or possibly a PA or musical instrument amp) you
need to consider the crossover distortion.

However, there is still a lower limit to absolute
detection threshold, regardless of what percentage
a component might be. If the system levels
(program, amp, speakers, listening position) are
set up such some signal component is below 0 dB
SPL, most people aren't going to hear it even if
it is the entire signal!

Best regards,




Bob Masta

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[Jon Kirwan]

On Fri, 05 Feb 2010 12:36:02 GMT, N0Spam@daqarta.com (Bob
Masta) wrote:

On Thu, 04 Feb 2010 12:07:16 -0800, Jon Kirwan
jonk@infinitefactors.org> wrote:

There's another question that comes to mind regarding the
output stage. A lot of talk seems to revolve around
"crossover distortion." Seems almost very first thing folks
talk about when discussing class of operation if not also at
other times.

Seems to me that in a three-rail power supply situation
without an output capacitor involved, the crossover takes
place near the midpoint (ground) voltage between the rails,
at a time when current into the speaker load is also near
zero. (I'm neglecting any thoughts about inductance in the
speaker and physical coupling into the air, for now.) In
other words, where power at the speaker is near zero. Is it
really that important to consider?

I was looking at that terrible large scale gain plot for the
quasicomplementary output stage on the web site recently
mentioned in the thread (the lower curve in Figure 4 on this
link):

http://www.embedded.com/design/206801065?printable=true

(It's not that terrible of a plot, as the variation is from
.96 to .98 with the "normal" middle at .97.)

What's experience say here? Is it really so terrible as to
worry too much about something that takes place near zero
voltage, anyway? I'm just questioning the concern, for now.
I have no understanding about it, at all. Just wondering.

Crossover distortion is a more-or-less fixed
amplitude, so at low signal levels it becomes a
large percentage of THD. Our ears are sensitive
to the relative amplitudes of components, so a
hypothetical fixed amplitude distortion component
might be totally inaudible when it is a low
percentage of a large fundamental, yet be
obnoxious as a larger percentage of a small
fundamental. (This is the same essential problem
as quantization distortion in digital circuits...
you don't hear it on the peaks, only on the quiet
parts.)

Okay. This is consistent with another way I was looking at
this. Since the crossover takes place near where power is
also close to zero, it's effect is fairly constant. If the
amplifier's swing is large scale (nearer the limits it was
designed to provide), it's not likely to be noticed buried
within that. If the amplifier's swing were small (nearer
that zero volt area) then the crossover distortion is quite a
bit more noticeable.

Makes sense, I think.

Unless you have some application that doesn't
involve soft passages in the signal (like a siren,
or possibly a PA or musical instrument amp) you
need to consider the crossover distortion.

Best I consider it, then. I'm looking to learn, not
specialize in bullhorns.

Interesting to me that you say that crossover distortion
might not be such an issue for a musical instrument
amplifier, though. I take it you must mean for stage work
where the power is going to be set pretty high, generally?

However, there is still a lower limit to absolute
detection threshold, regardless of what percentage
a component might be. If the system levels
(program, amp, speakers, listening position) are
set up such some signal component is below 0 dB
SPL, most people aren't going to hear it even if
it is the entire signal!

Hehe.

Best regards,

Bob Masta

Thanks,
Jon

 

[David Eather]

Jon Kirwan wrote:

On Thu, 04 Feb 2010 05:46:33 +1000, David Eather
eather@tpg.com.au> wrote:

A 30 volt CT transformer with 15% regulation and 7% mains over-voltage,
less voltage drop for the diode bride would give rails of +/- 25.

I'm not entirely familiar with the formal terms,

Four..mal terms? ....I don't think I have used one of those. Regulation
is just 1 - (no load AC voltage / full load AC voltage) expressed as a
percentage. A first order approximation of a real transformer would be
an ideal transformer with a resistor in series. PS pretty much nailed it
in his response.

There is no ripple to consider as there is no (significant) load on the
transformer and hence no (significant)load on the power supply. Also,
there is no (significant) current flowing through the diode bridge.

It was a 30 volt CT transformer with 15% regulation and 7% mains
over-voltage, less voltage drop for the diode bride would give rails of
ą 25.

Vsupply(max no load) = SQRT(2) * 15 * 1.15 * 1.07 - voltage lost from
the diode bridge. I rounded of the answer to +/- 25 volts. If I had
already selected a bridge I would have used whatever the data sheet
called the minimum voltage drop. That +/- 25 is the fully rectified and
filtered output of the power supply which is supplied by a relatively
cheap 30 volt CT transformer.

You use this voltage when you check that all transistors, caps etc are
adequately rated for the job and for the worst case situations you can
imagine - like that killer of speakers, the "1812 overture" driven into
heavy clipping.


but I figure

the 15% regulation you mentioned above must mean that the
ripple voltage goes from 100% to 85%. In general, the
equation for the angle would be something like:

angle = arcsin( 1 - Rf*(1-Vd/Vpk) )

This is with Rf being the ripple factor (not in percent
terms, obviously) and Vd being the sum of the diode drops
(full wave would be something like 2V) and Vpk being the
sqrt(2)*Vrms. At least, that's what the equation works out
for me on paper. (I can develop out here, if needed.)

The peak diode current happens just as the first moment of
conduction (which is neatly defined by the ripple factor, if
I understand you) and would be something like:

Ipk = 2*pi*f*C*Vpk*cos( angle )

Since the cos(arcsin(x)) is just sqrt(1-x^2), the computation
looks like:

Ipk = 2*pi*f*C*Vpk*SQRT(1-(1-Rf*(1-Vd/Vpk))^2)

There's probably some other adjustments to nail it, but that
probably gets somewhat close.

For a full wave bridge (I know, I think you were talking
about a half wave, but let's go with this for a moment) with
Vpk=25.2*SQRT(2) and Vd=2V (for two diodes in conduction in
the bridge) and Rf=.15 and f=60 (US-centric) and let's say a
C=2200uF, that gets something like Ipk = 15.2A.

None of this takes into account the average load current or
peak load currents. It just assumes that the ripple factor
is somehow known to be correct. I started out assuming that
the average load current would be defined entirely by the
droop (which is, of course, based upon the ripple factor
assumption) and the time from the peak of a previous cycle to
the point at which the above angle occurs and conduction
again begins. But it is complicated slightly by the fact
that the transformer supplies the entire current draw (if it
can) for a short part of the cycle _after_ the peak, as its
slope is less than the droop slope of the cap. I played with
accounting for all that but then decided that in most
practical cases the droop is hopefully not too excessive and
if not, then the angle after the peak isn't that much.. maybe
5 to 8 degrees or so... So I decided to ignore it and just
rely upon the capacitor's droop only:

Iave = C*Vpk*Rf/((pi/2+angle)/(2*pi*f))

So in the case just mentioned, I get Iave = 1.7A. I take it
that Ipk can easily be a factor of 9 or 10 greater. Also, I
note that it would probably be helpful to have nifty charts
of some kind to help pick off details like these.

Can you expand a little on what you were talking about,
though? Was that a half wave suggestion? I'm not sure I can
make sense of the rails, if so. If not, then I'd still
appreciate some of the calculations so that I can sure I
follow all of it.

I'm guessing that if a rail is to have a minimum of 25V on it
at the bottom of the ripple, and you are talking about 15%
regulation, the peak is going to be 29.4V -- not counting the
diode drops. Add 1V for that and it's 30.4V. Another 7% on
top would be 32.7V for the peak. RMS would be that figure
divided by sqrt(2), wouldn't it? Or 23.1Vrms or so?

No. 25 volt max. Under load less and also subtract the ripple.

So would that suggest two of the cheap 25.2Vrms transformers
and plan on rails still slightly higher?

