Discussing audio amplifier design -- BJT, discrete

[David Eather]

On 17/02/2010 2:28 PM, Jon Kirwan wrote:

On Thu, 11 Feb 2010 17:51:27 +1000, David Eather
eather@tpg.com.au> wrote:

snip
It is not a big stress. You can always use the junk-box transformer and
if it really isn't suitable replace it latter. For your consideration -
the RMS power of even compressed samples of music is only about 20% of
the peak.

There are a few variations on that figure. RCA did a lot of research in
the area and found that Radio broadcasts of compressed FM signals of
"Rock Music" - an undefined term, was the most demanding at 15%. Some
companies are trying to redefine that. IRF who call the same figure 1/8
of max power (12.5%) - which just happens to make their newest audio
mosfets look really good. It might be the other way around. They may
really believe it, and designed the mosfets to match. I forget where but
some group stated the 20% figure with respect to new modern music
styles. IIRC they were regarded as technically competent in the area and
had no axe to grind or wheelbarrow to push - so I filed the info away.
In any case an overestimate leads to a more conservative design and 5%
is not much. I'd be wary of definition of "modern music" too - badly
played organ music can be a stream of full amplitude waveforms that only
change in frequency at random intervals.

I'd use the junk box transformer and forget about allowing for the
electricity company slackness and just choose good sized caps that are a
reasonable price. I think a learning experience allows for a little
compromise.

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.)

OK. My bad.

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.

Yes. With a max of a ten watt sine wave output, the average long term
output would be between 1.25 watts and 2 watts. BUT that does depend on
what you intend the amp to be. The figures would be correct for
domestic, most PA and semi pro applications. IF the amp was intended for
musicians you would have to allow for the amp to be driven hard into
clipping - an output model more like a 12.7 volt square wave with short
circuit protection. In serious pro audio setups you might even feel it
is appropriate to design the amp to drive full time into short circuits
(not that it would be used that way, but during setup and the enevitable
modifications to the setup outputs can be shorted for long periods of
time).

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

Yes.

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


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?'

A ten watt amp delivers a sine wave producing 10 watts of output power
into a specified load. Ideally this would be 10 watts for an infinite
period of time but for audio amps, due to the nature of the signal, an
"infinite period of time" in practical terms may be as short as a few
seconds.

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?

40W output. I**2 x R. The power supply voltage is approximately constant.

But if class-A,

it's pretty much 10W no matter what?

If class A, power is 5 watts out with 4 ohms. Current is held constant.

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.

Your argument here is reasonable but ..... it is also the beginning of
the PMPO fiasco - since no advertising department could agree on what
constitutes "music" they used what ever figures looked best - and that
led to the PFPO (peak fantasy power output) fiasco where you just put
anything you like on the box.

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

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

Wait till you start talking about speakers!

Jon

 

[Jon Kirwan]

On Mon, 22 Feb 2010 22:56:22 +1000, David Eather
<eather@tpg.com.au> wrote:

On 17/02/2010 2:28 PM, Jon Kirwan wrote:
On Thu, 11 Feb 2010 17:51:27 +1000, David Eather
eather@tpg.com.au> wrote:

snip

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

Yes.

Okay. These concepts are slowly settling into my brain.

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

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

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

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

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

Okay, now I'm depressed. I go buy a 50W amplifier, stick a
sine wave signal generator on it and watch the thing toast
itself, bursting in fire soon enough?

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?'

A ten watt amp delivers a sine wave producing 10 watts of output power
into a specified load. Ideally this would be 10 watts for an infinite
period of time but for audio amps, due to the nature of the signal, an
"infinite period of time" in practical terms may be as short as a few
seconds.

Yeah. A few seconds. So... now I can go back with a
quasi-comp output stage and use a pair of those PN2222As for
it, without heat sinking! Nice little TO92 packages, too.

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?

40W output. I**2 x R. The power supply voltage is
approximately constant.

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

Now I'll do that.

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

Can you explain this 40W statement better for me?

But if class-A, it's pretty much 10W no matter what?

If class A, power is 5 watts out with 4 ohms. Current is held
constant.

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

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

I guess I need to delve into this a bit more to make sure I
understand. The class-B case seems easier for me to follow
(assuming I'm right, above, which I of course may not be.)

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.

Your argument here is reasonable but ..... it is also the beginning of
the PMPO fiasco - since no advertising department could agree on what
constitutes "music" they used what ever figures looked best - and that
led to the PFPO (peak fantasy power output) fiasco where you just put
anything you like on the box.

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

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

As a consumer, I've become a little better informed even if
all that means is I'm a lot more suspicious than before.

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

Wait till you start talking about speakers!

Hehe. Now I'm really scared.

Jon

 

[Jon Kirwan]

On Mon, 22 Feb 2010 08:41:39 -0800, I wrote:

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

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

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

Need to think more on that, as well.

Jon

 

[Jon Kirwan]

On Mon, 22 Feb 2010 10:03:41 -0800, I wrote:

On Mon, 22 Feb 2010 08:41:39 -0800, I wrote:

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

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

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

Need to think more on that, as well.

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

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

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

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

Jon

 

[Jon Kirwan]

On Tue, 23 Feb 2010 02:52:59 +0530, "pimpom"
<pimpom@invalid.invalid> wrote:

Jon Kirwan wrote:
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.

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


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.

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

No problem, at all. It means you are engaging yourself witgh
what I write, even if sometimes that means a mistake is made
in reading. And that means a great deal to me. Don't
apologize. I am honored even by the fact that you imagined a
problem there, because it means you scanned through it with
your educated eye.

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

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

Jon

 

[pimpom]

Jon Kirwan wrote:

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.

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

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.

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

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

 

[Paul E. Schoen]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:9lj5o5dicqhmln4on3uq9rlk6ejii12hq8@4ax.com...

On Mon, 22 Feb 2010 10:03:41 -0800, I wrote:

On Mon, 22 Feb 2010 08:41:39 -0800, I wrote:

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

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

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

Need to think more on that, as well.

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

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

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

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

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

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

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

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

Paul

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[Jon Kirwan]

On Mon, 22 Feb 2010 22:56:22 +1000, David Eather
<eather@tpg.com.au> wrote:

snip
Wait till you start talking about speakers!

