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Discussing audio amplifier design -- BJT, discrete

J

Jon Kirwan

Jan 1, 1970
0
At one time I had two reels of 5000 each of 1N4004 that I got surplus, but
I sold most of them. I also have a bag of about 1000 pieces of 1N4003. So I
have pretty much a lifetime supply. Either one is OK for 120 VAC mains and
perfect for lower voltage applications. But now my new designs are mostly
SMT. I was going to keep the thru holes and use the "free" parts I had, but
I figured that the labor cost of inserting, soldering, and clipping leads
on 6 diodes on 40 boards might be more than the $0.06 each for the S1G SMT
diodes. Once a commitment is made to SMT it is usually cost-effective to
use as many such parts as possible. I never fully analyzed it, though. I
figure about 2 minutes for the six diodes. At $60/hr, or $1/minute, I spend
$2/board for the leaded parts. The SMT assembly is probably $0.05 per part,
so I spend a total of $0.66 per board.


Some time ago I came up with a rule of thumb of 1000 uF per amp, and I
revised that to 2000 uF per amp. I used an RC time constant of 8 mSec
between peaks for a 37% discharge from peak which holds the approximate RMS
value, and for a typical 8 VDC power supply at 1 amp R=8 ohms. So C =
.008/8 = 1000 uF. But two time constants gives only 13% discharge so 2000
uF is much better. For a 16 VDC supply, 1000 uF is OK, and as the voltage
doubles the required capacitance is halved. So for most low voltage
applications, 1000 to 2000 uF per amp is reasonable, and easy to remember.
Okay.

Of course, if you enjoy mathematical analysis, you can spend time working
out effects of winding resistance and capacitor ESR and acceptable ripple.

I suppose what I don't enjoy is taking on faith "rules" or
"prepared charts" I'm handed. So I do the math once or
twice, just to verify and make sure I have a small sense of
understanding about the whys and wherefores. (And I enjoy
the math practice, from time to time.)
Or you can just use LTSpice.

Once I feel I grasp the theory I will use LTspice a lot and
not give it that much thought later on. But isn't it better
to make sure, at least once
But if I need a quick and dirty junkbox power
supply, 1000 uF/amp is good enough to grab and go.

And I think I understand the details why. (Unless someone
expresses an interest, I won't dump it out here.)
For example, using LTSpice, I find a 12.6 V transformer and I want to make
a 12 VDC power supply at 1 amp. Using a 1000 uF capacitor and a 12 ohm
load, my output is 13.3 V which has a peak of 16.1 V and drops to 10.4 V,
which is a 35% drop as predicted. With 2000 uF it drops to 12.6 VDC so my
output is high enough to provide the 12 VDC I wanted with a regulator. Of
course there are line variations and transformer regulation, but not bad
for a quick estimate.

As I just wrote, once I've done it in the "forward direction"
and feel I understand the details well enough, once or twice,
just selecting rules to follow after that make sense. If
something doesn't feel right in the simulation, you can
always return to the fundamentals on paper to double-check.

But I don't like using a tool in a fashion where I have no
clue whatsoever how to check the work on my own, should I
decide to do so. Doesn't feel right. (You are past that
point, of course, so no problem there.)
If I wanted 24 VDC, and I had a 25.2 V transformer, a 1000 uF capacitor
gives me a minimum of 26 VDC for a regulator with a little bit of headroom.

Now I actually add a simple emitter follower voltage regulator with a
2N3055 and two 12 V zeners and a diode in series, with 220 ohms and a 100
uF cap. I get an output of 24.18 VDC which varies from 23.99 VDC to 24.30
VDC. Adding the regulator improves the minimum voltage excursion on the
1000 uF main filter capacitor to 27.6 VDC.

Since I was originally designing for just such a regulated power supply, my
"grab-and-go" estimates for main filter capacitors seems to work out quite
well. And I found it more fun to build and test the circuit using LTSpice
rather than with math. Filter capacitors of this size are typically -20% /
+80% tolerance, so chances are the results will be even better than
expected.

Thanks. And I got it.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
I can't speak for pimpom but I don't think any offense has been taken.

I wasn't sure and I knew I didn't want any mistake there. It
doesn't hurt to clarify, just in case.
There are many ways to approach any problem and sometimes "quick and dirty"
is appropriate while other times a careful mathematical approach
considering all factors is required. There are some areas of mathematics
where my eyes glaze over and it becomes gobble-de-gook, while I can design
a circuit in my head and visualize currents and voltages and waveforms
which can then be verified and improved by using a tool such as LTSpice.
Previous to that I would rely on actual breadboard circuits and using test
equipment (with a good understanding of its limitations) to see how it
performs.

I study some new piece of mathematics as often as I'm able.
It's the language of science, so to speak. And it is very
difficult to "read" a science paper with good understanding
without it. I'm slogging through a book on atmospheric and
oceanic fluid dynamics -- can't read a single decent science
paper on the subject without getting some understanding of
the basics, which I'm finding 'hard.' But I'm slogging
through it. Only way to get to the other end.

...............
Also, I think this thread has about run its course, and it may be time to
start a new one.

I'm okay with that.
It has now morphed into power supply design (as it applies
to audio amps), and it seems to be more suited to sci.electronics.design.

I think I'm closing in on the end of that section, now. I
think I know mostly what parts to use, and why to use them. I
suppose there is always another consideration to take, that I
might have missed. But fuse placement, transorb use, and a
few other details aren't that hard.

What I need to think about a little more is organizing the
approach towards the end-point. I need to explore several
different types of output stages, for learning's purpose. Not
because I'll need all of them in the end. Just to make sure
I've got the salient details understood. Then, I make a
decision there. Finally, I need to figure out how I'm going
to include a microcontroller and external volume control
widget that meets my _real_ needs and move forward from
there.

I'm planning a year for getting there. No rush.

A project after that is re-designing the front panel of a
microwave oven. But that is yet another story.
You may consider yourself a beginner

I do.
but your theoretical knowledge and
mathematical analysis is beyond the range of basics.

It's just like being good at typing fast. Useless, if you
don't have any idea what you want to write about. Great, if
you do.

