Discussing audio amplifier design -- BJT, discrete

[Jon Kirwan]

On Wed, 17 Feb 2010 16:20:40 -0500, "Paul E. Schoen"
<paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:kvlmn5dhciv4c51a0au32qi52rjnl67sro@4ax.com...

Okay. I'm back to the power supply, again. (I'm convinced
that my junkbox unit will work fine -- I think it can hold
maybe 18V minimum under load on each rail. Which seems more
than enough headroom for 12.7V, plus output stage overhead.

I take a little issue with your use of terms in this phrase,
"RMS power of even compressed samples of music is only about
20% of the peak." Power is average and I don't think RMS
applies to power. Volts-to-power is a squared-phenomenon. So
are amps-to-power. RMS makes sense for those two. But power
is an average (integrated Joules divided by time.)

So I believe I have to interpret your meaning as suggesting
that the short-term power required (also an average of some
ill-defined kind, I suppose) when playing music can be a
factor of 5 times more than its long-term average power. You
also mentioned a figure as low as 12.5%, which would suggest
a factor of 8 used as a margin instead of 5.

But a requirement to support short-term power levels is
really just a compliance requirement on the power supply
rails, isn't it?

So put another way, if I wanted a long-term average of 10W
output and I wanted the extra margins required to support the
worst case estimate of a factor of 8 for short-term power
bursts, then I'd need to design rails that support a voltage
compliance level substantially higher. The parts would need
to withstand it, too. And because of the much higher rail
voltages that need to be dropped most of the time, the output
BJTs would need to have just that much more capacity to
dissipate.

Or put still another way, assuming that my output swing at
the output stage emitters cannot exceed a magnitude of 15V
and that everything is sized for dissipating 10W, does this
mean the amplifier is a 10W amplifier that can support a peak
of 14W=(15^2/(2*8))? (Which isn't so good, considering your
comments above regarding "music?")

What is meant when one says, '10 watts?'

This gets worse when I consider the class of operation,
doesn't it? I mean, class-B might be specified as 10W into 8
ohms, but wouldn't that be 20W into 4 ohms? But if class-A,
it's pretty much 10W no matter what?

I'm beginning to imagine amplifiers should be specified as to
their peak output voltage compliance into 8, 6, and 4 ohms;
instantaneous and sustained without damage to the unit. For
example, 35V into 8 ohms instantaneous, 15V sustained. Or
80W instantaneous, 15W sustained. That way, someone might
have some knowledge about how well it might handle _their_
music at, say, 15W average power. And could compare that
against another unit specified as 20V into 8 ohms, 15V
sustained.

How does one know what they are buying? What a headache.

Yes, as an extension of what (I think) Mark Twain said, there are lies,
damn lies, statistics, and specifications. Then there is the matter of
testing. An amplifier is a complex entity and its performance depends on
the power supply, the load, its components, environmental conditions, and
the nature of the signal being applied. So it may seem fair to level the
playing field by testing with a pure sine wave at certain frequencies and
determining that it maintains a certain level of maximum distortion without
overheating or shutting down over an extended period of time in a
controlled environment.

But in real life there are many more factors involved, and the actual
performance in an individual situation may vary widely. Power is indeed an
average function, but the ability to provide power involves efficiency and
a duty-cycle rated function of maximum temperature of components, and also
the ability of the power supply to maintain a certain voltage level for
long enough to "ride out" brief peaks in the signal of typical music.

The power that can be supplied to various loads depends largely on
impedance matching. But most solid state amplifiers are capable of
supplying a certain amount of current, so if it is optimized for eight
ohms, it may be able to provide even less continuous power at 4 ohms, but
possibly more peak power.

You have brought up some good points. But for most purposes, an amplifier
rated conservatively at 10W continuous power should be plenty for home
music listening. When pushed beyond its normal limits, much depends on how
the amplifier handles overloads, and your personal threshold of annoyance
when the inevitable distortion occurs.

Paul

It sure has been an education, so far. Now I am beginning to
understand the technical motivation for LOTS of rails and the
ability to select between them (perhaps automatically) in
those fancy-pants amplifier designs; dropping in (or out)
stacked BJTs as needed. Though I am loathe to even attempt
thinking more about them.