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

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

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

Jon

 

[Jon Kirwan]

On Tue, 23 Feb 2010 01:01:07 -0500, "Paul E. Schoen"
<paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:9lj5o5dicqhmln4on3uq9rlk6ejii12hq8@4ax.com...
On Mon, 22 Feb 2010 10:03:41 -0800, I wrote:

On Mon, 22 Feb 2010 08:41:39 -0800, I wrote:

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

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

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

Need to think more on that, as well.

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

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

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

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

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

Thanks. I'm seeing this well, now.

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

By this, do you mean several power rails?

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

An alternative is to create higher
rails for the driver stage, or to use a bootstrap approach.

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

I have attached a simple amplifier using MOSFETs

Saved and observed.

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

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

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

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

Also, high power FETs are "more expensive." At least, so far
as I've experienced. And I can consider paralleling BJTs
with emitter degeneration. I understand it and why it works.
I'm not sure I'd understand, at this point, how to parallel
FETs in linear operation.

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

Okay:

http://www.infinitefactors.org/misc/spice/behavioral%202-BJT%20output%2001.asc

Hopefully, you can pick that up okay.

Jon

 

[Jon Kirwan]

On Tue, 23 Feb 2010 05:13:15 -0800, I wrote:

http://www.infinitefactors.org/misc/spice/behavioral%202-BJT%20output%2001.asc

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

Jon

 

[David Eather]

Joh,

I'm leaving all the sniping to you because your the one who knows what
you want to investigate further.

Also, for today, I am using mostly using "perfect components" and
theoretical efficiencies etc to make my life easier - they will show the
point and I am talking about an amp spec'd for consumer audio.

On 23/02/2010 2:41 AM, Jon Kirwan wrote:

On Mon, 22 Feb 2010 22:56:22 +1000, David Eather
eather@tpg.com.au> wrote:

On 17/02/2010 2:28 PM, Jon Kirwan wrote:
On Thu, 11 Feb 2010 17:51:27 +1000, David Eather
eather@tpg.com.au> wrote:

snip

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

Yes.

Okay. These concepts are slowly settling into my brain.

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

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

No, that's the power of a 12.7 volt (peak) sine wave into an 8 ohm load.
An instantaneous peak power figure would be (Vmax**2)/R or 20watts.

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

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

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

Think of it this way:

You build an amplifier that puts out a 10 watt sine wave into 8 ohms
100% of the time. For a power transformer you will need something like a
30 volt CT rated at 40VA for this design (this one is a realistic not
theoretical estimate).

Use this amp in a consumer environment and the customer is happy about
everything except the cost.

Can you lower the cost without damaging the output quality?

Yes. As mentioned earlier the long term average output power of the
amplifier will be about 2 watts and any transformer will have a very
long thermal time constant compared to any other component in the
amplifier, so there is no danger of overheating during a peak in the
music output. 40VA x 0.2 = 8VA. An 8VA transformer is big enough (in
real life you would use a little bigger because of the increased I**2 x
R losses 10VA would probably be a good choice, (if you had more
information you could make a better choice but the result would be very
close to 10VA)

A 10VA transformer costs a whole lot less than 40VA and all you have
done is removed an unnecessary over-specification of a component and
that will have zero effect to the consumer.

You can do the same thing with the heatsink, but it is not so dramatic a
change, and it needs more care. On a small amp like this where the cost
of a heatsink is low I wouldn't bother - except as an exercise or if you
were making hundreds of them. You won't be able to proceed here until
you have a more finalized design.

Okay, now I'm depressed. I go buy a 50W amplifier, stick a
sine wave signal generator on it and watch the thing toast
itself, bursting in fire soon enough?

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?'

A ten watt amp delivers a sine wave producing 10 watts of output power
into a specified load. Ideally this would be 10 watts for an infinite
period of time but for audio amps, due to the nature of the signal, an
"infinite period of time" in practical terms may be as short as a few
seconds.

Yeah. A few seconds. So... now I can go back with a
quasi-comp output stage and use a pair of those PN2222As for
it, without heat sinking! Nice little TO92 packages, too.

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?

40W output. I**2 x R. The power supply voltage is
approximately constant.


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

Now I'll do that.

recheck everyone's figures

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

Can you explain this 40W statement better for me?

20 Watts - you were right. My figure wrong.

But if class-A, it's pretty much 10W no matter what?

If class A, power is 5 watts out with 4 ohms. Current is held
constant.

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

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

I guess I need to delve into this a bit more to make sure I
understand. The class-B case seems easier for me to follow
(assuming I'm right, above, which I of course may not be.)

You seem to be doing OK. Maybe we should brush up and compare notes on
the meanings of average and RMS but that's about it I think.

A class A amp say at 10 watts into 8 ohms will have an output stage with
a constant current sink (or source) set at 1.59 amps. If the speaker
load changes to 4 ohms the maximum current into and out of the speaker
is still 1.59 amps. How's the power now?

Self is talking about his practical results and if you dig around you
will see/find he believes in over-biasing the output stage current
source by 50% - 100% hence the apparent anomaly.

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.

Your argument here is reasonable but ..... it is also the beginning of
the PMPO fiasco - since no advertising department could agree on what
constitutes "music" they used what ever figures looked best - and that
led to the PFPO (peak fantasy power output) fiasco where you just put
anything you like on the box.

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

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

As a consumer, I've become a little better informed even if
all that means is I'm a lot more suspicious than before.

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

Wait till you start talking about speakers!

Hehe. Now I'm really scared.

In Sound Lounges no one can hear you scream....

Jon

 

[David Eather]

On 23/02/2010 4:03 AM, Jon Kirwan wrote:

On Mon, 22 Feb 2010 08:41:39 -0800, I wrote:

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

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

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

Need to think more on that, as well.

I'm not sure that is right, pure Class A doesn't degrade - does it have
a Vbe multiplier? Class AB can degrade into pure class B if the Vbe
multiplier is not set properly.