So I can type? I need to get to the point where I have some
story worth telling.
It seems that your lack of direct experience and practical
"knack" will soon pass as you build and test a hands-on circuit.

I can hope. I'm the kind that likes to "measure twice, cut
once," though.

When I designed my son's house, it was my _first_ ever
attempt. I had never studied a single architecture book,
never computed beam loading, etc. That was last spring.
Since then, I've discovered that there is ONLY ONE really
good book targeted squarely at people like me -- those
without prior, formal architecture training. (If interested,
I'd be glad to talk about that book.) I then did perform
those calculations and designed a gambrel roof I wanted for
the balloon framed structure and ran over to the planning
departments for an ear-full. I took into account the
required 80MPH side wind loads on the broadest side, with a
3' snow load at the same time, as well. And the rest. I
also read large sections of the NEC (adopted with slight
modifications by Oregon) and did my own Ufer ground for the
home (and a grounding well) and even had an interesting
discussion with the county electrician about what I
considered to be the stupidest part of the NEC -- the
allowance "as code" of a 20' #4 bare copper wire in the same
wet sponge cement as iron rebar. I chose NOT to do that. But
the funny part of the discussion was that the county official
broke protocol and started asking _me_ to help him understand
some problems he encountered!!! I was able to, by the way!

The result is fantastic. And worth the effort. And I
learned a lot, too.

But in no way did I want to build that darned thing twice!!
You better believe it! So I made VERY sure of every step and
asked for advice, even when people felt I should not be doing
this work myself. The gambrel roof is such a case, because
code required it to be signed off by a licensed professional.
Got that done, of course. But I did the work. All of it.
My main criticism would be that you tend to limit yourself too much by
using scavenged parts and freebies in a junkbox.

Well, this is part of a learning experience, right now. If
and when I decide to finalize and generate the final unit for
my daughter, in a year or so, I will use new parts and make
it more professional-looking. Right now, I'm not there.
I tend to do that myself,
and often wind up with an inferior design or one that acts abnormally
because perhaps a part is damaged or is not really the best choice given
the wide range of new devices available. And, unless your budget is
severely crimped, you can order new parts with guaranteed specs that will
result in a more predictable and satisfactory outcome, and if it is a
worthwhile design, others may use the same parts and benefit from your
work.

Hehe. Unless I get some of those "fake" parts that I saw
pimpom also discussing in a different thread! ;)

But of course I also take your point, too.
I have an old power supply right here that I built when I was in high
school and I've been itching to rebuild it to be more useful. But it has a
pair of 2N1540 transistors and a pair of 2N554 and two 450 uF 50 V metal
can capacitors and an RT-204 "Selenium Rectifier Type"

(I extracted some selenium rectifiers from an old WW II radar
set. VR-150 tubes, as well. Selsyn motors. Lots of very
interesting stuff. I very much know what they look like!)
transformer and a
1N2976B stud mount 12V zener, and the meters are 0-10 VDC and 0-3 Amps. I
no longer have the schematic and what I've been able to trace does not seem
to make much sense to me now. It is nicely packaged in a Bud Portacab but
I'd really like to have at least 0-15 VDC and more like 5 amps and better
regulation and current limiting rather than the crude 3 amp fuse it has
now. So should I use these old obsolete parts (those are Germanium
transistors!), and make compromises to get it working again or should I
design from scratch and make it do what I really want or do I just put it
back in the junk pile and buy what I'd like for a hundred bucks or so? If I
could just get it working OK in a few hours I could live with the limited
output, and maybe I could add a x2 switch so I can get 0-20V with the same
meter, or I could make a new scale and change the internal resistor, or...
so I wind up with one or two days work and I talk myself out of it again...

Hehe. Well, I think I explained my perspective for now. I'm
just learning, at this stage, so the random parts are fine.
Same with the amplifier. I think I can learn _about_
amplifiers plenty good enough even if I'm using poor quality
parts. In fact, I might learn some diagnostic tricks along
the way, using them. And maybe get lost a few times, too.
But what the heck? That's a good way to learn, too.

Jon
 
P

pimpom

Jan 1, 1970
0
Jon said:
Don't mistake me. All I meant to say is that I _am_ new and
therefore took a slower approach, not having developed the
well worn ruts from good experience as you have done. And
that I enjoyed seeing your way of cutting through it.


I think I clearly understood exactly that from your writing.


I didn't expect this and it all looks as though I may have
unintentionally implied something. If so, I hope you will
re-read what I wrote and understand that I'm merely
commenting upon my own painstaking processes, which are at
this point in time important steps for me to take, and in no
way commenting about anything you are saying (except perhaps
that I agree and otherwise like the way you thought about
it.) That's all there was there.

Thanks very much again,
Jon

If I seemed to be snapping at you, I apologise. I'd be less than
honest if I didn't admit that I was just a little bit irritated
at the time, but you weren't really the cause. I'd had a
frustrating day - no, make that a frustrating 2 weeks plus - from
being given the runaround by a component manufacturer. That and
the fact that I'm not a native user of English may have made me
sound more abrupt than I meant to be. Again, I apologise.

You and I are very similar in that I also like to dissect all the
little bits and pieces of any new ground I'm venturing into. But
I suspect that your keen desire to analyse everything in minute
detail, and the fact that you're probably better at that than a
lot of people here with more experience in applied electronics,
has put off more than one of the regulars in this NG.
 
P

pimpom

Jan 1, 1970
0
Paul said:
Some time ago I came up with a rule of thumb of 1000 uF per
amp, and I
revised that to 2000 uF per amp. I used an RC time constant of
8 mSec
between peaks for a 37% discharge from peak which holds the
approximate RMS value, and for a typical 8 VDC power supply at
1 amp
R=8 ohms. So C = .008/8 = 1000 uF. But two time constants gives
only
13% discharge so 2000 uF is much better. For a 16 VDC supply,
1000 uF
is OK, and as the voltage doubles the required capacitance is
halved.
So for most low voltage applications, 1000 to 2000 uF per amp
is
reasonable, and easy to remember.