No. 1 CT transformer might be OK. The PSU unloaded voltage might be

V supply(max no load) = SQRT(2) * 12.6 * 1.15 * 1.07 - bridge

which I would round off to +/- 21 volts. In my mind I rejected this
option because I thought it would be difficult to make work in a worst
case, such as when the mains is 7% under voltage and the transformer is
at it's rated load

V supply(max load) = SQRT(2) * 12.6 * 1.0 * 0.93 - bridge (say 2 volts
at 1.6 amps) equals 14.6 volts

For 10 watts of output the peak output voltage is 12.7 volts. You also
need some (configuration dependent) voltage to keep the output in linear
range and even before you make any allowance for ripple voltage you are
in trouble.

You could deal with this situation a few ways, like saying it is not
your fault if the mains voltage is under. Use a bigger 12.6 transformer
so it is never fully loaded and hence the voltage is somewhat higher,
try to pass of an 8 watt design as 10 watts music power (which it
probably is), use massive filter caps (infinite in this case) or use a
higher voltage transformer. My experiance suggests that no one
complains if their 10 watt amp can actually put out 11 watts, but lots
of them complain when their 10 watt amp only puts out 9 watts (even
though you really can't hear the difference)

********
(1).....Vout(to speaker) = sqrt(2*P*R)

With R as 8 ohms for a common speaker and 10 watts that is 12.7 volts -
actually ą 12.7 volts with a split power supply.

(2).....Imax = sqrt(2*P/R) This comes out to 1.6 amps.
*********

Oh, crap. The VA rating. That's another one to consider.
Later, I guess.

- always one more thing... later is OK. Just start thinking about how
the amp might be used (domestic music, PA, emergency announcement,
musician's band etc)

A dual 12.6 volt transformer would give a minimum (worst case with
transformer at full load and mains 7% under voltage) of 16.something
volts meaning big filter caps if you were serious at getting 10 watts.
One of the reasons to go PSU first I think. (Also I live in a tiny
jerk-water town where no one knows what a custom made transformer is let
alone where you can get one wound)

I suppose the old filament-type 12.6VAC transformers must be
common everywhere. I also see that Radio Shack (yes, I'm
holding my nose for a moment) still carries some "commodity"
type 25.2VAC CT 2.0A rated transformers for about US$10.
Their 12.6VAC CT 3.0A transformers are priced identically.

So what's considered to be generally available?

Job dependent. For a one off, whatever you can get. For high spec
commercial audio, whatever you can get custom made for the application.
For something laid out as a set of instructions what ever is widely
available - like 12.6v. In OZ we have a couple of hobbiest suppliers
that stock essentially the same range 6.3, 12.6, 15, 18, 20, 25, 30, 35..

Jon

 

[Jon Kirwan]

On Sat, 06 Feb 2010 06:22:37 +1000, David Eather
<eather@tpg.com.au> wrote:

Jon Kirwan wrote:
On Thu, 04 Feb 2010 05:46:33 +1000, David Eather
eather@tpg.com.au> wrote:

A 30 volt CT transformer with 15% regulation and 7% mains over-voltage,
less voltage drop for the diode bride would give rails of +/- 25.

I'm not entirely familiar with the formal terms,

Four..mal terms? ....I don't think I have used one of those.

I probably should have written, "I may not be familiar with
your use of this term, which may have a convention I'm
unaware of." I assumed your use had a formalized or
conventional meaning.

Regulation
is just 1 - (no load AC voltage / full load AC voltage) expressed as a
percentage. A first order approximation of a real transformer would be
an ideal transformer with a resistor in series. PS pretty much nailed it
in his response.

Yes, Paul helped me understand your meaning better.

There is no ripple to consider as there is no (significant) load on the
transformer and hence no (significant)load on the power supply. Also,
there is no (significant) current flowing through the diode bridge.

Well, I had been thinking about the loaded case perhaps too
much and hadn't broken out of that thinking. Seemed to me
that the unloaded situation never really occurs and that I
needed to maintain a fully loaded mindset at all times.

But I'd also wrongly imagined something that Paul caused to
change for the better. The bridge rectifier's peak current
occurs _before_ the top of the AC rise and has nearly gone
away by the time the rise hits the peak point. So the peak
voltage on the caps gets closer than I'd earlier guessed it
would. I knew that the slope (dv/dt) of the AC determines
the charging current on the caps and that the slope was zero
at the peak point, but a firm, simultaneous grasp hadn't yet
solidified in mind until he wrote.

I think this, taken together with your comment above, I must
also realize that the transformer itself will only be loaded
down for a small fraction of each cycle. If it has 15%
regulation, as you say, then for a short time while this peak
load is taking place the secondary voltage will be depressed.
But the fact remains that by the time the peak of the AC
arrives, the loading will be very light and thus the voltage
on the output will be at the higher, nearly unloaded value.
And thus, also, the caps (and BJTs) will experience this
value.

Did I get this about right?

It was a 30 volt CT transformer with 15% regulation and 7% mains
over-voltage, less voltage drop for the diode bride would give rails of
ą 25.

Vsupply(max no load) = SQRT(2) * 15 * 1.15 * 1.07 - voltage lost from
the diode bridge. I rounded of the answer to +/- 25 volts. If I had
already selected a bridge I would have used whatever the data sheet
called the minimum voltage drop. That +/- 25 is the fully rectified and
filtered output of the power supply which is supplied by a relatively
cheap 30 volt CT transformer.

Yes, I am better gathering the use of the term "regulation"
now. And maybe _why_ you think that way about it. You are,
I imagine, thinking about worst case voltage stresses that
need to be held off. The sqrt(2)*15 is the specified peak
stress, but one needs to be aware that under no load the
transformer will likely produce higher voltages on its output
which _will_ be passed along to the caps and BJTs.

You use this voltage when you check that all transistors, caps etc are
adequately rated for the job and for the worst case situations you can
imagine - like that killer of speakers, the "1812 overture" driven into
heavy clipping.

Yes, now that I read this it sounds almost like what I just
wrote. So maybe I am "getting a clue," now. Thanks very
much for taking the time for me.

but I figure
the 15% regulation you mentioned above must mean that the
ripple voltage goes from 100% to 85%. In general, the
equation for the angle would be something like:

angle = arcsin( 1 - Rf*(1-Vd/Vpk) )

This is with Rf being the ripple factor (not in percent
terms, obviously) and Vd being the sum of the diode drops
(full wave would be something like 2V) and Vpk being the
sqrt(2)*Vrms. At least, that's what the equation works out
for me on paper. (I can develop out here, if needed.)

The peak diode current happens just as the first moment of
conduction (which is neatly defined by the ripple factor, if
I understand you) and would be something like:

Ipk = 2*pi*f*C*Vpk*cos( angle )

Since the cos(arcsin(x)) is just sqrt(1-x^2), the computation
looks like:

Ipk = 2*pi*f*C*Vpk*SQRT(1-(1-Rf*(1-Vd/Vpk))^2)

There's probably some other adjustments to nail it, but that
probably gets somewhat close.

For a full wave bridge (I know, I think you were talking
about a half wave, but let's go with this for a moment) with
Vpk=25.2*SQRT(2) and Vd=2V (for two diodes in conduction in
the bridge) and Rf=.15 and f=60 (US-centric) and let's say a
C=2200uF, that gets something like Ipk = 15.2A.

None of this takes into account the average load current or
peak load currents. It just assumes that the ripple factor
is somehow known to be correct. I started out assuming that
the average load current would be defined entirely by the
droop (which is, of course, based upon the ripple factor
assumption) and the time from the peak of a previous cycle to
the point at which the above angle occurs and conduction
again begins. But it is complicated slightly by the fact
that the transformer supplies the entire current draw (if it
can) for a short part of the cycle _after_ the peak, as its
slope is less than the droop slope of the cap. I played with
accounting for all that but then decided that in most
practical cases the droop is hopefully not too excessive and
if not, then the angle after the peak isn't that much.. maybe
5 to 8 degrees or so... So I decided to ignore it and just
rely upon the capacitor's droop only:

Iave = C*Vpk*Rf/((pi/2+angle)/(2*pi*f))

So in the case just mentioned, I get Iave = 1.7A. I take it
that Ipk can easily be a factor of 9 or 10 greater. Also, I
note that it would probably be helpful to have nifty charts
of some kind to help pick off details like these.