.....

Now, I want axial leaded diodes for the bridge. From
simulating a load of 8 ohms, 1kHz, average power of 10W, and
my secondary winding resistance of 2.6 ohms, I'm finding that
each diode suffers under a quarter watt of dissipation. So,
any recommendations about diodes? Obviously, for a one-off,
cost is not really an issue. How important is 'fast
recovery'? (Outside of its impact on dissipation.) Seems
that anything with 100V or better for reverse voltage
standoff, 1/4 watt or better, should work. Leakage probably
isn't that important (except against as it may add to
dissipation.)

I'm expecting to use caps on the order of perhaps 2.2mF 50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad I
can make it while seeing what that means for the output. And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

Jon

 

[pimpom]

Jon Kirwan wrote:

On Wed, 17 Feb 2010 16:20:40 -0500, "Paul E. Schoen"
paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:kvlmn5dhciv4c51a0au32qi52rjnl67sro@4ax.com...

Okay. I'm back to the power supply, again. (I'm convinced
that my junkbox unit will work fine -- I think it can hold
maybe 18V minimum under load on each rail. Which seems more
than enough headroom for 12.7V, plus output stage overhead.

....
.............<snip>.............

Now, I want axial leaded diodes for the bridge. From
simulating a load of 8 ohms, 1kHz, average power of 10W, and
my secondary winding resistance of 2.6 ohms, I'm finding that
each diode suffers under a quarter watt of dissipation. So,
any recommendations about diodes? Obviously, for a one-off,
cost is not really an issue. How important is 'fast
recovery'? (Outside of its impact on dissipation.) Seems
that anything with 100V or better for reverse voltage
standoff, 1/4 watt or better, should work. Leakage probably
isn't that important (except against as it may add to
dissipation.)

Fast recovery is not important here since the diodes work at
mains frequency.

10W sinusoidal into 8 ohms = 1.58A peak = 0.503A dc average for
Class B. Add some mAs for the driver stages. That's slightly more
than 0.25A each for diodes in full-wave rectification. The
ubiquitous 1N4002 to 1N4007 rated for 1 Amp diodes will do fine.
They differ only in the maximum reverse voltage ratings and cost
almost the same. As a matter of convenience, I stock only the
1000-volt 1N4007. At less than 2 cents US each retail, I buy them
in batches of hundreds at a time.

I'm expecting to use caps on the order of perhaps 2.2mF 50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad I
can make it while seeing what that means for the output. And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of ripple

and load regulation. I divide the full supply dc voltage with the
current at maximum output. This gives the equivalent dc load as
seen by the power supply. In the sample design under
consideration, that's roughly 30 ohms on each side of the split
supply. Calculate the reactance of the filter capacitor at the
pulsating dc frequency which is twice the mains frequency for
full-wave. My rule of thumb is to get an R/Xc ratio of the order
of 50 for a medium quality amp. Your choice of 2200uF agrees well
with this.

 

[Jon Kirwan]

On Fri, 19 Feb 2010 02:20:29 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Wed, 17 Feb 2010 16:20:40 -0500, "Paul E. Schoen"
paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:kvlmn5dhciv4c51a0au32qi52rjnl67sro@4ax.com...

Okay. I'm back to the power supply, again. (I'm convinced
that my junkbox unit will work fine -- I think it can hold
maybe 18V minimum under load on each rail. Which seems more
than enough headroom for 12.7V, plus output stage overhead.

....
............<snip>.............

Now, I want axial leaded diodes for the bridge. From
simulating a load of 8 ohms, 1kHz, average power of 10W, and
my secondary winding resistance of 2.6 ohms, I'm finding that
each diode suffers under a quarter watt of dissipation. So,
any recommendations about diodes? Obviously, for a one-off,
cost is not really an issue. How important is 'fast
recovery'? (Outside of its impact on dissipation.) Seems
that anything with 100V or better for reverse voltage
standoff, 1/4 watt or better, should work. Leakage probably
isn't that important (except against as it may add to
dissipation.)