Jon

 

[David Eather]

On 23/02/2010 4:47 AM, Jon Kirwan wrote:

On Mon, 22 Feb 2010 10:03:41 -0800, I wrote:

On Mon, 22 Feb 2010 08:41:39 -0800, I wrote:

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

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

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

Need to think more on that, as well.

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

No problem for me.

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

Class-A appears to deliver the same thing as class-B, at
least using a TIP3055 and TIP2955 output pair, and using an
idealized voltage source between the bases to set the class
of operation. With a 4-ohm load and the exact same drive

I don't mind at all but your post gets a little weird from here on.

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From: HC<hboothe@gte.net
Newsgroups: sci.electronics.basics
Subject: Re: Stepper motor controller current limiting question
Date: Mon, 22 Feb 2010 14:50:10 -0800 (PST)
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On Feb 17, 7:19=A0am, N0S...@daqarta.com (Bob Masta) wrote:
On Tue, 16 Feb 2010 15:57:04 -0800 (PST), HC





hboo...@gte.net> wrote:
Hello, all. =A0I'm trying to build my first stepper motor controller and
I have two questions. =A0In trying to learn about stepper motors and
controllers I read the document AN907 from Microchip among other
documents and sites:http://ww1.microchip.com/downloads/en/AppNotes/00907=
a.pdf

In that document the author writes about current limiting a stepper
motor that is purposefully driven at a voltage that is higher than its
voltage (using again a voltage source as the VAS output), I
got 20.55 watts into 4 ohms with class-A operation and 10.31
watts into 8 ohms. (Which is not a 4X but 2X phenomenon.) In
class-B, this was 19.66 watts into 4 ohms and 10.16 watts
into 8. Again, 2X. (I think I might have been just slightly
into class-AB with that last test, but I got it close.)

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

Jon

 

[Jon Kirwan]

On Wed, 24 Feb 2010 15:03:18 +1000, David Eather
<eather@tpg.com.au> wrote:

I'm leaving all the sniping to you because your the one who knows what
you want to investigate further.

"Sniping" in the US has a negative connotation, which I'm not
sure you intended. I hope I'm not coming across in some
negative way. If so, I do apologize and will try for better.
I really do appreciate the time you've offered me.

Also, for today, I am using mostly using "perfect components" and
theoretical efficiencies etc to make my life easier - they will show the
point and I am talking about an amp spec'd for consumer audio.

Okay.

On 23/02/2010 2:41 AM, Jon Kirwan wrote:
On Mon, 22 Feb 2010 22:56:22 +1000, David Eather
eather@tpg.com.au> wrote:

On 17/02/2010 2:28 PM, Jon Kirwan wrote:
On Thu, 11 Feb 2010 17:51:27 +1000, David Eather
eather@tpg.com.au> wrote:

snip

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

Yes.

Okay. These concepts are slowly settling into my brain.

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

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

No, that's the power of a 12.7 volt (peak) sine wave into an 8 ohm load.
An instantaneous peak power figure would be (Vmax**2)/R or 20watts.

Clearly understood. I just wasn't thinking well at the
moment. I'm exactly with you on this.

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

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

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

Think of it this way:

You build an amplifier that puts out a 10 watt sine wave into 8 ohms
100% of the time. For a power transformer you will need something like a
30 volt CT rated at 40VA for this design (this one is a realistic not
theoretical estimate).

Use this amp in a consumer environment and the customer is happy about
everything except the cost.

Can you lower the cost without damaging the output quality?

Yes. As mentioned earlier the long term average output power of the
amplifier will be about 2 watts and any transformer will have a very
long thermal time constant compared to any other component in the
amplifier, so there is no danger of overheating during a peak in the
music output. 40VA x 0.2 = 8VA. An 8VA transformer is big enough (in
real life you would use a little bigger because of the increased I**2 x
R losses 10VA would probably be a good choice, (if you had more
information you could make a better choice but the result would be very
close to 10VA)

A 10VA transformer costs a whole lot less than 40VA and all you have
done is removed an unnecessary over-specification of a component and
that will have zero effect to the consumer.

You can do the same thing with the heatsink, but it is not so dramatic a
change, and it needs more care. On a small amp like this where the cost
of a heatsink is low I wouldn't bother - except as an exercise or if you
were making hundreds of them. You won't be able to proceed here until
you have a more finalized design.

I think I follow all of this. I guess my earlier writing was
about my own realizations and nothing else. I wrote more
strongly then because I'm just "seeing" a little better, is
all.

Okay, now I'm depressed. I go buy a 50W amplifier, stick a
sine wave signal generator on it and watch the thing toast
itself, bursting in fire soon enough?

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?'

A ten watt amp delivers a sine wave producing 10 watts of output power
into a specified load. Ideally this would be 10 watts for an infinite
period of time but for audio amps, due to the nature of the signal, an
"infinite period of time" in practical terms may be as short as a few
seconds.

Yeah. A few seconds. So... now I can go back with a
quasi-comp output stage and use a pair of those PN2222As for
it, without heat sinking! Nice little TO92 packages, too.

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?

40W output. I**2 x R. The power supply voltage is
approximately constant.

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

Now I'll do that.

recheck everyone's figures

Tentatively, I'm lumping his tabled results into effects I'm
less aware of, for now. Context will become clearer, later.

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

Can you explain this 40W statement better for me?

20 Watts - you were right. My figure wrong.

Thanks.

But if class-A, it's pretty much 10W no matter what?

If class A, power is 5 watts out with 4 ohms. Current is held
constant.

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

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

I guess I need to delve into this a bit more to make sure I
understand. The class-B case seems easier for me to follow
(assuming I'm right, above, which I of course may not be.)

You seem to be doing OK. Maybe we should brush up and compare notes on
the meanings of average and RMS but that's about it I think.

A class A amp say at 10 watts into 8 ohms will have an output stage with
a constant current sink (or source) set at 1.59 amps.

I know there are a number of structures, but I like to think
in terms of two BJTs, one NPN and one PNP, in a push-pull
arrangement to the rails -- for class-A -- with a Vbe
multiplier set to cause both BJTs to have at least some
emitter current at all times.

I know that there is also a single-ended arrangement. But I
never give that one more than a very slight nod. It's way,
way too inefficient to care about. I'm wondering if that is
what you are talking about here.