Can you expand a little on what you were talking about,
though? Was that a half wave suggestion? I'm not sure I can
make sense of the rails, if so. If not, then I'd still
appreciate some of the calculations so that I can sure I
follow all of it.




I'm guessing that if a rail is to have a minimum of 25V on it
at the bottom of the ripple, and you are talking about 15%
regulation, the peak is going to be 29.4V -- not counting the
diode drops. Add 1V for that and it's 30.4V. Another 7% on
top would be 32.7V for the peak. RMS would be that figure
divided by sqrt(2), wouldn't it? Or 23.1Vrms or so?

No. 25 volt max. Under load less and also subtract the ripple.

Right. I've been corrected and I think I may gather the why
of it.

So would that suggest two of the cheap 25.2Vrms transformers
and plan on rails still slightly higher?

No. 1 CT transformer might be OK. The PSU unloaded voltage might be

V supply(max no load) = SQRT(2) * 12.6 * 1.15 * 1.07 - bridge

which I would round off to +/- 21 volts. In my mind I rejected this
option because I thought it would be difficult to make work in a worst
case, such as when the mains is 7% under voltage and the transformer is
at it's rated load

V supply(max load) = SQRT(2) * 12.6 * 1.0 * 0.93 - bridge (say 2 volts
at 1.6 amps) equals 14.6 volts

For 10 watts of output the peak output voltage is 12.7 volts. You also
need some (configuration dependent) voltage to keep the output in linear
range and even before you make any allowance for ripple voltage you are
in trouble.

I see your thinking much better.

You could deal with this situation a few ways, like saying it is not
your fault if the mains voltage is under. Use a bigger 12.6 transformer
so it is never fully loaded and hence the voltage is somewhat higher,
try to pass of an 8 watt design as 10 watts music power (which it
probably is), use massive filter caps (infinite in this case) or use a
higher voltage transformer. My experiance suggests that no one
complains if their 10 watt amp can actually put out 11 watts, but lots
of them complain when their 10 watt amp only puts out 9 watts (even
though you really can't hear the difference)

Well, this is for learning and eventually for building and
testing here. Complaining to myself won't get far. But
I am very glad for the broader thinking process. Thanks.

********
(1).....Vout(to speaker) = sqrt(2*P*R)

With R as 8 ohms for a common speaker and 10 watts that is 12.7 volts -
actually ą 12.7 volts with a split power supply.

(2).....Imax = sqrt(2*P/R) This comes out to 1.6 amps.
*********

Yes, I think we've calculated that one a few times, already.

Oh, crap. The VA rating. That's another one to consider.
Later, I guess.

- always one more thing... later is OK. Just start thinking about how
the amp might be used (domestic music, PA, emergency announcement,
musician's band etc)

I'm going to hold off on that until I get more grasped well.
For now, I will imagine in my head that the main thing here
will be to look at the volts*amps right around the point
nearby the peak recharging current each cycle, except that
such a transformer would then be spec'd for a worst case peak
and would likely be overly large for realistic needs. Or
something like that.

A dual 12.6 volt transformer would give a minimum (worst case with
transformer at full load and mains 7% under voltage) of 16.something
volts meaning big filter caps if you were serious at getting 10 watts.
One of the reasons to go PSU first I think. (Also I live in a tiny
jerk-water town where no one knows what a custom made transformer is let
alone where you can get one wound)

I suppose the old filament-type 12.6VAC transformers must be
common everywhere. I also see that Radio Shack (yes, I'm
holding my nose for a moment) still carries some "commodity"
type 25.2VAC CT 2.0A rated transformers for about US$10.
Their 12.6VAC CT 3.0A transformers are priced identically.

So what's considered to be generally available?

Job dependent. For a one off, whatever you can get. For high spec
commercial audio, whatever you can get custom made for the application.
For something laid out as a set of instructions what ever is widely
available - like 12.6v. In OZ we have a couple of hobbiest suppliers
that stock essentially the same range 6.3, 12.6, 15, 18, 20, 25, 30, 35..

Okay. 25.2V is very common here, so if they are similarly
common elsewhere perhaps it's better to leave the final
output specification as an output of the end of the design
process, rather than an input at the front end of it. I
don't actually _need_ 10 watts. I just figured that would be
a good place to target to learn enough things to be useful.
In terms of Radio Shack, I've no idea what the regulation
spec on their transformers would be. But I have to imagine
that it's about as cheaply made as they can get away with.

I tracked down a very nice transformer in my box which may be
okay. It has two secondaries and was intended for 60Hz use.
It weighs in at 2.8 lbs (1.25 kg.)

Primary:
115VAC, 5.0 Ohms, 16 gauge
Secondaries: (Tested using 120.5VAC RMS on primary)
16VAC RMS CT, 0.05 Ohms, 14 gauge
30.4VAC RMS CT, 2.6 ohms, 22 gauge

The 16VAC RMS outer winding across a 56 ohm resistor yields
15.88VAC RMS. (I don't have a large wattage resistor with
lower values of resistance, so that needs to suffice.) Half
of the 30.4VAC RMS winding (CT to one side) yields 14.75VAC
RMS loaded with the same 56 ohm resistor. Across the entire
30.4VAC windings it is 28.9VAC RMS. (The poor thing is just
a 5W, so I didn't measure for longer than a few seconds.)

The 30.4VAC secondary looks reasonable, I think, for the two
amplifier rails and ground. The 16VAC might make another
supply for some other reason or, perhaps, provide another
pair of rails to use for a 2 ohm speaker?

I hadn't thought about that aspect, but as you earlier
pointed out the 25.2VAC CT standard transformer might be a
little light for a 10W amplifier... unless I spec'd a 4 ohm
speaker, I suppose. Then it might be fine.

Anyway, it looks like it may be a reasonable choice as
something I have available and ready from the junk box.

Jon

 

[Jon Kirwan]

On Fri, 05 Feb 2010 16:46:22 -0800, I wrote:

snip
I tracked down a very nice transformer in my box which may be
okay. It has two secondaries and was intended for 60Hz use.
It weighs in at 2.8 lbs (1.25 kg.)

Primary:
115VAC, 5.0 Ohms, 16 gauge
Secondaries: (Tested using 120.5VAC RMS on primary)
16VAC RMS CT, 0.05 Ohms, 14 gauge
30.4VAC RMS CT, 2.6 ohms, 22 gauge

The 16VAC RMS outer winding across a 56 ohm resistor yields
15.88VAC RMS. (I don't have a large wattage resistor with
lower values of resistance, so that needs to suffice.) Half
of the 30.4VAC RMS winding (CT to one side) yields 14.75VAC
RMS loaded with the same 56 ohm resistor. Across the entire
30.4VAC windings it is 28.9VAC RMS. (The poor thing is just
a 5W, so I didn't measure for longer than a few seconds.)

The 30.4VAC secondary looks reasonable, I think, for the two
amplifier rails and ground. The 16VAC might make another
supply for some other reason or, perhaps, provide another
pair of rails to use for a 2 ohm speaker?

I hadn't thought about that aspect, but as you earlier
pointed out the 25.2VAC CT standard transformer might be a
little light for a 10W amplifier... unless I spec'd a 4 ohm
speaker, I suppose. Then it might be fine.

Anyway, it looks like it may be a reasonable choice as
something I have available and ready from the junk box.