I have an even simpler rule of thumb for audio amps using the
standard series push-pull configuration. The ratio of rated
speaker impedance to the equivalent load resistance at full power
as seen by the power supply stays approximately the same no
matter what level of power and supply voltage. So, that rule of
thumb is 1000uF rail-to-rail for an amp with an 8-ohm load,
medium quality. 500uF where quality is not an important desing
criterion, and 2000uF where it is.

Jon may be interested in how those figures came about, but I
think he would rather deduce them himself instead of having them
handed to him on a platter.

Hint: Calculate the voltage and current swings needed for a given
power output. Add a few volts for transistor saturation and drops
across emitter resistors. Derive the average DC current from the
peak swing. That gives the equivalent load resistance as seen by
the power supply, and that resistance remains approximately the
same for a given speaker impedance, for a wide range of designed
power level.

In fact, if it weren't for the effects of transistor imperfection
and the need for emitter resistors and other compensation
techniques, the ratio of equivalent load resistance to speaker
impedance would be exactly the same for any power level.
 
J

Jon Kirwan

Jan 1, 1970
0
If I seemed to be snapping at you, I apologise. I'd be less than
honest if I didn't admit that I was just a little bit irritated
at the time, but you weren't really the cause. I'd had a
frustrating day - no, make that a frustrating 2 weeks plus - from
being given the runaround by a component manufacturer. That and
the fact that I'm not a native user of English may have made me
sound more abrupt than I meant to be. Again, I apologise.

No problem. I worried a little and just wanted to make sure
that I hadn't come across wrong. I had only meant that I was
merely taking extra pains myself and that I liked the way
you'd written, too. And I reserved out the possibility that
there were language/concept issues.

However, I had no clear idea either way, if English was your
native tongue. If my wife had asked, I'd have only said "I
don't know." You are just that good at it. Makes me wish I
weren't so provincial, myself.
You and I are very similar in that I also like to dissect all the
little bits and pieces of any new ground I'm venturing into.

I appreciate that.
But
I suspect that your keen desire to analyse everything in minute
detail,

I like to at least try my hand, at times. It's hard for me
to gain some creative insight without that work.
and the fact that you're probably better at that than a
lot of people here with more experience in applied electronics,
has put off more than one of the regulars in this NG.

I cannot say much there. They have to speak for themselves,
I suppose.

Jon
 
P

Paul E. Schoen

Jan 1, 1970
0
pimpom said:
I have an even simpler rule of thumb for audio amps using the standard
series push-pull configuration. The ratio of rated speaker impedance to
the equivalent load resistance at full power as seen by the power supply
stays approximately the same no matter what level of power and supply
voltage. So, that rule of thumb is 1000uF rail-to-rail for an amp with an
8-ohm load, medium quality. 500uF where quality is not an important
desing criterion, and 2000uF where it is.

Jon may be interested in how those figures came about, but I think he
would rather deduce them himself instead of having them handed to him on
a platter.

Hint: Calculate the voltage and current swings needed for a given power
output. Add a few volts for transistor saturation and drops across
emitter resistors. Derive the average DC current from the peak swing.
That gives the equivalent load resistance as seen by the power supply,
and that resistance remains approximately the same for a given speaker
impedance, for a wide range of designed power level.

In fact, if it weren't for the effects of transistor imperfection and the
need for emitter resistors and other compensation techniques, the ratio
of equivalent load resistance to speaker impedance would be exactly the
same for any power level.

Without going into my quick calculations, I came up with an impedance of
10.6 ohms for each rail, or about 21 ohms total. I think a perfect
amplifier would have an impedance that matched the 8 ohm output, so
considering possibly 40% efficiency that sounds about right. But for your
50:1 R/Xc ratio it seems like you need about 2500 uF.

I would expect the Xc/R ratio to be roughly the amount of ripple, and 2%
seems unnecessary. 1000 uF gives a ratio of 6.3%, which is still a
reasonable level of ripple. But then one must consider RMS versus P-P which
is a 3:1 ratio, and perhaps that is where the 2% comes from.

If the ratio of amplifier impedance to speaker impedance is roughly
constant, then it seems that one may only need to consider the ratio of Xc
to speaker impedance.

And these approximations are based on acceptable power line ripple, whereas
the amplifier needs to be able to draw on the power supply at audio
frequencies to 20 Hz, which may imply another factor of three for the
capacitance needed.

My rule-of-thumb values are based more on the requirements of a general
purpose power supply, and I was also assuming some sort of active
regulation. Still, the values come out within a couple multiples of each
other, so that shows validity of either method.

Paul
 
P

pimpom

Jan 1, 1970
0
Paul said:
Without going into my quick calculations, I came up with an
impedance
of 10.6 ohms for each rail, or about 21 ohms total. I think a
perfect
amplifier would have an impedance that matched the 8 ohm
output, so
considering possibly 40% efficiency that sounds about right.
But for
your 50:1 R/Xc ratio it seems like you need about 2500 uF.


This is how I go about it: Take the case of Jon's amp as an
example. 10W into 8 ohms is 12.65V, 1.581A (both sinusoidal
peak). That's 0.5A dc average. To get 12.65V swing, we'll need
about +/-16V Vcc. 16V/0.5A = 32 ohms for each rail, 64 ohms
total. 1000uF is 1.6 ohms at 100Hz, 1.326 at 120Hz. This gives
R/Xc of 40 and 48 at 100Hz and 120Hz respectively. (I'm sure you
knew that the 50:1 was a round figure). To get 1000uF rail to
rail, we'd need 2000uF for each rail, or 2200uF in practice.
I would expect the Xc/R ratio to be roughly the amount of
ripple, and
2% seems unnecessary. 1000 uF gives a ratio of 6.3%, which is
still a
reasonable level of ripple. But then one must consider RMS
versus P-P
which is a 3:1 ratio, and perhaps that is where the 2% comes
from.

I also assume reasonable values of equivalent source resistances,
contributed mainly by transformer losses. This equivalent source
resistance has a big influence on ripple and power supply current
and voltage ratios. These days I rely mostly on charts for those
ratios.
If the ratio of amplifier impedance to speaker impedance is
roughly
constant, then it seems that one may only need to consider the
ratio
of Xc to speaker impedance.

