J
Jon Kirwan
- Jan 1, 1970
- 0
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.
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.
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.
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.
20 Watts - you were right. My figure wrong.
Thanks.
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