Second thoughts. The 30.4VAC RMS CT secondary shows 2.6 ohms
and is 22 gauge. That's 1.3 ohms per half. I believe from
calculation that the peak diode current _might_ be 8-10 times
the load current in the ideal case (0 ohms.) Taking into
account the winding resistance, I may need to think more
closely about using this transformer in this application. The
winding resistance will limit the current and thus the energy
per unit time that can be transferred to the caps and that
will very likely lower the achievable rail voltage on the
other side of the bridge since the bridge itself simply won't
ever see the idealized peak voltage even right up to the
moment of peak where the dv/dt goes to zero. By the time
that happens, the cycle will already be on a decline again
while the resistance continues to limit inflow of charge.
Cripes.

Darn it. Back to monster caps to get a slight decent rail
voltage there.

Jon

 

[John Larkin]

On Fri, 05 Feb 2010 11:32:16 -0800, Jon Kirwan
<jonk@infinitefactors.org> wrote:

On Fri, 05 Feb 2010 12:36:02 GMT, N0Spam@daqarta.com (Bob
Masta) wrote:

On Thu, 04 Feb 2010 12:07:16 -0800, Jon Kirwan
jonk@infinitefactors.org> wrote:

There's another question that comes to mind regarding the
output stage. A lot of talk seems to revolve around
"crossover distortion." Seems almost very first thing folks
talk about when discussing class of operation if not also at
other times.

Seems to me that in a three-rail power supply situation
without an output capacitor involved, the crossover takes
place near the midpoint (ground) voltage between the rails,
at a time when current into the speaker load is also near
zero. (I'm neglecting any thoughts about inductance in the
speaker and physical coupling into the air, for now.) In
other words, where power at the speaker is near zero. Is it
really that important to consider?

I was looking at that terrible large scale gain plot for the
quasicomplementary output stage on the web site recently
mentioned in the thread (the lower curve in Figure 4 on this
link):

http://www.embedded.com/design/206801065?printable=true

(It's not that terrible of a plot, as the variation is from
.96 to .98 with the "normal" middle at .97.)

What's experience say here? Is it really so terrible as to
worry too much about something that takes place near zero
voltage, anyway? I'm just questioning the concern, for now.
I have no understanding about it, at all. Just wondering.

Crossover distortion is a more-or-less fixed
amplitude, so at low signal levels it becomes a
large percentage of THD. Our ears are sensitive
to the relative amplitudes of components, so a
hypothetical fixed amplitude distortion component
might be totally inaudible when it is a low
percentage of a large fundamental, yet be
obnoxious as a larger percentage of a small
fundamental. (This is the same essential problem
as quantization distortion in digital circuits...
you don't hear it on the peaks, only on the quiet
parts.)

Okay. This is consistent with another way I was looking at
this. Since the crossover takes place near where power is
also close to zero, it's effect is fairly constant. If the
amplifier's swing is large scale (nearer the limits it was
designed to provide), it's not likely to be noticed buried
within that. If the amplifier's swing were small (nearer
that zero volt area) then the crossover distortion is quite a
bit more noticeable.

Makes sense, I think.

Unless you have some application that doesn't
involve soft passages in the signal (like a siren,
or possibly a PA or musical instrument amp) you
need to consider the crossover distortion.

Best I consider it, then. I'm looking to learn, not
specialize in bullhorns.

Interesting to me that you say that crossover distortion
might not be such an issue for a musical instrument
amplifier, though. I take it you must mean for stage work
where the power is going to be set pretty high, generally?

However, there is still a lower limit to absolute
detection threshold, regardless of what percentage
a component might be. If the system levels
(program, amp, speakers, listening position) are
set up such some signal component is below 0 dB
SPL, most people aren't going to hear it even if
it is the entire signal!

Hehe.

Best regards,

Bob Masta

Thanks,
Jon

A few percent distortion at power levels is essentially inaudible.
Speakers do that already. Low-level crossover distortion is obvious
and obnoxious.

So: make sure the output stage transfer function is fast and
continuous, and apply a lot of feedback to straighten it out.

John

 

[Jon Kirwan]

On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 05 Feb 2010 11:32:16 -0800, Jon Kirwan
jonk@infinitefactors.org> wrote:

On Fri, 05 Feb 2010 12:36:02 GMT, N0Spam@daqarta.com (Bob
Masta) wrote:

snip

Crossover distortion is a more-or-less fixed
amplitude, so at low signal levels it becomes a
large percentage of THD. Our ears are sensitive
to the relative amplitudes of components, so a
hypothetical fixed amplitude distortion component
might be totally inaudible when it is a low
percentage of a large fundamental, yet be
obnoxious as a larger percentage of a small
fundamental. (This is the same essential problem
as quantization distortion in digital circuits...
you don't hear it on the peaks, only on the quiet
parts.)

Okay. This is consistent with another way I was looking at
this. Since the crossover takes place near where power is
also close to zero, it's effect is fairly constant. If the
amplifier's swing is large scale (nearer the limits it was
designed to provide), it's not likely to be noticed buried
within that. If the amplifier's swing were small (nearer
that zero volt area) then the crossover distortion is quite a
bit more noticeable.

Makes sense, I think.

Unless you have some application that doesn't
involve soft passages in the signal (like a siren,
or possibly a PA or musical instrument amp) you
need to consider the crossover distortion.

Best I consider it, then. I'm looking to learn, not
specialize in bullhorns.

Interesting to me that you say that crossover distortion
might not be such an issue for a musical instrument
amplifier, though. I take it you must mean for stage work
where the power is going to be set pretty high, generally?

However, there is still a lower limit to absolute
detection threshold, regardless of what percentage
a component might be. If the system levels
(program, amp, speakers, listening position) are
set up such some signal component is below 0 dB
SPL, most people aren't going to hear it even if
it is the entire signal!

Hehe.

Best regards,

Bob Masta

Thanks,
Jon

A few percent distortion at power levels is essentially inaudible.
Speakers do that already. Low-level crossover distortion is obvious
and obnoxious.

Yes, I think that's now much clearer to me now than it was
say two weeks ago -- without needing my ears to say so. Just
on understanding better _what_ crossover distortion is and
does.

So: make sure the output stage transfer function is fast and
continuous, and apply a lot of feedback to straighten it out.

Fast, I assume, to deal with high frequencies and still have
the needed gain to do it well; continuous in the sense that
the feedback isn't itself filled with potholes of some kind
or another; a lot because the more, the more linear you get.

Of course, that doesn't write equations for me or select
specific parts or topologies for me. But it's good so far as
it goes.

Jon

 

[pimpom]

Jon Kirwan wrote:

On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


A few percent distortion at power levels is essentially
inaudible.
Speakers do that already. Low-level crossover distortion is
obvious
and obnoxious.

Yes, I think that's now much clearer to me now than it was
say two weeks ago -- without needing my ears to say so. Just
on understanding better _what_ crossover distortion is and
does.

Jon, to see a graphical illustration of JL's point, see this

screenshot of some simple simulations I just ran:
http://img694.imageshack.us/img694/8967/crossoverdistortion.png

On the right, the complementary output stage is driven without
any bias, . The upper trace shows the output when the input
amplitude is +/-1V peak. The transistors are operating in Class C
and manage to conduct for less than half of each half cycle. Now
that's going to sound awful by any standard. I _know_ it sounds
awful because, when I was doing a lot of repairing work on
consumer products in the 80s, I came across some amps whose
biasing circuit had developed a fault.

Do a Fourier analysis and you'll get lots of harmonics. Reduce
the input amplitude even further and there won't be any output at
all below a certain amplitude.

The lower trace shows the output with +/-9V input. Crossover
distortion is much reduced, though still evident. This may or may
not be acceptable depending on the application. For anything that
needs good audio quality, any waveform distortion that can be
clearly seen in graphical form is still too high, especially in a
low-resolution bitmap trace like this.