That's true, and is really just another way of looking at the
rule of thumb of 1000uF for an amp with an 8-ohm speaker.
And these approximations are based on acceptable power line
ripple,
whereas the amplifier needs to be able to draw on the power
supply at
audio frequencies to 20 Hz, which may imply another factor of
three
for the capacitance needed.

Which is why I said earlier that I double my rule-of-thumb value
to 2000uF per 1/8ohms for a good quality amp, say 4700uF on each
supply rail. And 10,000uF where cost and size are not important
considerations. And double those values yet again for a 4-ohm
load.
 
P

pimpom

Jan 1, 1970
0
Jon said:
Hehe. What bothers me about having "simpler rules" is that
in the end it requires me to keep so many rules in mind.
Luckily, for a hobbyist type there aren't that many output
configurations.

I've been in electronics for 4 decades, but I still can't decide
whether I'm a hobbyist or a profressional! :) I earn a living
with it, but I also approach my work as if it was a hobby. I
waste a lot of time and earn only a fraction of what I could. I
do a lot of things for others for free or for a nominal fee, but
I also have much more fun.
(I've read about perhaps 20 sufficiently
different types that professionals might consider... but that
is happily not my problem.)

IMO, not even a budding pro should try to master everything at
once. I suggest you concentrate on the configuration presented in
its basic form in Wikipedia. What now look like very different
types may well turn out to be variants of a very few basic types.
This kind of writing confuses me a little. You start out
saying "at full power" then conclude "no matter what level of
power." But if I read this to mean that if the computation
is taken for the full power situation, that it will hold
regardless of the volume control setting or supply rail. But
if so, I have perhaps at least the following problem:

I may have been a bit careless with my wording. What I meant was
the ratio at full power output no matter what the designed
maximum power is. That is, the ratio holds equally
(approximately) for a 1W amp and a 100-watt amp.
Let's say "rated speaker impedance" is 8 ohms. Fixed. The
equivalent load resistance at full power as seen by the V+
power supply rail will be about the same as seen by the V-
rail, so I don't need to consider both in the ratio. For the
V+ rail, it's V(V+)/Ic_rms. The ratio is then:

8*Ic_rms/V(V+)

Which you claim is about the same regardless of power
setting, for example. However, while the power setting does
NOT affect V(V+) or the 8 ohm speaker, it does impact Ic_rms.
So the ratio does not seem to remain the same on that score.

In the case where V(V+) rises or falls, I'd imagine that
Ic_rms would similarly do so. So there, I think I follow
your point.


Let's hold on that point, for a moment.


Okay, let's do that.

Let's use P_out=10W average. We've already concluded that
this means V_peak=SQRT(2*P*R)=12.65V.

However, let's not assume even that much. In case anyone
needs to follow the details why, that equation comes from:

V0 = V_peak
w = 2*pi*f
V_t = V0 * SINE( 2*pi*f * t ) = V0 * SINE( w * t )
I_t = V_t / R
R = speaker impedance (say, 8 ohms?)
P = (Integral(V_t * I_t, dt) from 0 to t0) / t0
choose t0 where w*t0=2*pi, so t0=2*pi/w=1/f
= (Integral(V0^2*SINE^2(w*t)/R, dt) from 0 to 1/f) * f
move V0^2/R out:
= (Integral(SINE^2(w*t), dt) from 0 to 1/f) * f*V0^2/R
change of variable, x = w*t, gives dx = w*dt and dt = dx/w
= (Integral(SINE^2(x)/w, dx) from 0 to 2*pi) * f*V0^2/R
move 1/w out:
= (Integral(SINE^2(x), dx) from 0 to 2*pi) * f*V0^2/(w*R)
combine f/w at the end of the expression:
= (Integral(SINE^2(x), dx) from 0 to 2*pi) * V0^2/(2*pi*R)
solve integral:
= ((1/2x-1/4*SINE(2*x)) from 0 to 2*pi) * V0^2/(2*pi*R)
apply domain:
= (pi) * V0^2/(2*pi*R)
final solution:
= V0^2/(2*R)

Just for the record.

Since P=V_peak^2/(2*R), it follows that V_peak=SQRT(2*P*R).
And this documents the fact that we are talking about V_peak
and not V_p-p or V_rms or some other term. It's V_peak. So
for the p-p value, it's twice that.

That's nailed down, now. So I_peak will either be V_peak/R
or else SQRT(2*P/R), depending on how you like it stated.

V_peak = SQRT(2*P*R)
I_peak = SQRT(2*P/R)

Add a few volts transistors and so on and we get:

V_rail = V_peak + 2V

Derive average current from the peak swing? Hmm. This is
where I may get stuck.

We can compute peak values, no problem.

But now we need to consider class of operation and more.

For example, in class-B, each rail is used only _half the
time_, so to speak. So for half the time, it's zero. The
rest of the time looks like this:

I_avg = (Integral(I_t, dt) from 0 to t0) / t0
choose t0 where w*t0=pi, so t0=pi/w=1/(2*f)
= (Integral(V0*SIN(w*t)/R, dt) from 0 to 1/(2*f)) * 2*f
move V0/R out:
= (Integral(SIN(w*t), dt) from 0 to 1/(2*f)) * 2*f*V0/R
change of variable, x = w*t, gives dx = w*dt and dt = dx/w
= (Integral(SIN(x)/w, dx) from 0 to pi) * 2*f*V0/R
move 1/w out:
= (Integral(SIN(x), dx) from 0 to pi) * V0/(pi*R)
solve integral:
= (-COS(x)) from 0 to pi) * V0/(pi*R)
apply domain:
= (2) * V0/(pi*R)
final solution:
= 2*V0/(pi*R)

So I_avg so far is (2/pi) * V_peak/R. But that is only for
half the time. The other half, it is zero. So the actual
I_avg over the entire period of a full cycle from one rail,
in class-B, is half that or (1/pi) * V_peak/R.

The equivalent load resistance as seen by one rail is then:

R_equiv = pi * R * (V_rail/V_peak)
= pi * R * (V_rail/(R*I_peak))
= pi * V_rail / I_peak
= pi * V_rail / SQRT(2*P/R)

For a given R and V_rail and P, that does remain constant.