If so, then you'd indeed set Iq to be 1.59 amps, either to
the (+) or (-) rails, as I read you saying here. Because
then the single-ended BJT can either source/sink nothing
causing the -1.59 (or +1.59) amps to flow from speaker to
rail or the single-ended BJT can source/sink 3.18 amps, only
1.59 of which gets wasted via Iq and the rest going to the
speaker.

However, I don't think much about that arrangement and I'm
sure that Self wasn't talking about an amplifier designed
that way. I'm pretty sure he was discussing a push-pull
class-A amplifier.

If the speaker
load changes to 4 ohms the maximum current into and out of the speaker
is still 1.59 amps.

In the single-ended case, _very_ generally, yes. But let me
walk you through my single-ended thoughts. (I hate single
ended designs, so I hope I don't have to think about them
again!)

Let's assume we have (+), (-), and ground rails. Let's
arrange it so that the Iq current is a sink, as you stated.
So it goes from (-) to the speaker pin. There is a BJT that
goes from the speaker pin and up to (+). It's base is driven
by the VAS and the emitter simply follows that. The speaker
load goes to ground. That's what I'm imagining you are
talking about.

At 8 ohms and Iq = -1.59A, lets say that the emitter can rise
up to the point where Iload = +1.59A. (So about +12.7V at
the emitter, as discussed regularly.) This means the BJT's
emitter must be sourcing 3.18 amps, enough to supply both the
speaker and Iq. If the BJT effectively turns off and its
emitter current goes to about 0, then Iq causes Iload =
-1.59A.

A 4 ohm load would still "see" no lower than -1.59A, since
that is all that Iq can do. However, when the emitter rises
again to it's +12.7V (driven by the same exact signal at its
base, by assumption), then we will have 12.7V across 4 ohms
and +3.18A into the speaker. The BJT will not only have to
supply that, but also the 1.59A required by Iq. That's 4.77A
total. So it will operate in class-AB, now.

So I'm pretty sure I'm _not_ following you when you say the
output current is the same. Which I take to mean there is
something very wrong with the way I'm seeing this.

How's the power now?

Best to wait for your knock on my head about this. I would
have first preferred to talk about a push-pull class-A case,
which seems fundamentally different. But I'm still learning
and have to assume I am getting all of this wrong.

Self is talking about his practical results and if you dig around you
will see/find he believes in over-biasing the output stage current
source by 50% - 100% hence the apparent anomaly.
snip

I think he was NOT talking about single-ended class-A
designs, but instead push-pull class-A. I get your point, I
think, about over-biasing -- it is something I already feel
I'd want to do, too... though I'm not sure 50% is right and
even then I don't think he spent much of any time at all
talking about single-ended designs (for reasons I think I
agree with.)

I feel like I probably failed to get some point I should have
and so I'll stop here and wait.

Very much appreciated,
Jon

 

[Jon Kirwan]

On Wed, 24 Feb 2010 15:07:23 +1000, David Eather
<eather@tpg.com.au> wrote:

On 23/02/2010 4:03 AM, Jon Kirwan wrote:
On Mon, 22 Feb 2010 08:41:39 -0800, I wrote:

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

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

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

Need to think more on that, as well.

I'm not sure that is right, pure Class A doesn't degrade

In my mind it can. But I only think in terms of class-A push
pull class-A, not single-ended. Maybe I should be more
detailed when I write?

- does it have a Vbe multiplier?

It can. It doesn't have to have one, though. I was just
thinking about the case where there is one -- a push-pull
that looks a lot like a class-B stage but with higher Vbias.

Class AB can degrade into pure class B if the Vbe
multiplier is not set properly.

Right.

Jon

 

[David Eather]

On 24/02/2010 3:03 PM, David Eather wrote:

Joh,

I'm leaving all the sniping to you because your the one who knows what
you want to investigate further.

Also, for today, I am using mostly using "perfect components" and
theoretical efficiencies etc to make my life easier - they will show the
point and I am talking about an amp spec'd for consumer audio.

On 23/02/2010 2:41 AM, Jon Kirwan wrote:
On Mon, 22 Feb 2010 22:56:22 +1000, David Eather
eather@tpg.com.au> wrote:

On 17/02/2010 2:28 PM, Jon Kirwan wrote:
On Thu, 11 Feb 2010 17:51:27 +1000, David Eather
eather@tpg.com.au> wrote:

snip

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

Yes.

Okay. These concepts are slowly settling into my brain.

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

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

No, that's the power of a 12.7 volt (peak) sine wave into an 8 ohm load.
An instantaneous peak power figure would be (Vmax**2)/R or 20watts.


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

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

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


Think of it this way:

You build an amplifier that puts out a 10 watt sine wave into 8 ohms
100% of the time. For a power transformer you will need something like a
30 volt CT rated at 40VA for this design (this one is a realistic not
theoretical estimate).
It's been a bad day 40VA is way wrong. It should be more like 12VA

Use this amp in a consumer environment and the customer is happy about
everything except the cost.

Can you lower the cost without damaging the output quality?

Yes. As mentioned earlier the long term average output power of the
amplifier will be about 2 watts and any transformer will have a very
long thermal time constant compared to any other component in the
amplifier, so there is no danger of overheating during a peak in the
music output. 40VA x 0.2 = 8VA.

This 8VA transformer should be more like 2.3VA. My apologies - brain is
slowly dying.

An 8VA transformer is big enough (in

real life you would use a little bigger because of the increased I**2 x
R losses 10VA would probably be a good choice, (if you had more
information you could make a better choice but the result would be very
close to 10VA)

A 10VA transformer costs a whole lot less than 40VA and all you have
done is removed an unnecessary over-specification of a component and
that will have zero effect to the consumer.

You can do the same thing with the heatsink, but it is not so dramatic a
change, and it needs more care. On a small amp like this where the cost
of a heatsink is low I wouldn't bother - except as an exercise or if you
were making hundreds of them. You won't be able to proceed here until
you have a more finalized design.




Okay, now I'm depressed. I go buy a 50W amplifier, stick a
sine wave signal generator on it and watch the thing toast
itself, bursting in fire soon enough?