On the left, we have the same amp with diode biasing added.
Visible distortion of the waveshape has disappeared. The slight
irregularities in the sinusoidal shape are due to limitations of
the low-res, non-antialiased bitmap image.

 

[Bob Masta]

On Fri, 05 Feb 2010 11:32:16 -0800, Jon Kirwan
<jonk@infinitefactors.org> wrote:

<snip>

Interesting to me that you say that crossover distortion
might not be such an issue for a musical instrument
amplifier, though. I take it you must mean for stage work
where the power is going to be set pretty high, generally?

Yes. There might be a *very* quite venue
somewhere, where (say) the final note of a song
might trail away into audible crossover
distortion, but I'm skeptical.

However, note that the previously-mentioned
quantization distortion was first discovered as a
problem on early CDs where a final piano note
decayed into silence. When it got very soft, it
also got very "chunky" since it only used a few
active bits, and it sounded gritty just before it
became inaudible. The fix was to add dither
(noise) when recording, which effectively gave
PWM that miraculously eliminated the distortion at
the cost of a small increase in noise.

(You can get a fairly dramatic demo of this using
Daqarta. See <www.daqarta.com/dw_0gbb.htm>. This
is done using just the Generator, so it is totally
free.)

Alas, there is nothing like dither to "fix"
crossover distortion!

Another point to keep in mind about musical
instrument amps is that whatever distortion they
may have becomes part of the "color" or timbre of
the sound. Distortion may be added via dedicated
circuits, or obtained by use of tube output
stages. (Tubes naturally have a more-pleasant
distortion with stronger even harmonics, compared
to the harsh odd harmonics of crossover
distortion.)

Also, since they handle only one instrument at a
time, there is less problem with intermodulation.
As long as all the distortion components are
harmonics, the sound is just more reedy, or
richer, or whatever. IM distortion produces sum
and difference tones that are usually not
harmonically related to any fundamental, so it
sounds really bad.

Best regards,




Bob Masta

DAQARTA v5.00
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, Sound Level Meter
Frequency Counter, FREE Signal Generator
Pitch Track, Pitch-to-MIDI
DaqMusic - FREE MUSIC, Forever!
(Some assembly required)
Science (and fun!) with your sound card!

 

[pimpom]

pimpom wrote:

Jon Kirwan wrote:
On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


A few percent distortion at power levels is essentially
inaudible.
Speakers do that already. Low-level crossover distortion is
obvious
and obnoxious.

Yes, I think that's now much clearer to me now than it was
say two weeks ago -- without needing my ears to say so. Just
on understanding better _what_ crossover distortion is and
does.

Jon, to see a graphical illustration of JL's point, see this
screenshot of some simple simulations I just ran:
http://img694.imageshack.us/img694/8967/crossoverdistortion.png

On the right, the complementary output stage is driven without
any bias, . The upper trace shows the output when the input
amplitude is +/-1V peak. The transistors are operating in Class
C
and manage to conduct for less than half of each half cycle.
Now
that's going to sound awful by any standard. I _know_ it sounds
awful because, when I was doing a lot of repairing work on
consumer products in the 80s, I came across some amps whose
biasing circuit had developed a fault.

Do a Fourier analysis and you'll get lots of harmonics. Reduce
the input amplitude even further and there won't be any output
at
all below a certain amplitude.

The lower trace shows the output with +/-9V input. Crossover
distortion is much reduced, though still evident. This may or
may
not be acceptable depending on the application. For anything
that
needs good audio quality, any waveform distortion that can be
clearly seen in graphical form is still too high, especially in
a
low-resolution bitmap trace like this.

On the left, we have the same amp with diode biasing added.
Visible distortion of the waveshape has disappeared. The slight
irregularities in the sinusoidal shape are due to limitations
of
the low-res, non-antialiased bitmap image.

Correction: I interchanged left and right parts of the image in
my description. Sorry.

 

[Jon Kirwan]

On Sun, 7 Feb 2010 16:10:14 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


A few percent distortion at power levels is essentially
inaudible.
Speakers do that already. Low-level crossover distortion is
obvious
and obnoxious.

Yes, I think that's now much clearer to me now than it was
say two weeks ago -- without needing my ears to say so. Just
on understanding better _what_ crossover distortion is and
does.

Jon, to see a graphical illustration of JL's point, see this
screenshot of some simple simulations I just ran:
http://img694.imageshack.us/img694/8967/crossoverdistortion.png

One of the really nice things about imagining mathematical
functions in my own head is that I actually _saw_ this,
already. Exactly as it shows, in fact. The way I approached
it in my head was to realize that the gain effectively goes
to near zero around the zero-volt output area with the
crossover issue in play. I also realize that while it may
not go to zero in other cases where the bias is more usefully
set up, it will droop a little just the same. So it's not
far to go from there to those graphs.

By the way, I'd _also_ imagined the idea of driving the
center of the diode pair, rather than jacking up and slugging
around one side or the other. And I liked seeing that
schematic idea being applied in your PNG.

On the right, the complementary output stage is driven without
any bias, . The upper trace shows the output when the input
amplitude is +/-1V peak. The transistors are operating in Class C
and manage to conduct for less than half of each half cycle. Now
that's going to sound awful by any standard. I _know_ it sounds
awful because, when I was doing a lot of repairing work on
consumer products in the 80s, I came across some amps whose
biasing circuit had developed a fault.

Just as a note, you only include the output stage, which is
effectively a bipolar emitter follower. So your input
amplitude of +-1V peak is after any input stage and VAS. But
I get the point. It is noticeable, especially when the VAS
output peak is approaching the 1V level. Which, if I'm
gathering things okay may very well be the case more
especially for low-wattage amplifiers where that is more
nearly always the case.

Do a Fourier analysis and you'll get lots of harmonics. Reduce
the input amplitude even further and there won't be any output at
all below a certain amplitude.

Yes, I can _see_ that.

The lower trace shows the output with +/-9V input. Crossover
distortion is much reduced, though still evident. This may or may
not be acceptable depending on the application. For anything that
needs good audio quality, any waveform distortion that can be
clearly seen in graphical form is still too high, especially in a
low-resolution bitmap trace like this.

I have to believe you've made this subject so crystal clear
that anyone reading this thread must now understand it, if
not before.

On the left, we have the same amp with diode biasing added.
Visible distortion of the waveshape has disappeared. The slight
irregularities in the sinusoidal shape are due to limitations of
the low-res, non-antialiased bitmap image.

Neatly illustrated with clarity. Not only that, but even the
modification to the schematics is done, parsimoniously, which
aids understanding. The bias resistors are added in the
latter case, because of course they are needed. But no more
is required to make the point. I really like that approach
to teaching -- keeping the basics in view and simplifying,
but not oversimplifying. Cleanly handled.

Jon

 

[Jon Kirwan]

On Sun, 7 Feb 2010 19:28:34 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Correction: I interchanged left and right parts of the image in
my description. Sorry.

I had no problem following your points.

Jon

 

[pimpom]

Jon Kirwan wrote:

On Sun, 7 Feb 2010 16:10:14 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


A few percent distortion at power levels is essentially
inaudible.
Speakers do that already. Low-level crossover distortion is
obvious
and obnoxious.

Yes, I think that's now much clearer to me now than it was
say two weeks ago -- without needing my ears to say so. Just
on understanding better _what_ crossover distortion is and
does.

Jon, to see a graphical illustration of JL's point, see this
screenshot of some simple simulations I just ran:
http://img694.imageshack.us/img694/8967/crossoverdistortion.png

One of the really nice things about imagining mathematical
functions in my own head is that I actually _saw_ this,
already. Exactly as it shows, in fact. The way I approached
it in my head was to realize that the gain effectively goes
to near zero around the zero-volt output area with the
crossover issue in play. I also realize that while it may
not go to zero in other cases where the bias is more usefully
set up, it will droop a little just the same. So it's not
far to go from there to those graphs.