Let's pick some values, again. Just to make it concrete.
R=8, P=10, V_rail=18V, leads to just under 36 ohms for the
class-B case. That's from each rail and obviously includes
the speaker's R, itself.

I guess you enjoy reinventing the wheel all the time :)
Most people just use the long-established relation between peak,
rms and average values for known waveforms - sine in this case.
Your mathematical derivation was fundamentally correct, but where
you went astray was in that we are dealing with two half waves in
series because of the configuration. Therefore, the effective
average current drawn from each rail is Ipeak/pi, not 2Ipeak/pi.
If the two halves were drawn from the same rail, the total would
indeed be 2Ipeak/pi, but current for the two halves are drawn
from two separate rails in series, so the overall average is
still Ipeak/pi.

There's a once-common configuration (which you've probably come
across) in which two transistors take their power from the same
rail, and each transistor conducts on alternate half cycles. In
that configuration, the total average current drawn from the
power supply is indeed 2Ipeak/pi. That configuration has dropped
out of favor except for certain cases, mainly because it requires
an output transformer to combine the two output half waves. But
for the time being, let's not complicate things by going further
in that direction.
I'd guess that class-A would lead to different results due to
the fact that the half-cycle I set to zero would no longer be
zero.

In Class A, the output stage would be biased to a steady dc
current at least equal to the peak swing. At full drive in an
ideal Class A, the current will swing between 0 and twice the
steady value, and the average current will still be equal to that
without any output. Even with less-than-full drive, the swing is
still +/- around the bias value, and the average stays the same.
Hopefully, the above is about right. Let me know where I
screwed up.

_I_ screwed up again here because I failed to make it clear that
by "any power level", I meant amplifiers designed for any level
of full power output, NOT any level of output for the same
amplifier.
I'm now also thinking about something else.

The issue is peak power transfer, which usually is considered
to take place when R_src and R_load are equal.

This is a common misinterpretation of a law that's fundamentally
correct. The law applies when we try to draw as much power as
possible from a source and the limit is imposed by a series
resistance that we have no control over. That's not the case
here.

Let's take a hypothetical 6V battery with an internal resistance
of 1 ohm. Maximum power transfer to an external load occurs when
that external load is also 1 ohm. Half of that power will be
wasted inside the battery and efficiency is 50%. But that's often
not what we want. More often, we use the battery as a power
source and draw what we need from it. A load of 10 ohms results
in 3.27 watts of power, out of which 2.975W goes to the external
load, while the rest is wasted inside the battery. That's a
transfer effciency of about 91%.

If I managed to put that across clearly, I think you'll want to
revise the approach you use below.
 
J

Jon Kirwan

Jan 1, 1970
0
I've been in electronics for 4 decades, but I still can't decide
whether I'm a hobbyist or a profressional! :) I earn a living
with it, but I also approach my work as if it was a hobby. I
waste a lot of time and earn only a fraction of what I could. I
do a lot of things for others for free or for a nominal fee, but
I also have much more fun.

It's great when work and fun coincide. That's the way it is
with my profession, too, and I count myself very lucky for
that fact.
IMO, not even a budding pro should try to master everything at
once. I suggest you concentrate on the configuration presented in
its basic form in Wikipedia. What now look like very different
types may well turn out to be variants of a very few basic types.

It's nice to do a survey to get a feel for the variations
before drilling down on one or two. Partly, it's just about
being able to _read_ your comments and those of Paul's and
David's, for example. Sometimes, one of you might bring up
something and if I had no idea of the scope of possibilities
I might not have any chance to (a) interpret what I read or
(b) connect what I read with some other topologies to ask
further questions or (c) ask why still others aren't used.
It's not vital I know them, in detail. But it does help to
at least be able to "recognize" them when I see them.
I may have been a bit careless with my wording. What I meant was
the ratio at full power output no matter what the designed
maximum power is. That is, the ratio holds equally
(approximately) for a 1W amp and a 100-watt amp.

Okay. I think I'm following much better now.
I guess you enjoy reinventing the wheel all the time :)

What I don't like is not knowing _exact_ theory where certain
numbers/quantities come from. I think you already understood
this when you wrote: "I think he [me] would rather deduce
them himself instead of having them handed to him on a
platter." I guess I don't mind being told summaries, but I
also want to know how to get there entirely on my own, too.

I need to work on developing skills necessary to think for
myself, when faced with a novel situation.
Most people just use the long-established relation between peak,
rms and average values for known waveforms - sine in this case.

Since I'm at the front end of the learning curve, I guess
then I'm not most people.
Your mathematical derivation was fundamentally correct, but where
you went astray was in that we are dealing with two half waves in
series because of the configuration. Therefore, the effective
average current drawn from each rail is Ipeak/pi, not 2Ipeak/pi.

Perhaps you missed something in my writing. I wrote this
(also note above here, too):
: So I_avg so far is (2/pi) * V_peak/R. But that is only for
: half the time. The other half, it is zero. So the actual
: I_avg over the entire period of a full cycle from one rail,
: in class-B, is half that or (1/pi) * V_peak/R.

If you read that last equation, you see what amounts to
Ipeak/pi. Hopefully, anyway.

I think I _did_ follow and didn't make a mistake.
If the two halves were drawn from the same rail, the total would
indeed be 2Ipeak/pi, but current for the two halves are drawn
from two separate rails in series, so the overall average is
still Ipeak/pi.

Well, I came at the solution from about the same angle you do
here and come up with the same answer, too. So it must be
right. ;)
There's a once-common configuration (which you've probably come
across) in which two transistors take their power from the same
rail, and each transistor conducts on alternate half cycles. In
that configuration, the total average current drawn from the
power supply is indeed 2Ipeak/pi. That configuration has dropped
out of favor except for certain cases, mainly because it requires
an output transformer to combine the two output half waves. But
for the time being, let's not complicate things by going further
in that direction.
Okay.


In Class A, the output stage would be biased to a steady dc
current at least equal to the peak swing. At full drive in an
ideal Class A, the current will swing between 0 and twice the
steady value, and the average current will still be equal to that
without any output. Even with less-than-full drive, the swing is
still +/- around the bias value, and the average stays the same.