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?'

A ten watt amp delivers a sine wave producing 10 watts of output power
into a specified load. Ideally this would be 10 watts for an infinite
period of time but for audio amps, due to the nature of the signal, an
"infinite period of time" in practical terms may be as short as a few
seconds.

Yeah. A few seconds. So... now I can go back with a
quasi-comp output stage and use a pair of those PN2222As for
it, without heat sinking! Nice little TO92 packages, too.

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?

40W output. I**2 x R. The power supply voltage is
approximately constant.


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

Now I'll do that.

recheck everyone's figures


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

Can you explain this 40W statement better for me?

20 Watts - you were right. My figure wrong.


But if class-A, it's pretty much 10W no matter what?

If class A, power is 5 watts out with 4 ohms. Current is held
constant.

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

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

I guess I need to delve into this a bit more to make sure I
understand. The class-B case seems easier for me to follow
(assuming I'm right, above, which I of course may not be.)

You seem to be doing OK. Maybe we should brush up and compare notes on
the meanings of average and RMS but that's about it I think.

A class A amp say at 10 watts into 8 ohms will have an output stage with
a constant current sink (or source) set at 1.59 amps. If the speaker
load changes to 4 ohms the maximum current into and out of the speaker
is still 1.59 amps. How's the power now?

Self is talking about his practical results and if you dig around you
will see/find he believes in over-biasing the output stage current
source by 50% - 100% hence the apparent anomaly.


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.

Your argument here is reasonable but ..... it is also the beginning of
the PMPO fiasco - since no advertising department could agree on what
constitutes "music" they used what ever figures looked best - and that
led to the PFPO (peak fantasy power output) fiasco where you just put
anything you like on the box.

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

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

As a consumer, I've become a little better informed even if
all that means is I'm a lot more suspicious than before.

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

Wait till you start talking about speakers!

Hehe. Now I'm really scared.

In Sound Lounges no one can hear you scream....


Jon

 

[David Eather]

On 24/02/2010 6:06 PM, Jon Kirwan wrote:

On Wed, 24 Feb 2010 15:03:18 +1000, David Eather
eather@tpg.com.au> wrote:

I'm leaving all the sniping to you because your the one who knows what
you want to investigate further.

"Sniping" in the US has a negative connotation, which I'm not
sure you intended. I hope I'm not coming across in some
negative way. If so, I do apologize and will try for better.
I really do appreciate the time you've offered me.

Also, for today, I am using mostly using "perfect components" and
theoretical efficiencies etc to make my life easier - they will show the
point and I am talking about an amp spec'd for consumer audio.

Okay.

On 23/02/2010 2:41 AM, Jon Kirwan wrote:
On Mon, 22 Feb 2010 22:56:22 +1000, David Eather
eather@tpg.com.au> wrote:

On 17/02/2010 2:28 PM, Jon Kirwan wrote:
On Thu, 11 Feb 2010 17:51:27 +1000, David Eather
eather@tpg.com.au> wrote:

snip

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

Yes.

Okay. These concepts are slowly settling into my brain.

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

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

No, that's the power of a 12.7 volt (peak) sine wave into an 8 ohm load.
An instantaneous peak power figure would be (Vmax**2)/R or 20watts.

Clearly understood. I just wasn't thinking well at the
moment. I'm exactly with you on this.

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

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

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

Think of it this way:

You build an amplifier that puts out a 10 watt sine wave into 8 ohms
100% of the time. For a power transformer you will need something like a
30 volt CT rated at 40VA for this design (this one is a realistic not
theoretical estimate).

Use this amp in a consumer environment and the customer is happy about
everything except the cost.

Can you lower the cost without damaging the output quality?

Yes. As mentioned earlier the long term average output power of the
amplifier will be about 2 watts and any transformer will have a very
long thermal time constant compared to any other component in the
amplifier, so there is no danger of overheating during a peak in the
music output. 40VA x 0.2 = 8VA. An 8VA transformer is big enough (in
real life you would use a little bigger because of the increased I**2 x
R losses 10VA would probably be a good choice, (if you had more
information you could make a better choice but the result would be very
close to 10VA)

A 10VA transformer costs a whole lot less than 40VA and all you have
done is removed an unnecessary over-specification of a component and
that will have zero effect to the consumer.

You can do the same thing with the heatsink, but it is not so dramatic a
change, and it needs more care. On a small amp like this where the cost
of a heatsink is low I wouldn't bother - except as an exercise or if you
were making hundreds of them. You won't be able to proceed here until
you have a more finalized design.

I think I follow all of this. I guess my earlier writing was
about my own realizations and nothing else. I wrote more
strongly then because I'm just "seeing" a little better, is
all.

Okay, now I'm depressed. I go buy a 50W amplifier, stick a
sine wave signal generator on it and watch the thing toast
itself, bursting in fire soon enough?

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?'

A ten watt amp delivers a sine wave producing 10 watts of output power
into a specified load. Ideally this would be 10 watts for an infinite
period of time but for audio amps, due to the nature of the signal, an
"infinite period of time" in practical terms may be as short as a few
seconds.

Yeah. A few seconds. So... now I can go back with a
quasi-comp output stage and use a pair of those PN2222As for
it, without heat sinking! Nice little TO92 packages, too.

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?

40W output. I**2 x R. The power supply voltage is
approximately constant.

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

Now I'll do that.

recheck everyone's figures

Tentatively, I'm lumping his tabled results into effects I'm
less aware of, for now. Context will become clearer, later.

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

Can you explain this 40W statement better for me?

20 Watts - you were right. My figure wrong.

Thanks.

But if class-A, it's pretty much 10W no matter what?

If class A, power is 5 watts out with 4 ohms. Current is held
constant.

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

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

I guess I need to delve into this a bit more to make sure I
understand. The class-B case seems easier for me to follow
(assuming I'm right, above, which I of course may not be.)

You seem to be doing OK. Maybe we should brush up and compare notes on
the meanings of average and RMS but that's about it I think.

A class A amp say at 10 watts into 8 ohms will have an output stage with
a constant current sink (or source) set at 1.59 amps.