In my late teens almost 40 years ago, during the only time I ever
worked under someone else, my boss in a research lab once asked
me why I kept doing calculations in my head when I could use a
calculator. I said that it helps keep my brain sharp and also
lets me visualise the outcome even before I arrive at the final
figures. I do also use pen and paper, and a calculator for
complex sequences. But it's only very recently that I've started
using a calculator (mostly Windows' scientific calculator) for
routine work.

That said, I think it's also a good idea not to rely too much on
mental imaging. We do sometimes make mistakes, and what we
visualise may not at all be what actually happens.

In a way, I kind of miss the old days when the average EE and
tech didn't have access to computer simulation. Back then, when I
designed a tube amp for instance, I drew a load line on the tube
characteristics curves and marked the outputs for as many grid
voltages as possible. Then I plotted the input-output curve on a
separate graph paper, chose an operating point and marked the
appropriate points for a Fourier analysis of harmonics expected
without feedback.

I was less diligent with transistors because I felt that the much
wider tolerances made such a procedure less worthwhile.

By the way, I'd _also_ imagined the idea of driving the
center of the diode pair, rather than jacking up and slugging
around one side or the other. And I liked seeing that
schematic idea being applied in your PNG.

That was for the purpose of illustration with a symmetrical
drive. It's less convenient with a practical design that includes
all the other associated circuits.

On the right, the complementary output stage is driven without
any bias, . The upper trace shows the output when the input
amplitude is +/-1V peak. The transistors are operating in
Class C
and manage to conduct for less than half of each half cycle.
Now
that's going to sound awful by any standard. I _know_ it
sounds
awful because, when I was doing a lot of repairing work on
consumer products in the 80s, I came across some amps whose
biasing circuit had developed a fault.

Just as a note, you only include the output stage, which is
effectively a bipolar emitter follower. So your input
amplitude of +-1V peak is after any input stage and VAS. But
I get the point. It is noticeable, especially when the VAS
output peak is approaching the 1V level. Which, if I'm
gathering things okay may very well be the case more
especially for low-wattage amplifiers where that is more
nearly always the case.

The output stage is an emitter-follower in most designs.

 

[Jon Kirwan]

On Mon, 8 Feb 2010 02:53:51 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Sun, 7 Feb 2010 16:10:14 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Sat, 06 Feb 2010 10:48:16 -0800, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


A few percent distortion at power levels is essentially
inaudible.
Speakers do that already. Low-level crossover distortion is
obvious
and obnoxious.

Yes, I think that's now much clearer to me now than it was
say two weeks ago -- without needing my ears to say so. Just
on understanding better _what_ crossover distortion is and
does.

Jon, to see a graphical illustration of JL's point, see this
screenshot of some simple simulations I just ran:
http://img694.imageshack.us/img694/8967/crossoverdistortion.png

One of the really nice things about imagining mathematical
functions in my own head is that I actually _saw_ this,
already. Exactly as it shows, in fact. The way I approached
it in my head was to realize that the gain effectively goes
to near zero around the zero-volt output area with the
crossover issue in play. I also realize that while it may
not go to zero in other cases where the bias is more usefully
set up, it will droop a little just the same. So it's not
far to go from there to those graphs.

In my late teens almost 40 years ago, during the only time I ever
worked under someone else,

You sound a LOT like me... age and work history... unless I'm
projecting too much. I'm 54 now and I've been self-employed
almost my entire adult life.

my boss in a research lab once asked
me why I kept doing calculations in my head when I could use a
calculator. I said that it helps keep my brain sharp and also
lets me visualise the outcome even before I arrive at the final
figures. I do also use pen and paper, and a calculator for
complex sequences. But it's only very recently that I've started
using a calculator (mostly Windows' scientific calculator) for
routine work.

I feel comfortable with what you just wrote.

That said, I think it's also a good idea not to rely too much on
mental imaging. We do sometimes make mistakes, and what we
visualise may not at all be what actually happens.

The way I think about this, in agreeing, is that in our minds
it is far, far too easy to over-simplify the world around us
to the point of serious distortion if not wild inaccuracy.
When I _think_ about "building a house" there are so many
things that do NOT enter my thinking, so of course it seems
easier than it actually is to get right. (I'm using that
example because I recently forced myself to do just that for
my son, doing __all__ of the work myself including digging
the foundation, testing soils, leveling the surrounding area,
designing and placing foundation forms, rebar work and
clamps, designing and verifying quality of cement mixes, and
so on, all the way to doing the overall design for 3' snow
loads and 80 MPH side wind loads on the broadest side, for a
gambrel roof, balloon-framed two story.) When you put hands
to reality, you find yourself faced with far, far more than
your brain was able to hold in one spot at one time long
enough to figure things.

I suppose all this relates to the widely told, apocryphal
story that we cannot hold more than 3-7 concepts in mind at
one time. We simplify to think. We hope we don't over
simplify, but there's a phrase, "reality impinges," which
captures what happens when we do over-simplify.

In a way, I kind of miss the old days when the average EE and
tech didn't have access to computer simulation. Back then, when I
designed a tube amp for instance, I drew a load line on the tube
characteristics curves and marked the outputs for as many grid
voltages as possible. Then I plotted the input-output curve on a
separate graph paper, chose an operating point and marked the
appropriate points for a Fourier analysis of harmonics expected
without feedback.

I've used load lines for self-imposed homework. So I think I
know some of what you mean. Of course, I haven't designed
any realistic amplifier systems. So it goes only so far with
me. However, I very much liked the visualization it affords.

I was less diligent with transistors because I felt that the much
wider tolerances made such a procedure less worthwhile.

I considered trying that with BJTs, too, after having applied
my "homework" for tubes. But none of my books really went
that direction and so I left the idea and tried to follow
what _was_ being taught. By the way, the thing that really
lost me with tubes (and I was studying them as a teenager,
not since) was computing the grid leak resistor. I never did
find a satisfactory explanation to me, because no reference I
had access to back then helped me gain a quantitative
estimate of the grid current so I was left without knowing
how to compute a resistor value. I haven't returned to that
to fix that error in my understanding. I still imagine today
that it was a hit-and-miss affair, starting with values in
the 200k range and tinkering around from there.

By the way, I'd _also_ imagined the idea of driving the
center of the diode pair, rather than jacking up and slugging
around one side or the other. And I liked seeing that
schematic idea being applied in your PNG.

That was for the purpose of illustration with a symmetrical
drive. It's less convenient with a practical design that includes
all the other associated circuits.

It was nicely handled for the purpose.

On the right, the complementary output stage is driven without
any bias, . The upper trace shows the output when the input
amplitude is +/-1V peak. The transistors are operating in
Class C
and manage to conduct for less than half of each half cycle.
Now
that's going to sound awful by any standard. I _know_ it
sounds
awful because, when I was doing a lot of repairing work on
consumer products in the 80s, I came across some amps whose
biasing circuit had developed a fault.

Just as a note, you only include the output stage, which is
effectively a bipolar emitter follower. So your input
amplitude of +-1V peak is after any input stage and VAS. But
I get the point. It is noticeable, especially when the VAS
output peak is approaching the 1V level. Which, if I'm
gathering things okay may very well be the case more
especially for low-wattage amplifiers where that is more
nearly always the case.

The output stage is an emitter-follower in most designs.

Yes. I also considered the idea of using current mirrors on
both rails at the output _before_ I started this thread, not
knowing what someone might suggest and just stretching my
imagination along different lines. But I figure on leaving
such thoughts to the very end.

....

You triggered something in writing, "In a way, I kind of miss
the old days when the average EE and tech didn't have access
to computer simulation." Me, too. In the case of computer
software, where I specialize, the advent of cheap debuggers
has led, I think, to becoming intellectually "flabby" there.

