Qjhn.13494$Ab2.6638@newsfe23.iad...Like 13%. The magic active current source was contributing about half the
Your picture is what I was thinking about as a class A amp - a single
I don't get wound up very easily by this sort of thing - I know you
The current sink (or source) I was referring to was a matter of design a
Hope I'm not doing too badly.
You've got to love class A. The only thing better is to do class A withotherwise 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"
Lead on Jon!
Okay. At first, I wasn't understanding what you meant well.
In the above case, Vbias can be set about right for near
And at least one other one, as well.
Self says 12.5% is the best to hope for."Paul E. Schoen" <paul@peschoen.com> wrote in message
news
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.
Like 13%. The magic active current source was contributing about half the
power.
Sometimes, the fastest way from A to B is through some
No. Just keep going
One thing that's become pretty obvious is that audio
This company sells op amps to 450W:One thing that's become pretty obvious is that audio
amplifiers are roughly the same thing as big, monster opamps.
The main difference is that they hard-wire the power supply
design instead of offering connections, tend to require a
ground reference (though that isn't necessary, I'm seeing
that it can help a little bit), limit one to single-ended
drive and hard-wire their feedback.
Some of the thoughts about how to design in better CMRR in an
audio amplifier may very well be almost the same thoughts
used with designing some opamps, for example. So in ways
what I'm trying to learn is also applicable to opamps in near
equal measure.
Does anyone build and sell a giant 500W opamp?
This is actually very useful for me and not the least bit of
waste, I think. Glad to have started about a month back.
Glad for occasional company, too.
Wow. Thanks, Paul. It's really fun going through this."Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:qpubo5l2o2f14ev2jhi1bg2hhk60mmu9dp@4ax.com...
One thing that's become pretty obvious is that audio
amplifiers are roughly the same thing as big, monster opamps.
The main difference is that they hard-wire the power supply
design instead of offering connections, tend to require a
ground reference (though that isn't necessary, I'm seeing
that it can help a little bit), limit one to single-ended
drive and hard-wire their feedback.
Some of the thoughts about how to design in better CMRR in an
audio amplifier may very well be almost the same thoughts
used with designing some opamps, for example. So in ways
what I'm trying to learn is also applicable to opamps in near
equal measure.
Does anyone build and sell a giant 500W opamp?
This is actually very useful for me and not the least bit of
waste, I think. Glad to have started about a month back.
Glad for occasional company, too.
This company sells op amps to 450W:
http://www.powerampdesign.net/downloadfullcatalog.html
Apex has been around a long time. Here is a 400W amp (40A, 200V):
http://www.cirrus.com/en/products/pro/detail/P1150.html
A list of all their amps, including a 250W PWM device:
http://www.cirrus.com/en/products/pro/areas/PA139.html#PA142_open
National has had the LM12 for quite a while. It'll do 90W in a TO-3
package:
http://cache.national.com/an/AN/AN-446B.pdf
Around 1982 I used an RCA monolithic 100W power Op-Amp for the amplifier in
a frequency test set that had to produce 0-140 VAC at 45-450 Hz. It was in
a large square package with flying leads of perhaps #16 AWG. I still have
one that I cut open after it fried.
Thanks.
I got it in time, I guess.
Thanks. I scanned over it and will read it in more detail,
I've converted it to a single, 16-page PDF file. If you
Thanks for that. I'll look, a little later on. That will be
Here are the two, slightly more permanently posted up:
Some "IMO" on the RCA app note.