I know there are a number of structures, but I like to think
in terms of two BJTs, one NPN and one PNP, in a push-pull
arrangement to the rails -- for class-A -- with a Vbe
multiplier set to cause both BJTs to have at least some
emitter current at all times.

I know that there is also a single-ended arrangement. But I
never give that one more than a very slight nod. It's way,
way too inefficient to care about. I'm wondering if that is
what you are talking about here.

If so, then you'd indeed set Iq to be 1.59 amps, either to
the (+) or (-) rails, as I read you saying here. Because
then the single-ended BJT can either source/sink nothing
causing the -1.59 (or +1.59) amps to flow from speaker to
rail or the single-ended BJT can source/sink 3.18 amps, only
1.59 of which gets wasted via Iq and the rest going to the
speaker.

However, I don't think much about that arrangement and I'm
sure that Self wasn't talking about an amplifier designed
that way. I'm pretty sure he was discussing a push-pull
class-A amplifier.

If the speaker
load changes to 4 ohms the maximum current into and out of the speaker
is still 1.59 amps.

In the single-ended case, _very_ generally, yes. But let me
walk you through my single-ended thoughts. (I hate single
ended designs, so I hope I don't have to think about them
again!)

Let's assume we have (+), (-), and ground rails. Let's
arrange it so that the Iq current is a sink, as you stated.
So it goes from (-) to the speaker pin. There is a BJT that
goes from the speaker pin and up to (+). It's base is driven
by the VAS and the emitter simply follows that. The speaker
load goes to ground. That's what I'm imagining you are
talking about.

At 8 ohms and Iq = -1.59A, lets say that the emitter can rise
up to the point where Iload = +1.59A. (So about +12.7V at
the emitter, as discussed regularly.) This means the BJT's
emitter must be sourcing 3.18 amps, enough to supply both the
speaker and Iq. If the BJT effectively turns off and its
emitter current goes to about 0, then Iq causes Iload =
-1.59A.

A 4 ohm load would still "see" no lower than -1.59A, since
that is all that Iq can do. However, when the emitter rises
again to it's +12.7V (driven by the same exact signal at its
base, by assumption), then we will have 12.7V across 4 ohms
and +3.18A into the speaker. The BJT will not only have to
supply that, but also the 1.59A required by Iq. That's 4.77A
total. So it will operate in class-AB, now.

So I'm pretty sure I'm _not_ following you when you say the
output current is the same. Which I take to mean there is
something very wrong with the way I'm seeing this.

I am/was thinking of a single supply class A amp. It has a great big
output capacitor.

How's the power now?

Best to wait for your knock on my head about this. I would
have first preferred to talk about a push-pull class-A case,
which seems fundamentally different. But I'm still learning
and have to assume I am getting all of this wrong.

Self is talking about his practical results and if you dig around you
will see/find he believes in over-biasing the output stage current
source by 50% - 100% hence the apparent anomaly.
snip

I think he was NOT talking about single-ended class-A
designs, but instead push-pull class-A. I get your point, I
think, about over-biasing -- it is something I already feel
I'd want to do, too... though I'm not sure 50% is right and
even then I don't think he spent much of any time at all
talking about single-ended designs (for reasons I think I
agree with.)

I feel like I probably failed to get some point I should have
and so I'll stop here and wait.

Very much appreciated,
Jon

 

[Jon Kirwan]

On Wed, 24 Feb 2010 21:20:45 +1000, David Eather
<eather@tpg.com.au> wrote:

snip
I am/was thinking of a single supply class A amp. It has a great big
output capacitor.

Then I think class-AB remains the mode of operation when the
8 ohm is replaced with the 4 ohm output, per your question to
me about that. A single-sided class-A with a 1.59A sink and
an appropriately sized rails would barely work class-A with 8
ohms. And would move into AB, driving 4. I think.

I talked a bit about the topology I was considering, earlier,
so hopefully I didn't get that part wrong even if I did fail
to add the output cap to the description. Just to be clear,
here is what I'm imagining right now:

: V+
: |
: |
: |/c Q1
: VAS ----| TIP3055
: |>e
: | C1
: | || BIG
: +----||----,
: | || |
: | \
: | / R1
: / \ \ 8 or 4
: | I1 /
: v 1.59A |
: \ / |
: | gnd
: |
: gnd

Just an emitter follower feeding a sink and the speaker via a
cap. Maybe I'm getting that wrong, though.

Jon

 

[Paul E. Schoen]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:3rrao55usonbnvt12kv6iqg89sdeacc9no@4ax.com...

On Wed, 24 Feb 2010 21:20:45 +1000, David Eather
eather@tpg.com.au> wrote:

snip
I am/was thinking of a single supply class A amp. It has a great big
output capacitor.

Then I think class-AB remains the mode of operation when the
8 ohm is replaced with the 4 ohm output, per your question to
me about that. A single-sided class-A with a 1.59A sink and
an appropriately sized rails would barely work class-A with 8
ohms. And would move into AB, driving 4. I think.

I talked a bit about the topology I was considering, earlier,
so hopefully I didn't get that part wrong even if I did fail
to add the output cap to the description. Just to be clear,
here is what I'm imagining right now:

: V+
: |
: |
: |/c Q1
: VAS ----| TIP3055
: |>e
: | C1
: | || BIG
: +----||----,
: | || |
: | \
: | / R1
: / \ \ 8 or 4
: | I1 /
: v 1.59A |
: \ / |
: | gnd
: |
: gnd

Just an emitter follower feeding a sink and the speaker via a
cap. Maybe I'm getting that wrong, though.

My conception of a class A amplifier is one where instead of an active
current sink, there is just a resistor. It may be in the form of an emitter
follower as in this case with a unity gain, or the resistor may be in the
collector to obtain a voltage gain greater than one. But in these cases,
large signal linearity is not realized. So such a configuration is used for
very small signals that are at least an order of magnitude smaller than the
supply rails, and power levels in the milliwatt range. With a resistor
load, the maximum power output is where the output impedance equals the
resistor, and the maximum voltage that can be achieved is about half the
supply rails.