Fast recovery is not important here since the diodes work at
mains frequency.

Thanks. That had certainly crossed my mind as I was writing.
I just wanted to be sure I hadn't missed something important.

10W sinusoidal into 8 ohms = 1.58A peak = 0.503A dc average for
Class B. Add some mAs for the driver stages. That's slightly more
than 0.25A each for diodes in full-wave rectification. The
ubiquitous 1N4002 to 1N4007 rated for 1 Amp diodes will do fine.
They differ only in the maximum reverse voltage ratings and cost
almost the same. As a matter of convenience, I stock only the
1000-volt 1N4007. At less than 2 cents US each retail, I buy them
in batches of hundreds at a time.

I pull them out of CFL lamps before disposal. So they are
"free" to me. I've quite a few, now. 1200V PIV, I think.
Way overkill. But free. I'll use them.

I'm expecting to use caps on the order of perhaps 2.2mF 50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad I
can make it while seeing what that means for the output. And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of ripple
and load regulation. I divide the full supply dc voltage with the
current at maximum output. This gives the equivalent dc load as
seen by the power supply. In the sample design under
consideration, that's roughly 30 ohms on each side of the split
supply. Calculate the reactance of the filter capacitor at the
pulsating dc frequency which is twice the mains frequency for
full-wave. My rule of thumb is to get an R/Xc ratio of the order
of 50 for a medium quality amp. Your choice of 2200uF agrees well
with this.

Thanks for your thinking on this. I used more mathy stuff to
get there, but I like your practical slice through all that.
It is easy to follow.

So I'm settled on those particulars, now. The only thing I
don't have, right now, are the caps. Well, maybe. I just
found a 2.2mF, 35V cap. So that gives me one. I've got all
kinds of 200V caps, up to about 470uF. But still looking for
one more 'something close' on the order of 35-50V. I'll keep
looking through the junk box some more. Might turn up
another one.

If so, I then need to figure out all the mounting stuff for
the hardware. I have the AC plugs and grommets and fuse
holders. I can also pull a transorb out of the junk box
(from those CFL lamps, again.) The transformer doesn't have
any brackets or mounting holes in the laminated steel core so
I will have to fashion one from a simple strap of metal,
drilled out. Then to wire it all up and do the smoke test
and verify the output, with and without a load on it.

There and done, I'm ready to move on.

Thanks,
Jon

 

[pimpom]

Jon Kirwan wrote:

On Fri, 19 Feb 2010 02:20:29 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:

I'm expecting to use caps on the order of perhaps 2.2mF 50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad I
can make it while seeing what that means for the output. And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of
ripple
and load regulation. I divide the full supply dc voltage with
the
current at maximum output. This gives the equivalent dc load
as
seen by the power supply. In the sample design under
consideration, that's roughly 30 ohms on each side of the
split
supply. Calculate the reactance of the filter capacitor at the
pulsating dc frequency which is twice the mains frequency for
full-wave. My rule of thumb is to get an R/Xc ratio of the
order
of 50 for a medium quality amp. Your choice of 2200uF agrees
well
with this.

Thanks for your thinking on this. I used more mathy stuff to
get there, but I like your practical slice through all that.
It is easy to follow.

Rules of thumb are often based on previous mathematical
derivations, as it was in this case. However, after having done
umpteen calculations where absolute precision is not needed, the
novelty wears off after some time and one tends to be satisfied
with being able to intuitively predict the outcome within a per
cent or so without actually putting anything on paper. It's been
firmly etched in my mind for 40 years that 1000uF has a reactance
of 1.5815 ohms (usually taken as 1.6) at 100Hz (twice the mains
frequency here) and I quickly derive Xc for other values from
that within a second. Then I mentally divide the equivalent DC
resistance of a load (not necessarily an audio amplifier) with
that reactance and have a good idea of what to expect in terms of
ripple voltage amplitude, regulation, DC voltage, peak diode
current, rms transformer current, etc.

Heck, it's past 4 am over here. Time for bed. Bye.