I have watched as professional programmers "hack" their way
to solving bad code design, placing in if..else conditions in
what almost seems random locations to fix up design errors
rather than going back and crafting an appropriate design.
Debuggers _enable_ sloppy thinking, by making it more
successful to apply.

Having calculators and computers ever at the ready for "brute
force" analysis, while important in well-educated hands
because these tools _will_ be knowledgebly applied then, in
the hands of those lacking deeper understandings they can
lead to a false sense of confidence and, penultimately,
disasterous results.

Also, imagination can be developed and trained and that does
NOT happen when you are asking some program to do your work.
Daily exercise is important, here.

However, you also counter your own suggestion shortly later
by saying, "That said, I think it's also a good idea not to
rely too much on mental imaging." On this point, I can add:

Sloppy thinking also means that even excellent tools are
misapplied, ignorant of what is going on. A case in point
relates to an area I care about, which is the computation of
slopes of exponential decays. The k in Ae^(-kt)+C equation.
When taking measurements, there is noise. Most folks just
"linearize" the data and use a bog standard least squares on
the result. However, while equal treatment of the noise on
individual data points in the linear domain is assumed by the
usual lsq fit algorithm, it turns out that equal treatment of
the noise in the log domain is inappropriate to the problem
at hand and the method needs to be modified in a very
specific way to properly account for noise after the log has
been applied to the raw data.

It takes a clear __imagination__, and good understanding of
mathematics and the boundary conditions for its many tools,
to recognize the need to investigate further and to develop
an appropriate algorithm. So one cannot discount that need,
either.

A failure to _imagine_ is also what takes place when you try
and apply the usual standard deviation calculation using the
floating point system. Often, the computer programmer does
become aware of the dual-summation variety of it, but is
completely _unaware_ of the fact such variety of calculating
standard deviations takes, in the end, the difference between
two large, and similar values, sometimes leaving only a very
few bits of precision in the result. A failure to _imagine_
here could mean a calculation that carries significant
inaccuracy and, if relied upon for some automated process,
could result in occasional and unpredictable failures. A
solution would be to presort, before computing the sums, so
that smaller values have a chance to accumulate into middle
sized values before being truncated in the sum.

To cap this off, I recently discovered a way to self
calibrate gain and offset by using a _feature_ of the
measurement system projected into a particular mathematical
manifold. Without mathematical imagination skills, I can
assure you that no practitioner would have the chance of a
snowball in hades of uncovering the technique through
obseration and circuit experiments.

The point here is that imagination, while often unable to
fully account all the details of complex processes, also is
vitally important to use and exercise and not ignore. Just
like neither theory nor observation can ignore each other, I
suppose.

Saying all this reminds me of some lost arts, for which I'm
sure you can add piles to my own. I'll provide one beauty I
love, that has been "lost" to programmers and remains alive
in this one brain and, it seems, few others' -- conversion of
octal to decimal and back again. I wrote about this in
..basics back in 2003. Might be good to unearth it.

CONVERSION OF DECIMAL TO OCTAL

(0) Prefix the number with "0." Be sure to include
the radix point. It's an important marker.

(1) Double the value to the left side of the radix,
using octal rules, move the radix point one digit
rightward, and then place this doubled value
underneath the current value so that the radix
points align.

(2) If the moved radix point crosses over a digit
that is 8 or 9, convert it to 0 or 1 and add
the carry to the next leftward digit of the
current value.

(3) Add octally those digits to the left of the radix
and simply drop down those digits to the right,
without modification.

(4) If digits remain to the right of the radix, goto 1.

CONVERSION OF OCTAL TO DECIMAL

(0) Prefix the number with "0." Be sure to include
the radix point. It's an important marker.

(1) Double the value to the left side of the radix,
using decimal rules, move the radix point one digit
rightward, and then place this doubled value
underneath the current value so that the radix
points align.

(2) Subtract decimally those digits to the left of
the radix and simply drop down those digits to
the right, without modification.

(3) If digits remain to the right of the radix, goto 1.

For example,

0.4 9 1 8 decimal value
+0
---------
4.9 1 8
+1 0
--------
6 1.1 8
+1 4 2
--------
7 5 3.8
+1 7 2 6
--------
1 1 4 6 6. octal value

Let's convert it back:

0.1 1 4 6 6 octal value
-0
-----------
1.1 4 6 6
- 2
----------
9.4 6 6
- 1 8
----------
7 6.6 6
- 1 5 2
----------
6 1 4.6
- 1 2 2 8
----------
4 9 1 8. decimal value

There's also a very interesting way to visualize the _shape_
of the dynamics of systems that leads to better understanding
that I've never seen or read about anywhere, but which I've
uncovered through practice and trying to teach to high school
students without the use of calculus. When I showed the idea
to a physics teacher at the local high school, and started to
erase the chalk board, he stopped me. He said he needed to
keep it there and think about it more. So I left it.

I'd be happy to discuss the concept. But the point isn't
that discussion, but that imagination _is_ important. It's
just not the _only_ important thing. We need to develop it
continually and that is part of why I like to use things I
find an interest in (amplifier design, for example) as an
excuse to _also_ visualize and exercise thoughts. I'm not
merely trying to build an amplifier some day, though I want
to do that, too. I'm staying mentally in shape, while also
learning something, and at some point trying my hand building
it. It's all of a piece, I guess.

And I cannot express just how wonderful it is to have folks
to talk with like this. I _am_ learning to think better and
better also about practical details, too. But the discussion
is also just plain pleasant. As a hobbyist, it's very hard
to find it with local neighbors, you know? Community college
may immerse me and allow me some of this kind of talk, but it
also requires a regular pace that one cannot always afford
while taking care of a family, too.

Jon

 

[Jon Kirwan]

On Fri, 05 Feb 2010 17:06:54 -0800, I wrote:

On Fri, 05 Feb 2010 16:46:22 -0800, I wrote:

snip
I tracked down a very nice transformer in my box which may be
okay. It has two secondaries and was intended for 60Hz use.
It weighs in at 2.8 lbs (1.25 kg.)

Primary:
115VAC, 5.0 Ohms, 16 gauge
Secondaries: (Tested using 120.5VAC RMS on primary)
16VAC RMS CT, 0.05 Ohms, 14 gauge
30.4VAC RMS CT, 2.6 ohms, 22 gauge

The 16VAC RMS outer winding across a 56 ohm resistor yields
15.88VAC RMS. (I don't have a large wattage resistor with
lower values of resistance, so that needs to suffice.) Half
of the 30.4VAC RMS winding (CT to one side) yields 14.75VAC
RMS loaded with the same 56 ohm resistor. Across the entire
30.4VAC windings it is 28.9VAC RMS. (The poor thing is just
a 5W, so I didn't measure for longer than a few seconds.)

The 30.4VAC secondary looks reasonable, I think, for the two
amplifier rails and ground. The 16VAC might make another
supply for some other reason or, perhaps, provide another
pair of rails to use for a 2 ohm speaker?

I hadn't thought about that aspect, but as you earlier
pointed out the 25.2VAC CT standard transformer might be a
little light for a 10W amplifier... unless I spec'd a 4 ohm
speaker, I suppose. Then it might be fine.

Anyway, it looks like it may be a reasonable choice as
something I have available and ready from the junk box.

Second thoughts. The 30.4VAC RMS CT secondary shows 2.6 ohms
and is 22 gauge. That's 1.3 ohms per half. I believe from
calculation that the peak diode current _might_ be 8-10 times
the load current in the ideal case (0 ohms.) Taking into
account the winding resistance, I may need to think more
closely about using this transformer in this application. The
winding resistance will limit the current and thus the energy
per unit time that can be transferred to the caps and that
will very likely lower the achievable rail voltage on the
other side of the bridge since the bridge itself simply won't
ever see the idealized peak voltage even right up to the
moment of peak where the dv/dt goes to zero. By the time
that happens, the cycle will already be on a decline again
while the resistance continues to limit inflow of charge.
Cripes.