Okay. That makes sense.
_I_ screwed up again here because I failed to make it clear that
by "any power level", I meant amplifiers designed for any level
of full power output, NOT any level of output for the same
amplifier.

I think I got your point.
This is a common misinterpretation of a law that's fundamentally
correct. The law applies when we try to draw as much power as
possible from a source and the limit is imposed by a series
resistance that we have no control over. That's not the case
here.

Okay. I was just taking a silly thought, entirely out of
context, and "running it up the flag pole."
Let's take a hypothetical 6V battery with an internal resistance
of 1 ohm. Maximum power transfer to an external load occurs when
that external load is also 1 ohm. Half of that power will be
wasted inside the battery and efficiency is 50%. But that's often
not what we want. More often, we use the battery as a power
source and draw what we need from it. A load of 10 ohms results
in 3.27 watts of power, out of which 2.975W goes to the external
load, while the rest is wasted inside the battery. That's a
transfer effciency of about 91%.

So the "maximum power" metric would make sense if and only if
I were trying to get the most power possible out of some
given rail voltage and could adjust the load to any value I
wanted to achieve that. In other words, with your 9V battery
with 1 ohm internal resistance the external 1 ohm gets the
most out of the internal voltage source -- namely 9^2/2 or
40.5 watts, half of that or 20.25 watts occuring in the
external load (of course, I'm sure the 1 ohm internal
resistance model for a 9V battery would have long since
failed to be useful with that kind of external load.)

So maximum extractable power is 20.25 watts, math-wise. Your
example with 10 ohms requires quite a lot less and simply
doesn't meet the maximum power purpose. Which then allows it
to be more efficient, of course. Conversely, if the external
resistor were 0.2 ohms then total power would be 9^2/1.2 or
67.5 watts while the external power delivered to the load
would be 11.25 watts. Down from the maximum power criteria.

Applied to the audio amplifier case -- and let's assume for a
moment that the computed resistance comes from using the
average current and the voltage rail magnitude as already
discussed -- the "maximum power analogy" using a calculated
28 ohms in the driver section (removing the 8 ohms in the
speaker for a moment) would then suggest the best case would
occur when the external impedance was 28 ohms. With 18V
rails, maximum power with DC drive to the load would be
18^2/56 or half of 5.785 watts or about 2.9 watts. With 8
ohms in the load and _if_ the audio amplifier continued to
"look like" 28 ohms, this would then be 2 watts into the 8
ohms. Which is less than 2.9 watts.

However, the assumption that the amplifier is always
presenting 28 ohms fails in that case. So that is why the
whole idea of using that resistance calculation isn't any
good for the purpose to which I applied it. It does make
sense, broadly speaking, regarding capacitive loading
_because_ the integral I took, it's area, shows _charge_ that
must be delivered by the cap (time was included to compute a
current.. but the reality is that it was based upon finding
the total charge distributed, first.) Sizing a cap based
upon necessary charge storage at some voltage makes sense.
It's what they do and how they work in a circuit. It fails
for the purpose I then forced onto that idea because the
drivers don't present that resistance -- it's an artifact of
trying to deal with charge on the caps, not the reality of
the output driver BJTs.
 
P

pimpom

Jan 1, 1970
0
Jon Kirwan wrote:

.......<snipped for now>.........

Jon, it's getting very close to 3 am here and I've got to sign
off for now because I have to get up earlier than my usual 11-12
noon tomorrow.

Maybe I glanced through your calculation of Iav too cursorily and
wrongly concluded that you made a mistake where you didn't. I'll
give it a closer look and come back when I can.
 
J

Jon Kirwan

Jan 1, 1970
0
Jon Kirwan wrote:

......<snipped for now>.........

Jon, it's getting very close to 3 am here and I've got to sign
off for now because I have to get up earlier than my usual 11-12
noon tomorrow.

I see that you are +530 and I am -800, so we are either +1330
or -1030 apart (using my time as 0000) from each other --
roughly speaking at opposing corners of the day.

(Not sure how to apply the chemical analogy of para- to this.
Para-dies?)

11AM your time is 11+10.5 or 9:30PM my time. Which suggests
you are dead to me from 1PM to 10PM, my time. ;)
Maybe I glanced through your calculation of Iav too cursorily and
wrongly concluded that you made a mistake where you didn't. I'll
give it a closer look and come back when I can.

I very much appreciate any thoughts. So a lot of thanks go
to you for even taking a moment, at all. Whenever you feel
able and willing is nothing less than a fantastic boon to me.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0

Okay. These concepts are slowly settling into my brain.
The other ways around. The design will deliver ten watts maximum
(disregarding clipping) but the average power output will actually be
much lower - hence you can "skimp" a bit on the supply transformer and
heatsinks - which wrt overheating have very long time constants relative
the the peak output demands.

Cripes! Really? So a 10W amplifier isn't designed to
actually deliver a full 10W steadily into a load? That's the
peak power capability? Cripes.

Let me put this another way. I design a class-B output stage
with rails capable of 10W compliance into 8 ohms (roughly 13V
peak, so rails at maybe +/-17V or so?) With 10W into the 8
ohm load, let's say this means the upper power BJT is
handling about 4-5W and the lower BJT is handling 4-5W, as
well. Call it 10W total dissipation inside the amp while 10W
are dissipated in the speaker.

But I don't have to go find BJTs able to dissipate 4-5W,
because the 10W spec is just a max-unsustained case and the
real situation is more like 2W into the load, continuous? In
short, I need to find a BJT that only needs to dissipate 1W
for the high side and 1W for the low side? I could use two
PN2222As in parallel to do that!

I can _cheat_ like that and call it a 10W design? It doesn't
actually _have_ to sustain 10W without burning up?

Okay, now I'm depressed. I go buy a 50W amplifier, stick a
sine wave signal generator on it and watch the thing toast
itself, bursting in fire soon enough?
A ten watt amp delivers a sine wave producing 10 watts of output power
into a specified load. Ideally this would be 10 watts for an infinite
period of time but for audio amps, due to the nature of the signal, an
"infinite period of time" in practical terms may be as short as a few
seconds.