 

[Paul E. Schoen]

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hlk96n$6g8$1@news.albasani.net...

Jon Kirwan wrote:
On Wed, 17 Feb 2010 16:20:40 -0500, "Paul E. Schoen"
paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:kvlmn5dhciv4c51a0au32qi52rjnl67sro@4ax.com...

Okay. I'm back to the power supply, again. (I'm convinced
that my junkbox unit will work fine -- I think it can hold
maybe 18V minimum under load on each rail. Which seems more
than enough headroom for 12.7V, plus output stage overhead.

....
............<snip>.............

Now, I want axial leaded diodes for the bridge. From
simulating a load of 8 ohms, 1kHz, average power of 10W, and
my secondary winding resistance of 2.6 ohms, I'm finding that
each diode suffers under a quarter watt of dissipation. So,
any recommendations about diodes? Obviously, for a one-off,
cost is not really an issue. How important is 'fast
recovery'? (Outside of its impact on dissipation.) Seems
that anything with 100V or better for reverse voltage
standoff, 1/4 watt or better, should work. Leakage probably
isn't that important (except against as it may add to
dissipation.)


Fast recovery is not important here since the diodes work at mains
frequency.

10W sinusoidal into 8 ohms = 1.58A peak = 0.503A dc average for Class B.
Add some mAs for the driver stages. That's slightly more than 0.25A each
for diodes in full-wave rectification. The ubiquitous 1N4002 to 1N4007
rated for 1 Amp diodes will do fine. They differ only in the maximum
reverse voltage ratings and cost almost the same. As a matter of
convenience, I stock only the 1000-volt 1N4007. At less than 2 cents US
each retail, I buy them in batches of hundreds at a time.

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.

I'm expecting to use caps on the order of perhaps 2.2mF 50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad I
can make it while seeing what that means for the output. And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of ripple and load
regulation. I divide the full supply dc voltage with the current at
maximum output. This gives the equivalent dc load as seen by the power
supply. In the sample design under consideration, that's roughly 30 ohms
on each side of the split supply. Calculate the reactance of the filter
capacitor at the pulsating dc frequency which is twice the mains
frequency for full-wave. My rule of thumb is to get an R/Xc ratio of the
order of 50 for a medium quality amp. Your choice of 2200uF agrees well
with this.

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.

Of course, if you enjoy mathematical analysis, you can spend time working
out effects of winding resistance and capacitor ESR and acceptable ripple.
Or you can just use LTSpice. But if I need a quick and dirty junkbox power
supply, 1000 uF/amp is good enough to grab and go.

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.

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.

Paul

 

[Jon Kirwan]

On Fri, 19 Feb 2010 04:17:22 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Fri, 19 Feb 2010 02:20:29 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:

I'm expecting to use caps on the order of perhaps 2.2mF 50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad I
can make it while seeing what that means for the output. And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of
ripple
and load regulation. I divide the full supply dc voltage with
the
current at maximum output. This gives the equivalent dc load
as
seen by the power supply. In the sample design under
consideration, that's roughly 30 ohms on each side of the
split
supply. Calculate the reactance of the filter capacitor at the
pulsating dc frequency which is twice the mains frequency for
full-wave. My rule of thumb is to get an R/Xc ratio of the
order
of 50 for a medium quality amp. Your choice of 2200uF agrees
well
with this.

Thanks for your thinking on this. I used more mathy stuff to
get there, but I like your practical slice through all that.
It is easy to follow.

Rules of thumb are often based on previous mathematical
derivations, as it was in this case.

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.

However, after having done
umpteen calculations where absolute precision is not needed, the
novelty wears off after some time and one tends to be satisfied
with being able to intuitively predict the outcome within a per
cent or so without actually putting anything on paper.

I think I clearly understood exactly that from your writing.

It's been
firmly etched in my mind for 40 years that 1000uF has a reactance
of 1.5815 ohms (usually taken as 1.6) at 100Hz (twice the mains
frequency here) and I quickly derive Xc for other values from
that within a second. Then I mentally divide the equivalent DC
resistance of a load (not necessarily an audio amplifier) with
that reactance and have a good idea of what to expect in terms of
ripple voltage amplitude, regulation, DC voltage, peak diode
current, rms transformer current, etc.