Darn it. Back to monster caps to get a slight decent rail
voltage there.

Jon

An addition or two. The transformer mentioned above uses 18
gauge, not 16 gauge, for the primary. And since I'm still
wrestling with why there are 5 ohms, measured, I think it's
likely that is merely the wiring to the outside and that
perhaps even smaller diameter wire was used to wrap around
the core. So externally labeled gauge probably is NOT a
precise indication of what was used in the core.

I took a look at the weight of some similarly shaped 60Hz
power transformers, available from Stancor. It seems that
similar weights are on the order of 80 VA. (I've seen a few
rated 100 VA, but I'm betting less.)

For now, I'll assume that transformer is too small, not
because of my guessed-at VA rating but because of the
measured resistance in the 30.4VAC secondary.

I want to get back to the power supply design and finish
that.

I understant that at 10W, the 8 ohm speaker will experience
sqrt(2*10W*8 ohms) or 12.65Vrms and 1.58Arms. I could always
modify this to fit what I have available but rather than do
that, I think it's better to "stay on target" and see where
that goes.

So that's what is expected "at the speaker." The peak figure
required will be rounded up to 18V. To reach that peak, the
output BJTs will need some headroom of their own. Staying
out of saturation and assuming the output stage might use two
BJTs on either side, requiring perhaps two diode drops if
either of these quadrants uses both NPN or both PNP, I would
best figure another 4V of headroom on each side. So 22V
minimum there, __under load.__

And I begin to see why 25V isn't a bad target.

Which brings in the question about a linear regulator. It's
my vague feeling that there is NO need for one. I should be
able to arrange the circuity (current sources, etc.) so that
they are sufficiently immune to modest ripple that the 60Hz
(and other components due to loading causing cap voltage
changes, as well) can be rejected well enough. Besides, a
linear regulator would mean just that much more headroom and
wasted power/heat. So unless something very difficult is
shown to me, I'd like to take the position that a linear
regulator is a lot extra trouble without worthwhile payback.
(And dealing with the added poles/zeros would seem to make my
worries compounded, as if the rest weren't enough.)

The filter capacitors will probably have to be spec'd at 50V
given what I've read here. It seems 35V wouldn't be entirely
safe, given the comments about regulation at 15% and another
7% margin, as well. And something else that is bothering me.
Charging only takes place for short bursts and happen
_before_ the windings reach peak voltage. So there is a
small duty cycle during which usable energy is transferred.
Does this suggest that one might _under specify_ the VA
rating for the transformer to save cost and weight and get
away with it?

Jon

 

[Bob Masta]

On Mon, 8 Feb 2010 02:53:51 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

In my late teens almost 40 years ago, during the only time I ever
worked under someone else, my boss in a research lab once asked
me why I kept doing calculations in my head when I could use a
calculator. I said that it helps keep my brain sharp and also
lets me visualise the outcome even before I arrive at the final
figures. I do also use pen and paper, and a calculator for
complex sequences. But it's only very recently that I've started
using a calculator (mostly Windows' scientific calculator) for
routine work.

That said, I think it's also a good idea not to rely too much on
mental imaging. We do sometimes make mistakes, and what we
visualise may not at all be what actually happens.

In a way, I kind of miss the old days when the average EE and
tech didn't have access to computer simulation. Back then, when I
designed a tube amp for instance, I drew a load line on the tube
characteristics curves and marked the outputs for as many grid
voltages as possible. Then I plotted the input-output curve on a
separate graph paper, chose an operating point and marked the
appropriate points for a Fourier analysis of harmonics expected
without feedback.

One thing we had "back in the day" was slide
rules. For any whippersnappers out there who've
never used one, you only got answers to 2 or 3
decimal places, and you had to keep track of the
exponent yourself... you'd read 1.23 and have to
know whether that was 123000 or 0.000123 or
whatever.

That "shortcoming" of slide rules meant you had to
have a feel for the answer you were expecting,
with the slide rule just giving you more precision
on your guesstimate. I think that forced you to
stop and think a bit instead of blindly plugging
in numbers, and it meant that you never put blind
faith in your result, so you looked for other
validation, considered the problem from multiple
angles, etc. Not a bad thing.

Best regards,









Bob Masta

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[Jon Kirwan]

On Tue, 2 Feb 2010 22:30:31 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Sat, 30 Jan 2010 01:11:28 +0530, "pimpom" wrote:

snip
I've seen this as a modification. In ASCII form:

A
|
,---+---,
| |
| \
| / R3
\ \
/ R2 /
\ |
/ +--- C
| |
| |
| |/c Q1
+-----|
| |>e
\ |
/ R1 |
\ |
/ |
| |
'---+---'
|
B

We've already decided that R1 might be both a simple resistor
plus a variable pot to allow adjustment. The usual case I
see on the web does NOT include R3, though. However, I've
seen a few examples where R3 (small-valued) exists and one of
the two output BJTs' base is connected at C and not at A.

The above circuit is a somewhat different version of the Vbe
multiplier/rubber diode thing. The difference being R3,
which I'm still grappling with.

I've seen R3 used in that position too, but never gave it much
thought until you brought it up. Offhand I still can't see a
reason for it either. Maybe for stability against a local
oscillation? Perhaps taking some time to think about it will
bring some revelation. Or someone else can save us the trouble
and enlighten us.
snip

I had earlier said I thought you might be right about this R3
value. Now, I don't. I think it deals with something else
-- unregulated rail voltage variations.

In this thread, you've posted circuits with a resistor on one
side and the VAS on the other of this structure. The VAS
yanks one end around while the other side mostly follows it
around. However, the current through the resistor varies, of
course. Even if one of those BJTs+2 diodes thingies is used
for a current source instead of the resistor, which does
improve things, it still isn't very constant. Using the 2
BJT structure doesn't change that fact, though it does impact
variations. No matter how you arrange it, resistor or
current source, the fact is that the current into the Vbe
multiplier device changes around as the VAS yanks around one
side of it.

This variation means that Q1's Ic varies. To accomodate that
variation, Vbe varies. Since Vbe varies, so does the
multiplied value. And for no _other_ reason than variations
in signal. That changes the bias. Changing the bias changes
the quiescent current. Etc.

(Also, I suppose, the Early effect will add yet another
slight modification, since the Vce is slightly changed so is
the Ic for the same Vbe. The higher required Ic (assuming
the current source or resistor is supplying more current,
instead of less) requires a higher Vbe, as stated. So the
multiplied voltage at Vce is higher. But that multipled
voltage also slides over on the Vce axis for whatever Vbe
that has become and that suggests still more Ic due to Early
effect, so it is a positive feedback contributing to the
already existing problem, I think. I haven't tried to work
out just what percent it contributes, though.)

So a cludge fix for this is to insert a resistor in the
collector, which will act in the opposite direction to some
degreee. I'd imagine this would create a second degree poly
curve, with a maximum somewhere but gentle 'arms' outward,
which means less variation of the Vbe-multiplied value with a
tweakable peak point based upon a nominal Ic.

I imagine this is NOT nearly as important for class-A
operation, though, since it is already "biased up" and
variation at that point of operation probably isn't so
important. It _would_ matter, I think, in other classes of
operation.

Thinking as I am that I don't want to go with class-A, I am
trying to think of still better ways of replacing R3 with (or
adding) an active device to further improve it. Anyone have
a suggestion there?

Jon

 

[Jon Kirwan]

On Mon, 08 Feb 2010 11:54:42 -0800, I wrote:

I think it deals with something else
-- unregulated rail voltage variations.

__and__ variations due to the VAS driving the output stage.

Jon