When one adds an active current source or sink, it involves another
transistor and the circuit becomes essentially a half-bridge. For the
circuit shown above, with I1 = 0.75V and V+ = 12 VDC, a sine wave of 6V
amplitude will be reproduced across R1 = 8 ohms. With R1 = 4 ohms, the
current source must be set to 1.5A. The efficiency under these conditions
is 25%, and 4.5 W output. But this assumes that the current source can pull
the output below ground, which is not possible with any practical
component. And I did not factor in the power provided by the current
source, so actual efficiency will be lower.

If I use a 4 ohm resistor as the emitter load, and bias Q1 so that there is
equal clipping at the output, I can get an output of about 7.6 volts P-P,
or 1.8 W. The efficiency is about 8%. If I bias the resistor for 1/2 the
supply rail (6 V), I can get at most about 2.2 VRMS into 4 ohms, or 1.2 W.
Efficiency is 6.6%.

With a realizable current source made from a 2N3055 and a 0.2 ohm emitter
resistor, set at 1.55A, I can get about 3.9 watts into 4 ohms, and an
efficiency of 20%.

Now, I decided to see if I could get better efficiency by adding a variable
current sink. Essentially I am now making a push-pull circuit where the
lower half is not pulling so much when the upper half is pushing, and then
it pulls harder when it needs to do so during the negative excursions of
the signal. It simply required two additional components. My LTSpice
simulation shows an output of about 3.9 W into 4 ohms, with an efficiency
of about 27%.

This design is similar to class A in that it burns up about 15 watts with
low level signals. As such, maybe it is not so much and amplifier as an
"Apple-fryer"

And under those conditions the output stage is running 1.2A. But that is
better than running 1.5 A as was the case with the original design. There
does not appear to be any crossover distortion, and at high signal levels
you just get clipping, and that occurs within 1 volt of the supply rails.
As Scotty might say, "Cap'n, she just caint give ye no more!"

OK, I've played around enough. I've attached the ASC file if anyone wants
to play with it or criticize it. I just used a "shotgun" approach with
simplicity in mind. Maybe it's worth building, but I'm happy enough with
the usual Class B or AB amplifiers that don't function as space heaters
when they're just sitting there. And a class A power amplifier will never
win an Energy Star! Go green! Use PWM!

Paul

============================================================

Version 4
SHEET 1 880 680
WIRE 16 -80 -160 -80
WIRE 240 -80 16 -80
WIRE 240 -64 240 -80
WIRE 160 -16 -80 -16
WIRE 176 -16 160 -16
WIRE 16 0 16 -80
WIRE 16 96 16 80
WIRE -160 112 -160 -80
WIRE 240 128 240 32
WIRE 288 128 240 128
WIRE 416 128 352 128
WIRE 160 144 160 -16
WIRE 240 160 240 128
WIRE -80 208 -80 -16
WIRE 176 208 112 208
WIRE 416 208 416 128
WIRE 16 240 16 192
WIRE 112 240 112 208
WIRE 112 240 16 240
WIRE 112 256 112 240
WIRE 240 272 240 256
WIRE -160 352 -160 192
WIRE -80 352 -80 288
WIRE -80 352 -160 352
WIRE 112 352 112 336
WIRE 112 352 -80 352
WIRE 240 352 112 352
WIRE 416 352 416 288
WIRE 416 352 240 352
WIRE 416 368 416 352
FLAG 416 368 0
SYMBOL npn 176 -64 R0
SYMATTR InstName Q1
SYMATTR Value 2N3055
SYMBOL cap 352 112 R90
WINDOW 0 0 32 VBottom 0
WINDOW 3 32 32 VTop 0
SYMATTR InstName C1
SYMATTR Value 5000ľ
SYMBOL res 400 192 R0
SYMATTR InstName R1
SYMATTR Value 4
SYMBOL voltage -80 192 R0
WINDOW 3 -71 180 Left 0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value SINE(6.75 5.6 400 0 0 0 1000)
SYMBOL voltage -160 96 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 12
SYMBOL res 224 256 R0
SYMATTR InstName R2
SYMATTR Value .2
SYMBOL npn 176 160 R0
SYMATTR InstName Q2
SYMATTR Value 2N3055
SYMBOL res 96 240 R0
SYMATTR InstName R3
SYMATTR Value 15
SYMBOL pnp 80 192 R180
WINDOW 0 52 29 Left 0
WINDOW 3 60 68 Left 0
SYMATTR InstName Q3
SYMATTR Value 2N3906
SYMBOL res 0 -16 R0
SYMATTR InstName R4
SYMATTR Value 50
SYMBOL res 176 128 R90
WINDOW 0 0 56 VBottom 0
WINDOW 3 32 56 VTop 0
SYMATTR InstName R5
SYMATTR Value 100
TEXT -66 392 Left 0 !.tran .1 startup

 

[Jon Kirwan]

On Wed, 24 Feb 2010 19:50:15 -0500, "Paul E. Schoen"
<paul@peschoen.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:3rrao55usonbnvt12kv6iqg89sdeacc9no@4ax.com...
On Wed, 24 Feb 2010 21:20:45 +1000, David Eather
eather@tpg.com.au> wrote:

snip
I am/was thinking of a single supply class A amp. It has a great big
output capacitor.

Then I think class-AB remains the mode of operation when the
8 ohm is replaced with the 4 ohm output, per your question to
me about that. A single-sided class-A with a 1.59A sink and
an appropriately sized rails would barely work class-A with 8
ohms. And would move into AB, driving 4. I think.

I talked a bit about the topology I was considering, earlier,
so hopefully I didn't get that part wrong even if I did fail
to add the output cap to the description. Just to be clear,
here is what I'm imagining right now:

: V+
: |
: |
: |/c Q1
: VAS ----| TIP3055
: |>e
: | C1
: | || BIG
: +----||----,
: | || |
: | \
: | / R1
: / \ \ 8 or 4
: | I1 /
: v 1.59A |
: \ / |
: | gnd
: |
: gnd

Just an emitter follower feeding a sink and the speaker via a
cap. Maybe I'm getting that wrong, though.

My conception of a class A amplifier is one where instead of an active
current sink, there is just a resistor.

I first thought of that, as well. But for the purposes at
hand, it seemed a lot easier to plop a current sink in there.