Heck, it's past 4 am over here. Time for bed. Bye.

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

 

[Paul E. Schoen]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:buasn5pd2p5vs3uppunegtjcq8rfk8f3m1@4ax.com...

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,

I can't speak for pimpom but I don't think any offense has been taken.
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.

Also, I think this thread has about run its course, and it may be time to
start a new one. 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.
You may consider yourself a beginner but your theoretical knowledge and
mathematical analysis is beyond the range of basics. It seems that your
lack of direct experience and practical "knack" will soon pass as you build
and test a hands-on circuit.

My main criticism would be that you tend to limit yourself too much by
using scavenged parts and freebies in a junkbox. 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.

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" 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...

Paul

 

[Jon Kirwan]

On Thu, 18 Feb 2010 18:24:21 -0500, "Paul E. Schoen"
<paul@peschoen.com> wrote:

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hlk96n$6g8$1@news.albasani.net...
Jon Kirwan wrote:
On Wed, 17 Feb 2010 16:20:40 -0500, "Paul E. Schoen"
paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:kvlmn5dhciv4c51a0au32qi52rjnl67sro@4ax.com...

Okay. I'm back to the power supply, again. (I'm convinced
that my junkbox unit will work fine -- I think it can hold
maybe 18V minimum under load on each rail. Which seems more
than enough headroom for 12.7V, plus output stage overhead.

....
............<snip>.............

Now, I want axial leaded diodes for the bridge. From
simulating a load of 8 ohms, 1kHz, average power of 10W, and
my secondary winding resistance of 2.6 ohms, I'm finding that
each diode suffers under a quarter watt of dissipation. So,
any recommendations about diodes? Obviously, for a one-off,
cost is not really an issue. How important is 'fast
recovery'? (Outside of its impact on dissipation.) Seems
that anything with 100V or better for reverse voltage
standoff, 1/4 watt or better, should work. Leakage probably
isn't that important (except against as it may add to
dissipation.)


Fast recovery is not important here since the diodes work at mains
frequency.

10W sinusoidal into 8 ohms = 1.58A peak = 0.503A dc average for Class B.
Add some mAs for the driver stages. That's slightly more than 0.25A each
for diodes in full-wave rectification. The ubiquitous 1N4002 to 1N4007
rated for 1 Amp diodes will do fine. They differ only in the maximum
reverse voltage ratings and cost almost the same. As a matter of
convenience, I stock only the 1000-volt 1N4007. At less than 2 cents US
each retail, I buy them in batches of hundreds at a time.

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.


I'm expecting to use caps on the order of perhaps 2.2mF 50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad I
can make it while seeing what that means for the output. And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of ripple and load
regulation. I divide the full supply dc voltage with the current at
maximum output. This gives the equivalent dc load as seen by the power
supply. In the sample design under consideration, that's roughly 30 ohms
on each side of the split supply. Calculate the reactance of the filter
capacitor at the pulsating dc frequency which is twice the mains
frequency for full-wave. My rule of thumb is to get an R/Xc ratio of the
order of 50 for a medium quality amp. Your choice of 2200uF agrees well
with this.

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

 

[Jon Kirwan]

On Fri, 19 Feb 2010 02:43:12 -0500, "Paul E. Schoen"
<paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:buasn5pd2p5vs3uppunegtjcq8rfk8f3m1@4ax.com...

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,

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

 

[pimpom]

Jon Kirwan wrote:

On Fri, 19 Feb 2010 04:17:22 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Fri, 19 Feb 2010 02:20:29 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:

I'm expecting to use caps on the order of perhaps 2.2mF
50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad
I
can make it while seeing what that means for the output.
And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of
ripple
and load regulation. I divide the full supply dc voltage
with
the
current at maximum output. This gives the equivalent dc load
as
seen by the power supply. In the sample design under
consideration, that's roughly 30 ohms on each side of the
split
supply. Calculate the reactance of the filter capacitor at
the
pulsating dc frequency which is twice the mains frequency
for
full-wave. My rule of thumb is to get an R/Xc ratio of the
order
of 50 for a medium quality amp. Your choice of 2200uF agrees
well
with this.