Yeah. A few seconds. So... now I can go back with a
quasi-comp output stage and use a pair of those PN2222As for
it, without heat sinking! Nice little TO92 packages, too. ;)
40W output. I**2 x R. The power supply voltage is
approximately constant.

I was looking at some actual measurements taken by Mr. Self
on an actual class-B amplifier when I wrote that. I didn't
do a theory-based analysis. Just read off the figures when
he was comparing a class-A with a class-B into different
loads.

Now I'll do that.

I had then imagine it came from V^2/R and knowing that V^2
remains the same for a given amplifier and only the R changed
from 8 to 4. Which makes sense then that it would double,
not quadruple, the power output. From an I^2*R perspective,
I get the same estimate because a smaller load does double
the current, but the R divides in half, so the combination is
still just twice, not four-times.

Can you explain this 40W statement better for me?
If class A, power is 5 watts out with 4 ohms. Current is held
constant.

Again, looking at Self's chart (page 322 on his 5th edition)
I see a slight degregation into 4 ohms, going from about 20W
into 8 ohms to 15W into 4 ohms. I'm not entirely sure of
'theory' here, but I took this to suggest that at the higher
currents the drive circuitry's compliance coupled with the
likely somewhat lower gain caused by somewhat higher currents
now needed accounted for the droop.

But his chart certainly doesn't suggest 1/2 rated power.

I guess I need to delve into this a bit more to make sure I
understand. The class-B case seems easier for me to follow
(assuming I'm right, above, which I of course may not be.)
Your argument here is reasonable but ..... it is also the beginning of
the PMPO fiasco - since no advertising department could agree on what
constitutes "music" they used what ever figures looked best - and that
led to the PFPO (peak fantasy power output) fiasco where you just put
anything you like on the box.

For a short time some (better) manufactures used a figure they called
"headroom" which was the maximum possible instantaneous power output
when the power caps are fully charged divided by the long term power
output (10 watts in this case). It was always expressed in db - but was
confusing to the customer - so it disappeared.

Okay. Well, I can say one thing. I've learned that there
are output specs and there are output specs and what they
actually mean is yet another question, usually unanswered.

As a consumer, I've become a little better informed even if
all that means is I'm a lot more suspicious than before.
Wait till you start talking about speakers!

Hehe. Now I'm really scared. ;)

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
<snip>
Again, looking at Self's chart (page 322 on his 5th edition)
I see a slight degregation into 4 ohms, going from about 20W
into 8 ohms to 15W into 4 ohms. I'm not entirely sure of
'theory' here, but I took this to suggest that at the higher
currents the drive circuitry's compliance coupled with the
likely somewhat lower gain caused by somewhat higher currents
now needed accounted for the droop.

But his chart certainly doesn't suggest 1/2 rated power.
<snip>

Another thought crossed my mind, too. If the amplifier he
was testing used a Vbe multiplier to achieve class-A
operation, that won't be enough when faced with 4 ohms. If
so, it will degrade into class-AB operation. Not sure that
that means, yet.

Need to think more on that, as well.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
Another thought crossed my mind, too. If the amplifier he
was testing used a Vbe multiplier to achieve class-A
operation, that won't be enough when faced with 4 ohms. If
so, it will degrade into class-AB operation. Not sure that
that means, yet.

Need to think more on that, as well.

Sorry to keep responding to myself, but even more crosses my
mind, including VAS loading.

So I stopped letting things cross my mind and set up a spice
simulation to see what it tells me. (I hate doing this,
without applying theory, but I feel time is of the essence
and like cheating.. for now.)

Class-A appears to deliver the same thing as class-B, at
least using a TIP3055 and TIP2955 output pair, and using an
idealized voltage source between the bases to set the class
of operation. With a 4-ohm load and the exact same drive
voltage (using again a voltage source as the VAS output), I
got 20.55 watts into 4 ohms with class-A operation and 10.31
watts into 8 ohms. (Which is not a 4X but 2X phenomenon.) In
class-B, this was 19.66 watts into 4 ohms and 10.16 watts
into 8. Again, 2X. (I think I might have been just slightly
into class-AB with that last test, but I got it close.)

So it maybe doesn't matter about class of operation. But is
about the quiescent current flowing via the vbe multipler and
what is available to _drive_ the output BJTs and perhaps also
some estimation about output drive impedance of the VAS which
hauls the output section around in real amplifiers that
caused the table entry values I saw with Self's book.

Jon
 
P

pimpom

Jan 1, 1970
0
Jon said:
I see that you are +530 and I am -800, so we are either +1330
or -1030 apart (using my time as 0000) from each other --
roughly speaking at opposing corners of the day.

(Not sure how to apply the chemical analogy of para- to this.
Para-dies?)

11AM your time is 11+10.5 or 9:30PM my time. Which suggests
you are dead to me from 1PM to 10PM, my time. ;)

Yeah. Though, with your DST system, it's not easy to keep track
of the exact difference.
I very much appreciate any thoughts. So a lot of thanks go
to you for even taking a moment, at all. Whenever you feel
able and willing is nothing less than a fantastic boon to me.

I just had a quick look again and it seems I noticed only the
part where you got Iav as 2*Ipk/pi and missed the place where you
later corrected it to Ipk/pi for the series push-pull topology.
Sorry.

I'd like to chime in on your discussion of commercial power
ratings with David, but I'll have to leave that for later.
 
J

Jon Kirwan

Jan 1, 1970
0
Yeah. Though, with your DST system, it's not easy to keep track
of the exact difference.


I just had a quick look again and it seems I noticed only the
part where you got Iav as 2*Ipk/pi and missed the place where you
later corrected it to Ipk/pi for the series push-pull topology.
Sorry.

No problem, at all. It means you are engaging yourself witgh
what I write, even if sometimes that means a mistake is made
in reading. And that means a great deal to me. Don't
apologize. I am honored even by the fact that you imagined a
problem there, because it means you scanned through it with
your educated eye.
I'd like to chime in on your discussion of commercial power
ratings with David, but I'll have to leave that for later.

Understood, and thanks in advance for any thoughts when they
happen.

Jon
 
P

Paul E. Schoen

Jan 1, 1970
0
Jon Kirwan said:
Sorry to keep responding to myself, but even more crosses my
mind, including VAS loading.