David had written this to me:

"A class A amp say at 10 watts into 8 ohms will
have an output stage with a constant current
sink (or source) set at 1.59 amps. If the speaker
load changes to 4 ohms the maximum current into
and out of the speaker is still 1.59 amps. How's
the power now?"

I didn't want to wind up "getting corrected" for failing to
read well what he wrote, introducing some pre-conception of
mine.

That's why you see my schematic which _uses_ a current sink.
I'm trying to read David as accurately as I can and construct
from his words what I think he may be talking about. To do
otherwise would be to _change_ the subject on him and talk at
cross-purposes.

It may be in the form of an emitter
follower as in this case with a unity gain,

Yes, that's clear -- now that we are quickly moving to change
the subject.

or the resistor may be in the
collector to obtain a voltage gain greater than one.

Um. After moving the speaker/cap connection up there, too?
Right?

But in these cases,
large signal linearity is not realized.

You mean in the case where a resistor is used in the emitter
and where a collector resistor may (or may not) be used.
Right? In the case of the current sink I attempted, when
trying to follow David, it seems 'large signal linear' -ish
to me. (Speaking loosely. Except for Vbe variations on Ic
and maybe also the Early effect, anyway.)

So such a configuration is used for
very small signals that are at least an order of magnitude smaller than the
supply rails, and power levels in the milliwatt range. With a resistor
load, the maximum power output is where the output impedance equals the
resistor, and the maximum voltage that can be achieved is about half the
supply rails.

The configurations you now brought up? Or the one that I was
talking about, earlier, when trying to deal with David's
question to me?

When one adds an active current source or sink, it involves another
transistor

Yes, that's a given of sorts. And that is why I'd almost
certainly prefer to go with a push-pull style class-A of some
kind. It seems crazy to go single-ended under the
circumstances.

and the circuit becomes essentially a half-bridge.

A term I need to follow a little better, I suppose. I would
use it in the case of two diodes instead of four in a
full-wave, center-tapped PS with CT to ground and only one
other rail. You are using it differently than that, here.
Which makes me feel behind the terms-curve, still.

For the
circuit shown above, with I1 = 0.75V and V+ = 12 VDC,

Um... I1=.75A? Not 'V', right? (I assume we are getting
back to my ASCII schematic, now.)

a sine wave of 6V
amplitude will be reproduced across R1 = 8 ohms.

Yes, assuming as I know you must be that the drive is nicely
centered on +6V so that it goes from 0V to +12V -- which is
what I take you to mean here.

Actually, maybe 5.2V or so would be better, so that the
emitter can follow up and down well.

With R1 = 4 ohms, the
current source must be set to 1.5A.

Yes, this much I understand.

The question that David was asking me, if I understood him
accurately, didn't permit me to arbitrarily change the
current sink value. As such, the example case you are
bringing up would be a more accurate analogy to his question
if you kept the current source at .75A and changed R1 to 4
ohms. My reply, at least, was made on that basis.

The efficiency under these conditions
is 25%, and 4.5 W output. But this assumes that the current source can pull
the output below ground,

Two issues here. One with and one without the output cap
that David wisely mentioned in his response to me.

In the case without the output cap, the current sink needs
access to a rail _below_ that used by the speaker load's
other end. Otherwise, if they are common to each other, then
there is a DC bias current flowing through the speaker and
that's not really a good thing.

In the case with it, the cap provides the necessary 'most-
negative' side for the speaker and allows, after a few cycles
to pump up an equilibrium voltage on it, a DC center of 0A
for the speaker.

which is not possible with any practical
component.

Without the cap. With the cap, you are still right in that a
0V on the base of Q1 does not mean that the emitter can sink
to -0.8V or whatever, since there is no rail there for it.
(Unless some extra windings are added to the transformer to
get it, of course.)

And I did not factor in the power provided by the current
source, so actual efficiency will be lower.

Yup. Understood. I think Self says 12.5% is the best to be
hoped. I've not done my own double-check. But with your
estimate and adding in an equal amount for the sink, that
seems to get to about there.

If I use a 4 ohm resistor as the emitter load, and bias Q1 so that there is
equal clipping at the output, I can get an output of about 7.6 volts P-P,
or 1.8 W. The efficiency is about 8%. If I bias the resistor for 1/2 the
supply rail (6 V), I can get at most about 2.2 VRMS into 4 ohms, or 1.2 W.
Efficiency is 6.6%.

With a realizable current source made from a 2N3055 and a 0.2 ohm emitter
resistor, set at 1.55A, I can get about 3.9 watts into 4 ohms, and an
efficiency of 20%.

Now, I decided to see if I could get better efficiency by adding a variable
current sink. Essentially I am now making a push-pull circuit where the
lower half is not pulling so much when the upper half is pushing, and then
it pulls harder when it needs to do so during the negative excursions of
the signal. It simply required two additional components. My LTSpice
simulation shows an output of about 3.9 W into 4 ohms, with an efficiency
of about 27%.

Thanks for the circuit. I also ran it under LTspice.
Selecting from 100ms to 500ms as a range by which things have
settled out well, the resistor shows about 3.9 watts and 14.1
watts from the rail supply. Which gets to your number. As
you hoped, most of the 14 watts is in Q1, at about 6W. Q2
shows about 2.9W.

This design is similar to class A in that it burns up about 15 watts with
low level signals. As such, maybe it is not so much and amplifier as an
"Apple-fryer"

And under those conditions the output stage is running 1.2A. But that is
better than running 1.5 A as was the case with the original design. There
does not appear to be any crossover distortion, and at high signal levels
you just get clipping, and that occurs within 1 volt of the supply rails.
As Scotty might say, "Cap'n, she just caint give ye no more!"

OK, I've played around enough. I've attached the ASC file if anyone wants
to play with it or criticize it. I just used a "shotgun" approach with
simplicity in mind. Maybe it's worth building, but I'm happy enough with
the usual Class B or AB amplifiers that don't function as space heaters
when they're just sitting there. And a class A power amplifier will never
win an Energy Star! Go green! Use PWM!

Thanks, Paul. All discussion is most welcome to me. I
appreciate it very much.

Jon