Thanks for your thinking on this. I used more mathy stuff to
get there, but I like your practical slice through all that.
It is easy to follow.

Rules of thumb are often based on previous mathematical
derivations, as it was in this case.

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.

However, after having done
umpteen calculations where absolute precision is not needed,
the
novelty wears off after some time and one tends to be
satisfied
with being able to intuitively predict the outcome within a
per
cent or so without actually putting anything on paper.

I think I clearly understood exactly that from your writing.

It's been
firmly etched in my mind for 40 years that 1000uF has a
reactance
of 1.5815 ohms (usually taken as 1.6) at 100Hz (twice the
mains
frequency here) and I quickly derive Xc for other values from
that within a second. Then I mentally divide the equivalent DC
resistance of a load (not necessarily an audio amplifier) with
that reactance and have a good idea of what to expect in terms
of
ripple voltage amplitude, regulation, DC voltage, peak diode
current, rms transformer current, etc.

Heck, it's past 4 am over here. Time for bed. Bye.

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.

 

[pimpom]

Paul E. Schoen wrote:

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hlk96n$6g8$1@news.albasani.net...
Jon Kirwan wrote:


I'm expecting to use caps on the order of perhaps 2.2mF 50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad I
can make it while seeing what that means for the output. And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of
ripple and
load regulation. I divide the full supply dc voltage with the
current at maximum output. This gives the equivalent dc load
as seen
by the power supply. In the sample design under consideration,
that's roughly 30 ohms on each side of the split supply.
Calculate
the reactance of the filter capacitor at the pulsating dc
frequency
which is twice the mains frequency for full-wave. My rule of
thumb
is to get an R/Xc ratio of the order of 50 for a medium
quality amp.
Your choice of 2200uF agrees well with this.

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.

 

[Jon Kirwan]

On Fri, 19 Feb 2010 23:43:04 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Fri, 19 Feb 2010 04:17:22 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Fri, 19 Feb 2010 02:20:29 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:

I'm expecting to use caps on the order of perhaps 2.2mF
50V,
to be secure about the rails. But I expect to want to play
with that, once everything is working, to see just how bad
I
can make it while seeing what that means for the output.
And
then see if I can calculate a prediction that isn't too far
from those results, on paper.

I have my own rule of thumb here for acceptable levels of
ripple
and load regulation. I divide the full supply dc voltage
with
the
current at maximum output. This gives the equivalent dc load
as
seen by the power supply. In the sample design under
consideration, that's roughly 30 ohms on each side of the
split
supply. Calculate the reactance of the filter capacitor at
the
pulsating dc frequency which is twice the mains frequency
for
full-wave. My rule of thumb is to get an R/Xc ratio of the
order
of 50 for a medium quality amp. Your choice of 2200uF agrees
well
with this.

Thanks for your thinking on this. I used more mathy stuff to
get there, but I like your practical slice through all that.
It is easy to follow.

Rules of thumb are often based on previous mathematical
derivations, as it was in this case.

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.

However, after having done
umpteen calculations where absolute precision is not needed,
the
novelty wears off after some time and one tends to be
satisfied
with being able to intuitively predict the outcome within a
per
cent or so without actually putting anything on paper.

I think I clearly understood exactly that from your writing.

It's been
firmly etched in my mind for 40 years that 1000uF has a
reactance
of 1.5815 ohms (usually taken as 1.6) at 100Hz (twice the
mains
frequency here) and I quickly derive Xc for other values from
that within a second. Then I mentally divide the equivalent DC
resistance of a load (not necessarily an audio amplifier) with
that reactance and have a good idea of what to expect in terms
of
ripple voltage amplitude, regulation, DC voltage, peak diode
current, rms transformer current, etc.

Heck, it's past 4 am over here. Time for bed. Bye.

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.