So I stopped letting things cross my mind and set up a spice
simulation to see what it tells me. (I hate doing this,
without applying theory, but I feel time is of the essence
and like cheating.. for now.)

Class-A appears to deliver the same thing as class-B, at
least using a TIP3055 and TIP2955 output pair, and using an
idealized voltage source between the bases to set the class
of operation. With a 4-ohm load and the exact same drive
voltage (using again a voltage source as the VAS output), I
got 20.55 watts into 4 ohms with class-A operation and 10.31
watts into 8 ohms. (Which is not a 4X but 2X phenomenon.) In
class-B, this was 19.66 watts into 4 ohms and 10.16 watts
into 8. Again, 2X. (I think I might have been just slightly
into class-AB with that last test, but I got it close.)

So it maybe doesn't matter about class of operation. But is
about the quiescent current flowing via the vbe multipler and
what is available to _drive_ the output BJTs and perhaps also
some estimation about output drive impedance of the VAS which
hauls the output section around in real amplifiers that
caused the table entry values I saw with Self's book.

The most important distinction, I think, is the difference in efficiency at
power levels lower than maximum. Class A efficiency is about 35% to as low
as 15% at full power, and it drops to the point where it is essentially
just a heater when output is in the normal listening range of one or two
watts for a 10W amp. See http://sound.westhost.com/class-a.htm.

A class B amplifier should do about 69% efficiency at full power,
especially if you can drive the output stage with a source that is higher
than the rails for rail-to-rail output. An alternative is to create higher
rails for the driver stage, or to use a bootstrap approach.

I have attached a simple amplifier using MOSFETs that provides 14W into 8
ohms at 20 Hz - 20 kHz with a 35 VDC single supply, and it achieves 67%. It
has a quiescent power of about 2.2W. It is not a "practical" design,
however, as the biasing for the output MOSFETs depends on their Vto which
may not be stable. I added a simple resistor to adjust bias, and I used
logic level MOSFETs. The same basic circuit could be done with all bipolar
as well, but I like MOSFETs.

I'd like to see the ASC files for your LTSpice simulations. And it would be
interesting to see the results of a frequency sweep using AC analysis.

Paul

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

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J

Jon Kirwan

Jan 1, 1970
0
<snip>
Wait till you start talking about speakers!

Okay. Now you made me really worried. I just realized
(okay, so I "get it" for a moment once in a while and then
manage to forget it until the next time) again that speakers
can be nasty in terms of phase shifts and loads.

Cripes. Here I was about to embark on output stage design
and this rears its head, again. At certain slew rates, a
speaker can demand a lot from the amplifier's output and I
will need to think about 'protection' -- especially if that
takes place at low frequencies like 20Hz where the main
stretch of the slope from peak to valley can last as long as
maybe 20ms or so. Whatever the output stage is, it needs to
handle peak dissipations as well as peak currents, perhaps,
for that long. (I'm trying to keep in mind that current
phase may either lag or lead.)

Wait until I close the global NFB loop, too. And what about
oscillation in local FB, too? (If FETs, that might be more
[or less] a consideration... I have no idea, right now.)

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
The most important distinction, I think, is the difference in efficiency at
power levels lower than maximum. Class A efficiency is about 35% to as low
as 15% at full power, and it drops to the point where it is essentially
just a heater when output is in the normal listening range of one or two
watts for a 10W amp. See http://sound.westhost.com/class-a.htm.

Thanks. I'm seeing this well, now.
A class B amplifier should do about 69% efficiency at full power,
especially if you can drive the output stage with a source that is higher
than the rails for rail-to-rail output.

By this, do you mean several power rails?

(I'm trying to imagine a system that delays the output [equal
time for all frequencies of interest, or a group delay = 0...
haha] and "anticipates" the required voltages and uses an FFT
[applied to an inverse FFT 'filter' determined at startup] to
develop the appropriate signals that would automatically
generate the right voltages at the right times. That should
be 'fun', if it would work at all. Would have to be noisy as
all get-out, I think. But might be interesting.)
An alternative is to create higher
rails for the driver stage, or to use a bootstrap approach.

I know of the use of 'bootstrap' for other purposes, like
stiffening the apparent input impedance and I plan to use it
there. But what do you mean in this case?
I have attached a simple amplifier using MOSFETs

Saved and observed.
that provides 14W into 8
ohms at 20 Hz - 20 kHz with a 35 VDC single supply, and it achieves 67%. It
has a quiescent power of about 2.2W. It is not a "practical" design,
however, as the biasing for the output MOSFETs depends on their Vto which
may not be stable. I added a simple resistor to adjust bias, and I used
logic level MOSFETs. The same basic circuit could be done with all bipolar
as well, but I like MOSFETs.

I'm not yet 'used to' them, except as switches. And as I
understand things from reading people like Self, there are
tradeoffs -- neither BJT nor FET is a certain win over the
other.

I feel a little more ready to face the details of a BJT
output driver and probably need to get _one_ approach worked
out so that I understand it better before taking on the
other. (I'm assuming that you don't imagine using FETs all
the way through.)

In any case, I'll have some context, then, to fathom the pros
and cons, later on.

Also, high power FETs are "more expensive." At least, so far
as I've experienced. And I can consider paralleling BJTs
with emitter degeneration. I understand it and why it works.
I'm not sure I'd understand, at this point, how to parallel
FETs in linear operation.
I'd like to see the ASC files for your LTSpice simulations. And it would be
interesting to see the results of a frequency sweep using AC analysis.

Okay:

http://www.infinitefactors.org/misc/spice/behavioral 2-BJT output 01.asc

Hopefully, you can pick that up okay.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0

I modify Vb to set the class of operation, using larger
values (say, 2V, for a 4 ohm load for example) to make sure
it is in class-A. I use 1.3V or thereabouts for class-B. All
these depend upon load. Tweak as needed. I usually just
observe the two emitter currents to get a bead on the
operation mode and tweak the class-B to be just slightly on
the class-AB side. For class-A, I jack it up so that the
emitter currents don't show any visible "blunting" on their
sinusoidal shape.

Jon
 
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