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

 

[Paul E. Schoen]

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hlmn0i$aks$1@news.albasani.net...

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

 

[Jon Kirwan]

On Sat, 20 Feb 2010 00:28:22 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Paul E. Schoen wrote:

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.

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 read about perhaps 20 sufficiently
different types that professionals might consider... but that
is happily not my problem.)

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.

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:

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.

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.

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

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.

Exactly!

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.

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'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 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.

Hopefully, the above is about right. Let me know where I
screwed up.

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.

In the above example, with 36 ohms total, R_src would be
about 36-8 or 28 ohms (R_load is 8 ohms.) The ratio of power
delivered to the load, from that point of view, would be poor
wouldn't it? Bad on the order of 28/8 or 3.5:1 instead of an
optimal 1:1.

Wouldn't that suggest that if I dissipate 10W in R, that even
in class-B with 18V rails I'll be dissipating 35W across both
power BJTs? (Cutting it closer by using 15V rails would
yield more like 22/8 or 2.75:1 for 27.5W across both power
BJTs. But nothing very good there, either.)

Does this suggest a desire for an output transformer for
impedance matching and better transfer of power?

Thanks,
Jon

 

[pimpom]

Paul E. Schoen wrote:

"pimpom" <pimpom@invalid.invalid> wrote in message
news:hlmn0i$aks$1@news.albasani.net...

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.

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.

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

 

[pimpom]

Jon Kirwan wrote:

On Sat, 20 Feb 2010 00:28:22 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Paul E. Schoen wrote:

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.

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.

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.

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.

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.

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

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.

Exactly!

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.

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.

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.

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.

In the above example, with 36 ohms total, R_src would be
about 36-8 or 28 ohms (R_load is 8 ohms.) The ratio of power
delivered to the load, from that point of view, would be poor
wouldn't it? Bad on the order of 28/8 or 3.5:1 instead of an
optimal 1:1.

Wouldn't that suggest that if I dissipate 10W in R, that even
in class-B with 18V rails I'll be dissipating 35W across both
power BJTs? (Cutting it closer by using 15V rails would
yield more like 22/8 or 2.75:1 for 27.5W across both power
BJTs. But nothing very good there, either.)

Does this suggest a desire for an output transformer for
impedance matching and better transfer of power?

Thanks,
Jon

 

[Jon Kirwan]

On Sat, 20 Feb 2010 20:46:00 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
On Sat, 20 Feb 2010 00:28:22 +0530, "pimpom"
pimpom@invalid.invalid> wrote:

Paul E. Schoen wrote:

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.

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.

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.

(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.

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.

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.

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.

Okay. I think I'm following much better now.

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.

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.

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

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.

Exactly!

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.

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

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.

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.

Okay. That makes sense.

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.

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 think I got your point.

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.

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.

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

Yes.

Thanks,
Jon

In the above example, with 36 ohms total, R_src would be
about 36-8 or 28 ohms (R_load is 8 ohms.) The ratio of power
delivered to the load, from that point of view, would be poor
wouldn't it? Bad on the order of 28/8 or 3.5:1 instead of an
optimal 1:1.

Wouldn't that suggest that if I dissipate 10W in R, that even
in class-B with 18V rails I'll be dissipating 35W across both
power BJTs? (Cutting it closer by using 15V rails would
yield more like 22/8 or 2.75:1 for 27.5W across both power
BJTs. But nothing very good there, either.)

Does this suggest a desire for an output transformer for
impedance matching and better transfer of power?

Thanks,
Jon

 

[pimpom]

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.

 

[Jon Kirwan]

On Mon, 22 Feb 2010 02:58:40 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

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

 

[David Eather]

On 19/02/2010 8:00 PM, Jon Kirwan wrote:

On Fri, 19 Feb 2010 02:43:12 -0500, "Paul E. Schoen"
paul@peschoen.com> wrote:

"Jon Kirwan"<jonk@infinitefactors.org> wrote in message
news:buasn5pd2p5vs3uppunegtjcq8rfk8f3m1@4ax.com...

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,

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.
this bit - very easy.

***********

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